CBSE Class 12

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 7 The p-Block Elements.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 7 Important Extra Questions The p-Block Elements

The p-Block Elements Important Extra Questions Very Short Answer Type

Question 1.
Which one of PCl₄⁺ and PCl₄^– is not likely to exist and why? (CBSE Delhi 2012)
Answer:
PCl₄^– is not likely to exist because lone pair of PCl₃ can be donated to Cl⁺ and not to Cl^–.

Question 2.
Of PH₃ and H₂S which is more acidic and why? (CBSE Delhi 2012)
Answer:
H₂S is more acidic because of weak S-H bond. PH₃ behaves as a Lewis base because of the presence of lone pair of electrons on P.

Question 3.
Which is a stronger reducing agent, SbH₃ or BiH₃ and why? (CBSE Al 2012)
Answer:
BiH₃ is a stronger reducing agent than SbH₃. This is because BiH₃ is less stable than SbH₃ because of larger size of Bi than Sb.

Question 4.
What is the covalency of nitrogen in N₂O₅? (CBSE Delhi 2013)
Answer:
Four.

Question 5.
Name two poisonous gases which can be prepared from chlorine gas. (CBSE AI 2013)
Answer:

1. Phosgene (COCl₂)
2. Mustard gas (ClCH₂CH₂SCH₂CH₂Cl)

Question 6.
What is the basicity of H₃PO₃ and why? (CBSE AI 2013)
Answer:
H₃PO₃ contains two P-OH bonds and therefore can give 2H⁺ ions. Its basicity is two.

Question 7.
Why does ammonia act as a Lewis base? (CBSE 2014)
Answer:
Nitrogen atom in NH₃ has one lone pair of electrons which is available for donation. Therefore,it acts as a Lewis base.

Question 8.
What is the oxidation number of phosphorus in H₃PO₂ molecule? (CBSE Delhi 2010)
Answer:
+1

Question 9.
Out of white phosphorus and red phosphorus, which one is more reactive and why? (CBSE 2015)
Answer:
White phosphorus is more reactive because of angular strain in the P₄ molecules where the angles are only 60°.

Question 10.
What is the basicity of H₃PO₄? (CBSE Delhi 2015)
Answer:
Basicity of H₃PO₄ is three because it has three ionisable hydrogen atoms.

Question 11.
On heating Pb(NO₃)₂ a brown gas is evolved which undergoes dimerisation on cooling. Identify the gas. (CBSE 2016)
Answer:
NO₂.

2NO₂ → N₂O₄

Question 12.
Solid PCl₅ is ionic in nature. Give reason. (CBSE AI 2016)
Answer:
PCl₅ is ionic in the solid state because it exists as [PCl₄]⁺ [PCl₆]^– in which the cation is tetrahedral and anion is octahedral.

Question 13.
Write the structural difference between white P and red P. (CBSE Delhi 2014)
Answer:
White P consists of P₄ units in which four P atoms lie at the corners of a regular tetrahedron with ∠PPP = 60°. Red P also consists of P₄ tetrahedra units but it has polymeric structure consisting of P₄ tetrahedra linked together by covalent bonds.

Question 14.
What is the basicity of H₃PO₂ acid and why? (CBSE AI 2012)
Answer:
H₃PO₂ has the structure:

It has only one ionisable hydrogen and therefore, its basicity is one.

Question 15.
Name the promoter used In Haber’s process. (CBSE Sample Paper 2017-18)
Answer:
Molybdenum.

Question 16.
NF₃ is an exothermic compound whereas NCl₃ is not. Explain. (CBSE AI 2011, 2012)
Answer:
NF₃ is an exothermic compound while NCl₃ is not because:

• The bond dissociation enthalpy of F₂ is lower than that of Cl₂.
• Size of F is small as compared to Cl and therefore F forms stronger bonds with nitrogen releasing large amount of energy.

Therefore, overall energy is released during the formation of NF₃ and energy is absorbed during the formation of NCl₃.

Question 17.
N-N single bond is weaker than P-P single bond. (CBSE Delhi 2014)
Answer:
N-N single bond is weaker than P-P single bond because of high interelectronic repulsions of non-bonding electrons due to small bond length.

Question 18.
What happens when orthophosphorous acid is heated? (CBSE Sample Paper 2017-18)
Answer:
Phosphoric acid and phosphine are formed.

Question 19.
Write the order of thermal stability of the hydrides of group 16 elements. (CBSE Delhi 2017)
Answer:
The thermal stability of the hydrides of group 16 elements decreases from H₂O to H₂Te as:
H₂O > H₂S > H₂Se > H₂Te

Question 20.
The two O – O bond lengths in ozone molecule are equal. Why? (CBSE AI 2013, CBSE Delhi 2014)
Answer:
Ozone is a resonance hybrid of two structures and therefore, the two O-O bond lengths are equal.

Question 21.
SF₆ is not easily hydrolysed. (CBSE Delhi 2005)
OR
SF₆ is kinetically inert substance. Explain. (CBSE Al 2011, CBSE Delhi 2011)
Answer:
SF₆ is chemically inert and therefore, does not get hydrolysed. Its inert nature is due to the presence of stearically protected sulphur atom which does not allow thermodynamically favourable hydrolysis reaction.

Question 22.
Which of the following compounds has a lone pair of electrons at the central atom? (CBSE Sample Paper 2011)
H₂S₂O₈, H₂S₂O₇, H₂SO₃, H₂SO₄.
Answer:
H₂SO₃

Question 23.
Ozone is thermodynamically unstable. Explain. (CBSE Sample Paper 2011)
Answer:
Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (∆H is -ve) and increase in entropy (∆S is +ve). These two factors reinforce each other resulting negative ∆G (∆G = ∆H – T∆S) for its conversion to oxygen.

Question 24.
Which neutral molecule would be isoelectronic with ClO^–? Is that molecule a Lewis base? (CBSE Delhi 2008)
Answer:
ClO^– is isoelectronic with ClF. Yes, it is a Lewis base.

Question 25.
Which compound of xenon has distorted octahedral shape? (CBSE Sample Paper 2012)
Answer:
XeF₆ has distorted octahedral shape.

Question 26.
Iodide ions can be oxidised by oxygen in acidic medium. Give chemical equation to support this. (CBSE Sample Paper 2011)
Answer:
Iodide can be oxidised by oxygen in acidic medium:
4I^–(aq) + O₂(g) + 4H⁺(oq) > 2l₂(s) + 2H₂O(l).

Question 27.
What happens when XeF6 undergoes complete hydrolysis? (CBSE Sample Paper 2017-18)
Answer:
XeO₃ is formed.
XeF₆ + 3H₂O → XeO₃ + 6HF

Question 28.
Which xenon compound is isostructural with lCl₄^–? (CBSE Sample Paper 2011)
Answer:
XeF₄

Question 29.
Why conductivity of silicon increases on doping with phosphorus? (CBSE Delhi 2019)
Answer:
When silicon is doped with phosphorus, an extra electron is introduced after forming four covalent bonds. This extra electron gets delocalised and serves to conduct electricity. Hence, conductivity increases.

The p-Block Elements Important Extra Questions Short Answer Type

Question 1.
What inspired N. Bartlett for carrying out reaction between Xe and PtF₆? (CBSE Delhi 2013)
Answer:
In 1962, N. Bartlett noticed that platinum hexafluoride PtF₆, is a powerful oxidising agent which combines with molecular oxygen to form ionic compound, dioxygenyl hexafluoroplatinate (V), O₂⁺ [PtF₆]^–.
O₂(S) + PtF₆(g) → O₂⁺ [PtF₆]^–
This indicates that PtF₆ has oxidised O₂ to O₂⁺. Now, oxygen and xenon have some similarities:

• The first ionisation energy of xenon gas (1170 kJ mol^-1) is fairly close to that of oxygen (1177 kJ mol^-1).
• The molecular diameter of oxygen and atomic radius of xenon are similar (4Å). These prompted Bartlett to carry out the reaction between Xe and PtF₆.

Question 2.
Calculate the number of lone pairs on central atom in the following molecule and predict the geometry. (CBSE Sample Paper 2019)
Answer:
XeF₄ has square planar structure. It involves sp³d² hybridisation of Xenon in which two positions are occupied by lone pairs.

Structure of XeF₄ (Square planar)

Question 3.
How are interhalogen compounds formed? What general compositions can be assigned to them? (CBSE AI 2013)
Answer:
The compounds containing two or more halogen atoms are called interhalogen compounds. They may be assigned general composition as XX’_n, where X is halogen of larger size and X’ is of smaller size and X’ is more electronegative than X, e.g. ClF, ICl₃, BrF₅, IF₇, etc.

Question 4.
Though nitrogen exhibits +5 oxidation state, it does not form pentahalides. Give reason. (CBSE 2013, CBSE Delhi 2015)
Answer:
Nitrogen belongs to second period (n = 2) and has only s and p-orbitals. It does not have d-orbitals in its valence shell and therefore, it cannot extend its octet. That is why nitrogen does not form pentahalides.

Question 5.
Why does NO₂ dimerise? Explain. (CBSE 2014, CBSE Delhi 2014)
Answer:
NO₂ contains odd number of valence electrons. It behaves as a typical molecule. In the liquid and solid state, it dimerises to form stable N₂O₄ molecule, with even number of electrons. Therefore, NO₂ is paramagnetic, while N₂O₄ is diamagnetic in which two unpaired electrons get paired.

Question 6.
Draw the structure of O₃ molecule. (CBSE Delhi 2010)
Answer:
Ozone has angular structure as:

It is resonance hybrid of the following structures:

Question 7.
Complete the following equations: (CBSE 2014)
(i) Ag + PCl₅ →
(ii) CaF₂ + H₂SO₄ →
Answer:
(i) 2Ag + PCl₅ → 2AgCl + PCl₃
(ii) CaF₂ + H₂SO₄ → CaSO₄ + 2HF

Question 8.
Write balanced chemical equations for the following processes:
(i) XeF₂ undergoes hydrolysis.
(ii) MnO₂ is heated with cone. HCl.
OR
Arrange the following in order of property indicated for each set:
(i) H₂O, H₂S, H₂Se, H₂Te – increasing acidic character
(ii) HF, HCl, HBr, HI – decreasing bond enthalpy (CBSE Delhi 2019)
Answer:
(i) 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂
(ii) MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
OR
(i) H₂O < H₂S < H₂Se < H₂Te
(ii) HF > HCI > HBr > HI

Question 9.
Explain the following giving an appropriate reason in each case.
(i) O₂ and F₂ both stabilise higher oxidation states of metals but O₂ exceeds F₂ in doing so.
(ii) Structures of xenon fluorides cannot be explained by Valence Bond approach. (CBSE 2012)
Answer:
(i) This is because fluorine can show only -1 oxidation state, whereas oxygen can show -2 oxdiation state as well as positive oxidation states such as +2, +4 and +6.

(ii) Structures of xenon fluorides cannot be explained by Valence Bond approach because xenon is a noble gas and its octet is complete. Therefore, according to Valence Bond theory, xenon should not combine with fluorine to form fluorides due to its stable outermost shell configuration.

Question 10.
Draw the structures of the following molecules:
(i) XeOF₄
(ii) H₃PO₃ (CBSE 2013)
Answer:
(i)

Structure of XeOF₄

(ii)

Structure of H₃PO₃

Question 11.
How are interhalogen compounds formed? What general compositions can be assigned to them? (CBSE 2013)
Answer:
The compounds containing two or more halogen atoms are called interhalogen compounds. They may be assigned general composition as XX’_n, where X is halogen of larger size and X’ is of smaller size and X’ is more electronegative than X,
e.g. ClF, ICl₃, BrF₅, IF₇, etc.

Question 12.
Draw the structures of the following molecules: (CBSE 2013)
(i) N₂O₅
(ii) XeF₂
Answer:
(i)

(ii)

Structure of XeF₂ (Linear)

Question 13.
What happens when
(i) PCl₅ is heated?
(ii) H₃PO₃ is heated?
Write the reactions involved. (CBSE Delhi 2013)
Answer:
(i) On heating, PCl₅ sublimes and decomposes on strong heating:

(ii) On heating, H₃PO₃ decomposes into phosphoric acid and phosphine.
4H₃PO₃ → 3H₃PO₄ + PH₃

Question 14.
Complete the following chemical equations:
(i) Ca₃P₂ + H₂O →
(ii) Cu + H₂SO₄(conc.) →
OR
Arrange the following in the order of property indicated against each set:
(i) HF, HCl, HBr, HI – increasing bond dissociation enthalpy.
(ii) H₂O, H₂S, H₂Se, H₂Te – increasing acidic character. (CBSE Delhi 2014)
Answer:
(i) Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PH₃
(ii) Cu + 2H₂SO₂₄;(conc.) → CuSO₄
+ SO₂ + 2H₂O
OR
(i) HI < HBr < HCl < HF
(ii) H₂O < H₂S < H₂Se < H₂Te

Question 15.
Complete the following equations:
(i) P₄ + H₂O →
(ii) XeF₄ + O₂F₂ → (CBSE 2014)
Answer:
(i) P₄ + H₂O → no reaction
(ii) XeF₄ + O₂F₂ → XeF₆ + O₆

Question 16.
Draw the structures of the following:
(i) XeF₂
(ii) BrF₃
Answer:
(i)

Structure of XeF₂ (Linear)

(ii)

Structure of BrF₃ (T Shape)

Question 17.
(a) Draw the structure of XeF₄.
(b) What happens when CaF₂ reacts with cone. H₂SO₄? Write balanced chemical equation. (CBSE 2019C)
Answer:
(a)

(b) HF is formed.
CaF₂ + H₂SO₄ → 2HF + CaSO₄

Question 18.
(a) Draw the structure of H₂S₂O₇.
(b) What happens when carbon reacts with cone. H₂SO₄? Write balanced chemical equation. (CBSE 2019C)
Answer:
(a)

(b) Carbon dioxide is formed.
C + 2H₂SO₄(conc.) → CO₂ + 2 SO₂ + 2H₂O

Question 19.
Draw the structures of the following molecules: (CBSE 2014)
(i) XeO₂
(ii) H₂SO₄
Answer:
(i)

(ii)

Question 20.
Write the structure of the following: (HPO₃)₃ (CBSE 2016)
Answer:
(HPO₃)₃

Question 21.
Why is N₂ less reactive at room temperature? (CBSE Al 2015, H.P.S.B. 2015, 2016, Meghalaya S.B. 2017)
Answer:
In molecular nitrogen, there is a triple bond between two nitrogen atoms (NsN) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

Question 22.
Phosphine has lower boiling point than ammonia. Give reason. (CBSE Al 2008,
CBSE Delhi 2013)
Answer:
Ammonia exists as associated molecule due to its tendency to form hydrogen bonding. Therefore, it has high boiling point. Unlike NH₃, phosphine (PH₃) molecules are not associated through hydrogen bonding in liquid state.
This is because of low electronegativity of P than N. As a result, the boiling point of PH₃ is lower than that of NH₃.

Question 23.
Draw structures of the following:
(a) XeF₄
(b) S₂O₈^2- (CBSE AI 2019)
Answer:
(a)

(b)

Question 24.
(i) Draw the structure of phosphinic acid (H₃PO₂).
(ii) Write a chemical reaction for its use as reducing agent. (CBSE Sample Paper 2011)
Answer:
(i) Phosphinic acid, H₃PO₂.

(ii) H₃PO₂ reduces Ag⁺ ion to Ag which shows its reducing nature.
H₃PO₂ + 4AgNO₃ + 2H₂O → 4Ag + 4HNO₃ + H₃PO₄

Question 25.
Among the hydrides of Group 15 elements, which have the
(a) lowest boiling point?
(b) maximum basic character?
(c) highest bond angle?
(d) maximum reducing character? (CBSE AI 2018)
Answer:
(a) PH₃
(b) NH₃
(c) NH₃
(d) BiH₃

Question 26.
The bond angles (O-N-O) are not of the same value in NO₂^– and NO₂⁺. Give reason. (CBSE Delhi 2012)
Answer:
In NO₂ there is one electron on N. Therefore, in NO₂⁺, there is no electron on N atom and there is a lone pair of electrons on N atom in NO₂^–.

Question 27.
H₂PO₂ is a stronger reducing agent than H₃PO₃. Give reason. (CBSEAI 2014, 2016)
Answer:
The reducing character of the acid is due to H atoms bonded directly to P atom. In H₃PO₂, there are two P-H bonds whereas in H₃PO₃, there is one P-H bond. Therefore, H₃PO₂ is a stronger reducing agent than H₂PO₃.

Question 28.
Bi(V) is a stronger oxidising agent than Sb(V). Give reason. (CBSE Delhi 2014)
Answer:
On moving down the group, the stability of +5 oxidation state decreases while the stability of +3 oxidation state increases due to inert pair effect. Therefore, +5 oxidation state of Bi is less stable than +5 oxidation state of Sb. Thus, Bi(V) is a stronger oxidising agent than Sb(V).

Question 29.
Dioxygen is a gas while sulphur is a solid at room temperature. Why?
(CBSE AI 2013, 2018)
Answer:
Oxygen exists as a stable diatomic molecule and is, therefore, a gas. On the other hand, sulphur exists in solid state as S8 molecules and have puckered ring structure. The main reason for this different behaviour is that oxygen atom has strong tendency to form multiple bonds with itself and forms strong O=O bonds rather than O-O bonds.

On the other hand, sulphur-sulphur double bonds (S=S) are not very strong. As a result, catenated -O-O-O- chains are less stable as compared to O=O molecule while catenated -S-S-S- chains are more stable as compared to S = S molecule.
Therefore, oxygen exists as a diatomic gas and sulphur exists as S₈ solid.

Question 30.
Draw structures of the following:
(a) H₂S₂O₇
(b) HClO₃ (CBSE Al 2019)
Answer:
(a) H₂S₂O₇

(b) HClO₃

Question 31.
Complete and balance the following equations:
(a) C + H₂SO₄ (cone.) →
(b) XeF₂ + PF₅ →
OR
Write balanced chemical equations involved in the following reactions:
(a) Fluorine gas reacts with water.
(b) Phosphine gas is absorbed in copper sulphate solution. (CBSE AI 2019)
Answer:
(a)

(b)

OR

(a) 2F₂ + 2H₂O → 4HF + O₂

(b)

Question 32.
Sulphuric acid has low volatility. Give chemical reactions in support of this. (CBSE Sample Paper 2011)
Answer:
Sulphuric acid has low volatility and decomposes the salts of volatile acids forming its own salts:
2MX + H₂SO₄ → 2HX + M₂SO₄
(M = metal, X = F, Cl, NO3)
e.g.,NaCl + H₂SO₄ → NaHSO₄ + HCl
KNO₃ + H₂SO₄ → KHSO₄ + HNO₃.

Question 33.
Electron gain enthalpies of halogens are largely negative. Why? (CBSE AI 2017)
Answer:
The halogens have the smallest size in their respective periods and therefore, high effective nuclear charge. Moreover, they have only one electron less than the stable noble gas configuration (ns²np⁶).

Therefore, they have strong tendency to accept one electron to acquire noble gas electronic configurations and hence have largely negative electron gain enthalpies.

Question 34.
Why does R₃P = 0 exist but R₃N = 0 does not (R = alkyl group). (CBSE AI 2014)
Answer:
R₃N = 0 does not exist because nitrogen cannot have covalency more than four. Moreover, R₃P = 0 exists because phosphorus can extend its covalency more than 4 as well as it can form dπ-pπ bond whereas nitrogen cannot form dπ-pπ bond.

Question 35.
Write any two oxoacids of sulphur and draw their structures. (CBSE Al 2019)
Answer:
(i) Sulphuric acid

(ii) Peroxodisulphuric acid

Question 36.
Complete and balance the following equations:
(a) NH₃ (excess) + Cl₂ →
(b) XeF₆ + 2H₂O →
OR
Write balanced chemical equations involved in the following reactions:
(a) XeF₄ reacts with SbF₅.
(b) Ag is heated with PCl₅. (CBSE AI 2019)
Answer:
(a)
8NH₃ + 3Cl₂ → 6NH₄Cl + N₂
excess
(b) XeF₆ + 2H₂O → XeO₂F₂ + 4HF

OR

(a) XeF₄ + SbF₅ → [XeF₃]⁺[SbF₆]^–
(b) 2Ag + PCl₅ → 2AgCl + PCl₃

Question 37.
Complete and balance the following equations:
(a) S + H₂SO₄ (cone.) →
(b) PCl₃ + H₂O →
Write balanced chemical equations involved in the following reactions:
(a) Chlorine gas reacts with cold and dilute NaOH
(b) Calcium phosphide Is dissolved in water. (CBSE AI 2019)
Answer:
(a) S + 2H₂SO₄(conc.) → 3SO₂ + 2H₂O
(b) PCl₃ +3H₂O → H₃PO₃ + 3HCl
OR
(a) Cl₂ + ZNaOH → NaCl + NaOCl + H₂O
(b) Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PH₃

Question 38.
Bond enthalpy of fluorine Is lower than that of chlorine. Why? (CBSE AI 2018)
Answer:
Fluorine atoms are smalL and internuclear distance between two F atoms In F₂ molecule is also small (1 .43Å). As a result, electron-electron repulsions among the lone pairs on two fluorine atoms in F₂ molecule are large. These Llarge electron-electron repulsions weaken the F-F bond. However, the electron-electron repulsions among lone pairs in Cl₂ molecule are less because of Larger Cl atoms and larger Cl-Cl distance (1.99Å). Therefore, the bond enthalpy of fluorine is lower than that of chlorine.

The p-Block Elements Important Extra Questions Long Answer Type

Question 1.
Give reasons for the following:
(a) Dioxygen is a gas but sulphur a solid.
(b) NO (g) released by jet aeroplanes is slowly depleting the ozone layer.
(c) Interhalogens are more reactive than pure halogens. (CBSE Al 2019)
Answer:
(a) Due to small size and high electronegativity, oxygen atom forms pπ-pπ double bond, O = O. The intermolecular forces in oxygen are weak van der Waal’s forces and therefore, oxygen exists as a gas. On the other hand, sulphur does not form stable pπ-pπ bonds and does not exist as S₂. It is linked by single bonds and forms polyatomic complex molecules having eight atoms per moledule (S₈) and has puckered ring structure. Therefore, S atoms are strongly held together and it exists as a solid.

(b) Nitrogen oxide (NO) gas emitted from the exhaust of jet aeroplanes readily combines with ozone to form nitrogen dioxide and diatomic oxygen.
NO + O₃ → NO₂ + O₂
Since jet aeroplanes fly in the stratosphere, near ozone layer, these are responsible for the depletion of ozone layer.

(c) This is because covalent bond between dissimilar atoms in interhalogens (X- Y) is polar and weaker than between similar atoms (X-X and Y-Y). This is due to the fact that the overlapping of orbitals of dissimilar atoms is less effective than the overlapping between similar atoms.

Question 2.
Give reasons for the following:
(i) Where R is an alkyl group, R₃P = 0 exists but R₃N = 0 does not.
(ii) PbCl₄ is more covalent than PbCl₂.
(iii) At room temperature, N₂ is much less reactive. (CBSE AI 2013)
Answer:
(i) R₃N = 0 does not exist because nitrogen cannot have covalency more than four. But R₃P = 0 exists because phosphorus can extend its covalency more than 4 as well as it can form dπ-pπ bonds whereas nitrogen cannot form dπ-pπ bond.

(ii) Pb⁴⁺ has smaller size than Pb²⁺. As a result, it can polarise the Cl^– ion to greater extent than Pb²⁺. Consequently, PbCl₄ is more covalent than PbCl₂.

(iii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

Question 3.
Give reasons for the following:
(i) Oxygen is gas but sulphur is a solid,
(ii) O₃ acts as a powerful oxidising agent.
(iii) BiH₃ is the strongest reducing agent amongst all the hydrides of Group 15 elements. (CBSE Al 2013)
Answer:
(i) Oxygen exists as a stable diatomic molecule and is, therefore, a gas.
On the other hand, sulphur exists in solid state as S₈ molecules and have puckered ring structure. The main reason for this different behaviour is that oxygen atom has good tendency to form multiple bonds with itself and forms strong o = o bonds than o – o bonds. On the other hand, sulphur- sulphur double bonds (S = S) are not very strong. As a result, catenated -O-O-O- chains are less stable as compared to O = O molecule while catenated -S-S-S- chains are more stable as compared to S = S molecule. Therefore, oxygen exists as a diatomic gas and sulphur exists as S₈ solid.

