CBSE Class 11

NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrogen

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrogen

Question 1.
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.
Answer:
Position of Hydrogen in the Periodic Table

In the periodic table, hydrogen has been placed at the top of the alkali metals in group 1 although it is not a member of the group. However, its position is not properly justified simply because of its electronic configuration (Is¹). It can be placed along with the alkali metals as they too have ns¹ configuration. However, it can also be placed along with halogens in group 17 since just like halogens, it also requires any one electron to have the configuration of the nearest noble gas element helium. Let us compare its characteristics with the members of both the families

Resemblance with Alkali metals
Hydrogen and alkali metals resemble in the following respects.
Electronic configuration. Hydrogen has one electron in its valence shell like the alkali metals. For example,
Element                                                 H                   Li                         Na
Atomic No.                                            1                     3                           11
Electronic configuration                      Is¹               1s²2s’               ls²2s²2p⁶3s¹

Question 2.
Write the names of the three isotopes of hydrogen. What is the mass ratio of these isotopes ?
Answer:
The three isotopes of hydrogen are : protium (\( _{ 1 }^{ 1 }{ H }\)) deuterium( \( _{ 1 }^{ 2 }{ H }\)or D) and tritium ( \( _{ 1 }^{ 3 }{ H }\) or T). Mass ratio of the three isotopes is :

Question 3.
Why does hydrogen occur in diatomic form rather than in monoatomic form under normal conditions ?
Answer:
In monoatomic form, hydrogen atom has only one electron in K-shell (Is¹) while in diatomic form, the K-shell is complete (Is²). This means that in diatomic form, hydrogen (H₂) has acquired the configuration of noble gas helium. It is therefore, quite stable.

Question 4.
How can the production of dihydrogen from ‘coal gasification’ be increased ?
Answer:
In coal gasification reaction, steam vapours are passed through coke to form mixture of CO and H₂ also called syn gas.

The production of dihydrogen can be increased by reacting CO(g) present in syn gas with steam in the presence of iron chromate catalyst With the removal of CO(g) or CO₂(g), the reaction shifts in the forward direction. This is known as water gas shift reaction. As a result, the production of dihydrogen will increase

Question 5.
Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an electrolyte in the process ?
Answer:
Dihydrogen is formed mainly from water by carrying its electrolysis in the presence of suitable electrolytes i.e., small amount of an acid or alkali. Actually, water as such is a poor conductor of electricity. Addition of small amount of electrolyte increases its conductivity.

Question 6.
Complete the following

Answer:


Question 7.
Describe the consequences of high bond enthalpy of H-H bond in terms of chemical reactivity of dihydrogen.
Answer:
The bond enthalpy or bond dissociation enthalpy of dihydrogen (H₂) is very high (AH = 435 -9 kJ mol^-1). This is on account of its small atomic size and also small bond length (74 pm) of H—H bond. Consequently, the bond cleavage is extremely difficult and molecular hydrogen or dihydrogen takes part in chemical reactions only under specific conditions.

Question 8.
What do you understand by
(1) electron deficient
(2) electron precise and
(3) electron rich compounds of hydrogen ? Provide justification with suitable examples.
Answer:
(1) Electron deficient hydrides : These hydrides have lesser number of electrons available for writing their conventional structures according to Lewis concept. One popular example of the hydrides belonging to this class is of diborane (B₂Hg). No electrons are available to account for the bonding between the two atoms of boron. We shall discuss the details about the structure of diborane in unit-11 onp-block elements. Most of the hydrides of elements of group 13 fall under this category. This electron deficient hydrides may be regarded as Lewis acids and are electron acceptors.

(2) Electron precise hydrides : These are the covalent hydrides in which the atoms have the required number of electrons which are permissible according to Lewis concept of bond formation. Hydrocarbons like methane (CH₄), ethane (C₂Hg), ethene (C₂H₄) and ethyne (C₂H₂) etc. are electron precise hydrides. We shall discuss them in unit-13.

(3) Electron rich hydrides : In these hydrides, the participating elements donot have the required number of hydrogen atoms. As a result, they have one or more lone pairs of electrons present. The elements generally belong to groups 15, 16 and  17 A few important of out of these are

They have one, two and three lone pairs of electrons present one N, O and F atoms respectively. Due to the presence of lone electron pairs, these are expected to behave as Lewis bases. However, except for NH₃, both H₂O and HF hardly exhibit basic nature of oxygen and fluorine. However, all the three hydrides which are listed, participate in intermolecular hydrogen bonding and get associated due to strong polar character.

Question 9.
What characteristics do you expect from an electron deficient hydride with respect to its structure and chemical reaction ?
Answer:
It is expected to be a Lewis acid. For example, diborane B₂H₆ (dimer of BH₃) forms a complex with LiH which gives H^– ion (Lewis base)

Similarly it also combines with trimethylamine (organic compound) and carbon monoxide which act as Lewis base to form addition compounds.

Question 10.
Do you expect the carbon hydride of the type (C_nH_2n+2) to act as Lewis acid or base ? Justify you answer.
Answer:
It is neither a Lewis acid nor a Lewis base. In all the hydrides belonging to this type, the carbon atoms have complete octet. Therefore, these hydrides behave as normal covalent hydrides also called saturated hydrocarbons or alkanes, (e.g., CH₄, C₂H₆, C₃H₈ etc.). These are also called electron precise hydrides. We shall study in details about these hydrides at a later stage in unit 13.

Question 11.
What do you understand by the term non-stoichiometric hydrides ? Do you expect these types of hydrides to be formed from alkali metals ? Justify your answer.
Answer:
Metallic or Non-stiochjometric (or Interstitial) Hydrides
Many transition and inner transition metals absorb hydrogen into the interstices of their lattices to yield metal like hydrides also called interstitial hydrides. The hydrogen is present in the atomic form in these hydrides. It is interesting to note that the transition metals of groups 3, 4 arid 5 form metallic hydrides. In group 6, chromium alone has a tendency to form hydride CrH. Then there is a gap and the metals of groups 7, 8 and 9 donot form hydrides. This is known as hydride gap. Latest studies have shown that only the hydrides of Ni, Pd, Ce and Ac are interstitial in nature which means that hydrogen atoms occupy the interstitial sites.

The rest of the hydrides have different lattices and it is not proper to call them interstitial hydrides. The elements of/-block form limited number of hydrides. The hydrides are generally non-stoichiometric and their composition vary with temperature and pressure. For example, TiH_1.73, CeH_2.7, LaH_2.8 etc. These hydrides have metallic appearance and their properties are closely related to those of the parent metal. Most of them are strong reducing agents probably due to the presence of free hydrogen atoms in metal lattice. Metallic hydrides can be used for storing hydrogen Alkali metals do not form these type of hydrides because in their crystal lattices, atoms of hydrogen, which are very small in size do not fit in.

Question 12.
How would you expect the metallic hydrides to be useful for hydrogen storage ? Explain.
Answer:
Some metals like palladium (Pd), platinum (Pt) etc. have a capacity to adsorb large volume of hydrogen on their surface forming hydrides. In fact, hydrogen dissociates on the surface of metal as H atoms which are adsorbed. In order to accommodate these atoms, the metal lattice expands and becomes rather unstable upon heating. The hydride releases hydrogen and changes back to the metallic state. The hydrogen evolved in this manner can be used as a fuel. The metals listed above (belonging to transition metals) can be used to store as well as transport hydrogen which is to be used as a fuel. Thus, metal hydrides have a very useful role in hydrogen economy.

Question 13.
How does atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes ? Explain.
Answer:
Atomic hydrogen torch.When molecular hydrogen is passed through tungsten electric arc (2000 – 3000°C), at low pressure, it dissociates to form atoms of hydrogen, known as atomic hydrogen.

Atomic hydrogen is extremely unstable (life period is about 0-3 sec). Therefore, hydrogen atoms immediately unite to form molecular hydrogen again accompanied by the liberation of a large amount of heat energy. The temperature rises to 4000 – 5000°C. This is the principle of atomic hydrogen torch which is used for welding purposes. oxy-hydrogen torch.

When hydrogen is burnt in oxygen, the reaction is highly exothermic in nature. A temperature ranging between 2800^c to 4000°C is generated. This temperature can be employed for welding purpose in the form of oxy-hydrogen torch.

Question 14.
Among NH₃, H₂O and HF, which would you expect to have highest magnitude of hydrogen bonding and Why ?
Answer:
Hydrogen bonding represents dipolar attraction. Its strength depends upon the magnitude of the polarity of the bond. The electro negativities of three non-metals F, O and N involved in these compounds are 4, 3-5 and 3 respectively. This means that H—F bond is maximum polar and as a result, highest magnitude of hydrogen bonding is HF molecules.

Question 15.
Saline hydrides are known to react violently with water producing fire. Can CO₂, a well known extinguisher, be used in this case ? Explain.
Answer:
Whenever a saline hydride (NaH or CaH₂) reacts with water, the reaction is so highly exothermic that the hydrogen evolved catches fire. For example,

NaH(s) + H₂O(aq) → NaOH(aq) + H₂(g) + heat
CaH₂(s) + 2H₂O(aq) → Ca(OH)₂(aq) + 2H₂(g) + heat

CO₂ which is normally used as fire extinguisher cannot be used in this case because it will react with the hydroxide formed in the reaction to form a carbonate. This will increase the rate of the forward reaction in which heat is evolved.