(ii) Ozone acts as a powerful oxidising agent because it has higher energy content and decomposes readily to give atomic oxygen as:
O₃ → O₂ + O
Therefore, ozone can oxidise a number of non-metals and other compounds. For example,
PbS (s) + 4O₃ (8) → PbSO₄ (s) + 4O₂ (g)
2I^– (aq) + H₂O(l) + O₃(g) → 2OH^– (aq) + I₃₂(s) + O₂(g)

(iii) The stability of group 15 hydrides decreases from NH₃ to BiH₃. Therefore, BiH₃ is most unstable hydride among group 15 hydrides because of lowest M-H bond strength. As a result, BiH₃ is the strongest reducing agent.

Question 4.
Give reasons for the following:
(i) Though nitrogen exhibits +5 oxidation state, it does not form pentahalide.
(ii) Electron gain enthalphy with negative sign of fluorine is less than that of chlorine.
(iii) The two oxygen-oxygen bonds lengths in ozone molecule are identical. (CBSE Al 2013)
Answer:
(i) Nitrogen does not have vacant d-orbitals in its valence shell. Therefore, it cannot extend its valency beyond 3. Thus, it cannot form pentahalides.

(ii) The less negative electron gain enthalpy of fluorine as compared to chlorine is due to very small size (72 pm) of fluorine atom. As result, there are strong inter- electronic repulsions in the relatively small 2p sub-shell of fluorine and therefore, the incoming electron does not feel much attraction. Thus, its electron gain enthalpy is small.

(iii) Ozone is resonance hybrid of two structures and therefore, the two oxygen-oxygen bonds are identical.

Question 5.
How would you account for the following:
(i) H₂S is more acidic than H₂O.
(ii) The N-O bond in NO₂^– is shorter than the N-O in NO₃^–.
(iii) Both O₂ and F₂ stabilise higher oxidation states but the ability of oxygen to stabilise the higher oxidation states exceeds that of fluorine. (CBSE 2011)
Answer:
(i) Since size of S is more than that of O, the distance between central atom and hydrogen atoms is more in H₂S than in H₂O. Therefore, bond dissociation enthalpy of H₂S is less and bond cleavage is more easy than in H₂O. Therefore, H₂S is more acidic than H₂O.

(ii) In the resonance structures of NO₂^–, two bonds are sharing a double bond, while in NO₃^–, three bonds are sharing a double bond.

As a result, the bond in NO₂^– will be shorter than in NO₃^–.

(iii) The larger tendency of oxygen to stabilise the higher oxidation state is because of the tendency of oxygen to form multiple bonds with metal.

Question 6.
How would you account for the following:
(i) NF₃ is an exothermic compound but NCl₃ is not.
(ii) The acidic strength of compounds increases in the order:
PH₃ < H₂S < HCl
(iii) SF₆ is kinetically inert. (CBSE 2011)
Answer:
(i) NF₃ is an exothermic compound while NCl₃ is an endothermic compound because
(a) The bond dissociation enthalpy of F₂ is lower as compared to Cl₂.
(b) Size of F is small as compared to Cl and therefore, F forms stronger bond with N releasing large amount of energy.
Therefore, overall energy is released during the formation of NF₃ and energy is absorbed during the formation of NCl₃.

(ii) The bond dissociation enthalpy decreases in the order:
P-H > S-H > Cl-H
Therefore, acidic character follows the order:
PH₃ < H₂S < HCl

(iii) SF₆ is kinetically inert because sulphur is sterically protected by six F atoms and it is a coordinatively saturated compound. Therefore, it does not undergo thermodynamically favourable hydrolysis reaction.

Question 7.
(a) Draw the structures of the following molecules:
(i) XeOF₄
(ii) H₂SO₄
(b) Write the structural difference between white phosphorus and red phosphorus. (CBSE Delhi 2014)
Answer:
(a) (i) XeOF₄

(ii) H₂SO₄

(b) White phosphorus consists of P₄ units in which four P atoms lie at the corners of a regular tetrahedron with ΔPPP = 60°.
Answer:
Red phosphorus also consists of P₄ tetrahedra units, but it has polymeric structure consisting of P₄ tetrahedra linked together by covalent bond.

Question 8.
Account for the following:
(i) Bi(V) is a stronger oxidising agent than Sb(V).
(ii) N-N single bond is weaker than P-P single bond.
(iii) Noble gases have very low boiling points. (CBSE Delhi 2014)
Answer:
(i) On moving down the group, the stability of +5 oxidation state decreases while the stability of +3 oxidation state increases due to inert pair effect. Therefore, +5 oxidation state of Bi is less stable than +5 oxidation state of Sb. Thus, Bi(V) is a stronger oxidising agent than Sb(V).

(ii) N-N single bond is weaker than single P-P bond because of high interelectronic repulsions of non-bonding electrons due to small bond length.

(iii) Noble gases are monoatomic gases and are held together by weak vander Waals forces. Therefore, these are liquified at very low temperatures. Hence they have low boiling points.

Question 9.
Give reasons for the following: (CBSE 2014)
(i) (CH₃)₃ P = 0 exists but (CH₃)₃ N = 0 does not.
(ii) Oxygen has less electron gain enthalpy with negative sign than sulphur.
(iii) H₃PO₂ is a stronger reducing agent than H₃PO₃.
Answer:
(i) (CH₃)₃ N = 0 does not exist because
nitrogen cannot have covalency more than four. Moreover, (CH₃)₃ P = 0 exists because phosphorus can extend its covalency more than 4 as well as it can form dπ – pπ bonds whereas nitrogen cannot form dπ – pπ bonds.

(ii) This is due to small size of oxygen atom so that its electron cloud is distributed over a small region of space and therefore, it repels the incoming electron. Hence the electron gain enthalpy of oxygen is less negative than sulphur.

(iii) In H₃PO₃, there are two P-H bonds whereas in H₃PO₃, there is one P-H bond. Therefore, H₃PO₃ is stronger reducing agent than H₃PO₃.

Question 10.
Assign reason for the following:
(i) H₃PO₂ is a stronger reducing agent than H₃PO₄.
(ii) Sulphur shows more tendency for catenation than oxygen.
(iii) Reducing character increases from HF to HI. (CBSE 2016)
Answer:
(i) The reducing character of the acid is due to H atoms bonded directly to P atom. In H₃PO₂, there are two P-H bonds whereas in H₃PO₃, there is one P-H bond. Therefore, H₃PO₂ is a stronger reducing agent than H₃PO₃.

(ii) Sulphur shows more tendency for catenation than oxygen because of stronger S-S bonds than O-O bonds.

(iii) Due to decrease in bond enthalpy down the group, the thermal stability decreases from HF to HI.
As a result of decrease in stability from HF to HI, the reducing character increases down the group as:
HF < HCl < HBr < HI

Question 11.
What happens when
(i) (NH₄)₂Cr₂O₇ is heated?
(ii) PCl₅ is heated?
(iii) H₃PO₃ is heated? Write the equations involved. (CBSE AI 2015, CBSE Delhi 2008, 2013, 2017)
Answer:
(i) On heating (NH₄)₂Cr₂O₇, nitrogen gas is evolved.

(ii) On heating, PCl₅ first sublimes and then decomposes on strong heating:

(iii) On heating, H₃PO₃ disproportionates to give orthophosphoric acid and phosphine.

Question 12.
(a) Suggest a quantitative method for estimation of the gas which protects us from U.V. rays of the sun.
(b) Nitrogen oxides emitted from the exhaust system of supersonic jet aeroplanes slowly deplete the concentration of ozone layer in upper atmosphere. Comment. (CBSE Sample Paper 2011)
Answer:
(a) The gas which protects us from U.V. rays of the sun is ozone. It reacts with I^– ions to give iodine as:
O₃ + 2I^– + H₂O → O₂ + l₂ + 2OH^–
l₂ liberated is titrated against sodium thiosulphate solution and amount of O₃ can be estimated.

(b) The release of nitrogen oxides (NO_x) into stratosphere by the exhaust system of supersonic jet aeroplanes deplete the concentration of O₂ because NO reacts with O₃ to give O₂.
NO (g) + O₂ (g) → NO₂ (S) + O₂ (s)
Therefore, NO is slowly depleting the concentration of ozone.

Question 13.
Give reasons for the following:
(a) Acidic character decreases from N₂O₃ to Bi₂O₃.
(b) All the P-Cl bonds in PCl₅ are not equivalent.
(c) HF is a weaker acid than HCl in an aqueous solution. (CBSE Al 2019)
Answer:
(a) On moving down the group, metallic character increases. Therefore acidic character of oxides decreases from nitrogen to bismuth.

(b) PCl₅ has trigonalbipyramidal geometry.

In this geometry, axial bonds are larger than equatorial bonds because of large repulsions at axial positions. Therefore, all P-Cl bonds in PCl₅ are not equivalent.

(c) The H-F bond has high bond dissociation enthalpy than H-Cl bond. Therefore, H-F bond is stronger and can be dissociated into H⁺ and F^– ions with difficulty. Hence, it is weaker acid than HCl.

Question 14.
Give reasons for the following:
(a) O-O single bond is weaker than S-S single bond.
(b) Tendency to show -3 oxidation state decreases from Nitrogen (N) to Bismuth (Bi).
(c) Cl₂ acts as a bleaching agent. (CBSE AI 2019)
Answer:
(a) O-O single bond is weaker than S-S single bond because oxygen has smaller size and has high inter-electronic repulsion.

(b) Tendency to show -3 oxidation state decreases because metallic character and size increase down the group.

(c) Cl₂ acts as a bleaching agent in the presence of moisture
Cl₂ + H₂O → 2HCl + [O]
Coloured matter + [O] → Colourless matter
The nascent oxygen is responsible for the bleaching action of Cl₂.

Question 15.
(a) Draw the structures of the following molecules:
(i) XeOF₄
(ii) H₂SO₄
(b) Write the structural difference between white phosphorus and red phosphorus. (CBSE Delhi 2014)
Answer:
(a) (i) XeOF₄

(ii) H₂SO₄

(b) White phosphorus consists of P₄ units in which four P atoms lie at the corners of a regular tetrahedron with ΔPPP = 60°.
Red phosphorus also consists of P₄ tetrahedra units, but it has polymeric structure consisting of P₄ tetrahedra linked together by covalent bond.

Question 16.
Account for the following:
(i) PCl₅ is more covalent than PCl₃.
(ii) Iron on reaction with HCl forms FeCl₂ and not FeCl₃.
(iii) The two O-O bond lengths in the ozone molecule are equal. (CBSE Delhi 2014)
Answer:
(i) In PCl₅, the oxidation state of P is +5 while in PCl₃, it is +3. As a result of higher oxidation state of the central atom, PCl₅ has larger polarising power and can polarise the chloride ion (Cl^–) to a greater extent than in the corresponding PCl₃. Since larger the polarisation, larger is the covalent character; PCl₅ is more covalent than PCl₃.

(ii) Iron reacts with HCl to form ferrous chloride with the evolution of hydrogen gas.
Fe + 2HCl → FeCl₂ + H₂

(iii) The two O-O bond lengths in ozone molecule are equal and are intermediate between single and double bonds. This is because of resonance between two structures (a) and (b) and actual molecule is resonance hybrid of these two structures:

Question 17.
Account for the following:
(a) Moist SO₂ decolourises KMnO₄ solution.
(b) In general, interhalogen compounds are more reactive than halogens (except fluorine).
(c) Ozone acts as a powerful oxidising agent. (CBSE Sample Paper 2019)
Answer:
Moist sulphur dioxide behaves as a reducing agent. It reduces MnO₄^– to Mn²⁺.
5 SO₂ + 2MnO₄^– + 2H₂O → 5 SO₄^2- + 4H⁺ + 2Mn²⁺

(b) X-X’ bond in inter halogens is weaker than X-X bond in halogens except F-F bond. This is due to the fact that the overlapping of orbitals of two dissimilar atoms is less effective than the overlapping of orbitals of similar atoms.

(c) Due to the ease with which it liberates atoms of nascent oxygen, Ozone acts as a powerful oxidising agent.
O₃ → O₂ + O

Question 18.
(a) What happens when
(i) chlorine gas is passed through a hot concentrated solution of NaOH?
(ii) sulphur dioxide gas is passed through an aqueous solution of Fe(lll) salt?
Answer:
(a) (i) Sodium chlorate and sodium chloride are formed.

(ii) Fe (III) salt is reduced to Fe (II) salt.
2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^{2- + 4H+}

(b) Answer the following:
(i) What is the basicity of H₃PO₃ and why?
(ii) Why does fluorine not play the role of a central atom in interhalogen compounds?
(iii) Why do noble gases have very low boiling points? (CBSE Delhi 2011)
Answer:
(i) H₃PO₃ is dibasic and has basicity of
two because it has two P-OH bonds which are ionisable. The third H atom is linked to P and is non-ionisable.

H₃PO₃ ⇌ HPO₃^2- + 2H⁺

(ii) Fluorine does not play the role of a central atom in interhalogen compounds because it is highly electronegative. Moreover, it has only one electron less than the octet and does not have vacant d-orbitals in its valence shell. Therefore, it can form only one bond with other halogen atoms and cannot act as central atom in interhalogen compounds.

(iii) Noble gases have very low boiling points because only weak van der Waals’ forces are present between the atoms of the noble gases in the liquid state.

Question 19.
(a) Give reasons for the following:
(i) Bond enthalpy of F₂ is lower than that of Cl₂.
(ii) PH₃ has lower boiling point than NH₃.
Answer:
(a) (i) Due to small size of F atom, there are strong repulsions between the non-bonding electrons of F atoms in the small sized F₂ molecule. Therefore, bond enthalpy of F₂ is lower than relatively larger Cl₂ molecule in which repulsions between non-bonding electrons are less.

(ii) Ammonia exists as associated molecules due to its tendency to form hydrogen bonding. Therefore, it has high boiling point. Unlike NH₃, phosphine (PH₃) molecules are not associated through hydrogen bonding in liquid state. This is because of low electronegativity of P than N. As a result, the boiling point of PH₃ is lower than that of NH₃.

(b) Draw the structures of the following molecules:
(i) BrF₃
(ii) (HPO₃)₃
(iii) XeF₄
Answer:
(i)

BrF₃ (T shape)

(ii)

(HPO₃)₃

(iii)

XeF₄

OR

(a) Account for the following:
(i) Helium is used in diving apparatus.
(ii) Fluorine does not exhibit positive oxidation state.
(iii) Oxygen shows catenation behaviour less than sulphur.
Answer:
(a) (i) Helium alongwith oxygen (helium- oxygen mixture) is used by deep sea divers in preference to nitrogen-oxygen mixture because of its low solubility in blood.

(ii) Fluorine is the most electronegative element and therefore, it shows an oxidation state of -1 only. It does not show any positive oxidation state.

(iii) Sulphur has strong tendency for catenation, i.e. forming bonds with itself in comparison to oxygen. This is because of stronger S-S single bond than O-O single bond.

(b) Draw the structures of the following molecules.
(i) XeF₂
(ii) H₂S₂O₈ (CBSE Delhi 2013)
Answer:
(i)

XeF₂ (Linear)

(ii)

Question 20.
(a) Give reasons for the following:
(i) Sulphur in vapour state shows paramagnetic behaviour.
(ii) N-N bond is weaker than P-P bond.
(iii) Ozone is thermodynamically less stable than oxygen.
Answer:
(a) (i) In the vapour state, sulphur exists as S2 molecules. S2 molecule has two unpaired electrons in anti-bonding molecular orbitals (π^x and π^y) and hence shows paramagnetism.

(ii) The N-N bond is weaker than P-P bond because of high interelectronic repulsions of non-bonding electrons due to small N-N bond length (109 pm).

(iii) Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (ΔH is negative) and increase in entropy (ΔS positive). These two factors make large negative ΔG for its conversion to oxygen.

(b) Write the name of gas released when Cu is added to
(i) dilute HNO₃ and
(ii) conc. HNO₃
Answer:
(i) With dilute HNO₃; NO (nitric oxide)
3Cu + 8HNO₃(dil) → 3Cu(NO₃)₂ + 2NO + 4H₂O

(ii) With cone. HNO₃: NO₂ (nitrogen dioxide)
Cu + 4HNO₃(conc) → Cu(NO₃)₂ + 2NO₂ + 2H₂O

OR

(a) (i) Write the disproportionation reaction of H₃PO₃.
(ii) Draw the structure of XeF₄.
Answer:
(i) 4H₃PO₃ → 3H₃PO₄ + PH₃

(ii)

(b) Account for the following:
(i) Although fluorine has less negative electron gain enthalpy, yet F₂ is strong oxidising agent.
(ii) Acidic character decreases from N₂O₃ to Bi₂O₃ in group 15.
Answer:
(i) Fluorine is the strongest oxidising agent, although it has less negative electron gain enthalpy. This is because of its smaller size, large bond dissociation enthalpy of F₂ and high exothermic hydration enthalpy of small F^– ion.

(ii) On moving down the group, the metallic character increases. Therefore, the acidic character of oxides decreases down the group from N₂O₃ to Bi₂O₃

(c) Write a chemical reaction to test sulphur dioxide gas. Write chemical equation involved. (CBSE Delhi 2019)
Answer:
SO₂ decolourises pink violet colour of acidified potassium permanganate solution.
2KMnO₄ + 5SO₂ + 2H₂O → K₂SO₄ + 2MnSO₄ + 2H₂SO₄

Question 21.
(a) Complete the following chemical reaction equations:
(i) P₄ + SO₂Cl₄ →
(ii) XeF₆ + H₄O →
Answer:
(i) P₄ + 10SO₂Cl₂ → 4PCl₅ + 10SO₂

(ii) XeF₆ + H₂O → XeOF₄ + 2HF

(b) Predict the shape and the asked angle (90° or more or less) in each of the following cases:
(i) SO₃^2- and the angle O – S – O
(ii) ClF₃ and the angle F – Cl – F
(iii) XeF₂ and the angle F – Xe – F
Answer:
(i) SO₃^2-

The O-S-O angle is more than 90°.

(ii) ClF₃

The F-Cl-F bond angle is equal to 90°.

(iii)
XeF₂

F-Xe-F bond angle is equal to 180° (greater than 90°).

OR

(a) Complete the following chemical equations:
(i) NaOH + Cl₂ →
(hot and cone.)
(ii) XeF₄ + O₂F₂ →
Answer:
(i) 6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O
(Hot and conc.)

(ii) XeF₄ + O₂F₂ → XeF₆ + O₂

(b) Draw the structures of the following molecules:
(i) H₃PO₂
(ii) H₂S₂O₇ (CBSE 2012)
Answer:
(i) H₃PO₂

(ii) H₂S₂O₇

Question 22.
(a) Draw the molecular structures of following compounds:
(i) XeF₆
(ii) H₂S₂O₈
Answer:
(i)

(ii)

(b) Explain the following observations:
(i) The molecules NH₃ and NF₃ have dipole moments which are of opposite directions.
(ii) All the bonds in PCl₅ molecule are not equivalent.
(iii) Sulphur in vapour state exhibits paramagnetism.
Answer:
(i) Both NH₃ and NF₃ have pyramidal shape with one lone pair on N atom.

The lone pair on N is in opposite direction to the N-F bond moments and therefore, it has very low dipole moment (about 0.234 D). But ammonia has high dipole moment because its lone pair is in the same direction as the N-H bond moments.

(ii) PCl₅ has trigonal bipyramidal structure in which there are three P-Cl equatorial bonds and two P-Cl axial bonds. The two axial bonds are being repelled by three bond pairs at 90° while the three equatorial bonds are being repelled by two bond pairs at 90°. Therefore, axial bonds are repelled more by bond pairs than equatorial bonds and hence are larger (219 pm).

(iii) In vapour state, sulphur partly exists as S₂ molecule and S₂ molecule like O₂ has two unpaired electrons in anti-bonding π* molecular orbitals. Therefore, it is paramagnetic.

OR

(a) Complete the following chemical equations:
(i) XeF₄ + SbF₅ →
(ii) Cl₂ + F₂ (excess ) →
Answer:
(i) XeF₄ + SbF₅ → XeF₄.SbF₅ → [XeF₃]⁺ [SbF₆]^–

(ii)

(b) Explain each of the following:
(i) Nitrogen is much less reactive than phosphorus.
(ii) The stability of +5 oxidation state decreases down group 15.
(iii) The bond angles (O – N – O) are not of the same value in NO₂^– and NO₂⁺. (CBSE Delhi 2012)
Answer:
(i) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is non-polar in character.
Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(ii) The stablity of +5 oxidation state decreases down the group because of inert pair effect. Therefore, the +5 oxidation state of Bi is less stable than that of Sb.

(iii) In NO₂, there is one electron on N while in NO₂⁺ there is no electron and in NO₂⁺, there is a lone pair of electrons on N as shown below:

NO₂⁺ molecule is linear and has bond angle of 180°. The repulsion by single electron is less as compared to repulsion by a lone pair of electrons. Therefore, bond pairs in NO₂^– are forces more closer than in NO₂.

Question 23.
(a) Draw the structure of the following molecule: H₃PO₂
Answer:
(a)

(b) Explain the following observations:
(i) Nitrogen is much less reactive than phosphorus.
(ii) Despite having greater polarity, hydrogen fluoride boils at a lower temperature than water.
(iii) Sulphur has greater tendency for catenation than oxygen in the same group.
Answer:
(i) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is nonpolar in character.
Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(ii) Though electronegativity of F is more than O, yet water forms more extensive hydrogen bonding than HF. In H₂O, each oxygen is tetrahedrally surrounded by two covalent bonds and two hydrogen bonds, (-O….H). In HF there is only one hydrogen bond (-F….H).

(iii) Sulphur has strong tendency for catenation, i.e. forming bonds with itself in comparison to oxygen. This is because of stronger S-S single bond than O-O single bond.

OR

(a) Draw the structures of the following molecules:
(i) N₂O₅
(ii) HClO₄
Answer:

(a)
(i)

(ii)

(b) Explain the following observations:
(i) H₂S is more acidic than H₂O.
(ii) Fluorine does not exhibit any positive oxidation state.
(iii) Helium forms no real chemical compound. (CBSE 2012)
Answer:
(i) The size of S is more than that of O. Therefore, the distance between S and H, i.e. S-H bond length, is more than O-H bond length. As a result, the bond dissociation enthalpy of S-H will be less and it will be easier to break the bond in H₂S than O-H bond in water. Therefore, H₂S will be more acidic than H₂O.

(ii) Fluorine is the most electronegative element and therefore, it shows oxidation state of – 1 only. It does not show any positive oxidation state.

(iii) Helium does not form compounds because it has very high ionisation enthalpy and smallest size. Its electron gain enthalpy is also almost zero.

Question 24.
Write balanced equations for the following reactions:
(a) P₄ + NaOH + H₂O → CBSE Delhi 2009)
(b) As₄ + Cl₂(excess) →
(c) P₄O₁₀ + H₂O →
(d) Ca₃P₂ + H₂O → (CBSE Delhi 2008, 2014)
(e) POCl₃ + H₂O →
(f) HgCl₂ + PH₃ → (CBSE AI 2010)
(g) Ag + PCl₅ → (CBSE AI 2014)
Answer:
(a) P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂
(b) As₄ + 10Cl₄(excess) → 4AsCl₅
(c) P₄O₁₀ + 6H₄O → 4H₃PO₄
(d) Ca₃P₂ + 6H₂O → 2PH₃ + 3 Ca(OH)₂
(e) POCl₃ + 3H₂O → H₃PO₄ + 3HCl
(f) 3HgCl₂ + 2PH₃ → Hg₃P₂ + 6HCl
(g) 2Ag + PCl₅ → 2 AgCl + PCl₃

Question 25.
Complete the following reactions
(i) NaOH + Cl₂ →
hot and conc.
(ii) NH₃ + Cl₂ (excess) → (CBSE Delhi 2017)
(iii) NaNO₂ + HCl →
(iv) F₂(g) + H₂O(l) → (CBSE Delhi 2008)
(v) K₂CO₃ + HCl →
(vi) Cl₂ + H₂O → (CBSE Delhi 2017)
(vii) F₂ + 2Cl^– → (CBSE Delhi 2017)
Answer:
(i) 6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O
hot and conc.
(ii) NH₃ + 3Cl₂ (excess) → NCl₃ + 3HCl
(iii) 2NaNO₂ + 2HCl → 2NaCl + H₂O + NO₂ + NO
(iv) 2F₂(g) + 2H₂O(l) → 4H⁺(aq) + 4F^– (aq) + O₂(g)
(v) K₂CO₃ + 2HCl → 2KCl + CO₂ + H₂O
(vi) 2Cl₂ + 2H₂O → 4HCl + O₂
(vii) F₂ + 2Cl^– → 2F^– + Cl₂

Question 26.
Complete the following reactions:
(i) XeF₄ + SbF₅ → (CBSE Delhi 2012)
(ii)

(CBSE Delhi 2012, CBSE Al 2012, 2014)
(iii) XeF₂ + H₂O →
(iv) XeF₄ + H₂O →
(v) XeF₆ + H₂O → (CBSE Delhi 2012)
(vi) XeF₆ + 2H₂O → (CfiSE Delhi 2017)
(vii) XeF₆ + 3H₂O → (CBSE Delhi 2017)
Answer:
(i) XeF₄ + SbF₅ → [XeF₃]⁺ [SbF₆]^–
(ii)

(iii) XeF₂ + H₂O → 2Xe + 4HF + O₂
(iv) XeF₄ + H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂
(v) XeF₆ + H₂O → XeOF₄ + 2HF
(vi) XeF₆ + 2H₂O → XeO₂F₂ + 4HF
(vii) XeF₆ + 3H₂O → XeO₃ + 6HF

Question 27.
(a) Account for the following:
(i) Ozone is thermodynamically unstable.
(ii) Solid PCl₅ is ionic in nature.
(iii) Fluorine forms only one oxoacid HOF.
Answer:
(a) (i) Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (ΔH is -ve) and increase in entropy ΔS is +ve). These two factors reinforce each other resulting in negative ΔG (ΔG = ΔH – TΔS) for its conversion to oxygen.

(ii) PCl₅ has trigonal bipyramidal structure and is not very stable. It splits up into more stable tetrahedral and octahedral structures which are stable as
PCl₅ ⇌ [PCl₄]⁺ [PCl₆]^–
Therefore, it exists as ionic.

(iii) Due to small size and high electronegativity, fluorine cannot act as central atom in higher oxoacids.

(b) Draw the structure of
(i) BrF₅
(ii) XeF₄
Answer:
(i)

Square Pyramidal

(ii)

OR

(i) Compare the oxidising action of F₂ and Cl₂ by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
Answer:
F₂ is stronger oxidising agent than Cl₂. This can be explained on the basis of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
The process of oxidising behaviour may be expressed as:

The overall tendency for the change (i.e. oxidising behaviour) depends upon the net effect of three steps. As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy change for this step is positive. On the other hand, energy is released in steps (II) and (III), therefore, enthalpy change for these steps is negative.

Now although fluorine has less negative electron gain enthalpy, yet it is stronger oxidising agent because of low enthalpy of dissociation and very high enthalpy of hydration. In other words, large amount of energy released in step (III) and lesser amount of energy required in step (I) overweigh the smaller energy released in step (II) for fluorine. As a result, the AH overall is more negative for F₂ than for Cl₂. Hence, F₂ is stronger oxidising agent than Cl₂.

(ii) Write the conditions to maximise the yield of H₂SO₄ by contact process.
Answer:
Conditions for maximum yield of H₂SO₄ by Contact Process:
(a) Low temperature (optimum temperature 720 K)
(b) High pressure (optimum pressure 2 bar)
(c) Presence of catalyst (V₂O₅ catalyst).

(iii) Arrange the following in the increasing order of property mentioned:
(a) H₃PO₃, H₃PO₄, H₃PO₂ (Reducing character)
(b) NH₃, PH₃, ASH₃, SbH₃, BiH₃ (Base strength) (CBSE 2016)
Answer:
(a) H₃PO₂ > H₃PO₃ > H₃PO₄
(b) NH₃ > PH₃ > ASH₃ > SbH₃ > BiH₃

Question 28.
(a) Account for the following:
(i) Acidic character increases from HF to HI.
(ii) There is large difference between the melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
Answer:
(a) (i) In gaseous state, hydrogen halides are covalent. But in aqueous solution, they ionise and behave as acids. The acidic strength of these acids decreases in the order:
HI > HBr > HCl > HF
Thus, HF is the weakest acid and HI is the strongest acid among these hydrogen halides.

The above order of acidic strength is reverse of that expected on the basis of electronegativity. Fluorine is the most electronegative halogen, therefore, the electronegativity difference will be maximum in HF and should decrease gradually as we move towards iodine through chlorine and bromine.

Thus, HF should be most ionic in nature and consequently it should be strongest acid. Although many factors contribute towards the relative acidic strengths, the major factor is the bond dissociation energy. The bond dissociation energy decreases from HF to HI so that HF has maximum bond dissociation energy and HI has the lowest value.

Since H-I bond is weakest, it can be dissociated into H⁺ and I^– ions readily while HF can be dissociated with maximum difficulty. Thus, HI is the strongest acid while HF is the weakest acid among the hydrogen halides.

(ii) Oxygen molecule is held by weak van der Waals forces because of the small size and high electronegativity of oxygen. On the other hand, sulphur molecules do not exist as S₂ but form polyatomic molecules having eight atoms per molecule (S₈) linked by single bonds. Therefore, S atoms are strongly held together by intermolecular forces and its melting point is higher than that of oxygen. Hence, there is large difference in melting and boiling points of oxygen and sulphur.

(iii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N ≡ N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation enthalpy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(b) Draw the structures of the following:
(i) CIF₃
(ii) XeF₄
Answer:
(i)

T – Shaped molecule

(ii)

Square planar molecule

OR

(i) Which allotrope of phosphorus is more reactive and why?
Answer:
White phosphorus is most reactive of all the allotropes because it is unstable due to angular strain on P₄ molecule with bond angle of 60°.

(ii) How are the supersonic jet aeroplanes responsible for the depletion of ozone layers?
Answer:
Nitrogen oxide emitted from exhausts of supersonic jet aeroplanes readily combines with ozone to form nitrogen dioxide and diatomic oxygen. Since supersonic jets fly in the stratosphere near ozone layer, they are responsible for the depletion of ozone layer.

(iii) F₂ has lower bond dissociation enthalpy than Cl₂. Why?
Answer:
Because of small size of fluorineatoms, there are strong electron-electron repulsions between the lone pairs of electrons on F atoms. Hence, bond dissociation enthalpy of F₂ is lower than that of Cl₂.

(iv) Which noble gas is used in filling balloons for meteorological observations?
Answer:
Helium is used for filling balloons for meteorological observations because it is non-inflammable.

(v) Complete the equation: (CBSE 2015)
XeF₂ + PF₅ →
Answer:
XeF₂ + PF₅ → [XeF]⁺ [PF₆]^–

Question 29.
(a) Account for the following:
(i) Interhalogens are more reactive than pure halogens.
(ii) N₂ is less reactive at room temperature.
(iii) Reducing character increases from NH₃ to BiH₃.
Answer:
(i) Interhalogen compounds are more reactive than component halogens.
This is because covalent bond between dissimilar atoms in interhalogen compounds is polar and weaker than between similar atoms in halogens (except F-F). This is due to the fact that the overlapping of orbitals of dissimilar atoms is less effective than the overlapping of orbitals of similar atoms.

(ii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N ≡ N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(iii) The reducing character of hydrides of group 15 depends upon the stability of the hydride. On going down the group, the size of the central atom increases and therefore, its tendency to form stable covalent bond with small hydrogen atom decreases.

The greater unstability of a hydride, the greater is its reducing character. Since the stability of the group 15 hydrides decreases from NH₃ to BiH₃, hence reducing character increases.

(b) Draw the structures of the following:
(i) H₄P₂O₇ (Pyrophosphoric acid)
(ii) XeF₄
Answer:
(i)

(ii)

OR

(a) Which poisonous gas is evolved when white phosphorus is heated with conc. NaOH solution? Write the chemical equation involved.
Answer:
Phosphine, PH₃
P₄ + 3NaOH + 3H₂O → 3NaH₂PO₃ + PH₃

(b) Which noble gas has the lowest boiling point?
Answer:
Helium

(c) Fluorine is a stronger oxidising agent than chlorine. Why?
Answer:
Fluorine has lower bond dissociation enthalpy of F-F bond than Cl-Cl bond of Cl₂ and high enthalpy of hydration because of smaller size of F ion. As a result it has greater tendency to accept electron in solution and is stronger oxidising agent than chlorine.

(d) What happens when H₃PO₃ is heated?
Answer:

(e) Complete the equation:
PbS + O₃ → (CBSE 2015)
Answer:
PbS + 4O₃ → PbSO₄ + 4O₂

Question 30.
(a) Account for the following observations:
(i) SF₄ is easily hydrolysed whereas SF₆ is not easily hydrolysed.
(ii) Chlorine water is a powerful bleaching agent.
(iii) Bi(V) is a stronger oxidising agent than Sb(V).
Answer:
(a) (f) S atom in SF₄ is not sterically protected as it is surrounded by only four F atoms, so attack of water molecules can take place easily. In contrast, S atom in SF₆ is protected by six F atoms. Thus attack by water molecules cannot take place easily.

(ii) Chlorine water produces nascent oxygen (causes oxidation) which is responsible for bleaching action.
Cl₂ + H₂O → 2HCl + O

(iii) Due to inert pair effect Bi(V) can accept a pair of electrons to form more stable Bi (III) (+3 oxidation state of Bi is more stable than its +5 oxidation state).

(b) What happens when
(i) White phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂.
(ii) XeF₆ undergoes partial hydrolysis.
(Give the chemical equations involved).
Answer:
(i) Phosphorus undergoes disproportionation reaction to form phosphine gas.
P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂

(ii) On partial hydrolysis, XeF6 gives oxyfluoride XeOF4 and HF.
XeF₆ + H₂O → XeOF₄ + 2HF

OR

(a) What inspired N.Bartlett for carrying out reaction between Xe and PtF₆?
Answer:
N. Bartlett first prepared a red compound O₂⁺PtF₆^–. He then realised that the first ionisation enthalpy of molecular oxygen was almost identical with Xenon. So he carried out reaction between Xe and PtF₆.

(b) Arrange the following in the order of property indicated against each set:
(i) F₂, I₂, Br₂, Cl₂ (increasing bond dissociation enthalpy)
(ii) NH₃, ASH₃, SbH₃, BiH₃, PH₃ (decreasing base strength)
Answer:
(i) I₂ < F₂ < Br₂ < Cl₂
(ii) NH₃ > PH₃ > ASH₃ > SbH₃ > BiH₃

(c) Complete the following equations:
(i) Cl₂ + NaOH (cold and dilute) →
(ii) Fe³⁺ + SO₂ + H₂O →
(CBSE Sample Paper 2018)
Answer:
(i) 2NaOH + Cl₂ → NaCl + NaOCl + H₂O
(ii) 2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^2- + 4H⁺

Question 31.
(a) Give reasons:
(i) H₃PO₃ undergoes disproportionation reaction but H₃PO₄ does not.
(ii) When Cl₂ reacts with excess of F₂, ClF₃ is formed and not FCl₃.
(iii) Dioxygen is a gas while Sulphur is a solid at room temperature.
Answer:
(i) In +3 oxidation state phosphorus tends to disproportionate to higher and lower oxidation states / Oxidation state of P in H₃PO₃ is +3 so it undergoes disproportionation but in H₃PO₄ it is +5 which is the highest oxidation state.

(ii) F cannot show positive oxidation state as it has highest electronegativity/ Because Fluorine cannot expand its covalency / As Fluorine is a small sized atom, it cannot pack three large sized Cl atoms around it.

(iii) Oxygen has multiple bonding due to pπ-pπ bonding whereas sulphur shows catenation. Oxygen is diatomic therefore held by weak intermolecular force while sulphur is polyatomic and held by strong intermolecular forces.

(b) Draw the structures of the following:
(i) XeF₄
(ii) HClO₃
Answer:
(i)

(ii)

OR

(a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas Intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into a colourless solid (B).
(i) Identify (A) and (B).
(ii) Write the structures of (A) and (B).
(iii) Why does gas (A) change to solid on cooling?
Answer:
(i) A = NO₂, B = N₂O₄
(ii)

(iii) NO₂ contains an odd electron. So it dimerises to give N₂O₄.

(b) Arrange the following In the decreasing order of their reducing character:
HF, HCl, HBr, HI
Answer:
HI > HBr > HCl > HF

Complete the following reaction:
XeF₄ + SbF₅ →, (CBSE 2018)
Answer:
(C) XeF₄ + SbF₅ → [XeF₃] [SbF₆]

Question 32.
(i) What happens when
(a) chlorine gas reacts with cold and dilute solution of NaOH?
(b) XeF₂ undergoes hydrolysis?
Answer:
(a) 2NaOH + Cl₂ → NaCl + NaOCl + H₂O
(cold and dilute)
(b) 2XeF₂(s) + 2H₂O (l) → 2Xe (g) + 4HF(aq) + O₂(S)

(ii) Assign suitable reasons for the following:
(a) SF₆ Is inert towards hydrolysis.
(b) H₃PO₃ is diprotic.
(C) Out of noble gases only Xenon is known to form established chemical compounds.
Answer:
(a) Sulphur is sterically protected by six F atoms, hence does not allow the water molecules to attack.

(b) It contains only two ionisable H-atoms which are present as -OH groups, thus behaves as dibasic acid.

(c) Xe has least ionisation energy among the noble gases and hence it forms chemical compounds particularly with O₂ and F₂.

OR

(i) Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F₂ and Cl₂.
Answer:
F₂ is stronger oxidising agent than Cl₂. This can be explained on the basis of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
The process of oxidising behaviour may be expressed as:

The overall tendency for the change (i.e. oxidising behaviour) depends upon the net effect of three steps. As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy change for this step is positive. On the other hand, energy is released in steps (II) and (III), therefore, enthalpy change for these steps is negative.

Now although fluorine has less negative electron gain enthalpy, yet it is stronger oxidising agent because of low enthalpy of dissociation and very high enthalpy of hydration. In other words, large amount of energy released in step (III) and lesser amount of energy required in step (I) overweigh the smaller energy released in step (II) for fluorine. As a result, the AH overall is more negative for F₂ than for Cl₂.
Hence, F₂ is stronger oxidising agent than Cl₂.

(ii) Complete the following reactions :
(a) Cu + HNO₃(dilute) →
(b) Fe³⁺ + SO₂ + H₂O →
(c) XeF₄ + O₂F₂ → (CBSE 2018)
Answer:
(a) 3Cu + 8 HNO₃ (dilute) → 3Cu(NO₃)₂ + 2NO + 4H₃O
(b) 2Fe³⁺ + SO₃ + 2H₃O → 2 Fe²⁺ + SO₄^2- + 4H⁺
(c) XeF₄ + O₂F₂ → XeF₆ + O₂

Question 33.
A crystalline solid ‘A’ burns in air to form a gas ‘B’ which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified KMnO₄ (aq.) solution and reduces Fe³⁺ to Fe²⁺. Identify ‘A’ and ‘B’ and write the reactions involved. (CBSE 2019C)
Answer:
A = S₈ / Sulphur
S₈ + 8 O₂ → 8SO₂ / S + O₂ → SO₂
B = SO₂

Decolourises KMnO₄
2KMnO₄ + 5 SO₂ + 2H₄O → 2H₂SO₄ + 2MnSO₄ + K₂SO₄ / 2MnO₄^– + 5SO₂ + 2H₂O → 4H⁺ + 2Mn²⁺ + 5SO₄^2-
Reduces Fe³⁺ to Fe²⁺
2Fe³⁺ + SO₂ + 2 H₂O → 2 Fe²⁺ + SO₄^2- + 4H⁺

OR

Answer the following:
(a) Arrange the following hydrides of Group 16 elements in the decreasing order of their acidic strength:
H₂O, H₂S, H₂Se, H₂Te
Answer:
H₂Te > H₂Se > H₂S > H₂O

(b) Which one of PCl₄⁺ and PCl₄^– is not likely to exist and why?
Answer:
PCl^4- is not likely to exist because lone pair on P in PCl₃ can be donated to Cl⁺ and not to Cl^–. Phosphorus has 10e^– which cannot be accommodated in sp³ orbitals.

(c) Which aliotrope of sulphur is thermally stable at room temperature?
Answer:
Rhombic sulphur is thermally stable at room temperature. Its melting point is 385.8 K. All other varieties of sulphur change into this form on standing. It has low thermal and electrical conductivity.

(d) Write the formula of a compound of phosphorus which is obtained when cone. HNO₃ oxidises P₄.
Answer:
HNO₃ oxidises phosphorus to phosphoric acid. H₃PO₄ is formed
P₄ + 2OHNO₃ → 4H₃PO₄ + 2ONO₂ + 4H₂O

(e) Why does PCl₃ fume in moisture?
Answer:
PCl₃ fumes in moist air and reacts with water violently to form phosphorus acid. It hydrolyses in the presence of moisture to give fumes of HCl.
PCl₃ + 3H₂O → H₃PO₃ + 3HCl

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 6 Electromagnetic Induction. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 6 Important Extra Questions Electromagnetic Induction

Electromagnetic Induction Important Extra Questions Very Short Answer Type

Question 1.
What is the function of a step-up transformer? (CBSE AI 2011C)
Answer:
The function of a step-up transformer is to step-up the alternating voltage.

Question 2.
State Lenz’s law. (CBSE AI 2012C)
Answer:
It states that the direction of induced emf is such that it opposes the cause of its production.

Question 3.
How can the self-inductance of a given coil having ‘N’ number of turns, area of cross-section of ‘A’ and length T be increased? (CBSE AI 2012C)
Answer:
By inserting a core of high permeability inside the coil.

Question 4.
How does the mutual inductance of a pair of coils change when
(a) the distance between the coils is increased and
(b) the number of turns in the coils is increased? (CBSE AI 2013)
Answer:
(a) decreases
(b) increases.

Question 5.
The motion of the copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping? (CBSE AI 2013)
Answer:
Production of eddy current.

Question 6.
Why is the core of a transformer laminated? (CBSE Delhi 2013C)
Answer:
To reduce the effects of eddy currents.

Question 7.
A metallic piece gets hot when surrounded by a coil carrying a high-frequency alternating current. Why? (CBSE Delhi 2014C)
Answer:
Due to the production of eddy current which generates heat.

Question 8.
Name any two applications where eddy currents are used to advantage. (CBSE Delhi 2016C)
Answer:

1. Electromagnetic damping
2. Induction furnace.

Question 9.
A long straight current-carrying wire passes normally through the centre of the circular loop. If the current through the wire increases, will there be an induced emf in the loop? Justify. (CBSE Delhi 2017)
Answer:
Yes, as there will be a change in magnetic flux.

Question 10.
Predict the polarity of the capacitor in the situation described below. (CBSE AI 2017)

Answer:
The upper plate will be positive with respect to the lower plate in the capacitor.

Question 11.
In the figure given, mark the polarity of plates A and B of a capacitor when the magnets are quickly moved towards the coil. (CBSE AI 2017C)

Answer:
Plate A will be positive with respect to plate B in the capacitor.

Question 12.
A long straight current-carrying wire passes normally through the centre of the circular loop. If the current through the wire increases, will there be an induced emf in the loop? Justify. (CBSE Delhi 2017)
Answer:
Yes, as there will be a change in magnetic flux.

Question 13.
An air-cored solenoid has self-inductance 2.8 H. When the core is removed, the self-inductance becomes 2 mH. What is the relative permeability of the core used? (CBSE Delhi 2017C)
Answer:
μ_r = 2.8 / 2 × 10^-3 = 1.4 × 10³

Question 14.
A choke and a bulb are in series to a dc source. The bulb shines brightly. How does its brightness change when an iron core is inserted inside the choke coil?
Answer:
There is no change in the brightness as the inductive reactance is zero for dc.

Question 15.
Why does the acceleration of a magnet falling through a long solenoid decrease?
Answer:
It decreases because of the opposing induced emf produced in the soLenoid due to the rate of change of magnetic flux.

Question 16.
Why is the core of a transformer laminated?
Answer:
It is done to reduce the effect of Eddy Currents.

Question 17.
A vertical metallic pole falls down through the plane of the magnetic meridian. Will any emf be Induced between Its ends?
Answer:
No emf will be induced because the pote neither intercepts the vertical component nor the horizontal component of the earth’s magnetic field.

Question 18.
A magnet Is moved towards a coil and an electric charge is induced in it. If the resistance of the coil is increased, how will the induced charge change?
Answer:
On increasing the resistance of the colt, the magnitude of induced charge decreases.

Question 19.
Can a transformer be used In a dc circuit?
Answer:
No, because there is no change in magnetic flux.

Question 20.
Why does a metallic piece become very hot when it is surrounded by a coil carrying high-frequency alternating current?
Answer:
The high-frequency coil induces eddy currents in the metallic piece. These eddy currents produce heat hence the metalLic piece becomes hot.

Question 21.
The figure shows a horizontal solenoid PQ connected to a battery and a switch. A copper ring R is placed on a frictionless track, the axis of the ring being along the axis of the solenoid. What would happen to the ring as the switch S is closed.

Answer:
The ring WilL be repelled due to opposing induced emf produced in it.

Question 22.
An air-core solenoid is connected to an ac source and a bulb. If an iron core Is inserted in the solenoid, how does the brightness of the bulb change? Give reasons for your answer.
Answer:
Insertion of an iron core In the solenoid increases its inductance. This in turn increases the value of inductive reactance. This decreases the current and hence the brightness of the bulb.

Question 23.
A magnet is moved in the direction indicated by an arrow between two coils AB and CD as shown in the figure. Suggest the direction of current in each coil,

Answer:
In coil AB induced current flows from A to B and in colt CD current flows from C to D.

Question 24.
Predict the directions of induced currents in metal rings 1 and 2 lying in the same place where current I in the wire is increasing steadily. (CBSE Delhi 2012)

Answer:
1 -clockwise, 2-anticlockwise.

Question 25.
The electric current flowing in a wire In the direction from B to A is decreasing. Find out the direction of the induced current in the metallic loop kept above the wire as shown. (CBSE AI 2014)
Answer:
Clockwise.

Question 26.
Predict the polarity of plate A of the capacitor, when a magnet is moved towards it, as is shown In the figure. (CBSE AI 2014C)

Answer:
Positive.

Question 27.
The figure below shows two positions of a loop PQR in a perpendicular uniform magnetic field. In which position of the coil is there induced emf?

Answer:
In position b.

Question 28.
If the self-inductance of an air-core inductor increases from 0.01 mH to 10 mH on introducing an iron core into it, what is the relative permeability of the core used?
Answer:
We know that µ = \(\frac{L}{L_{0}}=\frac{10}{0.01}\) = 1000

Electromagnetic Induction Important Extra Questions Short Answer Type

Question 1.
An induced current has no direction of its own, comment.
Answer:
Yes, it is perfectly correct to say that an induced current has no fixed direction of its own. The direction of induced current depends upon the change in magnetic flux because in accordance with Lenz’s law the induced current always opposes the change in magnetic flux.

Question 2.
How are eddy currents produced? Mention two applications of eddy currents?
Answer:
Eddy currents are the currents induced in the body of a thick conductor when the magnetic flux linked with the conductor changes. When a thick conductor is moved in a magnetic field, magnetic flux linked with it changes. In situations like these, we can have induced currents that circulate throughout the volume of a material.

Because their flow patterns resemble swirling eddies in a river, therefore they are called eddy currents.

• Electromagnetic braking, and
• Induction furnace.

Question 3.
Name and define the unit used for measuring the coefficient of mutual inductance. State the relation of this unit with the units of magnetic flux and electric current.
Answer:
In SI the unit of mutual inductance is henry (H). Now from the expression
ε = – \(\frac{d \phi}{d t}\) = – M \(\frac{d l}{d t}\)
we have M = ε l \(\frac{d l}{d t}\).

Let ε = 1 volt and dl/dt = 1 As^-1, then
M = 1 volt/1 As^-1 = 1 henry.

The mutual-inductance of a coil is said to be 1 henry if a rate of change of current of 1 ampere per sec in the neighbouring coil induces in at an emf of 1 volt.

Question 4.
What are eddy currents? Write any two applications of eddy currents. (CBSE A! 2011)
Answer:
Eddy currents are the currents induced in the body of a thick conductor when the magnetic flux linked with a bulk piece of conductor changes.

1. Dead Beat Galvanometer, and
2. Induction furnace.

Question 5.
(a) Obtain the expression for the magnetic energy stored in a solenoid in terms of the magnetic field B, area A and length l of the solenoid.
(b) How is this magnetic energy per unit volume compared with the electrostatic energy per unit volume stored in a parallel plate capacitor? (CBSE Delhi 2011C)
Answer:
The magnetic field stored in a solenoid is given by the expression U = – \(\frac{1}{2}\)Ll².

But for a solenoid B = μ₀nl
or
l = B / μ₀ n

Substituting in the above expression we have
U = \(\frac{1}{2}\) × (μ₀n²Al)\(\left(\frac{B}{\mu_{0} n}\right)^{2}\) as L = μ₀ n² A l

U = \(\frac{1}{2}\)\(\frac{B^{2} A l}{\mu_{0}}\)

We know that the energy stored per unit volume in a parallel plate capacitor is
U_E = \(\frac{1}{2}\)ε₀E²

It is clear that in both cases the energy stored per unit volume is proportional to the square of the field intensity.