2 NaOH(aq) + CO₂(g) →Na₂CO₃(aq) + H₂O(aq).

Question 16.
Arrade the following:
(1) IH, NaH and CsH in order of increasing ionic character.
(2) Hlr-H, D—D, and F—F in order of increasing bond dissociation enthalpy.
(3) N₄₁, MgH₂ and H₂O in order of increasing reducing power.
Answer:
(1) Increasing ionic character: LiH < Nail < CsH.
Reason : The ionisation enthaiphy decreases in the order Li > Na> Cs. This influences the ionic character adversely which increases as shown
(2) Increasing bond dissociation enthalpy : F—F < H—H < D—D.
Reason : The bond dissociation enthalphy of :F—F: fluorine is very small (242-6 kJ mol^-1) due to the repulsion in the lone pairs of electrons present on the two F atoms. Out of H₂ and D₂, the bond dissociation enthalphy of H—H (435-88 kJ mol^-1) is less than that of D—D (443-35 kJ mol^-1).
(3)  Increasing reducing power : H₂O < MgH₂ < NaH.
Reason : NaH being ionic in nature is the strongest reducing agent. Both H₂O and MgH₂ are covalent in nature but bond dissociation enthalpy of H₂O is higher. Therefore, it is a weaker reducing agent than MgH₂.

Question 17.
Compare the structures of H₂O and H₂O₂
Answer:
The structure of H₂O is angular while that of H₂O₂ is non-linear just like two opposite open pages of a book. For actual structure of H₂O,
Structure of water (H₂O).

H₂O is a covalent molecule in which the two hydrogen atoms are linked to the oxygen atoms by single covalent bonds. The oxygen atom is surrounded by two shared pairs with hydrogen atoms and it has also two lone pairs of electrons. To have a minimum force of repulsion in the four electron pairs around the oxygen atom, the structure of H₂O molecule is expected to be tetrahedral.

However, the two lone pairs distort its geometry and its bond angle (104-5°) is less than the bond angle of regular tetrahedron (109-28°). As the oxygen atom is more electronegative (3-5) than the hydrogen atom (2-1), therefore, both the O—H bonds are polar in nature. Because of the unsymmetrical nature of H₂O molecule, the bond polarities do not cancel. Therefore, it is polar with dipole moment (ju) equal to 1 -84 D.

Structure of Hydrogen Peroxide
Hydrogen peroxide is a dihydroxy compound (H-O-O-H) and the 0-0 linkage is known as a peroxide linkage. It is a non­linear molecule as the two 0-H bonds are in different planes. The interplanar (dihedral) angle is 111-5° in the gaseous phase but is reduced to 9p-2° in the crystalline state because of hydrogen bonding. The molecular dimensions of hydrogen peroxide in the gas and solid phases are shown.

Question 18.
What do you understand by ‘auto-protolysis’ of water ? What is its significance ?
Answer:
Auto-protolysis of water means ‘self-ionisation’ which proceeds as follows :

Because of auto-protolysis, water behaves both as a Bronsted acid and Bronsted base i.e. it is amphoteric in nature

Question 19.
Consider the reaction of water with F₂ and suggest in terms of oxidation and reduction which species are oxidised/reduced.
Answer:

• Water (H₂O) is oxidised to O₂
  
• Fluorine (F₂) is reduced to F^– ions or HF

Question 20.
Complete the following :
(1) PbSfs) + H₂O₂(aq)→
(2) MnO₄(dq) + H₂O₂(aq) + H⁺(aq) →
(3) CaO(s) + H₂O(g)→
(4) AlCl₃(s) + H₂O(l) →
(5) Ca₃N₂(s) + H₂O(l) →
Answer:

1. PbS(s)+ 4H₂O₂(aq) →PbSO₄(.s) + 4U₂Oaq)
2. 2MnO^–₄ (aq) + 6iA⁺(aq) + 5R₂O₂(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 5O₂(g)
3. CaO(s) + H₂O(g) → Ca(OH)₂(s)
4. AlCl₃(s) + 3H₂O(l) → Al(OH)₃(s) + 3HCl (l)
5. Ca₃N_2(s) + 6H₂O(l) → 3Ca(OH)₂(s) + 2NH₃(g).

Question 21.
Describe the structure of common form of ice.
Answer:
1. Ice fleets over water. In ice (solid state), the H₂O molecules are arranged tetrahedrally ini space. The oxygen atom in each H₂O molecule is linked to two hydrogen atoms by covalent bonds and at the same time its forms hydrogen bonds with the hydrogen atoms of the neighbouring H₂O molecules. This leads to a cage like structure as shown in the Figure 9.5. The structure is also porous because of the voids or empty spaces left.