Question 6.
State Lenz’s Law.
A metallic rod held horizontally along the east-west direction is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer. (CBSE Delhi 2013)
Answer:
Lenz’s law states that the polarity of the induced emf is such that it tends to oppose the cause of its production.

Yes, as it will cut the horizontal component of the earth’s magnetic field.

Question 7.
Starting from the expression for the energy W = 1/2Ll², stored in a solenoid of self-inductance L to build up the current l, obtain the expression for the magnetic energy in terms of the magnetic field B, area A and length l of the solenoid having n number of turns per unit length. Hence show that the energy density is given by 8z/2m0. (CBSE Delhi 2013C)
(i) The magnetic energy is
U = \(\frac{1}{2}\)LI² = \(\frac{1}{2}\)L\(\left(\frac{B}{\mu_{0} n}\right)^{2}\) since B = μ₀nl

Now L = μ₀n² Al, therefore we have
U_B = \(\frac{1}{2}\)(μ0n2 Al)\(\left(\frac{B}{\mu_{0} n}\right)^{2}\) = \(\frac{1}{2 \mu_{0}} B^{2} A l\)

(ii) The magnetic energy per unit volume is
U_B = \(\frac{U_{B}}{V}=\frac{U_{B}}{A l}=\frac{B^{2}}{2 \mu_{0}}\)

Question 8.
Define mutual inductance. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? (CBSE Delhi 2016)
Answer:
Mutual inductance is numerically equal to the magnetic flux linked with a coil when the unit current passes through the neighbouring coil.
Given M = 1.5 H, dl = 20 – 0 = 20 A,
dt = 0.5 s, Φ = ?
Φ = – M\(\frac{d l}{d t}\)
or
Φ = – 1.5 × \(\frac{20}{0.5}\) = – 60 Wb

Question 9.
Explain the principle on which the metal detector is used at airports for security reasons.
Answer:
The metal detectors used in airport security checkpoints operate by detecting eddy currents induced in metallic objects. The detector generates an alternating magnetic field. This induces eddy currents in the conduction object carried through the detector. The eddy currents in turn produce an alternating magnetic field. This field induces a current in the detectors receiver coil.

Question 10.
A current is induced in coil C₁, due to the motion of current-carrying coil C₂.
(i) Write any two ways by which a large deflection can be obtained in the galvanometer G.
(ii) Suggest an alternative device to demonstrate the induced current in place of a galvanometer. (CBSE Delhi 2011)

Answer:
(ii) The two ways are
(a) Passing a large current through coil C₂ and
(b) Moving coil C₂ quickly towards the coil.
(ii) A magnetic compass can be placed at the centre of coil C₁. Whenever current will be induced it will show a deflection.

Question 11.
Consider a cube EFGHIJKL of side ‘a’ placed in a magnetic field B acting perpendicular to the face FJKG as shown in the figure. Write magnetic flux through the following faces (a) EFGH (b) EFJI (c) EILH (d) FJKG

Answer:
The magnetic flux depends upon the angle (q) between the magnetic field and area vector.
(a) Here θ = 90°, therefore magnetic flux Φ = BA cos θ = BA cos 90° = 0
(b) Here θ = 90°, therefore magnetic flux Φ = BA cos θ = BA cos 90° = 0
(c) Here θ = 180°, therefore magnetic flux Φ = BA cos 180° = -BA
(d) Here θ = 0°, therefore magnetic flux Φ = BA cos 0° = BA

Question 12.
A metallic rod of ‘L’ length is rotated with an angular frequency of ‘ω’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring. (CBSE Delhi 2012, 2013)
Answer:
Let the rod move from OP to OQ through a small sector of angle θ.

The small area covered by the rod is
dA= πL² × \(\frac{\theta}{2 \pi}=\frac{1}{2}\)L²θ

where L is the radius of the circle in which the rod rotates. Hence the induced emf is
ε = \(\frac{d \phi}{d t}=\frac{d B A}{d t}=\frac{B d A}{d t}=B \frac{d}{d t}\left[\frac{1}{2} L^{2} \theta\right]\)

ε = – \(\frac{1}{2}\)L2B\(\frac{d \theta}{d t}=\frac{1}{2}\)BωL²

Question 13.
A metallic rod of length l length is rotated with a frequency ω, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius R = l, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf induced in the rod. If r is the resistance of the rod and the metallic ring has negligible resistance, obtain the expression for the power generated. (CBSE AI 2013C)
Answer:
For the first part refer above question.
The power generated is given by
P = \(\frac{\varepsilon^{2}}{r}=\frac{B^{2} \omega^{2} l^{4}}{2 r}\)

Question 14.
A rectangular loop PQMN with movable ‘ arm PQ of length 10 cm and resistance 4 Ω is placed in a uniform magnetic field of 0.25 T acting perpendicular to the plane of the loop as is shown in the figure. The resistances of the arms MN, NP and MQare negligible.
Calculate the
(i) emf induced in the arm PQand
(ii) current induced in the loop when arm PQ is moved with velocity 20 m s^-1. (CBSE Delhi 2014C)

Answer:
Given L = 10 cm = 0.1 m, B = 0.25 T, v = 20 m s^-1
(a) ε = BLv = 0.25 × 0.1 × 20 = 0.5 V
(b) l = ε/R = 0.5/4 = 0.125 A

Question 15.
In the given diagram a coil B is connected to a low voltage bulb L and placed parallel to another coil A as shown. Explain the following observations
(i) Bulb lights, and
(ii) Bulb gets dimmer if coil B is moved upwards.

Answer:
(a) When ac is applied across coil A an induced emf is produced in coil B due to mutual induction between the two coils. This makes the lamp light up.
(b) When coil B is moved upwards the mutual induction and hence induced emf in coil B decreases. This makes the lamp dimmer.

Question 16.
A horizontal straight wire of length L extending from east to west is falling with speed v at right angles to the horizontal component of Earth’s magnetic field B.
(a) Write the expression for the instantaneous value of the emf induced in the wire.
Answer:
ε = BLv

(b) What is the direction of the emf?
Answer:
west to east

(c) Which end of the wire is at the higher potential? (CBSEAI2011)
Answer:
East.

Question 17.
Two concentric circular coils one of small radius r₁ and the other of large radius r₂, such that r₁ << r₂, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. (NCERT)
Answer:
Let a current l₂ flow through the outer circular coil. The field at the centre of the coil is B₂ = \(\frac{\mu_{0} I_{2}}{2 r_{2}}\). Since the other coaxially placed coil has a very small radius, B2 may be considered constant over its cross¬sectional area. Hence,
Φ = πr₁²B2 = \(\frac{\mu_{0} \pi r_{1}^{2} l_{2}}{2 r_{2}}\) = M₁₂l₂

Thus M₁₂ = \(\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}}\)

But M₁₂ = M₂₁, therefore M₁₂ = M₂₁ = \(\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}}\)

Note that we calculated M₁₂ from an approximate value of Φ₁ assuming the magnetic field B₂ to be uniform over the area πr₁². However, we can accept this value because r₁ << r₂.

Question 18.
Consider a magnet surrounded by a wire with an on/off switch S (figure). If the switch is thrown from the off position (open circuit) to the on position (closed circuit), will a current flow in the circuit? Explain. (NCERT Exemplar)

Answer:
There is no relative motion between the magnet and the coil. This means that there is no change in magnetic flux, hence no electromotive force is produced and hence no current will flow in the circuit.

Question 19.
A wire in the form of a tightly wound solenoid is connected to a DC source and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil, will the current increase or decrease? Explain. (NCERT Exemplar)
Answer:
The current will increase. As the wires are pulled apart the flux will leak through the gaps. Lenz’s law demands that induced e.m.f. resist this decrease, which can be done by an increase in current.

Question 20.
A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain. (NCERT Exemplar)
Answer:
The current will decrease. As the iron core is inserted in the solenoid, the magnetic field increases and the flux increases. Lenz’s law implies that induced e.m.f. should resist this increase, which can be achieved by a decrease in current.

Electromagnetic Induction Important Extra Questions Long Answer Type

Question 1.
11 kW of electric power can be transmitted to a distant station at (i) 220 V or (ii) 22,000 V. Which of the two modes of transmission should be preferred and why? Support your answer with possible calculations.
Answer:
1. Consider that 11000 watt of energy has to be transmitted. First at 220 V and then at 22000 V. When the power is transmitted at 220 V then the current flowing through the wires is 11000/220 = 50 A

2. When power is transmitted at 22000 V then the current through the wires is 11000 / 22000 = 0.5 A. If R is the resistance of the line wire then the energy dissipated in the two cases is 2500R joule per sec and 0.25R. joule per sec.

This shows that if energy is transmitted at low voltages there is more loss in energy than when it is transmitted at high voltages. Furthermore, if power is to be transmitted at low voltage then the resistance of the line wire should be low, as such thick wires will be required. To support these thick wires strong poles situated close to each other will be needed. This will increase the cost of transmission. But at high voltages, even thin wires will do.

Question 2.
A coil A is connected to a voltmeter V and the other coil B to an alternating current source D. If a large copper sheet C is placed between the two coils, how does the induced emf in coil A change due to current in coil B. Justify your answer.

Answer:
In the absence of sheet C, an induced emf is set up in coil due to mutual induction phenomenon when an alternating current is passed through coil B.

However, when induced copper sheet C is placed, eddy currents are set up in the sheet due to a change in flux.

Thus, now coil A has a positive effect due to coil B and a negative effect due to eddy currents in C. Consequently, the flux of coil A and hence the induced emf in coil A is decreased, i.e. the reading of voltmeter V is reduced.

Question 3.
A bar magnet is dropped so that it falls vertically through coil C. The graph obtained for the voltage produced across the coil versus time is as shown in figure (b).
(i) Explain the shape of the graph and
(ii) why is the negative peak longer than the positive peak?

Answer:
(a) As the magnet approaches the coil, an emf is induced in it. As the magnet approaches the coil the magnetic flux linked with the coil increases. As a result the induced emf increases. When the magnet enters the coil, the change in magnetic flux linked with the coil begins to decrease and becomes zero when the magnet is completely inside the coil. This starts decreasing the emf and makes it zero.

When the magnet comes out of the coil the direction of induced emf changes direction and begins to increase in the opposite direction. When the magnet moves far away from the coil the induced emf becomes zero.

As the magnet comes out of the coil with a speed greater than the speed at which it approaches the coil, therefore the induced emf is more in the second case. Hence the longer negative peak.

Question 4.
(a) State Lenz’s law: Using this law indicate the direction of the current in a closed loop when a bar magnet with the North Pole is brought close to it. Explain briefly how the direction of the current predicted wrongly results in the violation of the law of conservation of energy.
(b) A rectangular loop and circular loop are moving out of a uniform magnetic field region with a constant velocity v as shown in the figure. In which loop do you expect the induced emf to be constant during the passage out of the field region? The field is normal to the loops. (CBSE AI 2011C)

Answer:
(a) Lenz’s law states that the direction of the induced emf is such that it opposes the cause of its production. When the north pole of the magnet is brought near the coil, the upper end of the coil will acquire north polarity so as to oppose the approaching North Pole (by repelling). This means that the direction of current must be anticlockwise as seen from the side of the magnet.

If the current is wrongly predicted as clockwise, then the upper face will acquire south polarity which will attract the North Pole. This means the current is being produced without doing any work. This leads to the violation of the law of conservation of energy.

(b) It is expected that the induced emf will be constant in the rectangular coil. In the case of the rectangular coil, when pulled out of the magnetic field, the rate of change of magnetic flux will be constant because the rate of change of area is constant. This is not so in the case of the circular coil.

Question 5.
The current through two inductors of self-inductance 15 mH and 25 mH is increasing with time at the same rate. Draw graphs showing the variation of the
(a) emf induced with the rate of change of current
(b) energy stored in each inductor with the current flowing through it. Compare the energy stored in the coils, if the power dissipated in the coils is the same. (CBSE AI 2017C)
Answer:
Given L₁ = 15 mH, L₂ = 25 mH.
(a) The emf induced across an inductor is given by the expression ε = – L\(\frac{d i}{d t}\) since di/dt is the same for both coils therefore induced emf will depend upon the value of inductance. Hence the graphs are as shown

(b) The energy stored in an inductor is given by U = \(\frac{1}{2}\)Ll²

(c) The power dissipated P = l₂X_L
or
l₂ = Pl XL and U = \(\frac{1}{2}\)Ll¹ = \(\frac{1}{2} \frac{L P}{X_{L}}=\frac{1}{2} \frac{P L}{\omega L}=\frac{P}{2 \omega}\)

Since power dissipated is same and co is also same therefore energy stored in the coils will also be the same.

Question 6.
A rectangular frame of wire is placed in a uniform magnetic field directed outwards, normal to the paper. AB is connected to a spring which is stretched to AB and then released at time t = 0. Explain qualitatively how induced e.m.f. in the coil would vary with time. (Neglect damping of oscillations of spring) (CBSE AI 2018C)

Answer:
As the spring is released AB is pulled out of the field. This increases the area of the loop inside the magnetic field. This increases flux and hence an induced emf is produced. The portion AB does not stop at Ab but moves outwards. Now the spring will push AB inwards. This will decrease the area of the loop thereby decreasing the induced emf. This continues and hence the emf increases and decreases periodically.

Question 7.
Define mutual inductance between a pair of coils. Derive an expression for the mutual inductance of two long coaxial solenoids of the same length wound one over the other. (CBSE AI 2017)
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil.

As shown in the figure consider two long co-axial solenoids S₁ and S₂ with S₂ wound over S₁.

Let l = length of each solenoid r₁, r₂ = radii of the two solenoids

A = πr₁² = area of cross-section of inner solenoid S₁
N₁, N₂ = number of turns in the two solenoids First we pass a time-varying current l₂ through S₂. The magnet field set up inside S₂ due to l₂ is
B₂ = μ_o n₂ l₂

where n₂ = N₂/l = the number of turns per unit length of S₂.

Total magnetic flux linked with the inner solenoid S₁ is
Φ₁ = B₂AN₁ = μ0 n₂ l₂ × AN₁

Mutual inductance of coil 1 with respect to coil 2 is
M₁₂ = \(\frac{\phi_{1}}{l_{2}}\) = μ0n2AN1 = \(\frac{\mu_{0} A N_{1} N_{2}}{l}=\frac{\mu_{0} \pi r^{2} N_{1} N_{2}}{l}\)

The mutual inductance of any two coils is always proportional to the product N₁ N₂ of their number of turns. This is termed the reciprocity theorem.

Question 8.
Define mutual inductance and write its SI unit.
A square loop of side ‘a’ carrying a current l₂ is kept at distance x from an infinitely long straight wire carrying a current l1 as shown in the figure. Obtain the expression for the resultant force acting on the loop. (CBSE Delhi 2019)

Answer:
(a) Mutual inductance equals the magnetic flux associated with a coil when unit current flows in its neighbouring coil. (b) Force per unit length between two parallel straight conductors
\(\frac{F}{L}=\frac{\mu_{o} I_{1} I_{2}}{2 \pi r}\)

Force on the part of the loop which is parallel to infinite straight wire and at a distance x from it F = \(\frac{\mu_{o} l_{1} l_{2} a}{2 \pi x}\)(away from the infinitely 2KX long wire)

Force on the part of the loop which is at a distance (x + a) from it
F₂ = \(\frac{\mu_{0} I_{1} I_{2} a}{2 \pi(x+a)}\) (towards the infinite long wire)

Net force F = F₁ – F₂
F = \(\frac{\mu_{o} l_{1} l_{2} a^{2}}{2 \pi(x+a)}\) (away from the infinite straight line)

Question 9.
What are the possible causes of energy loss in a transformer? How are these minimised?
Answer:
The possible causes of energy losses in transformers are:
(a) Flux leakage: There is always some flux leakage. It can be reduced by winding the primary and secondary coils one over the other.
(b) Copper: The copper wires used for the windings have some resistance and hence some energy is lost due to heat produced in the wire. It can be minimised by taking thick wire.
(c) Eddy currents: The alternating magnetic flux induces eddy currents in the iron core, which results in loss of electrical energy. To minimise it we use a laminated iron core.
(d) Hysteresis loss: As the magnetisation cycle of the iron core is repeated again and again some loss of energy takes place due to magnetic hysteresis. To minimise it we prefer a soft iron core for which hysteresis loss is less.

Question 10.
Discuss how Faraday’s law of e.m.f induction is applied in an ac generator for converting mechanical energy into electrical energy. Draw graphs to show the ’phase relationship’ between the instantaneous
(a) magnetic flux (Φ) linked with the coil and
(b) induced emf (ε) in the coil. (CBSE Delhi 2016C)
Answer:
In an ac generator, the change in magnetic flux is brought about by rotating the coil in a magnetic field. According to Faraday’s law, induced emf is set up in the coil on changing the magnetic flux linked with it. Hence mechanical energy, which is supplied to rotate the coil, gets converted into electrical energy.

Let the coil be rotated with a constant angular speed, co, the angle 0 between magnetic field 8, and area vector A of the coil at any instant is θ = ωt

Therefore magnetic flux at any instant is Φ = BA cos ωt

From Faraday’s law, induced emf is
ε = – n\(\frac{d \phi}{d t}\) = -nBA\(\frac{d}{d t}\)cos ωt
ε = nBAa sin ωt

Magnetic flux is given by Φ = BA cos ωt, therefore the graph will be a cosine curve as shown

The induced emf is given by ε = nBA sin ωt, therefore the graph will be a sine curve as shown.

Question 11.
(a) Define self-inductance of a coil. Obtain an expression for the energy stored in a solenoid of self-inductance ‘L’ when the current through it grows from zero to l.
(b) A square loop MNOP of the side 20 cm is placed horizontally in a uniform magnetic field acting vertically downwards as shown in the figure. The loop is pulled with a constant velocity of 20 cm s^-1 till it goes out of the field.

(c) Depict the direction of the induced current in the loop as it goes out of the field. For how long would the current in the loop persist?
(d) Plot a graph showing the variation of magnetic flux and induced emf as a function of time. (CBSE AI 2015)
Answer:
(a) Let i be the current at some instant through a pure inductor. If di / dt is the rate of change of current through the inductor then the voltage between the terminals of the inductor at this instant is
V=L di / dt

Therefore instantaneous power in the inductor is
P = V_ab i = Li \(\frac{d i}{d t}\)

The energy dU supplied to the inductor during an infinitesimal time interval dt is
dU = P dt, so dU = Li di

The total amount of energy supplied while the current increases from zero to a final value l is
U = L ∫_o¹ i di = \(\frac{1}{2}\)L l²

(b) (i) Direction of induced current – clockwise (MNOP), the duration of induced current is 1s.
(ii) The graph is as shown.

Question 12.
(a) Define mutual inductance and write its SI units.
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil. It is measured in henry.

(b) Derive an expression for the mutual inductance of two long co-axial solenoids of the same length wound one over the other.
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil.

As shown in the figure consider two long co-axial solenoids S₁ and S₂ with S₂ wound over S₁.

Let l = length of each solenoid r₁, r₂ = radii of the two solenoids

A = πr₁² = area of cross-section of inner solenoid S₁
N₁, N₂ = number of turns in the two solenoids First we pass a time-varying current l₂ through S₂. The magnet field set up inside S₂ due to l₂ is
B₂ = μ_o n₂ l₂

where n₂ = N₂/l = the number of turns per unit length of S₂.

Total magnetic flux linked with the inner solenoid S₁ is
Φ₁ = B₂AN₁ = μ_o n₂ l₂ × AN₁

Mutual inductance of coil 1 with respect to coil 2 is
M₁₂ = \(\frac{\phi_{1}}{l_{2}}\) = μ0n2AN1 = \(\frac{\mu_{0} A N_{1} N_{2}}{l}=\frac{\mu_{0} \pi r^{2} N_{1} N_{2}}{l}\)

The mutual inductance of any two coils is always proportional to the product N1 N2 of their number of turns. This is termed the reciprocity theorem.

(c) In an experiment, two coils c₁ and c₂ are placed close to each other. Find out the expression for the emf induced in the coil c₁ due to a change in the current through the coil c₂. (CBSE Delhi 2015)
Answer:
Consider the experimental set up as shown.

When the key K is pressed in coil C₁ a magnetic flux linked with C₂ changes.

The magnetic flux linked with coil C₂ is given by
f₂ = M l₁ where M is the mutual inductance between the two coils.

But by Faraday’s flux rule we have
ε₂ = – \(\frac{d \phi}{d t}\) therefore we have
ε₂ = – \(\frac{d M l_{1}}{d t}\) = – M \(\frac{d l_{1}}{d t}\)

Question 13.
(a) When a bar magnet is pushed towards (or away) from the coil connected to a galvanometer, the pointer in the galvanometer deflects. Identify the phenomenon causing this deflection and write the factors on which the amount and direction of the deflection depend. State the laws describing this phenomenon.
(b) Sketch the change In flux, emf and force when a conducting rod PQ of resistance R and length I moves freely to and fro between A and C with speed v on a rectangular conductor placed in the uniform magnetic field as shown in the figure. (CBSE AI 2016)

Answer:
(a) Phenomenon: electromagnetic induction

Factors:

• Strength of the magnetic field of the magnet
• speed of motion of bar magnet.

Direction depends upon

• the motion of magnet whether inward or outward
• the direction of the north/south pole.

Law: The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.

(b) The sketch is as shown below.

Question 14.
What is induced emf? Write Faraday’s laws of electromagnetic induction. Express it mathematically. A conducting rod of length ‘L’ with one end pivoted is rotated with a uniform angular velocity ‘ω’ in a vertical plane, normal to a uniform magnetic field ‘B’. Deduce an expression for the emf induced in this rod.
Answer:
It is the emf induced when the magnetic flux linked with a coil changes.

Faraday put forward the following laws called Faraday’s laws of electromagnetic induction.

• Law 1. Whenever the magnetic flux linked with a coil changes an induced emf is produced.
• Law 2. The induced emf lasts as long as the change in magnetic flux continues.

The expression for induced emf in rotating rod: Let the rod move from OP to OQ through a small sector of angle θ.

The small area covered by the rod is
dA = πL² × \(\frac{\theta}{2 \pi}=\frac{1}{2}\)L2θ

where L is the radius of the circle in which the rod rotates. Hence the induced emf is
ε = \(\frac{d \phi}{d t}=\frac{d B A}{d t}=\frac{B d A}{d t}=B \frac{d}{d t}\left[\frac{1}{2} L^{2} \theta\right]\)
Or
ε = \(\frac{1}{2} L^{2} B \frac{d \theta}{d t}=\frac{1}{2}\)BωL²

Question 15.
How is the mutual inductance of a pair of coils affected when:
(a) the separation between the coils is increased?
Answer:
When the separation between the two coils is increased the mutual inductance decreases. It is because the flux linked with the secondary due to a current in the primary decreases.

(b) the number of turns of each coil is increased?
Answer:
When the number of turns of each coil is increased the mutual inductance increases because M₁₂ (= M₂₁) ∝ N₁N₂,

i.e. mutual inductance is directly proportional to the number of turns. The linkage of flux increases with an increase in the number of turns in the coils.

(c) a thin iron sheet is placed between the two coils, other factors remaining the same? Explain your answer in each case.
Answer:
When a thin iron sheet is placed between the coils the permeability of the medium between the coil increases. As mutual inductance is directly proportional to the permeability, therefore the mutual inductance increases.

Question 16.
A rectangular coil of area A, having a number of turns N, is rotated at f revolutions per second in a uniform magnetic field B, the field being perpendicular to the coil. Prove that the maximum emf induced in the coil is 2π f N B A.
Answer:
Consider an armature of the ac generator having n turns and placed in a uniform magnetic field B.

Suppose at any instant t the normal to the plane of the coil makes an angle 0 with the direction of the magnetic field. If r_o is the uniform angular velocity with which, the coil rotates then θ = ωt.

The flux through the loop equals its area A multiplied by B_⊥ = Bcos θ, the component of magnetic field B perpendicular to the area, hence

Φ = n B A cos Φ = n B A cos ω t …(1)
where is the number of turns in the armature?

By Faraday’s flux rule,
ε = – \(\frac{d \phi}{d t}=-\frac{d}{d t}\) n B A cos ω t
= – n B A \(\frac{d}{d \phi}\)cos ω t = – n B A (- ω sin ω t) ….(2)
Or
ε = n B A ω sin ω t

The induced emf is maximum when sin ωt = maximum = 1, therefore the maximum induced emf is given by
ε₀ = n B A ω sin ω t …(3)

Since ω = 2πf, therefore equation (3) becomes ε₀ = 2πf nBA

Question 17.
State Lenz’s law. Explain by giving examples that Lenz’s law is a consequence of the law of conservation of energy. (CBSE Delhi 2017C)
Answer:
It states that the direction of induced emf in a coil is such that it opposes the cause of its production. Lenz’s law is a consequence of the law of conservation of energy. To show it, let us consider a bar magnet pushed towards a conducting loop. When N-pole moves towards the loop, the face of the loop facing the North Pole develops north polarity as per Lenz’s law so as to oppose the motion of the magnet.

Again, when N-pole moves away from the loop, the nearby face develops south polarity, thus opposing the motion of the magnet away from the loop. It means that the motion of the magnet is automatically opposed every time. Hence, some work is to be done on the magnet to move it and this mechanical work is transformed into electrical energy. Thus, the conservation law of energy is followed.

Question 18.
Define the term ‘self-inductance’ and write its SI unit.
Obtain the expression for the mutual inductance of two long co-axial solenoids S₁ and S₂ wound one over the other, each of length L and radii r₁ and r₂ and n₁ and n₂ number of turns per unit length when a current I is set up in the outer solenoid S₂. (CBSE Delhi 2017)
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil.

As shown in the figure consider two long co-axial solenoids S₁ and S₂ with S₂ wound over S₁.

Let l = length of each solenoid r₁, r₂ = radii of the two solenoids

A = πr₁² = area of cross-section of inner solenoid S₁
N₁, N₂ = number of turns in the two solenoids First we pass a time-varying current l₂ through S₂. The magnet field set up inside S₂ due to l₂ is
B₂ = μ_o n₂ l₂

where n₂ = N₂/l = the number of turns per unit length of S₂.

Total magnetic flux linked with the inner solenoid S₁ is
Φ₁ = B₂AN₁ = μ_o n₂ l₂ × AN₁

Mutual inductance of coil 1 with respect to coil 2 is
M₁₂ = \(\frac{\phi_{1}}{l_{2}}\) = μ_on₂AN₁ = \(\frac{\mu_{0} A N_{1} N_{2}}{l}=\frac{\mu_{0} \pi r^{2} N_{1} N_{2}}{l}\)

The mutual inductance of any two coils is always proportional to the product N₁ N₂ of their number of turns. This is termed the reciprocity theorem.

Question 19.
Obtain the expression for the magnetic energy stored in an ideal inductor of self-inductance L when a current l passes through it. Hence obtain the expression for the energy density of magnetic field B produced in the inductor. (CBSE Delhi 2016C)
Answer:
Let i be the current at some instant through a pure inductor. If di / dt is the rate of change of current through the inductor then the voltage between the terminals of the inductor at this instant is
V=L di / dt

Therefore instantaneous power in the inductor is
P = V_ab i = Li\(\frac{d i}{d t}\)

The energy di supplied to the inductor during an infinitesimal time interval dt is
dU = P dt, so dU = Li di

The total amount of energy supplied while the current increases from zero to a final value l is

U = L ∫₀¹ di = \(\frac{1}{2}\)L l²

This gives the expression for the energy stored in an inductor.

The magnetic energy is

The magnetic energy per unit volume is
U_B = \(\frac{U_{B}}{V}=\frac{U_{B}}{A l}=\frac{B^{2}}{2 \mu_{0}}\)

Question 20.
Derive the expression for the self-inductance of a solenoid.
Answer:
Consider a uniformly wound solenoid with N turns and length l. Let the length of the solenoid be large as compared to its radius and let the core of the solenoid have air. Due to this, we can take the interior field uniform. Therefore the magnetic field in the interior of the solenoid is given by

B = μ₀ n l = μ₀ \(\frac{N}{l}\) l …(1)

where n is the number of turns per unit length. The flux through each turn is given by
Φ_m = B A = μ₀ \(\frac{N A}{l}\) l …(2)

where A Is the area of cross-section of the solenoid. But
L = \(\frac{N \Phi_{\mathrm{m}}}{I}\) …(3)

Therefore from equations (2) and (3) we have
L = \(\frac{N \Phi_{m}}{l}=\frac{\mu_{0} N^{2} A}{l}\) ….(4)

This shows that L depends on the geometric factors and is proportional to the square of the number of turns. Since N N = n l, we can therefore express the above result as
L = µ₀\(\frac{(n l)^{2}}{l}\)A = µ₀ n² A l …(5)

Question 21.
(a) State Faraday’s laws of electromagnetic Induction.
Answer:
Faraday’s Laws of electromagnetic induction
First law: Whenever the magnetic flux (inked with a circuit (or coil) changes, an emf is induced in the circuit. The induced emf Lasts so Long as the change in the magnetic flux continues.

Second law:
The magnitude of induced emf in the circuit (or coil) is directly proportional to the rate of change of magnetic flux linked with the circuit.
i.e. e = – \(\frac{d \phi}{d t}\)

(b) Derive an expression for the emf induced across the ends of a straight conductor of length I moving at right angles to a uniform magnetic field B with a uniform speedy.
Answer:
Let a conductor ab of Length l be placed in a magnetic field \(\vec{B}\) shown by (x) directed towards the reader. When the conductor moves with velocity v perpendicular to B, the force on any free electron of this conductor is given

F = e vB sin 90° = evß

When this electron moves across the Length of the conductor, the work done by the electron
W = F × l = evBl

After some time when all the free electrons are shifted towards the end b, the end b becomes negative and end a becomes positively charged.

Hence the net induced emf in the conductor
e = \(\frac{W}{q}=\frac{e v B l}{e}=v B l\)
e = Blv

(c) Obtain the expression for the magnetic energy stored in a solenoid in terms of the magnetic field B, area A and length I of the solenoid through which a current j is passed. (CBSE 2019C)
Answer:
Energy stored in an inductor Considers an inductor of inductance L. Let I will be instantaneous current in the inductor and dl/dt be the rate of growth of current.

Induced emf, E = L\(\frac{dl}{dt}\) (in magnitude)

If the source sends a charge dq in time dt, then
dq = ldt

Small amount of work done by the source
dW = edq = eldt = L\(\frac{dl}{dt}\). ldt
or
dW = Ll dl

Total work was done during the growth of current in an Inductor by the external source
W = ∫dW =L∫₀¹ ldl
= L \(\left[\frac{l^{2}}{2}\right]_{0}=\frac{1}{2}\)Ll²
W = \(\frac{1}{2}\) Ll² …(1)

This work done is stored in the form of magnetic energy.
In a solenoid,
L = μ₀n²Al
And B = μ₀nl
∴ l = \(\frac{B}{\mu_{0} n}\)

So Eq. (1) becomes
W = \(\frac{1}{2}\)μ0n2Al\(\left(\frac{B}{\mu_{0} n}\right)^{2}=\frac{B^{2} A l}{2 \mu_{0}}\)

Question 22.
(a) A metallic rod of length ‘l’ and resistance ‘R’ is rotated with a frequency ‘v’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius ‘l’, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field ‘B’ parallel to the axis is present everywhere.
(i) Derive the expression for the induced emf and the current in the rod.
(ii) Due to the presence of current in the rod and of the magnetic field, find the expression for the magnitude and direction of the force acting on this rod.
(iii) Hence, obtain an expression for the power required to rotate the rod.
(b) A copper coil l is taken out of a magnetic field with a fixed velocity. Will it be easy to remove it from the same field if its ohmic resistance is increased?
OR
(a) A rectangular coil rotates in a uniform magnetic field. Obtain an expression for induced emf and current at any instant. Also find their peak values. Show the variation of induced emf versus angle of rotation (ωt) on a graph.
(b) An iron bar falling through the hollow region of a thick cylindrical shell made of copper experiences a retarding force. What can you conclude about the nature of the iron bar? Explain. (CBSEAI 2019)
Answer:
(a) (/) Let the rod moves from OP to OQ through a small sector of angle θ.

The small area covered by the rod is
dA = πl² × \(\frac{\theta}{2 \pi}=\frac{1}{2}\)l²θ

where L is the radius of the circle in which the rod rotates. Hence the induced emf is

Now induced current is given by
i = \(\frac{\varepsilon}{R}=\frac{B \omega l^{2}}{2 R}\)

(ii) \(\vec{F}=i(\vec{l} \times \vec{B})\)
F = \(\frac{l^{3} \omega B}{2 R}\)

Direction of \(\vec{F}\) is perpendicular to both \(\vec{i}\) and \(\vec{B}\),

(iii) P = i²R
= \(\left(\frac{B l^{2} \omega}{2 R}\right)^{2}\)R = \(\frac{B^{2} l^{4} \omega^{2}}{4 R}\)

(b) Yes, since induced current will reduce, it will be a little easier to remove the coil.
Or
(a) Consider an armature of the ac generator having n turns placed in a uniform magnetic field B.

Suppose at any instant t the normal to the plane of the coil makes an angle θ with the direction of the magnetic field. If ω is the uniform angular velocity with which the coil rotates then θ = ωt.

The flux through the loop is Φ = n BA cos Φ = n BA cos ω t
where n is the number of turns in the armature.

By Faraday’s flux rule,

Now,i₀ = \(\frac{\varepsilon_{0}}{R}=\frac{n B A \omega}{R}\)

The graph is as shown.

(b) The bar is magnetic in nature. It experiences retardation in accordance with Lenz’s law.

Question 23.
State Lenz’s law. The energy f required to build up a steady current l, in a given coil, varies with l in the manner as shown. Calculate the self-inductance of the coil.

A circular coil of radius r, is placed coaxially with another coil of radius R (R >> r) with the centre of the two coils coinciding with each other. Obtain an expression for the mutual inductance of the two coils.
Answer:
Lenz’s law states that the direction of induced emf is such that it opposes the cause of its production.
The energy stored in an inductor is given by the relation
E = – \(\frac{1}{2}\)Ll²

From the graph for a current of 200 mA the energy is 4 mJ, therefore self-inductance of the coil.
\(\frac{2 E}{l^{2}}\) = L
or
\(\frac{2 \times 4 \times 10^{-3}}{\left(200 \times 10^{-3}\right)^{2}}\) = 0.2 H
or
L = 200 mH

Consider the two coils of radius r and R placed coaxially, as shown in the figure.

Let n₁ and n₂ be the number of turns per unit length in them.

If a current l₂ is passed through the outer coil, there will a magnetic field
B_o = μ_o n₁ l₁ produced in it.

The flux associated with the inner coil will undergo variation as the field B0 grows.
The effective magnetic flux
Φ₁ = B_o × A_eff = \(\frac{\mu_{0} l_{2}}{2 R}\) × πr²

Therefore mutual inductance of the coils is
M = \(\frac{\phi_{1}}{l_{2}}=\frac{\mu_{0} \pi r^{2}}{2 R}\)

Question 24.
(a) Describe a simple experiment (or activity) to show that the polarity of emf induced in a coil is always such that it tends to produce a current which opposes the change of magnetic flux that produces it.
(b) The current flowing through an inductor of self-inductance L is continuously increasing. Plot a graph showing the variation of
(i) Magnetic flux versus the current
(ii) induced emf versus dl/dt
(iii) Magnetic potential energy stored versus the current. (CBSE Delhi 2014)
Answer:
(a) Consider a coil C connected to a galvanometer. When the North Pole of a bar magnet is pushed towards the coil, the pointer in the galvanometer deflects, indicating the presence of electric current in the coil. When the magnet is pulled away from the coil, the galvanometer shows deflection in the opposite direction, which indicates a reversal of the current’s direction.

We see that the North Pole of a bar magnet is being pushed towards the closed coil. As the North Pole of the bar magnet moves towards the coil, the magnetic flux through the coil increases. Hence current is induced in the coil in such a direction that it opposes the increase in flux.

This is possible only if the current in the coil is in a counter-clockwise direction with respect to an observer situated on the side of the magnet. Similarly, if the North Pole of the magnet is being withdrawn from the coil, the magnetic flux through the coil will decrease. To counter this decrease in magnetic flux, the induced current in the coil flows in a clockwise direction and the South Pole faces the receding North Pole of the bar magnet. This would result in an attractive force that opposes the motion of the magnet and the corresponding decrease in flux.

(b) The graphs are as shown.
(i) Φ = Ll
Therefore the graph is

(ii) ε = – L\(\frac{dl}{dt}\)
Therefore the graph is

(iii) U = \(\frac{1}{2}\)Ll²
Therefore the graph is

Numerical Problems:

Formulae for solving numerical problems

• Induced emf ε = –\(\frac{d \phi}{d t}\)
  or
  ε = \(\frac{-\left(\phi_{2}-\phi_{1}\right)}{t_{2}-t_{1}}\)
• Emf induced in a rod of Length L is ε = Blv.
• When a conducting rod of length L kept perpendicular to a uniform magnetic field B is rotated about one of its ends with uniform angular velocity, the emf induced between its ends has a magnitude
  \(\frac{1}{2}\)BωL² = BπvL².
• When a current in a coil changes, it induces a back emf in the same coil. The self-induced emf is given by
  ε = – \(\frac{d \phi}{d t}\) = – L \(\frac{d l}{d t}\) inductance of the coil. It is a measure of the inertia of the coil against the change of current through it. Also Φ = L l
• ε = – M\(\frac{d l}{d t}\) . Also Φ = Ml for mutual induction.
• \(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}=\frac{l_{p}}{l_{s}}\) = k.
• Average power η = \(\frac{\varepsilon_{s} l_{s}}{\varepsilon_{p} l_{p}}\)

Question 1.
A square loop of side 10 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s^-1 in the positive X-direction containing a magnetic field in the positive Z-direction. The field is non-uniform and has a gradient of 10^-3 T cm^-1 along the negative X-direction (i.e. it increases by 10^-3 T cm^-1 as one move in the negative X-direction). Calculate the emf induced. (CBSE 2019C)
Answer:

Question 2.
In a ceiling fan, each blade rotates in a circle of radius 0.5 m. If the fan makes 2 rotations per second and the vertical component of the earth’s magnetic field is 8 × 10^-5 T, calculate the emf induced between the inner and outer ends of each blade. (CBSE2019C)
Answer:

Question 3.
A circular coil of radius 10 cm, 500 turns and resistance 2 Ω are placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. Estimate the magnitudes of the emf and the current induced in the coil. The horizontal component of the earth’s magnetic field at the place is 3.0 × 10^-5 T. (NCERT)
Answer:
Initial flux through the coil
Φ_in = BA cos θ = 3.0 × 10^-5 × π × 10^-2 × cos 0°
= 3π × 10^-7 Wb

Final flux after the rotation
Φ_f = BA cos θ = 3.0 × 10^-5 × π × 10^-2 × cos 180° = – 3π × 10^-7 Wb

Therefore estimated value of induced emf is
ε = n\(\frac{d \phi}{d t}\) = 500 × (6π × 10^-7)/ 0.25 = 3.8 × 10^-3 V dt

Hence the current induced is
l = ε/R = 3.8 × 10^-3 / 2 = 1.9 × 10^-3 A

Question 4.
Kamla peddles a stationary bicycle. The pedals of the bicycle are attached to a 100 turn coil of area 0.10 m². The coil rotates at half a revolution per second and is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil? (NCERT)
Answer:
Given n = 100, A = 0.10 m2, B = 0.01 T, f = 0.5 rps
Using the expression ε₀ = nBAω

we have ε₀ = 100 × 0.01 × 0.10 × 2 × 3.14 × 0.5 = 0.314 V

Question 5.
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of the uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s^-1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? (NCERT)
Answer:
Given L = 8 cm = 8 × 10^-2 m, b = 2 cm = 2 × 10^-2 m, 8 = 0.3 T, v = 1 cm s^-1 = 0.01 m s^-1 ε = ?
(a) Using the expression ε = B L v
ε = 0.3 × 8 × 10^-2 × 0.01 = 2.4 × 10^-4 V

This emf will last till the loop comes out of the magnetic field. Since the shorter side is moving out, therefore it will take 2 s to cover 2 cm with a velocity of 1 cm s^-1.

(b) Using the expression e = B L v
ε = 0.3 × 2 × 10^-2 × 0.01 = 0.6 × 10^-4 V

This emf will last till the loop comes out of the magnetic field. Since the longer side is moving out, therefore it will take 8 s to cover 8 cm with a velocity of 1 cm s^-1.

Question 6.
A 1.0 m long metallic rod is rotated with an angular frequency 400 rad s-1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. (NCERT)
Answer:
Given L = 1.0 m, ω = 400 rad s^-1, B = 0.5 T,

ε = ? Using the relation ε = \(\frac{1}{2}\)BωL²

we have
ε = \(\frac{1}{2}\)0.5 × 400 × 1² =100 V

Question 7.
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s^-1 in a uniform horizontal magnetic field of magnitude 3.0 × 10^-2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from? (NCERT)
Answer:
Given r = 8.0 cm = 8.0 × 10^-2 m, n = 20, ω = 50 rad s-1 B = 3.0 × 10^-2 T, ε₀ = ?, ε_a = ? R= 10 Ω l = ?, P = ?
Using the relation
ε0 = nBAω
= 20 × 3 × 10^-2 × 3.14 × (8 × 10^-2)² × 50 = 0.603 V

The average induced emf is given by
εa = 0 over one cycle.

Also l₀ = ε₀ / R
= 0.603 / l₀
= 0.0603 A

Now P = 1 /2 ε₀ × l₀ = 1 /2 × 0.603 × 0.0603
= 0.018 W

The induced current causes a torque opposing the rotation of the coil. An external agent (rotor) must supply torque to counter this torque in order to keep the coil rotating uniformly. Thus the source of the power dissipated as heat in the coil is the external rotor.

Question 8.
Current in a circuit falls from 5.0 A to 0. 0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit (NCERT)
Answer:
Given dl = 5.0 – 0.0 = 5.0 A, dt = 0.1 s, ε = 200 V, L = ?

Using the relation ε = – L \(\frac{dl}{dt}\)
or
L = \(\frac{\varepsilon}{d l / d t}\)

L = \(\frac{200}{5 / 0.1}=\frac{200 \times 0.1}{5}\) = 4 H

Question 9.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? (NCERT)
Answer:
Given dl = 20- 0.0 = 20 A, dt = 0.5 s, ε = ?,
M = 1.5 H,
Using the relation ε = – \(\frac{d \phi}{d t}\) = -M\(\frac{dl}{dt}\)
or
dΦ = Mdl = 1.5 × 20 = 30 Wb

Question 10.
A jet plane is travelling towards the west at a speed of 1800 km h^-1. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the earth’s magnetic field at the location has a magnitude of 5 × 10^-4 T and the dip angle is 30°? (NCERT)
Answer:
Given v= 1800 km h-1, ε = ?, L = 25 m,
B = 5 × 10^-4 T, δ = 30°
Now vertical component of earth’s magnetic field is
B_V = B × sin 30°
= 5 × 10^-4 × 0.5
= 2.5 × 10^-4T
Now using the expression ε = B L v
ε = 2.5 × 10^-4 × 25 × 500 = 3.125 V

Question 11.
1 MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of a radius of 0.5 cm for this purpose. Calculate the fraction of ohmic losses to the power transmitted if
(a) power is transmitted at 220 V. Comment on the feasibility of doing this.
(b) a step-up transformer is used to boost the voltage to 11000 V, power transmitted, then a step-down transformer is used to bring the voltage to 220 V. ρ_cu = 1.7 × 10^-8 SI unit. (NCERT Exemplar)
Answer:
Given P = 1 MW, 2L = 20 km = 2 × 104 m, r = 0.5 cm = 0.5 × 10^-2 m
(a) Resistance of the Cu wire used
R = \(\frac{\rho L}{A}=\frac{1.7 \times 10^{-8} \times 2 \times 10^{4}}{\pi \times\left(0.5 \times 10^{-2}\right)^{2}}\) = 4 Ω

Now current at 220 V is

l = \(\frac{P}{V}=\frac{10^{6}}{220}\) = 0.45 × 10⁴ A

Therefore power loss
= l²R = (0.45 × 10⁴)² × 4 = 10⁶W

This is a huge loss; therefore this method is not feasible.

(b) Now P = Vl’ or
l’ = \(\frac{P}{V^{\prime}}=\frac{10^{6}}{11000}=\frac{1}{1.1}\) × 10² A

Hence power loss
= l’²R = \(\frac{1}{1.21}\) × 4 × 10⁴ = 3.3 × 10⁴ W

Hence fraction of power loss
\(\frac{3.3 \times 10^{4}}{10^{6}}\) = 3.3%

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 6 General Principles and Processes of Isolation of Elements. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 6 Important Extra Questions General Principles and Processes of Isolation of Elements

General Principles and Processes of Isolation of Elements Important Extra Questions Very Short Answer Type

Question 1.
What is the composition of copper matte? (CBSE Delhi 2013)
Answer:
Copper matte contains cuprous sulphide (Cu₂S) and iron sulphide (FeS).

Question 2.
Name the method used for refining of
(i) Nickel
(ii) Zirconium (CBSE Sample Paper 2007, 2014)
Answer:
(i) Mond’s process
(ii) Van Arkel method

Question 3.
Write the overall reaction taking place in the process used for the electrolysis of alumina by Hall-Heroult process. (CBSE Sample Paper 2011)
Answer:
2Al₂O₃ + 3C → 4Al + 3CO₂

Question 4.
Write a non-exothermic reaction taking place in the blast furnace during extraction of iron. (CBSE Sample Paper 2011)
Answer:

Question 5.
Name the method of refining of metals such as germanium. (CBSE Delhi 2016)
Answer:
Zone refining.

Question 6.
In the extraction of Al, impure Al₂O₃ is dissolved in cone. NaOH to form sodium aluminate and leaving impurities behind. What is the name of the process? (CBSE Delhi 2016)
Answer:
Al₂O₃ + 2NaOH + 3H₂O → 2Na [Al(OH)₄]
This process is called leaching.

Question 7.
Name the method of refining which is based on the principle of adsorption.
(CBSE AI 2016)
Answer:
Chromatography.

Question 8.
Out of PbS and PbCO₃ (ores of lead), which one is concentrated by froth floatation process preferably? (CBSE Delhi 2017)
Answer:
PbS

Question 9.
What is the role of depressants in the froth floatation process? (CBSE Al 2017)
Answer:
The depressants in the froth floatation process selectively prevent one of the sulphide ores from forming the froth with air bubbles.

Question 10.
What is the role of collector and froth stabilizer in froth floatation process?
(CBSE AI 2012)
Answer:

• Collector enhances non-wettability of the mineral particles.
• Froth stabilisers stabilise the froth.

Question 11.
Name the method used for refining of copper. (CBSE AI 2013)
Answer:
Electrolytic refining

Question 12.
What is the role of graphite in the electrometallurgy of aluminium? (CBSE Delhi 2012)
Answer:
The graphite rod is useful in the electrometallurgy of aluminium for reduction of alumina to aluminium.
2Al₂O₃ + 3C → 4Al + 3CO₂

Question 13.
Why is it that only sulphide ores are concentrated by ‘froth floatation process’? (CBSE Delhi 2011)
Answer:
This is because the sulphide ore particles are preferentially wetted by oil and the gangue particles by water.

Question 14.
What is the role of collectors in Froth- Floatation process? (CBSE AI 2012, 2017)
Answer:
The collectors such as pine oil, eucalyptus oil and fatty acids enhance the non – wettability of the mineral particles in froth – floatation process.

Question 15.
Name the substance used as depressant in the separation of two sulphide ores in froth floatation method. (CBSE Sample Paper 2019)
Answer:
Sodium cyanide.

Question 16.
Which reducing agent is employed to get copper from the leached low grade copper ore? (CBSE Delhi 2014)
Answer:
Iron scraps.

Question 17.
What is the role of zinc metal in the extraction of silver? (CBSE 2014)
Answer:
Zinc acts as a reducing agent.

General Principles and Processes of Isolation of Elements Important Extra Questions Short Answer Type

Question 1.
Out of C and CO which is a better reducing agent for FeO?
(i) In the lower part of blast furnace (Higher temperature)
(ii) In the upper part of blast furnace (Lower temperature) (CBSE Sample Paper 2011)
Answer:
(i) C is better reducing agent at higher temperature (lower part of blast furnace).
(ii) CO is better reducing agent at lower temperature (higher part of blast furnace).

Question 2.
What is the role of limestone in the extraction of iron from its oxides? (CBSE Al 2016)
Answer:
Limestone decomposes to form lime (CaO) and carbon dioxide (CO₂). The lime thus produced acts as a flux and combines with silica (present as impurity in oxides of iron) to produce slag.

Question 3.
How is copper extracted from a low grade ore of it? (CBSE Al 2012)
Answer:
Copper is extracted by hydrometallurgy from low grade ores. It is leached out using acid or bacteria. The solution containing copper ions (Cu²⁺) is treated with scrap iron or H₂ as:
Cu²⁺(aq) + H₂(g) → Cu(s) + 2H⁺(aq)
In this way, copper is obtained.