Now, as ice mbits to form liquid water, the heat energy supplied tends to break some of the hydrogen bonds in H₂O molecules. As a result, the tetrahedral arrangements start collapsing and the H₂O molecules in water come closer to one another. The number of vacant spaces also start decreasing. Therefore, the density of water is more than that of ice. This means that ice always floats over water.

2. Water has maximum density at 4°C. At 273 K (or 0°C), both ice and water coexist. As the temperature is increased, the heat energy supplied will further break the tetrahedral arrangements because of decrease in hydrogen bonding. Therefore, the density of water is expected to increase. But the rise in temperature is also expected to increase the average kinetic energy of the H₂O molecules leading to increase in volume (or “decrease in density). But this effect is negligible upto 4°C and the density of water, therefore, increases. However, if the temperature is increased beyond 4°C, then the effect of the increase in kinetic energy will be more as compared to the effect of increase in density. This is likely to increase the volume of water and decrease its density. Thus, we conclude that upto 4°C, the density of water increases and then decreases after this temperature. In other words, water has maximum density and minimum volume at 4°C.

This property of water is extremely helpful for the animals living under sea water. In severe cold, the surface of the sea almost freezes. But below the surface, there is water at a temperature of about 4°C. The aquatic animals can safely live in water at this temperature.

Question 22.
What causes temporary and permanent hardness of water ?
Answer:
1. Temporary hardness is due to the presence of hydrogen carbonates of calcium and magnesium dissolved in water. It is called temporary because it can be easily removed by simply boiling hard water.

2. Permanent hardness is due to the presence of chlorides and sulphates of calcium and magnesium dissolved in water. It is called permanent hardness as it cannot be removed on boiling hard water

Question 23.
Discuss the principle and method of softening of hard water by synthetic ion-exchange method.
Answer:
Softening of hard water. Softening of water means the removal of the ions from water which make it hard.

Question 24.
Write chemical reactions to show amphoteric nature of water.
Answer:
water covers nearly three-fourth of the earth’s surface in the form of snow over mountains as liquid in the rivers, lakes, springs and oceans. Next to oxygen, water is the most important for human life. One can live without food for a number of days but not without water. It is also used as a solvent for many substances and takes part in a variety of chemical reactions. Water is also a source of heavy water (D₂O) which is extremely essential for nuclear reactors to control the speed of neutrons.

Question 25.
Hydrogen peroxide can act both as an oxidising as well as reducing agent. Explain.
Answer:
The Change in oxidation number accompanying the decomposition of H₂O₂ is as follows :

Since oxygen atom in H₂O₂ can undergo an increase as well as decrease in oxidation number therefore, it can act both as reducing as well as oxidising agent. This is supported by the following reactions.
As reducing agent:

In this reaction, H₂O₂ acts as reducing agent and has reduced Ag₂O K to metallic Ag.
As oxidising agent:

In this reaction, H₂O₂ has oxidised PbS to PbSO₄.

Question 26.
What is meant by demineralised water ? How is it obtained ?
Answer:
Water free from cations (Ca²⁺, Mg²⁺ etc.) and anions (Cl^–, SO^–₄, HCO₃ etc.) responsible for hardness is known as de-ionised or demineralised water. It is formed by passing water repeatedly through cation and anion exchangers.

Now let us try to investigate as to why hard water does not form lather with soap readily. Actually soaps are water soluble sodium (or potassium salts) of higher fatty acids with formula NaX [where X may represent C₁₇H₄₅COO^– (palmitate), C₁₇H₄₃COO^– (oleate), or C₁₇H₃₅COO^–(stereate) ions]. When soap is added to the hard water, the Na⁺ ions are replaced by Ca²⁺ and Mg²⁺ ions present in hard water to form the corresponding calcium and magnesium salts of the acids which get precipitated.

Thus, a lot of soap is wasted due to formation of curdy white precipitate and hard water is not useful in lauhdary. In addition, it cannot be employed in boilers for raising steam. Over a period of a time, the salts present in hard water get deposited inside the walls of the boiler in the form of hard scale. The boiler scale does not allow the flow of heat since it is a poor conductor. Sometimes, it also damages the boiler in case it cracks. Therefore, removal of these salts from water is necessary. This is called softening of water.

Question 27.
Is dimineralised or distilled water always useful for drinking purposes ? If not, how can it be made
useful ?
Answer:
No, dimineralised water or distilled water is not always useful for drinking purposes. It is usually tasteless. Moreover,some ions such as Na⁺, K⁺ and Mg²⁺ etc. are essential to the body. In order to make dimineralised water more useful, some useful salts of sodium and potassium etc. must be dissolved in it.