Question 4.
(i) Which solution is used for the leaching of silver metal in the presence of air in the metallurgy of silver?
(ii) Out of C and CO, which is a better reducing agent at the lower temperature range in the blast furnace to extract iron from the oxide ore? (CBSE Delhi 2013)
Answer:
(i) Dilute solution (0.5%) of NaCN or KCN in the presence of air.
(ii) CO

Question 5.
(i) Which of the following ores can be concentrated by froth floatation method and why?
Fe₂O₃, ZnS, Al₂O₃
(ii) What is the role of silica in the metallurgy of copper? (CBSE Delhi 2013)
Answer:
(i) ZnS, because in sulphide ores, the sulphide ore particles are preferentially wetted by oil and gangue particles by water.
(ii) During roasting, the copper pyrites are converted into a mixture of FeO and Cu₂O.

To remove FeO (basic), the roasted ore is mixed with silica and heated. Silica acts as a flux and combines with ferrous oxide present to form fusible slag of iron silicate.

The slag being lighter floats and forms the upper layer and is removed through slag hole. Therefore, silica helps to remove FeO in the metallurgy of copper.

Question 6.
(i) Name the method used for removing gangue from sulphide ores.
(ii) How is wrought iron different from steel? (CBSE Al 2013)
Answer:
(i) Froth floatation method.
(ii) Wrought iron is the pure form of iron and contains carbon and other impurities not more than 0.5%.
Steel contains 0.5 to 1.5% carbon along with small amounts of other elements such as Mn, Cr, Ni, etc. and other impurities.

Question 7.
Outline the principles behind the refining of metals by the following methods: (CBSE Delhi 2014)
(i) Zone refining method
(ii) Chromatographic method
Answer:
(i) Zone refining method. It is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.
(ii) Chromatographic method. This is based on the principle that different components of a mixture are differently adsorbed on an adsorbent.

Question 8.
Write the principle behind the froth floatation process. What is the role of collectors in this process? (CBSE 2014)
Answer:
The froth floatation process is based on the principle of difference in the wetting properties of the ore and gangue particles with water and oil. The collectors such as pine oil, eucalyptus oil, fatty acids, etc. are used to enhance the non-wettability of the mineral particles.

Question 9.
What is meant by vapour phase refining? Write any one example of the process which illustrates this technique, giving the chemical equations involved.
OR
Write and explain the reactions involved in the extraction of gold. (CBSE Sample Paper 2019)
Answer:
Vapour phase refining: This method is based on the fact that certain metals are converted to their volatile compounds while the impurities are not affected during compound formation. The compound formed decomposes on heating to give pure metal. Thus, the two requirements are:

• The metal should form a suitable compound with a suitable reagent.
• The volatile compound should be easily decomposable so that the metal can be easily recovered.

For example, nickel is refined by this technique and the method is known as Mond process. In this method, nickel is heated in a stream of carbon monoxide to form volatile nickel tetracarbonyl, Ni(CO)₄, complex.
The carbonyl vapours when subjected to higher temperature (450-470 K) undergo thermal decomposition giving pure nickel.

OR
The extraction of gold involves leaching of metal present in the ore with CN^– ions. This is an oxidation-reaction because during the leaching process, Au is oxidised to Au⁺ which then combines with CN^– ions to form their respective soluble complexes.

The metal is then recovered from these complexes by reduction or displacement method using a more electropositive zinc metal.
2 [Au (CN)₂]^– (aq) + Zn(s) → 2 Au (s) + [Zn (CN)₄]^2- (aq)
Zinc acts as a reducing agent.
This process is called hydro metallurgy.

General Principles and Processes of Isolation of Elements Important Extra Questions Long Answer Type

Question 1.
(i) Give one example of each the following:
(a) Acidic flux
(b) Basic flux
(ii) What happens when
(a) Cu₂O undergoes self reduction in a silica line converter
(b) Haematite oxidises carbon to carbon monoxide? (CBSE Sample Paper 2012)
Answer:
(i) (a) Acidic flux: SiO₂
(b) Basic flux: CaO

(ii) (a) Cu₂O undergoes self reduction to form blister copper as:

(b) Haematite oxidises carbon to carbon monoxide forming iron.
Fe₂O₃ + 3C → 3CO + 2Fe

Question 2.
What is a flux? What is the role of flux in the metallurgy of iron and copper? (CBSE Sample Paper 2011)
Answer:
Flux is a substance which combines with gangue which may still be present in the calcined or roasted ore to form an easily fusible material called the slag.
Flux + Gangue → Slag (fusible)
In the metallurgy of copper, most of the ferrous sulphide present as impurity gets oxidised to ferrous oxide which combines with silica (flux) to form fusible slag.
FeO + SiO₂ → FeSiO₃
The slag being lighter floats and forms the upper layer which is removed from the slag hole from time to time.
In the blast furnace for metallurgy of iron, lime acts as a flux and combines with silica present as impurity to form slag.

Question 3.
Describe the principle involved in each of the following processes.
(i) Mond process for refining of Nickel.
(ii) Column chromatography for purification of rare elements. (CBSE 2012)
Answer:
(i) Mond’s process for refining of nickel. Impure nickel is heated in a stream of carbon monoxide forming volatile complex, tetra carbonylnickel (0).

The volatile complex is decomposed at high temperature to give pure metal

(ii) Column chromatography for purification of rare elements. This technique is based on the differences in the adsorbing capacities of the metal and its impurities. The metal and the impurities are differently adsorbed. Impure metal is put in a liquid or gaseous medium (called moving phase) which is moved through an adsorbent (stationary phase).
Metal and impurities are adsorbed at different levels in column.

Question 4.
Which methods are usually employed for purifying the following metals?
(i) Nickel
(ii) Germanium
Mention the principle behind each one of them. (CBSE 2012)
Answer:
(i) Nickel. By vapour phase refining. This method is based on the principle that certain metals are converted to their volatile compounds while the impurities are not affected during compound formation. The compound formed decomposes on heating to give pure metal.

For example, nickel is refined by this technique and the method is known as Mond process. In this method, nickel is heated in a steam of carbon monoxide to form volatile nickel carbonyl Ni(CO)₄.
The carbonyl vapours when subjected to higher temperature (450-470 K) undergo thermal decomposition giving pure nickel.

(ii) Germanium. By Zone refining method. This method is based on the principle that the impurities are more soluble in melt than in the solid state of the metal. Therefore, an impure metal on solidification will deposit pure metal and the impurities will remain behind in the molten part of the metal.

Question 5.
Explain the principle of the method of electrolytic refining of metals. Give one example. (CBSE 2014)
Answer:
In electrolytic refining method, the impure metal is made to act as anode. A strip of the same metal in pure form is used as cathode. Both anode and cathode are placed in a suitable electrolytic bath containing soluble salt of the same metal. On passing the electrical current, metal ions from the electrolyte are deposited at the cathode in the form of pure metal, while equivalent amount of metal dissolves from the anode into the elctrolyte in the form of metal ions. The impurities fall down below the anode as anode mud. The reactions occurring at the electrodes are:
At cathode: Mⁿ⁺ + ne^– → M
At anode: M → Mⁿ⁺ + ne^–
Copper is refined using electrolytic refining method.

Question 6.
(i) Name the method used for the refining of zirconium. (CBSE 2015)
(ii) What is the role of CO in the extraction of iron?
(iii) Reduction of metal oxide to metal becomes easier if the metal obtained is in liquid state. Why?
Answer:
(i) Van Arkel method
(ii) CO acts as reducing agent. It reduces oxides of iron to iron.
FeO + CO → Fe + CO₂
3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂
Fe₃O₄ + 4CO → 3Fe + 4CO₂
Fe₂O₃ + CO → 2FeO + CO₂

(iii) This is because if the metal is in liquid state, its entropy is higher than when it is in solid state. Therefore, the value of entropy change (ΔS) of the reduction process is more on +ve side when the metal formed is in the liquid state and the metal oxide being reduced is in solid state. As a result, the value of ΔG° becomes more on negative side and hence the reduction becomes easier.

Question 7.
(i) Name the method of refining which is based on the principle of adsorption. (CBSE 2008, 2016)
(ii) What is the role of depressant in froth floatation process?
(iii) What is the role of limestone in the extraction of iron from its oxides?
Answer:
(i) Chromatography

(ii) The depressants are used to prevent certain types of particles from forming the froth with bubbles in froth floatation process. This helps to separate two sulphide ores. For example, in case of an ore containing zinc sulphide (ZnS) and lead sulphide (PbS), sodium cyanide (NaCN) is used as a depressant. It forms a layer of zinc complex Na₂[Zn(CN)₄] with ZnS on the surface of ZnS and therefore, prevents it from forming the froth. Therefore, it acts as a depressant.

However, NaCN does not prevent PbS from forming the froth and allows it to come with the froth.

(iii) Limestone decomposes to form lime (CaO) and carbon dioxide (CO₂). The lime thus produced acts as a flux and combines with the silica (present as impurity in oxides of iron) to produce slag.

Question 8.
(i) Name the method of refining of metals such as germanium. (CBSE Delhi 2016)
(ii) In the extraction of Al, impure Al₂O₃ is dissolved in cone. NaOH to form sodium aluminate and leaving impurities behind. What is the name of this process?
(ii) What is the role of coke in the extraction of iron from its oxides?
Answer:
(i) Zone refining
(ii) Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄]
This process is called leaching.
(iii) The coke serves as a fuel as well as a reducing agent. It burns to produce CO₂ and heat which raises the temperature to about 2200 K (in combustion zone).
C + O₂ → CO₂, ΔH = -393.4 kJ
It also reduces Fe₂O₃ to iron.
Fe₂O₃ + 3C → 2Fe + 3CO + Heat

Question 9.
(i) Indicate the principle behind the method used for the refining of zinc.
(ii) What is the role of silica in the extraction of copper?
(iii) Which form of the iron is the purest form of commercial iron? (CBSE Delhi 2015)
Answer:
(i) Zinc is refined by electrolytic refining:
In this method, the impure zinc is made anode, while a plate of pure zinc is made the cathode. These are placed in a suitable electrolytic bath containing suitable salt of the same metal (e.g. zinc sulphate) containing a small amount of dilute sulphuric acid. On passing current, zinc from the solution is deposited at the cathode, while an equivalent amount of zinc from anode goes into the electrolytic solution. Therefore, pure zinc is obtained at cathode.

(ii) During roasting, the copper pyrites are converted into a mixture of FeO and Cu₂O.

To remove FeO (basic), the roasted ore is mixed with silica and heated. Silica acts as a flux and combines with ferrous oxide present to form fusible slag of iron silicate.

The slag being lighter floats and forms the upper layer and is removed through slag hole. Therefore, silica helps to remove FeO in the metallurgy of copper.

(iii) Wrought iron is the purest form of commercial iron.

Question 10.
(a) What is the difference between calamine and malachite?
(b) Why is zinc used instead of Cu for recovery of Ag from [Ag(CN)₂]?
(c) What is the role of cryolite in metallurgy of Al? (CBSE 2019C)
OR
(a) Give two points of differences between pig iron and cast iron.
(b) Outline the principle of zone refining.
Answer:
(a) Calamine is an ore of Zn while malachite is an ore of copper. Calamine is ZnCO₃ while malachite is CuCO₃.Cu(OH)₂

(b) Zinc is more electropositive than silver and therefore, zinc displaces silver from its solution.
2[Ag(CN)₂]^– + Zn → [Zn(CN)₄]^2- + 2Ag
On the other hand, copper is less electropositive than silver and therefore, cannot displace silver from its solution. Zn is more reactive than Cu, so reduction will be faster in case of Zn.

(c) Fused alumina is a bad conductor of electricity. Purified alumina is mixed with molten cryolite (Na₃AlF₆) and is electrolysed in an iron tank lined inside with carbon. The molten cryolite decreases the melting point and increases the electrical conductivity. It acts as a solvent.

OR

(a)

(b) Zone refining – It is based on the principle that impurities are more soluble in the melt than the solid state of the metal. Therefore, an impure metal on solidification will deposit crystals of pure metal and the impurities will remain behind in the molten part of the metal.

Question 11.
How will you convert the following:
(i) Impure Nickel to pure Nickel
(ii) Zinc blende to Zinc metal
(iii) [Ag(CN)₂]^– to Ag (CBSE Delhi 2019)
Answer:
(i) Impure nickel is heated in a stream of carbon monoxide forming volatile complex tetracarbonylnickel (0). The volatile complex is decomposed at high tempreature to give pure nickel.

This is called Mond’s process.

(ii) The zinc blende ore is concentrated by froth floatation process and then roasted in the presence of excess air at about 1200 K.
2ZnS + 3O₂ → 2ZnO + 2SO₂
Zinc oxide formed is reduced to zinc by heating with crushed coke at 1673 K in vertical fire clay retorts.

(iii) [Ag(CN)₂]^– complex is heated with zinc to get silver.
2[Ag(CN)₂]^– + Zn → [Zn(CN)₄]^2- + 2Ag

Question 12.
(a) Name the method of refining which is
(i) used to obtain semiconductor of high purity,
(ii) used to obtain low boiling metal.
(b) Write chemical reactions taking place in the extraction of copper from Cu₂S. (CBSE Delhi 2019)
Answer:
(a) (i) Zone refining
(ii) Distillation

(b) Cu₂S is first roasted and converted to oxide
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
The oxide is then easily reduced to metallic copper with coke.
Cu₂O + C → 2Cu + CO

Question 13.
Describe how the following steps can be carried out.
(i) Recovery of Gold from leached gold metal complex.
(ii) Conversion of Zirconium iodide to pure Zirconium.
(iii) Formation of slag in the extraction of copper.
(Write the chemical equations also for the reactions involved)
OR
Explain the use of the following:
(i) NaCN in Froth Floatation Method.
(ii) Carbon monoxide in Mond process.
(iii) Coke in the extraction of Zinc from Zinc Oxide.
Answer:
(i) Leached gold complex is treated with Zinc and gold is recovered by displacement method.
2Au[(CN)₂]r(aq) + Zn(s) → 2Au(s) + [Zn(CN)₄]^2-(aq)

(ii) Zirconium iodide is decomposed on a tungsten filament, electrically heated to 1800 K. Pure Zr metal is deposited on the filament.
ZrI₄ → Zr + I₂

(iii) Silica is added to the ore and heated. It helps to slag off iron oxide as iron silicate.
FeO + SiO₂ → FeSiO₃ (slag)

OR

(i) NaCN is used as depressants to separate two sulphide ores (ZnS and PbS) in Froth Floatation Method.
(ii) Carbon monoxide forms a volatile complex of nickel, nickel tetracarbonyl.
(iii) Coke is used as a reducing agent to reduce zinc oxide to zinc.

Question 14.
Write the role of
(i) NaCN in the extraction of gold from its ore.
(ii) Cryolite in the extraction of aluminium from pure alumina.
(iii) CO in the purification of Nickel. (CBSE 2018)
Answer:
(i) Gold is leached out in the form of a complex with dil. solution of NaCN in the presence of air. Here NaCN acts as leaching agent.
(ii) It lowers the melting point of alumina and makes it a good conductor of electricity.
(iii) CO forms a volatile complex with nickel which is further decomposed to give pure Ni metal.

Question 15.
(i) Write the role of ‘CO’ in the purification of nickel.
(ii) What is the role of silica in the extraction of copper?
(iii) What type of metals are generally extracted by electrolytic method? (CBSE Delhi 2019)
Answer:
(i) Carbon monoxide forms a volatile complex, Ni(CO)₄, on heating nickel in a stream of CO while the impurities remain unaffected. The compound on further heating decomposes to give pure nickel.

Thus, CO helps to purify nickel.

(ii) During roasting, the copper pyrites are converted into a mixture of Cu₂O and FeO.

To remove FeO (basic), the roasted ore is mixed with silica and heated. Silica acts as a flux and combines with FeO to form fusible slag of iron silicate.
FeO + SiO₂ → FeSiO₃
The slag being lighter floats and forms the upper layer and is removed. Thus, silica helps to remove FeO.

(iii) Very reactive metals such as sodium, potassium, calcium, aluminium, etc. are extracted by electrolytic methods. Certain less reactive metals such as copper are also purified by using electro-refining method.

Question 16.
Write down the reactions taking place in blast furnace related to the metallurgy of iron in the temperature range 500 K to 800 K. What is the role of limestone in the metallurgy of iron?
OR
What happens when
(a) Silver is leached with NaCN in the presence of air?
(b) Copper matte is charged into silica-lined converter and hot air blast is blown?
(c) NaCN is added in an ore containing PbS and ZnS during concentration by froth floatation method? (CBSE Delhi Al 2019)
Answer:
Reactions taking place in blast furnace at 500K to 800K: The oxides of iron are reduced by CO.
FeO + CO → Fe + CO₂
3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂
Fe₃O₄ + 4CO → 3Fe + 4CO₂
Fe₂O₃ + CO → 2FeO + CO₂
Role of Limestone: It acts as a flux which decomposes to calcium oxide.
CaCO₃ → CaO + CO₂
Limestone
CaO combines with impurity (e.g. Si02) to form slag which is then removed.
CaO + SiO₂ → CaSiO₃

OR

(a) Silver is leached with dilute solution (0.5%) of NaCN in the presence of atmospheric oxygen. The metal dissolves forming the complex.
4Ag + 8CN^– + 2H₂O + O₂ → 4[Ag(CN)₂]^– + 4OH^–
The metal is extracted from water-soluble complex by zinc metal.
2[Ag(CN)₂]^– + Zn → [Zn(CN)₄]^2- + 2Ag

(b) The copper matte containing Cu₂S and FeS is put in silica-lined convertor and hot air blast is blown to convert remaining FeS to FeO, which is removed as slag with silica.
2FeS + 3O₂ → 2FeO + 2SO₂
FeO + SiO₂ → FeSiO₃
Cu2S or CuO gets converted to copper
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2Cu₂O + Cu₂S → 6Cu + SO₂

(c) NaCN acts as a depressant in preventing ZnS from forming the froth. If forms a layer of zinc complex Na₂[Zn(CN)₄] on the surface of ZnS and thereby prevents it from forming the froth.
4NaCN + ZnS → Na₂[Zn(CN)₄] + Na₂S

Long Answer Type Questions (LA-II)

Question 1.
Explain the following:
(i) Although thermodynamically feasible, in practice magnesium metal is not used for the reduction of alumina in the metallurgy of aluminium. Why?
(ii) Why is zinc and not copper used for the recovery of silver from the complex [Ag(CN)₂]^–?
(iii) The extraction of Au by leaching with NaCN involves both oxidation and reduction. Justify giving equations. (CBSE Sample Paper 2011)
(iv) Lime-stone is used in the manufacture of pig iron from haematite. Why?
Answer:
(i) Inspection of Ellingham diagram shows that ∆G vs T curves for Al₂O₃ and MgO intersect at a point corresponding to very high temperature of the order of 2000 K. This means above this temperature, ∆G for the reaction:
Al₂O₃ + 3Mg → 2Al + 3Mgo
would become negative and hence reduction will be feasible. However, this temperature is very high so that the process is uneconomical and technologically difficult.

(ii) Zinc is a stronger reducing agent (E° = – 0.76 V) and more electropositive than copper (E° = + 0.34 V). Therefore, zinc is used for recovery of Ag from [Ag(CN₂)]^– complex.

(iii) During leaching process, gold (Au) is first oxidised by O₂ of the air to Au⁺ which then combines with CN^– ions to form the soluble complex, dicyanoaurate (I).

Gold is then extracted from this complex by displacement method by using a more electropositive zinc metal. In this method, zinc acts as a reducing agent and it reduces Au⁺ to Au. Zinc itself gets oxidised to Zn²⁺ ions which combine with CN^– ions to form soluble complex, sodium tetracyanozincate (II).

Thus, extraction of gold by leaching with NaCN involves both oxidation and reduction.

(iv) Haematite is an ore of iron and contains silica (SiO₂) as the main impurity. The purpose of limestone is to remove SiO₂ as calcium silicate (CaSiO₃) slag.

Question 2.
Name the principal ore of aluminium. Explain the significance of leaching in the extraction of aluminium. (CBSE AI 2013)
Answer:
The principal ore of aluminium is bauxite, Al₂O₃.2H₂O. Leaching process is used to concentrate the ore of aluminium, bauxite, which is contaminated with impurities of silica (SiO₂), iron oxide (Fe₂O₃), titanium oxide (TiO₂), etc. Leaching is done by treating the powdered ore with hot cone. (45%) solution of NaOH at about 473-523 K and 35-36 bar pressure.
Al₂O₃.2H₂O(s) + 2NaOH(aq) + H₂O(l) → 2Na [Al(OH)₄](aq)
The impurities of ferric oxide and silica are insoluble and are removed by filtration.
The solution containing sodium aluminate is neutralised by passing C02 gas and hydrated alumina is precipitated.
2Na [Al(OH)₄] (aq) + CO₂ (g) → Al₂O₃. x H₂O (s) + 2Na HCO₃(s)
Hydrated alumina is heated to 1473 K to get pure alumina.

Question 3.
Describe the principle controlling each of the following processes:
(i) Vapour phase refining of titanium metal.
(ii) Froth floatation method of concentration of sulphide ore.
(iii) Recovery of silver after silver ore was leached with NaCN. (CBSE AI 2011, CBSE Delhi 2011)
Answer:
(i) The principle of vapour phase refining is that certain metals are converted to their volatile compounds while the impurities are not affected during compound formation. The compound formed decomposes on heating to give pure metal. During refining of titanium metal, the titanium metal is heated with l₂ to form a volatile compound Til₄, which on further heating at higher temperature decomposes to give pure titanium metal.

(ii) This is based upon the different wetting properties of the ore and the gangue particles with oil and water. The mineral particles become wet by oil and the gangue particles by water. When the ore is mixed with water containing small quantities of pine oil and then by agitating the water by blowing air violently, the froth is formed. The froth carries the lighter ore particles to the surface and the heavier impurities sink to the bottom.

(iii) This is a chemical method and is useful for the ores which are soluble in certain suitable solvents but the impurities are not soluble. The impurities left undissolved are removed by filtration. The ore of silver, Ag₂S (argentite), reacts with NaCN as

Sodium sulphide thus formed is oxidised to sodium sulphate by blowing air into the solution. This helps the reaction to occur in the forward direction.
4Na₂S + 5O₂ + 2H₂O → 2Na₂SO₄ + 4NaOH + 2S

The above solution after filtration and removing insoluble impurities is heated with zinc to get silver.

Question 4.
Describe the role of the following:
(i) NaCN in the extraction of silver from a silver ore
(ii) Iodine in the refining of titanium
(iii) Cryolite in the metallurgy of aluminium (CBSE 2010)
Answer:
(i) The role of NaCN in the extraction of silver is to do the leaching of silver ore in the presence of atmospheric air from which silver is obtained later by replacement with zinc as:

(ii) Iodine is used to form a volatile unstable compound of the metal, which is decomposed to get the pure metal. For example, titanium is heated with iodine to 523 K forming unstable titanium iodide which is decomposed to Ti as

(iii) Cryolite is added to bauxite ore before electrolysis because of the following reasons:
(a) It acts as a solvent.
(b) It lowers the melting point of alumina to about 1173 K.
(c) Addition of cryolite to alumina increases the electrical conductivity.

Question 5.
Describe the role of the following:
(i) SiO₂ in the extraction of copper from copper matte
(ii) NaCN in froth floatation process
Answer:
(i) The copper matte containing Cu₂S and FeS is put in silica lined converter. Some silica is also added and hot air blast is blown to convert remaining FeS to FeO, which is removed as slag with silica.
2FeS + O₂ → 2FeO + 2SO₂
FeO + Si0₂ → FeSiO₃
Cu₂S or CuO gets converted to copper.
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2Cu₂O + Cu₂S → 6Cu + SO₂

(ii) NaCN in froth floatation process is used to separate one sulphide ore from another and is known as depressant.
For example, sodium cyanide can be used as a depressant in the separation of zinc sulphide ore (ZnS) and lead sulphide ore (PbS). Sodium cyanide forms a layer of zinc complex, Na₂[Zn(CN)₄], on the surface of ZnS and therefore, prevents it from forming the froth. Therefore, it acts as a depressant.

However, NaCN does not prevent PbS from forming the froth. Thus, it selectively prevents ZnS from coming to the froth but allows PbS to come with the froth. Thus, the two ores can be separated by the use of a depressant.

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 5 Surface. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 5 Important Extra Questions Surface Chemistry

Surface Chemistry Important Extra Questions Very Short Answer Type

Question 1.
Give an example of a shape selective catalyst. (CBSE Delhi 2010, 2011)
Answer:
Zeolite catalyst known as ZSM-5.

Question 2.
Why are medicines more effective in colloidal state? (CBSE Delhi 2019)
Answer:
Medicines are most effective in colloidal state because colloids have large surface area and hence the medicines can be easily adsorbed and assimilated.