Question 28.
What properties of water make it useful as a solvent ? What type of compounds can it dissolve and hydrolyse ?
Answer:
Water is a useful, rather excellent solvent due to the following properties :

1. It has high enthalpy of vaporisation and heat capacity.
2. It is a liquid over a wide range of temperature (0° to 100°C)
3. It is polar in nature and high dielectric constant (78-39).
   It can dissolve polar substances and also some organic compounds due to hydrogen bonding.Water can hydrolyse oxides, halides, phosphides, nitrides etc.

Question 29.
Describe be the usefulness of water in biosphere and biological systems.
Answer:

• Water is a liquid with freezing point of 273-2 K and boiling point of 373-2 K.
• Water has the maximum density at 277 K (4°C) e. 1 gm cm^-3.
• In water, the molecules are hydrogen bonded and the hydrogen bonding influences all the physical properties such as state, heat of fusion, heat of vaporisation, melting point and boiling point etc. For example, H₂O is the hydride of the first member of group 16 e. oxygen. The rest of the members of the group are sulphur, selenium, tellurium and polonium. The hydrides of these elements do not show any hydrogen bonding and their physical properties are quite different from H₂O. For example, H₂O is a liquid while H₂S is a gas at room temperature.
• Water is of polar nature. As a result, most of the inorganic substances and also many organic substances having polar bonds in their molecules are water soluble because df inter-molecular hydrogen bonding. In addition to this, water has high heat of vaporisation and it exists in the liquid form over a wide range of temperature (0 to 100°C). Therefore, water is regarded as an excellent solvent and is quite often knoy/n as universal solvent.
• Because of polar nature,’‘there are strong intermolecular forces of attraction and hydrogen bonding present in the H₂O molecules. As a results water has higher specific heat, thermal conductivity, surface tension, dipole moment as compared to other liquids. ‘These properties enable water to play a major role in biosphere. It is responsible for maintaining the body temperature of living beings and also for moderating climate.

Question 30.
Knowing the properties of H₂O and D₂O, do you think that D₂O can be used for drinking purposes ?
Answer:
Heavy water (D₂O) is quite injurious to living beings, plants and animals since it slows down the rates of reactions which occur in them. It fails to supporfelife and has no utility in biosphere.

Question 31.
Hydrolysis is different from hydration. Elaborate.
Answer:
Hydrolysis is a chemical reaction in which a substance reacts with water under neutral, acidic or alkaline conditions. For example, aluminium chloride is hydrolysed by water as follows:
AlCl₃ + 3H₂O—>Al(OH)₃ + 3HCl
Hydration on the other hand is the property of a chemical compound to take up molecules of water of crystallisation and get hydrated. For example, anhydrous copper sulphate (CuSO₄) which is white in colour takes up five molecules of H₂O to form hydrated copper sulphate (CuSO₄ 5H₂O) which has blue colour.

Question 32.
Saline hydrides are generally used to remove traces of water from organic coin pounds. Explain.
Answer:
Saline hydrides (e.g., NaH, CaH₂ etc.) absorb water and react with it as follows :
NaH(s) + H₂O(l) → NaOH (ag) + H₂(g)
CaU₂(s) + 2H₂O(l) →Ca(OH)₂(aq) + H₂(g)
The hydrpgen escapes leaving behind metal hydroxides. Thus, these hydrides can be used to remove traces of water from the organic compounds.

Question 33.
What do you expect the nature of hydrides if formed by the elements of atomic numbers 15, 19, 23 and 44 with dihydrogem?
Answer:

• The elefpent with atomic number (Z) = 15 is P and the corresponding hydride is phosphine [PH₃]. It is covalent in nature.
• The element with atomic number (Z) = 19 is K and the corresponding hydride is potassium hydride (K⁺H“). It is ionic in nature.
• The element with atomic number (Z) = 23 is V(vanadium). It is a transition metal. It forms an interstitial hydride (VH_x.g)
• The element with atomic number (Z) = 44 is Rh (Ruthenium). It is also a transition metal but does not form any hydride due to hydride gap (as it is present in group 8).Out of the hydrides listed, only the ionic hydrides react with water to evolve hydrogen gas.
  2K⁺H-(s) + 2H₂O(l)→ 2KOH(aq) + H₂(g)

Question 34.
Do you are expect different products in solution when aluminum(III) chloride and potassium chloride are treated separately with (1) normal water (2) acidified water and (3) alkaline water ?   (P.I.S.A. Based)
Answer:
Both the compounds are salts and they react in different manner with water of different nature.
(a) Aluminium (III) chloride or AlCl₃ will react with water as follows :
AlCl₃ + 3H₂O →Al(OH)₃ + 3HCl.
The reaction is known as hydrolysis.