Question 3.
What is difference between an emulsion and a gel? (CBSE Delhi 2019)
Answer:
Emulsions are colloidal solutions in which both the dispersed phase and the dispersion medium are liquids. Gel is a colloidal system in which a liquid is dispersed in a solid.

Question 4.
Define ‘peptization’. (CBSE2012)
Answer:
The process of converting a freshly prepared precipitate into colloidal form by addition of a suitable electrolyte is called peptization.

Question 5.
What is meant by ‘shape-selective catalysis? (CBSE 2012)
Answer:
The catalytic reaction which depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape selective catalysis.

Question 6.
What happens when a freshly precipitated
Fe(OH)₃ is shaken with little amount of dilute solution of FeCl₃? (CBSE 2013)
Answer:
A reddish brown colloidal solution of Fe(OH)₃ is obtained. This process is called peptization.

Question 7.
What is especially observed when a beam of light is passed through a colloidal solution? (CBSE 2013)
Answer:
Tyndall effect

Question 8.
Give one example each of sol and gel. (CBSE Delhi 2014)
Answer:
Sol: As₂S₃ sol.
Gel: Cheese.

Question 9.
Give one example each of ‘oil in water’ and ‘water in oil’ emulsion. (CBSE 2014)
Answer:
Oil in water emulsion:
Milk Water in oil emulsion: Butter

Question 10.
Write a method by which lyophobic colloids can be coagulated. (CBSE 2015)
Answer:
By electrophoresis lyophobic colloids can be coagulated. Other methods are by mixing two oppositely charged sols or by boiling or by addition of electrolyte.

Question 11.
Write the main reason for the stability of colloidal sols. (CBSE Delhi 2016)
Answer:
The stability of colloidal sols is because of the presence of electrical charge on the particles which prevents the particles to come close together.

Question 12.
What are lyophobic colloids? Give one example for them. (CBSE Delhi 2011)
Answer:
The colloidal solutions in which the particles of the dispersed phase have no affinity for the dispersion medium are called lyophobic colloids. For example, metal sulphides like As₂S₃.

Question 13.
Write one similarity between physisorption and chemisorption. (CBSE Delhi 2017, CBSE AI 2017)
Answer:
Both are surface phenomena and increase with increase in surface area.

Question 14.
Why are lyophilic colloidal sols more stable than lyophobic colloidal sols? (CBSE Delhi 2015)
Answer:
The lyophilic colloidal sols are more stable because they are highly hydrated in solution.

Question 15.
What is the difference between a sol and a gel? (CBSE Delhi 2017)
Answer:
In a sol, dispersion medium is liquid and dispersed phase is solid. On the other hand, in a gel, dispersion medium is solid and dispersed phase is liquid.

Question 16.
What type of colloid is formed when a liquid is dispersed in a solid? Give an example. (CBSE AI 2017)
Answer:
Gel
For example: Cheese

Question 17.
Which of the following is most effective electrolyte in the coagulation of Fe₂O₃.x H₂O/Fe³⁺ sol?
KCl, AlCl₃, MgCl₂, K₄[Fe(CN)₆] (CBSE Sample Paper 2011)
Answer:
Since Fe(OH)₃ sol is positively charged, the anion having highest charge will be most effective, i.e. [Fe(CN)₆]⁴.

Question 18.
Why is colloidal gold used for intramuscular injection? (CBSE Sample Paper 2011)
Answer:
Colloidal gold is more effective because of larger surface area and therefore, is easily assimilated with blood which is colloidal.

Question 19.
What is colloidion? (CBSE Sample Paper 2011)
Answer:
4% solution of nitrocellulose in a mixture of alcohol and ether.

Question 20.
Leather gets hardened after tanning. Why? (CBSE Delhi 2015)
Answer:
Animal hide is colloidal in nature and has positively charged particles. When it is soaked in tannin which has negatively charged colloidal particles, it results in mutual coagulation. This results in the hardening of leather.

Question 21.
It is necessary to remove CO when ammonia is prepared by Haber’s process. Explain. (CBSE Delhi 2015)
Answer:
Carbon monoxide acts as a poison for the catalyst in Haber’s process and therefore, it will lower the activity of the catalyst. Thus, CO must be removed when ammonia is obtained by Haber’s process.

Question 22.
Addition of alum purifies water. Why? (CBSE AI 2015)
Answer:
Alum coagulates the impurities present in water by neutralising the charge.

Question 23.
Out of sulphur sol and proteins, which one forms multimolecular colloids? (CBSE Delhi 2016)
Answer:
Proteins are macromolecules which cannot form multimolecular colloids while sulphur sol has smaller S₈ molecules which can coagulate to form multimolecular colloids.

Question 24.
Write the chemical method by which Fe(OH)₃ sol is prepared from FeCl₃. (CBSE AI 2017)
Answer:

Question 25.
Out of MgCl₂ and AlCl₃ which one is more effective in causing coagulation of negatively charged sol and why? (CBSE Delhi 2016)
Answer:
According to Hardy-Schulze rule, for negatively charged sol, greater the valency of the positive ion of the electrolyte added, greater is its coagulating power. Thus, AlCl₃ (Al³⁺ ions) is more effective in causing coagulation of negatively charged sol than MgCl₂ (Mg²⁺ ions).

Question 26.
Give one example each of lyophobic sol and lyophilic sol. (CBSE Delhi 2014)
Answer:
Lyophobic : As₂S₃, Lyophilic : Gelatin

Question 27.
Out of BaCl₂ and KCl which one is more effective in causing coagulation of negatively charged colloidal sol. Give reason. (CBSE Delhi 2015)
Answer:
BaCl₂ because greater the valency of the coagulating ion (positive ion), greater is its tendency to coagulate.

Question 28.
What are the dispersed phase and . dispersion medium in milk? (CBSE AI 2014)
Answer:
Liquid (dispersed phase), Liquid (dispersion medium)

Question 29.
What type of colloid is formed when a solid is dispersed in liquid? Give an example. (CBSE AI 2017)
Answer:
Sol, paints

Question 30.
CO (g) and H₂(g) react to give different products in the presence of different catalysts. Which ability of the catalyst is shown by these reactions? (CBSE AI 2018)
Answer:
Selectivity of a catalyst. It is the ability of a catalyst to selectively form a particular product.

Surface Chemistry Important Extra Questions Short Answer Type

Question 1.
Describe a conspicuous change observed when
(i) a solution of NaCl is added to a sol of hydrated ferric oxide.
(ii) a beam of light is passed through a solution of NaCl and then through a sol. (C.B.S.E. Delhi 2012)
Answer:
1. Hydrated ferric oxide is positively charged. When an electrolyte such as NaCl is added, the excess Cl^– ions neutralise positive charge and cause coagulation of the sol.

2. When a beam of light is passed through a colloidal solution, the path of the light becomes visible when viewed from the direction at right angle to that of incident beam. This phenomenon is called Tyndall effect. This is because of scattering of light by colloidal particles. The particles first absorb the incident light and then a part of it gets scattered by them and therefore the part becomes visible when seen at right angle to the direction of incident beam.

Question 2.
Name the two groups into which phenomenon of catalysis can be divided. Give an example of each group with the chemical equation involved. (CBSE Delhi 2012)
Answer:
Catalysis can be divided into two groups
(i) Homogeneous catalysis
(ii) Heterogeneous catalysis

(i) Homogenous catalysis: When the catalyst is present in the same phase as the reactants and products, it is called homogeneous catalysis. For example,

SO₂(g) is oxidised to SO₃(g) in the presence of gaseous nitric oxide as catalyst.

(ii) Heterogeneous catalysis: When the catalyst is in different phase than the
reactants, it is called heterogeneous catalysis. For example,

Question 3.
Write the dispersed phase and dispersion medium of the following colloidal systems: (CBSE Delhi 2013)
(i) Smoke
(ii) Milk
Answer:
Colloidal system: Dispersed phase/Dispersion medium
(i) Smoke: Solid/Gas
(ii) Milk: Liquid/Liquid

Question 4.
What are lyophilic and lyophobic colloids? Which of these sols can be easily coagulated on the addition of small amounts of electrolytes? (CBSE Delhi 2013)
Answer:
The colloidal solutions in which the particles of the dispersed phase have great affinity for the dispersion medium are called lyophilic sols. For example, glue, gelatin, starch, proteins, etc.

The colloidal solutions in which there is no affinity between the particles of the dispersed phase and the dispersion medium are called lyophobic sols. For example, solutions of . metals like Ag, Au, Al(OH)₃, Fe(OH)₃, etc.

Lyophobic sols can be easily coagulated on the addition of small amounts of electrolytes.

Question 5.
What is the difference between oil/water (O/W) type and water/oil (W/O) type emulsions? Give an example of each type. (CBSE Delhi 2013)
Answer:
1. Oil-in-water emulsion (o/w). In this case, oil acts as the dispersed phase (small amount) and water as the dispersion medium (excess), e.g. milk is an emulsion of soluble fats in water and here casein acts as an emulsifier. Vanishing cream is another example of this class. Such emulsions are called aqueous emulsions.

2. Water-in-oil emulsion (w/o). In this case, water acts as the dispersed phase while the oil behaves as the dispersion medium, e.g. butter, cod liver oil, cold cream, etc. Such types of emulsions are called oily emulsions.

Question 6.
Peptizing agent is added to convert precipitate into colloidal solution. Explain. (CBSE Sample Paper 2011)
Answer:
Peptization is a process of converting a freshly prepared precipitate into colloidal form by the addition of an electrolyte called peptizing agent. The suitable ions from the peptizing agent (electrolyte) are adsorbed by the particles of the precipitate giving it positive or negative charge. The charged particles repel one another and break up the precipitate into smaller particles of the size of the colloid. Therefore, it results into the formation of colloid.

For example, on treating a precipitate of iron (III) oxide with a small amount of FeCl₃ solution gives a reddish brown coloured colloidal solution.

Question 7.
Cottrell’s smoke precipitator is fitted at the mouth of chimney used in factories. Give reasons. (CBSE Sample Paper 2011)
Answer:
Smoke coming out of chimney of a factory is a colloidal solution of solid carbon particles which are charged in nature.

The mouth of the chimneys used in factories is fitted with Cottrell smoke precipitator. In this method, the smoke is allowed to pass through a chamber having a series of plates charged to very high potential (20,000 to 70,000 V).

Charged particles of smoke get attracted by charged plates, get precipitated and the gases coming out of chimney become free of charged carbon and dust particles.

Question 8.
Differentiate between peptization and coagulation.
(CBSE Sample Paper 2011, CBSE AI 2017)
Answer:
Peptization is the process of converting a freshly prepared precipitate into collodial form by the the addition of a suitable electrolyte. The electrolytes used for the purpose are called peptizing agents. On the other hand, coagulation is the phenomenon of precipitation of a collodial solution by the addition of excess of an electrolyte.

Question 9.
Explain:
(i) Sky appears blue in colour.
(ii) A freshly formed precipitate of ferric hydroxide can be converted to a colloidal sol by shaking it with a small quantity of ferric chloride. (CBSE Sample Paper 2011)
Answer:
1. Dust particles along with water suspended in air scatter blue light which reaches our eyes and therefore, sky looks blue to us.

2. When we add FeCl₃ to a freshly formed precipitate of Fe(OH)₃, peptization occurs. The Fe³⁺ ions are adsorbed on the surface of the precipitate which ‘ ultimately breaks down into smaller particles of colloidal size.

Question 10.
What happens when
(i) a freshly prepared precipitate of Fe(OH)₃ is shaken with a small amount of FeCl₃ solution?
(ii) persistent dialysis of a colloidal solution is carried out?
(iii) an emulsion is centrifuged? (CBSE 2018)
Answer:
(i) Peptization occurs / Colloidal solution of Fe(OH)₃ is formed.
(ii) Coagulation occurs
(iii) Demulsification or breaks into constituent liquids

Surface Chemistry Important Extra Questions Long Answer Type

Question 1.
Classify colloids where the dispersion medium is water. State their characteristics and write one example of each of these classes.
OR
Explain what is observed when
(i) an electric current is passed through a sol.
(ii) a beam of light is passed through a sol.
(iii) an electrolyte (say NaCl) is added to ferric hydroxide sol. (CBSE 2011)
Answer:
These colloids are of two types:
(i) Hydrophilic
(ii) Hydrophobic

Characteristics: In lyophilic colloids, there is affinity between dispersed phase and the dispersion medium while in the lyophobic colloids, there is no affinity rather hatred between the phases.

The main points of distinction between lyophilic and lyophobic sols are summarized below:

Example: Hydrophilic: Starch,
Hydrophobic : Metal sulphide

OR

(i) On passing electric current through colloidal solution, the colloidal particles move towards the oppositely charged electrodes, where they lose their charge and get coagulated. This is electrophoresis process.

(ii) When a beam of light is passed through a colloidal solution, the path of the light becomes visible when viewed from the direction at right angle to that of incident beam. This phenomenon is called Tyndall effect. This is because of scattering of tight by colloidal particles. The particles first absorb the incident light and then a part of it gets scattered by them and therefore the part becomes visible when seen at right angle to the direction of incident beam.

(iii) Hydrated ferric oxide is positively charged. When an electrolyte such as NaCl is added, the excess Cl^– ions neutralise positive charge and cause coagulation of the sol.

Question 2.
Explain how the phenomenon of adsorption finds application in each of the following processes:
(i) Production of vacuum
(ii) Heterogeneous catalysis
(iii) Froth floatation process (CBSE Delhi 2011)
Answer:
(i) Production of vacuum: The adsorption of air in liquid air helps to create a high vacuum in a vessel. This process is used in high vacuum instruments such as Dewar flask for storage of liquid air or liquid hydrogen. The remaining traces of air can be adsorbed by charcoal from the vessel evacuated by a vacuum pump to give a very high vacuum.

(ii) Heterogeneous catalysis: The phenomenon of adsorption is useful in the heterogeneous catalysis. Adsorption of reactants on the solid surface of catalysts increases the rate of reaction. The metals such as Fe, Ni, Pt, Pd, etc. are used in the manufacturing processes. Manufacture of ammonia using iron as catalyst (Haber process), manufacture of sulphuric acid by Contact process and use of finely divided nickel in the hydrogenation of oils are excellent examples of heterogeneous catalysis. Its use is based upon the phenomenon of adsorption.

(iii) In froth floatation process: A low grade sulphide ore is concentrated by separating it from silica and other earthly matter by adsorption using pine oil and frothing agent.

Question 3.
What is meant by coagulation of a colloidal solution? Describe briefly any three methods by which coagulation of lyophobic sols can be carried out. (CBSE 2012)
Answer:
The phenomenon of precipitation of a colloidal sol by the addition of excess of an electrolyte is called coagulation of colloidal sol. Coagulation of lyophobic sols can be carried out by the following methods:

1. By addition of an electrolyte: When an electrolyte is added to the sol, colloidal particles take up ions carrying opposite charge from the electrolyte and get neutralised resulting in coagulation.

2. By mutual precipitation: When two oppositely charged sols are mixed in equimolar proportions, they mutually neutralise their charge and both get coagulated.

3. By electrophoresis: When electrophoresis is carried out for a long time, the particles of dispersed phase move towards oppositely charged electrodes, lose their charge and get coagulated.

Question 4.
Explain the following terms giving a suitable example for each:
(i) Aerosol
(ii) Emulsion
(iii) Micelle (CBSE 2012)
Answer:
(i) Aerosol: It is a colloidal dispersion of a liquid in a gas. For example, fog, mist, cloud.

(ii) Emulsion. Emulsions are the colloidal solutions of two immiscible liquids in which the liquid acts as the dispersed phase as well as the dispersion medium. Types of emulsion. These are of two types:

(a) Oil-in-water emulsion. In this case, oil acts as the dispersed phase (small amount) and water as the dispersion medium (excess), e.g. milk is an emulsion of soluble fats in water and here casein acts as an emulsifier. Vanishing cream is another example of this class. Such emulsions are called aqueous emulsions.

(b) Water-in-oil emulsion. In this case, water acts as the dispersed phase while the oil behaves as the dispersion medium, e.g. butter, cod liver oil, cold cream, etc. Such types of emulsions are called oily emulsions.

(iii) Micelle: The particles of colloidal size formed due to aggregation of several ions or molecules with lyophobic as well as lyophilic particles are called micelles. The common micelles system is soap. The micelles behave as normal electrolytes at low concentrations but as colloids at higher concentrations.

Question 5.
Write three distinct features of chemisorption which are not found in physisorption. (CBSE 2012)
Answer:
Characteristics of Chemisorption:

1. The forces between the adsorbate molecules and the adsorbent are strong chemical forces similar to chemical bonds. But no chemical bonds exist in physisorption.
2. Chemisorption is irreversible in nature but physisorption is reversible.
3. Chemisorption forms mono – molecular layers, but in physisorption multimolecular layers are present.

Question 6.
What are the characteristics of the following colloids? Give one example of each.
(i) Multimolecular colloids
(ii) Lyophobic sols
(iii) Emulsions (CBSE 2013)
Answer:
(i) The colloidal solutions in which atoms or smaller molecules of substances (having diameter less than 1 nm) aggregate together to form particles of colloidal dimensions. The particles thus formed are called multimolecular colloids. For example, sols of gold atoms and sulphur (S₈) molecules.

(ii) The colloidal solutions in which there in no affinity (or love rather they have hatred) between the particles of the dispersed phase and the dispersion medium are called lyophobic sols.
For example, metal sulphides like As₂S₃.

(iii) The colloidal solutions in which both the dispersed phase and the dispersion medium are liquids are called emulsions. For example, milk.

Question 7.
Write the dispersed phase and dispersion medium of the following colloidal systems:
(i) Smoke
(ii) Milk
OR
What are lyophilic and lyophobic colloids? Which of these sols can be easily coagulated on the addition of small amounts of electrolytes? (CBSE Delhi 2013)
Answer:

OR
The colloidal solutions in which the particles of the dispersed phase have great affinity for the dispersion medium are called lyophilic sols. For example, glue, gelatin, starch, proteins, etc.

The colloidal solutions in which there is no affinity between the particles of the dispersed phase and the dispersion medium are called lyophobic sols. For example, ‘ solutions of metals like Ag, Au, Al(OH)₃, Fe(OH)₃, etc.
Lyophobic sols can be easily coagulated on the addition of small amounts of electrolytes.

Question 8.
What are emulsions? What are their different types? Give one example of each type. (CBSE 2014)
Answer:
Emulsions are the colloidal solutions of two immiscible liquids in which the liquid acts as the dispersed phase as well as the dispersion medium.

Types of emulsions. These are of two types:
1. Oil-in-water emulsion. In this case, oil acts as the dispersed phase (small amount) and water as the dispersion medium (excess), e.g. milk is an emulsion of soluble fats in water and here casein acts as an emulsifier. Vanishing cream is another example of this class. Such emulsions are called aqueous emulsions.

(ii) Water-in-oil emulsion. In this case, water acts as the dispersed phase while the oil behaves as the dispersion medium, e.g. butter, cod liver oil, cold cream, etc. Such types of emulsions are called oily emulsions.

Question 9.
(i) In reference to Freundlich adsorption isotherm, write the expression for adsorption of gases on solids in the form of an equation.
(ii) Write an important characteristic of lyophilic sols.
(iff) Based on type of particles of dispersed phase, give one example each of associated colloid and multimolecular colloid. (CBSE Delhi 2014)
Answer:
(i) \(\frac{x}{m}\) = kp^1/n
where \(\frac{x}{m}\) is extent of adsorption, x is the mass of adsorbate, m is the mass of adsorbent, p is pressure at a particular temperature, k is a constant and n takes any whole number value.

(ii) in lyophilic sols, the particles of dispersed phase have a great affinity for the dispersion medium and are reversible in nature.

(iii) Associated colloids: Soap (e.g. sodium stearate).
Multimolecular colloids: Sol of gold (atoms).

Question 10.
Give reasons for the following observations:
(i) Leather gets hardened after tanning.
(ii) Lyophilic sol is more stable than lyophobic sol.
(iii) It is necessary to remove CO when ammonia is prepared by Haber’s process. (CBSE Delhi 2015)
Answer:
(i) Animal hide is colloidal in nature and has positively charged particles. When it is soaked in tannin which has negatively charged colloidal particles, it results in mutual coagulation. This results in the hardening of leather.

(ii) Lyophilic sol is more stable than lyophobic sol because it is highly hydrated or solvated in solution.

(iii) Carbon monoxide acts as a poison for the catalyst in Haber’s process and therefore, it will lower the activity of the catalyst.
Thus, CO must be removed when ammonia is obtained by Haber’s process.

Question 11.
Give reasons for the following observations:
(i) Physisorption decreases with increase in temperature.
(ii) Addition of alum purifies the water.
(iii) Brownian movement provides stability to the colloidal solution. (CBSE 2015)
Answer:
(i) In physisorption, the attractive forces between adsorbent and adsorbate molecules are weak vander Waals forces. When temperature is increased, the kinetic energy of the molecules of the gas increases and they can easily leave the surface of adsorbent.
Therefore, physisorption decreases with increase in temperature.

(ii) Alum coagulates the impurities present in water by neutralising the charge.

(iii) The Brownian movement is due to the unbalanced bombardment of the particles by the molecules of the dispersion medium. This results in zigzag motion. Because of brownian movement, the particles do not settle down and hence is responsible for the stability of the sols.

Question 12.
Define the following terms:
(i) O/W Emulsion
(ii) Zeta potential
(iii) Multimolecular colloids (CBSE 2016)
Answer:
(i) O/W Emulsion: It is the emulsion in which oil is the dispersed phase and water is the dispersion medium. For example, milk is an emulsion of soluble fats in water.

(ii) Zeta potential: The potential difference between the fixed layer and the diffused layer of opposite charges is called Zeta potential.

(iii) Multimolecular colloids: When a dissolution, atoms or smaller molecules of substances (having diameter less than 1 nm) aggregate together to form particles of colloidal dimension, the particles thus formed are called multimolecular colloids.

Question 13.
(i) Differentiate between adsorption and absorption.
(ii) Out of MgCl₂ and AlCl₂, which one is more effective in causing coagulation of negatively charged sol and why?
(iii) Out of sulphur sol and proteins, which one forms multimolecular colloids? (CBSE Delhi 2016)
Answer:
(i)

(ii) According to Hardy-Schulze rule, for negatively charged sol, greater the valency of the positive ion of the electrolyte added, greater is its coagulating power. Thus AlCl₃ (Al³⁺ ion) is more effective in causing coagulation of negatively charged sol than MgCl₂ (Mg²⁺ ions).

(iii) Proteins are macromolecules which cannot form multimolecular colloids while sulphur sol has smaller S₈ molecules which can coagulate to form multimolecular colloids.

Question 14.
What is meant by coagulation of a colloidal solution? Describe briefly any three methods by which coagulation of lyophobic sols can be carried out.
(CBSE Delhi 2013)
Answer:
The phenomenon of precipitation of a colloidal sol by the addition of excess of an electrolyte is called coagulation of colloidal sol.
Coagulation of lyophobic sols can be carried out by the following methods:
(i) By addition of an electrolyte: When an electrolyte is added to the sol, colloidal particles take up ions carrying opposite charge from the electrolyte and get neutralised resulting in coagulation.

(ii) By mutual precipitation: When two oppositely charged sols are mixed in equimolar proportions, they mutually neutralise their charge and both get coagulated.

(iii) By electrophoresis: When electrophoresis is carried out for a long time, the particles of dispersed phase move towards oppositely charged electrodes, lose their charge and get coagulated.

Question 15.
(i) What is the role of activated charcoal in gas mask?
(ii) A colloidal sol is prepared by the given method in figure. What is the charge on hydrated ferric oxide colloidal particles formed in the test tube? How is the sol represented?

(iii) How does chemisorption vary with temperature? (CBSE Delhi 2019) Answer:
(i) Activated charcoalin gas mask adsorbs the poisonous or toxic gases from air. Therefore, these masks help to purify the air for breathing.

(ii) Negatively charged sol
Representation: Fe₂O₃. xH₂O/OH^–

(iii) In chemisorption, adsorption initially increases with rise in temperature and then decreases.

Question 16.
Define the following terms with an example in each case:
(i) Macromolecular sol
(ii) Peptization
(iii) Emulsion (CBSE Al 2013)
Answer:
(i) Macromolecular sol: In this type, the particles of the dispersed phase are sufficiently big in size (macro) to be of colloidal dimensions. In this case, a large number of small molecules are joined together through their primary valencies to form giant molecules. These molecules are called macromolecules and each macromolecule may consist of hundreds or thousands of simple molecules. The solutions of such molecules are called macromolecular solutions. For example, colloidal solution of starch, cellulose, etc.