• In normal water, Both Al(OH)₃ and HCl will be present.
• In alkaline water, HCl will be neutralised by the alkali added. Therefore, it will contain mainly Al(OH)₃ and will be alkaline in nature.
• In acidic water. In this medium, Al(OH)₃ will be neutralised by the acid added. Therefore, water will contain mainly HCl and will be acidic in nature.

(b) Potassium chloride is a salt of strong acid and strong base. It will remain as such under all the conditions and will not undergo any chemical reaction.

Question 35.
How does H₂O₂ behave as a bleaching agent ?
Answer:
Bleaching nature of hydrogen peroxide is due to oxidation

Question 36.
What do you understand by the terms :
(1) hydrogen economy
(2) hydrogenation
(3) syn gas
(4) water-gas shift reaction
(5) fuel cell ?
Answer:
1. Hydrogen Economy. The basic principle of hydrogen economy is the transportation and storage of energy in the form of liquid or gaseous dihydrogen. In India, dihydrogen was used for the first time in October 2005 for running automobiles like four- wheelers. Initially 5% dihydrogen was mixed with C.N.G. which is commonly used. The experiment is successful till to-day. Efforts are on to increase the percentage of dihydrogen gradually.

Hydrogen oxygen fuel cells can also be used to generate electricity to run electric powered cars. A fuel cell produces electricity from a chemical reaction just as in a battery and can work as long as hydrogen and oxygen are present. One such fuel­ cell made by German motors was on display at Sydney Olympics. A number of companies are working on these projects because of the limited availability and very high cost of petrol. A number of cars were launched in the year 2004 and it is predicted that by 2025, automobiles employing hydrogen as fuel will capture about 25% of the world market.

In these fuel cells, porous carbon electrodes are placed in a concentrated aqueous solution of KOH or NaOH. The temperature is maintained to about 400 K when the gases react to form water accompanied by the release of electrical energy. We will study in detail the working of fuel cells in next class under Electrochemistry .

2. Hydrogenation : Hydrogenation of vegetable oils. It is done by passing hydrogen gas through edible oils (groundnut oil, cotton seed oil etc.) in the presence of Ni at 473 K. As a result, the oils are converted into solid fats, also called vegetable ghees. Actually, oils are unsaturated due to the presence C=C bond. On hydrogenation, the bond changes into C—C bond and as a result, unsaturated oils change into saturated fats.

3.Syn gas : The mixture of CO and H₂ was previously known as water gas. These days, it is known as synthesis gas or syn gas. In addition to hydrocarbons and coke, syn gas’ can also be produced from some other materials containing carbon and hydrogen, e.g., wood scrap, saw dust, newspapers, sewage etc. The process of preparing ‘syn gas’ from coke, coal and other materials is known as coal gasification.

4. Water-gas shift reaction :In place of iron chromate, a mixture of ferric oxide (Fe₂O₃) and chromium oxide (Cr₂O₃) can be used. This reaction is known as Water-gas shift reaction. From the gaseous mixture, carbon dioxide can be removed by either passing into water under pressure or by scrubbing with sodium arsenite solution.

5. Fuel cell : It is a cell which converts chemical energy of fuel directly into electrical energy.

We hope the NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrogen, help you. If you have any query regarding NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrogen, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions.

Question 1.
Assign oxidation numbers to the underlined elements in each of the following species ;
(a) NaH₂PO₄
(b) NaHSO₄
(c) H₄P₂O₇
(d) K₂MnO₄
(e) CaO₂
(f) NaBH₄
(g) H₂S₂O₇
(h) KAl(SO₄)_2.12H₂O
Answer:

Question 2.
What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results ?|
(a) KI₃ (b) H₂S₄O_e (c) Fe₃O₄ (d) CH₃CH₂OH (e) CH₃COOH         (H P. Board 2015)
Answer:
(a) O.N. of iodine in KI₃ : By conventional method, the oxidation number of iodine in the I₃^_ ion may be expressed as :

Explanation : The oxidation number of iodine (I) comes out to be fractional which does not seem to be possible. Let us consider the structure of I₃~ ion. In it, two atoms of iodine are linked to each other by covalent bond (I—I). The iodide ion (I~) is linked to the molecule by co-ordinate bond [I—I <—I]“. The molecule may be represented as K⁺[I—I <—1| . Now, in the anion, the oxidation number of the two I atoms is zero while I^– ion has – 1 oxidation number. The overall oxidation number of L“ ion is :

(b) O.N. of S in H₂S₄O₆ : By conventional method, the oxidation number of sulphur may be calculated as :
H⁺¹₂  S^x₄ O^-2₆
2 + 4x + 6 (- 2) =0 or 4x = 12 – 2 = 10 or x = 5/2
Explanation : In order to account for the fractional value, let us assign oxidation numbers to different sulphur atoms in the structural formula of the acid. The oxidation numbers of the two middle sulphur atoms is zero while the two atoms at the terminal positions have + 5 oxidation number.