(ii) Peptization: The process of converting a freshly prepared precipitate into colloidal form by addition of a suitable electrolyte is called peptization. The electrolytes used for the purpose are called peptizing agents.

(iii) Emulsions: They are the colloidal solutions of two immiscible liquids in which the liquid acts as the dispersed phase as well as the dispersion medium. Normally, these are obtained by mixing an oil with water. Since the two do not mix well, the emulsion is generally unstable and is stabilised by adding a suitable reagent called emulsifier or emulsifying agent. The substances that are commonly employed for the purpose are gum, soap, glass powder, etc.

Question 17.
Define the following with a suitable example of each:
(a) Coagulation
(b) Multimolecular colloid
(c) Gel
Answer:
(a) Coagulation. The phenomenon of precipitation of a colloidal solution by the addition of excess of an . electrolyte is called coagulation. For
example, Fe(OH)₃ colloidal solution is coagulated by adding HCl.

(b) Multimolecular colloid. When on dilution, atoms or smaller molecules of substances (having diameter less than 1 nm) aggregate together to form particles of colloidal dimensions, then the particles thus formed are called multimolecular colloids, for example, sols of gold atoms.

(c) Gel. A gel is a colloidal system in which a liquid is dispersed in a solid. For example, jellies, cheese.

OR

(a) Out of starch and ferric hydroxide sol, which one can easily be coagulated and why?
(b) What is observed when an emulsion is centrifuged?
(c) What is the role of promoters and poisons in catalysis? (CBSE Al 2019)
Answer:
(a) Ferric hydroxide sol is easier to coagulate because it is a lyophobic sol.
(b) De-emulsification occurs.
(c) Promoters increase the efficiency of catalyst whereas poisons inhibit the efficiency of catalyst.

Question 18.
Give reason for the following observations:
(i) When silver nitrate solution is added to potassium iodide solution, a negatively charged colloidal solution is formed.
(ii) Finely divided substance is more effective as an adsorbent.
(iii) Lyophilic colloids are also called reversible sols.
Answer:
(i) The precipitated silver iodide adsorbs iodide ions from the dispersion medium resulting in the negatively charged colloidal solution.
(ii) Due to large surface area
(iii) If the dispersion medium is separated from the dispersed phase, the sol can be reconstituted by simply remixing with the dispersion medium. That is why these sols are also called reversible sols.

Question 19.
Answer the following questions:
(a) Which of the following electrolytes is most effective for the coagulation of Agl/Ag⁺ sol?
MgCl₂, K₂SO₄, K₄[Fe(CN)₆]
(b) What happens when a freshly precipitated Fe(OH)₃ is shaken with a little amount of dilute solution of FeCl₃?
(c) Out of sulphur sol and proteins, which one forms macromolecular colloids? (CBSE Sample Paper 2019)
Answer:
(a) K₄[Fe(CN)₆]
Hardy Schulze rule: The coagulation tendency of different electrolytes is different. It depends upon the valency of the active ion called flocculatins ion, which is the ion-carrying charge opposite to the charge on the colloidal particles. According to Hardy Schulze rule, greater the valency of the active ion or flocculating ion, greater will be its coagulating power. For example, to coagulate negative sol of AS₂S₃, the coagulating power of different cations has been found to decrease in the order as:
Al³⁺ > Mg²⁺ > Na⁺
Similarly, to coagulate a positive sol such as Fe(OH)₃, the coagulating power of different anions has been found to decrease in the order:
[Fe(CN)₆]4^– > PO₄^3- > SO₄^2- > Cl^–

(b) Due to the phenomenon of peptization, a reddish brown positively charged sol of ferric hydroxide is obtained when a freshly precipitated Fe(OH)₃ is shaken with a little amount of dilute solution of FeCl₃. It involves the adsorption of suitable ions from the electrolyte by . the particles of the precipitate. The charged particles repel one another and form a colloidal sol. Ferric ions from ferric chloride are adsorbed by Fe(OH)₃, precipitate and get converted into colloidal sol.

(c) Proteins are macromolecules which cannot form multi-molecular colloids while sulphur sol has smaller S₈ molecules which can coagulate to form multi-molecular colloids.

Question 20.
Answer the following:
(a) Why is ester hydrolysis slow in the beginning and then becomes faster after some time?
(b) In a solution of methylene blue, animal charcoal is added, the solution is then well shaken. What will be observed and why?
(c) Give an example of oil in water emulsion.
Answer:
(a) This is because of the process of autocatalysis. Ester on hydrolysis gives an acid which starts acting as a catalyst after sometime and therefore, the reaction becomes fast.

(b) Adsorption of a dye by charcoal. When animal charcoal is shaken with a solution of an organic dye such as methylene blue it is observed that the solution turns colourless. The discharge of the colour is due to the fact that the coloured component (generally an organic dye) gets adsorbed on the surface of animal charcoal. Therefore, animal charcoal is used for decolourising a number of organic substances in the form of their solutions.
(c) Milk, vanishing cream.

OR

Define the following:
(a) Associated colloids
(b) Electrophoresis
(c) Zeta potential (CBSE 2019C)
Answer:
(a) Associated colloids: These are the substances which when dissolved
in a medium behave as normal electrolytes at low concentration but behave as colloidal particles at higher concentration due to the formation of aggregated particles. The aggregate particles thus formed are called micelles. For example, in aqueous solution, soap (sodium stearate) ionises as:

In concentrated solution, these ions get associated to form an aggregate of colloidal size.

(b) The phenomenon of movement of colloidal particles under an applied electric field is called electrophoresis. If the particles accumulate near the negative electrode, the charge on the particles is positive. On the other hand, if the sol particles accumulate near the positive electrode, the charge on the particles is negative.

(c) The combination of two layers of opposite charges around the colloidal particle is called Helmholtz electrical double layer. The first layer of ions is firmly held and is termed fixed layer while the second layer is mobile which is termed diffused layer. The charges of opposite signs on the fixed and diffused parts of the double layer result in a difference in potential between these layers. The potential difference between the fixed layer and the diffused layer of opposite charges is called the electrokinetic potential or zeta potential.

Question 21.
Write the differences between physisorption and chemisorption With respect to the following:
(i) Specificity
(ii) Temperature dependence
(iii) Reversibility and
(iv) Enthalpy change (CBSE Delhi 2013)
Answer:

Question 22.
Describe some features of catalysis by zeolites.
Answer:
Zeolites are microporous aluminosilicates of the general formula M_x/n [(AlO₂)_x (SiO₂)y] mH₂O. These are most important oxide catalysts. These are used in petrochemical industries for cracking of hydrocarbons and isomerisation. The reactions in zeolites depend upon the size of the cavities (cages) or pores (apertures) present in them.

The most remarkable feature of zeolite catalysis is the shape selectivity. Therefore, the selectivity of catalyst depends on the pores structure. It has been observed that the pore size in zeolites generally varies between 260 pm and 740 pm. Depending upon the size of the reactants and products compared to the size of the cages or pores of zeolite, reactions proceed in specific manner.

A zeolite catalyst called ZSM-5 converts alcohols to gasoline, by first dehydrating the alcohol by loss of water.

Question 23.
What do you understand by adsorption isotherm? Discuss briefly Freundlich adsorption isotherm.
Answer:
Adsorption isotherm. A graph between amount of adsorption and gas pressure keeping temperature constant is called an adsorption isotherm.

Freundlich adsorption isotherm. The behaviour of adsorption with pressure can be expressed by adsorption isotherm as shown in figure.

The inspection of the graph reveals the following facts:
(i) At low pressure, the graph is almost straight line which indicates that x/m is directly proportional to the pressure. This may be expressed as:
\(\frac{x}{m}\) ∝ P or \(\frac{x}{m}\) = kP ……. (i)
where k is constant.

(ii) At high pressure. The graph becomes almost constant which means that x/m becomes independent of pressure. This may be expressed as:

(iii) Thus, in the intermediate range of pressure, x/m will depend upon the power of pressure which lies between 0 and 1, i.e. fractional power of pressure. This may be expressed as:

where n can take any value between 1 and a larger integer depending upon the pressure. The above relationship is also called Freundlich’s adsorption isotherm.

Question 24.
Explain the following in connection with colloids:
(i) Hardy-Schulze rule
(ii) Dialysis.
Answer:
(i) Hardy-Schulze rule: This rule states that greater the valency of the active ion or flocculating ion, greater will be its coagulating power. According to this rule:

(a) The ions carrying the charge opposite to that of sol particles are effective in causing coagulation of the sol.

(b) Coagulation power of an electrolyte is directly proportional to the valency of the active ions.
For example, for negative As₂S₃ sol, the coagulating power of cations decreases as Al³⁺ > Mg²⁺ > K⁺
Similarly, for positively charged sol such as Fe (OH)₃, the coagulating power decreases as
[Fe(CN)₃]^4- > PO₄^3- > SO₄^2- > Cr^–

(ii) Dialysis. The method used to separate the impurities from the colloidal solution is called dialysis. Its principle is based upon the fact that colloidal particles cannot pass through a parchment or cellophane membrane while the ions of the electrolyte can pass through it. The colloidal solution is taken in a bag made of cellophane or parchment. The bag is suspended in fresh water. The impurities slowly diffuse out of the bag leaving behind pure colloidal solution. Dialysis can be used for removing HCl from the ferric hydroxide sol. The method is shown in figure.

The ordinary process of dialysis is slow. To increase the process of purification, the dialysis is carried out by applying electric field. This process is called electrodialysis.

Question 25.
Explain the following terms:
(i) Electrodialysis
(ii) Tyndall effect.
Answer:
(i) Electrodialysis. It is the process of separating the impurities from the colloidal solution by applying electric field. It is based on the fact that colloidal particles cannot pass through a parchment or cellophane membrane while the ions of the electrolyte can pass through it. The colloidal solution is taken in a bag made of cellophane or parchment and is suspended in a fresh water and electric current is applied. The impurities slowly diffuse out of the bag leaving behind pure colloidal solution.

(ii) Tyndall effect. When a strong beam of light is passed through a true solution placed in a beaker, in a dark room, the path of the light does not become visible. However, if the light is passed through a sol, placed in the same room, the path of the light becomes visible when viewed from a direction at right angle to that of the incident beam.

This phenomenon was studied for the first time by Tyndall and therefore it is called Tyndall effect. The cause to Tyndall effect is the scattering of light by colloidal particles, i.e. these particles first absorb the incident light and then a part of it gets scattered by them. Since the intensity of the scattered light is at right angle to the plane of the incident light, the path becomes visible only when seen in that direction. The particles in true solution are too small in size to cause any scattering, i.e. the Tyndall effect is not noticed in true solutions.

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 5 Magnetism and Matter. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 5 Important Extra Questions Magnetism and Matter

Magnetism and Matter Important Extra Questions Very Short Answer Type

Question 1.
A small magnetic needle pivoted at the center is free to rotate In a magnetic meridian. At what place will the needle
be vertical?
Answer:
At the potes

Question 2.
What is the angle of dip at a place where the horizontal and vertical components of the earth’s magnetic field are equal?
Answer:
450

Question 3.
How does the intensity of a paramagnetic sample vary with temperature?
Answer:
it decreases with the increase in temperature.

Question 4.
What should be the orientation of a magnetic dipole in a uniform magnetic field so that its potential energy is maximum?
Answer:
It should be anti-parallel to the applied magnetic field.

Question 5.
What is the value of angle of dip at a place on the surface of the earth where the ratio of the vertical component to the horizontal component of the earth’s magnetic field is 1/\(\sqrt{3}\)?
Answer:
Using the expression tan δ = \(\frac{B_{V}}{B_{H}}=\frac{1}{\sqrt{3}}\)
Therefore, δ = 30°

Question 6.
Where on the surface of the earth is the angle of dip 90°? (CBSE Al 2011)
Answer:
Poles.

Question 7.
Where on the surface of the earth is the angle dip zero? (CBSE Al 2011)
Answer:
Magnetic equator

Question 8.
What are permanent magnets? Give one example. (CBSE Delhi 2013)
Answer:
It is an arrangement that has a permanent dipole moment, e.g. bar magnet.

Question 9.
At a place, the horizontal component of the earth’s magnetic field is B, and the angle of dip is 60°. What is the value of the horizontal component of the earth’s magnetic field at the equator? (CBSE Delhi 2017)
Answer:
Zero.

Question 10.
Is the steady electric current the only source of the magnetic field? Justify your answer. (CBSE Delhi 2013C)
Answer:
No, the magnetic field is also produced by alternating current.

Question 11.
Write one important property of a paramagnetic material. (CBSEAI2019)
OR
Do the diamagnetic substances have a resultant magnetic moment in an atom in the absence of an external magnetic field?
Answer:
It moves from the weaker to the stronger regions of the magnetic field.
OR
No

Question 12.
Steel is preferred for making permanent magnets, whereas soft iron is preferred for making electromagnets. Give one reason.
Answer:

1. The Retentivity of steel is more than that of soft iron.
2. Soft iron has a smaller value of coercivity than steel.

Question 13.
Define angle of inclination at a given place due to the earth’s magnetic field. (CBSE 2019C)
Answer:
The angle of inclination is the angle at which the earth’s magnetic field at a given place makes with the horizontal line in the magnetic meridian.

Question 14.
Relative permeability of a material p = 0.5. Identify the nature of the magnetic material and write its relation to magnetic susceptibility. (CBSE Delhi 2014C)
Answer:
Paramagnetic.
Susceptibility is small and greater than one.

Magnetism and Matter Important Extra Questions Short Answer Type

Question 1.
(a) Define the term magnetic susceptibility and write its relation in terms of relative magnetic permeability.
Answer:
It refers to the ease with which a substance can be magnetized. It is defined as the ratio of the intensity of magnetization to the magnetizing field. The required relation is µ_r = 1 + χ_m

(b) Two magnetic materials A and B have relative magnetic permeabilities of 0. 96 and 500. Identify the magnetic materials A and B. (CBSE Al, Delhi 2018C)
Answer:
A: Paramagnetic,
B: Ferromagnetic

Question 2.
A magnetic needle free to rotate in a vertical position orients itself with its axis vertical at a certain place on the earth. What are the values of
(a) the angle of dip and
(b) the horizontal component of the earth’s magnetic field at this place? Where will this place be on the earth?
Answer:
The angle of dip is 90° and the horizontal component of the earth’s magnetic field is zero. This place is the magnetic pole of the earth.

Question 3.
Out of the two magnetic materials ‘A’ has relative permeability slightly greater than unity while ‘B’ has less than unity. Identify the nature of the materials ‘A’ and ‘B’. Will their susceptibilities be positive or negative? (CBSE Delhi 2014)
Answer:

• ‘A’ is paramagnetic and ‘B’ is diamagnetic.
• ‘A’ will have positive susceptibility while
• ‘B’ will have negative susceptibility.

Question 4.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its northern tip down at 60° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place. (CBSE Delhi 2011)
Answer:
Given δ = 30°, B_H = 0.4 G, B = ?
Using the expression
B_H = Bcos δ we have
B = \(\frac{B_{H}}{\cos \delta}=\frac{0.4}{\cos 30^{\circ}}=\frac{0.4}{\sqrt{3} / 2}=\frac{0.8}{\sqrt{3}}\) G

Question 5.
The susceptibility of a magnetic material is -0.085. Identify the type of magnetic material. A specimen of this material is kept in a non-uniform magnetic field. Draw the modified field pattern.
Answer:
The material is a diamagnetic material as diamagnetic materials have negative susceptibility. The modified field pattern is as shown below.

Question 6.
A uniform magnetic field gets modified as shown below when two specimens X and Y are placed in it.

(a) Identify the two specimens X and Y.
Answer:
X is a diamagnetic substance and Y is a paramagnetic substance.

(b) State the reason for the behavior of the field lines in X and Y.
Answer:
This is because the permeability of a diamagnetic substance is less than one and that of a paramagnetic substance is greater than one.

Question 7.
Three identical specimens of magnetic materials nickel, antimony, and aluminum are kept in a non-uniform magnetic field. Draw the modification in the field lines in each case. Justify your answer.
Answer:
Nickel is ferromagnetic, antimony is diamagnetic and aluminium is paramagnetic. Therefore, they will show the behavior as shown in the following figures.

Question 8.
Define neutral point. Draw lines of force when two identical magnets are placed at a finite distance apart with their N-poles facing each other. Locate the neutral points.
Answer:
It is a point near a magnet where the magnetic field of the earth is completely balanced by the magnetic field of the magnet. The figure is as shown below.

The cross indicates the neutral point.

Question 9.
The susceptibility of a magnetic material is -2.6 x 10^-5. Identify the type of magnetic material and state its two properties. (CBSE Delhi 2012)
Answer:
Diamagnetic:

1. Very small and negative susceptibility
2. Permeability is less than one.

Magnetism and Matter Important Extra Questions Long Answer Type

Question 1.
Write the expression for the magnetic dipole moment for a closed current loop. Give its SI unit. Derive an expression for the torque experienced by a magnetic dipole in a uniform magnetic field.
Answer:
The required expression is m = nIA.
It is measured in A m².

Consider a uniform magnetic field of strength B. Let a magnetic dipole be suspended in it such that its axis makes an angle 6 with the field as shown in the figure below. If ‘m’ is the strength of each pole, the two poles experience two equal and opposite force ‘B’ each. These forces constitute a couple that tends to rotate the dipole. Suppose the couple exerts a torque of magnitude τ.

Then

τ = either force × arm of the couple
= mB × AN = mB × 2 L sin θ
or
Since m × 2L is the magnetic dipole moment of the magnet.
Therefore τ = MB sin θ in vector form

we have \(\vec{τ}\) = \(\vec{M}\) × \(\vec{B}\)

Question 2.
(a) State Gauss’s law for magnetism. Explain Its significance.
Answer:
(a) Gauss’s Law for magnetism states that “The total flux of the magnetic field, through any closed surface, is aLways
zero, i.e. ∮\(\vec{B}\).\(\vec{dL}\) = 0

This law implies that magnetic monopoles do not exist” or magnetic field lines form closed loops.

(b) Write the four Important properties of the magnetic field lines due to a bar magnet. (CBSE Delhi 2019)
Answer:
Four properties of magnetic field lines are as follows:

1. Magnetic field lines always form continuous closed loops.
2. The tangent to the magnetic field line at a given point represents the direction of the net magnetic field at that point.
3. The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field.
4. Magnetic field lines do not intersect.

Question 3.
A short bar magnet placed with its axis at 30° with an external field of 800 G experiences a torque of 0.016 Nm.
(a) What is the magnetic moment of the magnet?
Answer:
(a) From equation τ = mB sin θ,
For θ = 30°, sin θ = 0.5

Thus, 0.016 = m × 800 × 10^-4 × 0.5 or m = 160 × 2/800 = 0.40 Am²

(b) What is the work done in moving it from its most stable to a most unstable position?
Answer:
From equation U= -mB cos θ, the most stable position is for θ = 0° and the most unstable position is for θ = 180°.

Work done is given by
W = U_m (θ = 180°) – Um for (θ = 0°)
= 2 mB = 2 × 0.40 × 800 × 10^-4
= 0.064 J

(c) The bar magnet is replaced by a solenoid of cross-sectional area 2 × 10^-4 m² and 1000 turns but of the same magnetic moment. Determine the current flowing through the solenoid. (NCERT Delhi 2005C)
Answer:
From part (i), m = 0.40 Am² Using equation m = NIA.
0.40 = 1000 × l × 2 × 10^-4
l = 2 A

Question 4.
(a) What happens if a bar magnet is cut into two pieces: (i) transverse to its length and (ii) along its length?
Answer:
(a) In either case one gets two magnets each with a north pole and a south pole.

(b) A magnetized needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to torque. Why?
Answer:
The needle experiences no net force because the field is uniform.

The iron nail experiences a non¬uniform field due to the bar magnet. There is an induced magnetic moment in the nail; therefore, it experiences both force and torque.

The net force is attractive because the induced south pole (say) in the nail is closer to the north pole of the magnet than the induced north pole.

Question 5.
Name the elements of the earth’s magnetic field at a place. Explain their meaning.
Answer:
The elements required to completely specify the earth’s magnetic field are called magnetic elements. These elements are:
(a) Declination (θ)
(b) Dip or inclination (δ)
(c) Horizontal component of the earth’s magnetic field (B_H).

(a) Declination (θ): Declination at a place is defined as the acute angle between the magnetic meridian and the geographic meridian.
In the figure above ∠BAB’ = θ is called the angle of declination. There are several periodic variations in the value of declination for a given location.

(b) Dip (δ): Dip at a place is defined as the angle between the direction of the total intensity of the earth’s magnetic field and a horizontal line in the magnetic meridian at that place. It is denoted by 8.

Suppose a freely suspended magnet aligns itself along the direction AB as shown in the figure, then this gives the direction of the earth’s total magnetic field at that place. Then the angle ∠BAO = δ is called the angle of dip or simply dip at that point.

(c) Horizontal component of the earth’s magnetic field (BH): It is the component of the total intensity of the earth’s magnetic field along a horizontal line in the magnetic meridian. It is denoted by BH or H.

Question 6.
Name the three types of magnetic materials which behave differently when placed in a non-uniform magnetic field. Give two properties for each of them.
Answer:
The three types of magnetic materials are
(a) Diamagnetic
(b) Paramagnetic and
(c) Ferromagnetic

Question 7.
The susceptibility of a magnetic material is 0-9853. Identify the type of magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field. (CBSEAI, Delhi 2018)
Answer:
We know that the susceptibility of a paramagnetic substance is positive but small. Hence, the given magnetic material is paramagnetic in nature.

When a specimen of a paramagnetic substance is placed in a magnetizing field, the magnetic field lines prefer to pass through the specimen rather than through air. Thus, magnetic induction 6 inside the sample is more than the magnetic induction B0 outside the sample.

Question 8.
(a) Draw the magnetic field lines due to a circular loop of area \(\vec{A}\) carrying current l. Show that it acts as a bar magnet of magnetic moment \(\vec{m}\) = I \(\vec{A}\).
Answer:
(a) The magnetic field lines are as shown.

Magnetic field due to circular loop on its axis at far off points B = \(\frac{\mu_{0}}{4 \pi} \frac{2 l A}{x^{3}}\)

Magnetic field due to a bar magnet at its axial point is B = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{x^{3}}\)
Comparing the above two we have m = IA

(b) Derive the expression for the magnetic field due to a solenoid of length ‘21’, radius ‘a’ having ‘n’ number of turns per unit length and carrying a steady current T at a point on the axial line, distant ‘r’ from the center of the solenoid. How does this expression compare with the axial magnetic field due to a bar magnet of the magnetic moment ‘m’? (CBSEAI 2015)
Answer:
(b) By the-Biot-Savarts law, the magnetic field at point P on the axis of the solenoid due to a small element of length dx at a distance ‘x’ from the center of the solenoid is

dB = \(\frac{\mu_{0} n d x / a^{2}}{2\left[(r-x)^{2}+a^{2}\right]^{3 / 2}}\)

Hence total magnetic field is given by
B = ∫dB = \(\frac{\mu_{0} / a^{2} n^{+L}}{2} \int_{-L}^{d x} \frac{d x}{\left[(r-x)^{2}+a^{2}\right]^{3 / 2}}\)

For r >> a (R >> L)
We have
B = ∫dB = \(\frac{\mu_{0} / a^{2} n^{+L}}{2 r^{3}} \int_{-L}^{+L} d x=\frac{\mu_{0} \ln }{2} \frac{2 L a^{2}}{r^{3}}\)

Magnetic moment of a solenoid
m = (n × 2L) I (πa²)

Therefore,
B = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}\) as that of a bar magnet.

Numerical Problems :

• Torque on a magnetic dipole τ = MB sin θ
• Potential energy of a magnetic dipole U = – MB cos θ
• \(B_{H}^{2}+B_{v}^{2}=B^{2}\)
• tan θ = \(\frac{B_{V}}{B_{H}}\)
• H = B cos δ
• V = B sin δ

Question 1.
A bar magnet of the magnetic moment 6 J T¹ is aligned at 60° with a uniform external magnetic field of 0 – 44 T. Calculate
(a) the work is done in turning the magnet to align its magnetic moment (i) normal to the magnetic field and (ii) opposite to the magnetic field and
Answer:

(b) the torque on the magnet in the final orientation in case (ii). (CBSEAI, Delhi 2018)
Answer:
τ = M × B sin θ₂
= 6 × 0.44 × sin 180° = 0

Question 2.
A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 x 10^-2 J. What is the magnitude of the magnetic moment of the magnet?
Answer:
Given θ = 30°, B = 0.25 T, τ = 4.5 × 10^-2 J,M = ?
Using the expression τ = MB sin θ we have
M = \(\frac{\tau}{B \sin \theta}=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}\) = 0.36 J T^-1