Average O.N. of S = —1/4 [5+ 0 + 0 +5] = 5/2

(c) O. N. of Fe in Fe₃O₄ : By conventional method, oxidation number of Fe may be calculated as :
Fe₃^x ₃ 0₃₄
3x + 4(- 2) =0 or x = 8/3
Explanation : Fe₃O₄ is a mixed oxide and is an equimolar mixture of Fe⁺² O^-2 and Fe⁺³₂ O^-2₃
Average O.N. of Fe = 1/3(2 + 2 x 3) =8/3

(d) O.N. of C in CH₃CH₂OH : By conventional method, the oxidation number of carbon in ethanol molecule may be calculated as :
CH₃cH₂OH or C^x₂ H⁺¹₆ O^-2
2x + 6(+ 1) + (- 2) = 0 or 2x + 6 – 2 = 0 or 2x = – 4 or x = – 2
Explanation : Let us calculate the oxidation number of both C₁ and C₂ atoms.
C₂ : The carbon atom is attached to three H atoms (less electronegative than carbon) and one CH₂OH group (more electronegative than carbon).
∴ O.N. of C₂ = 3(+ 1) +x + (- 1) = O or x = -2
C1 : The carbon atom is attached to one OH group (O.N. = – 1), two H atoms (O.N. = + 1) and one CH₃ group (O.N. = + 1)
O.N. of C1 = ( + 1) + x + 2 (+ 1) + 1(- 1) =0 or x = – 2 Average O.N. of C = l/2[ – 2 + (- 2)] = – 2

(e) O.N. of C in CH₃COOH : By conventional method the O.N. of carbon may be calculated as:
CH3COOH or c₂ H₄ 62
2x + 4(+ 1) + 2(— 2) = 0 or x — 0
Explanation : Let us calculate the oxidation number of both C₂ and C1 atoms.
C1 : The carbon atom is attached to three H atoms (less electronegative than carbon) and one COOH group (more electronegative than carbon).
∴ O.N. of C₂ = 3 (+ 1) + x + 1(- 1) = 0 or x = – 2
C₂ : The carbon atom is attached to one OH group (O.N. = – 1) one oxygen atom by double bond
(O.N. = – 2) and one CH₃ group (O.N. = + 1)
O.N. of C1 = 1( + 1) + Jt + 1 (-2) + 1(- 1) = O or x = + 2
∴ Average O.N. of C = 1/2 [+ 2 + (- 2)] = 0

Question 3.
How will you justify that the following reactions are redox reactions in nature ?
(a)  CuO(s) + H₂(g)  → Cu(s) + H₂O(g)
(b) Fe₂O₃(s) + 3CO(g) →2Fe(s) + 3CO₂(g)
(c) 2K(s) + F₂(g)→2K⁺F^–(s)
(d) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
Answer:
A chemical reaction may be regarded as redox reaction if one of the reacting species undergoes increase in O.N. (oxidation) and the other decrease in O.N. (reduction). Based upon this, let us try to justify that the reactions under study are redox reactions in nature.

Question 4.
Fluorine reacts with ice as follows :
H₂O(s) + F₂(g) — HF(g) + HOF(g)
Justify that this reaction is a redox reaction.
Answer:

In this reaction, the F₂ molecule has undergone increase in O.N. in changing to HOF and decrease in O.N. in change to HF. It is therefore, a redox reaction. Since the same reacting species has undergone increase as well as decrease in O.N., it is also called disproportionation reaction.

Question 5.
Calculate the O.N. of sulphur, chromium and nitrogen in H₂SO₅, CrO₅ and NO^–₃ ion. Suggest structure of these compounds. Account for the fallacy if any.
Answer:
(1) Oxidation number of S in H₂SO₅
By conventional method, O.N. of sulphur may be calculated as :
2 x 1 + x + 5 X (-2)= 0 or 2 + x – 10 = 0 or x = + 8 (wrong)
But this cannot be true as maximum oxidation number for sulphur cannot exceed +6 as it has only six valence electrons. The oxidation number of the element can be calculated from the structure of the persulphuric acid.
The sulphur atom is linked one OH group (O.N. = – 1) two oxygen atoms (O.N. = – 2) and one peroxide (- O – O -) linkage (O.N. = – 1)
O.N. of S = 1 (- 1) + x + 2(- 2) + 1( – 1) = 0
    or 
x = + 6

(2) Oxidation number of Cr in CrO₅
By conventional method, O.N. of chromium may be calculated as :
x + 5 (-2) = 0 or x = + 10 (wrong)

This is wrong because maximum oxidation number of Cr cannot be more than +6, since it has only six electrons (3d⁵4.s¹) to take part in bond formation. The oxidation number of the metal atom can be calculated by taking into account its structure.
The Cr atom is linked with one oxygen atom by double bond (O. N. = – 2) and four oxygen atoms by peroxide linkages (O.N. = -1)
.’. O.N. of Cr = 4(- 1) \x + 1(- 2) = 0 or x = + 6

(3) Oxidation number N in NO^–₃ ion
By conventional method, O.N. of N atom in the ion may be calculated as : x + 3 (- 2) = – 1 or x = +5.This can be further made clear from the structure of the ion. In it, the nitrogen atom is linked to one single bonded oxygen atom (O.N. = – 2), one double bonded oxygen atom
(O.N.= – 2) and one oxygen atom by dative bond (O.N. = – 2).
O.N. of N = 1 (- 2) + x + 1( – 2) + 1 (- 2) = – 1 or x = + 5

Question 6.
Write the formulas of the following compounds :
(a) Mercury (II) chloride (b) Nickel (II) sulphate (c) Tin (IV) oxide (d) Thallium (I) sulphate (e) Iron (III) sulphate (f) Chromium (III) oxide.
Answer:
The formulas have been shown by stock notations. The Roman numerals given in parenthesis represent the oxidation state of the metal atom. Taking into account the oxidation state of the anions, the chemical formulas of the compounds may be written as :
(a) HgCl₂ (b) NiSO₄ (c) SnO₂ (d) T1₂SO₄ (e) Fe₂(SO₄)₃ (f) Cr₂O₃

Question 7.
Suggest a list of substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.
Answer:
Variable oxidation states of carbon :

Question 8.
While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid can act only as oxidising agents. Why ?
Answer:
In sulphur dioxide (SO₂) and hydrogen peroxide (H₂O₂), the oxidation states of sulphur and oxygen are +4 and -1 respectively. Since they can increase as well as decrease when these compounds take part in chemical reactions, they can act as oxidising as well as reducing agents. For example


In ozone (O₃), the oxidation state of oxygen is zero while in nitric acid (HNO₃), the oxidation state is nitrogen is +5. Since both them can undergo decrease in oxidation state and not an increase in its value, they can act only as oxidising agents and not as reducing agents.

Question 9.
Consider the reactions :
(a) 6CO ₂(g) + 6H₂O (l) → C₆H₁₂O₆(s) + 6O₂(g)
(b) O₃(g) + H₂O₂(l) → H₂O (l) + 2O₂(g)
Why it is more appropriate to write these reactions as :
(a) 6CO₂(g) + 12H₂O(l)→ C₆H₁₂O₆(s) + 6H₂O(l) + 6O₂(g)
(b) 6₃(g) + H₂O₂(l)→ H₂O(l) + O₂(g) + O₂(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
Answer:
(a)  6CO₂(g) + 6H₂O(l) → C₆H₆O₆(s) + 6O₂(g)
Let us try to balance the equation by ion-electron method.


Note:
(1) In this reaction O₃ acts as an O.A. and H₂O₂ acts as a R.A.
(2) If a co-ordinate bond exist between two similar atoms, the donor atom acquires + 2 O.N. and the acceptor gets -2 O.N.
This explains the path of the reaction i.e., how electrons are lost and gained as well as and why it is more appropriate to write the equation as above.

Question 10.
The compound AgF₂ is an unstable compound. However if formed, the compound acts as a very strong oxidising agent. Why ?
Answer:
In AgFL the oxidation state of silver is + 2 i.e., it exists as Ag²⁺ ion. This is quite unstable since the normal stable oxidation state of the metal is + 1 or it exists as Ag⁺ ion. This means that if the compound AgF₂ be formed, it will undergo reduction by the gain of electrons. It is therefore, a very strong oxidising agent
Ag²⁺ + e^– — Ag⁺

Question 11.
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations
Answer:

Question 12.
(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene, we use alcoholic potassium permanganate as an oxidant, why ? Write a balanced redox equation for the reaction.

(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get a colourless pungent smelling gas HC1. However, if mixture contains bromide, then red vapours of bromine are evolved. Why ?
Answer:

In the manufacture of benzoic acid from toluene, neutral medium is preferred over acidic or alkaline medium due to following reasons.

•  In the neutral medium, neither acid or base is to be added externally. This will definitely reduce the cost of the manufacturing process.
• Alcohol if used as solvent will help in the formation of a homogeneous mixture between toluene (non-polar) and KMnO₄ (ionic). Actually alcohol has non-polar alkyl group as well as polar OH group.

(b)
A chloride such as NaCl reacts with concentrated sulphuric acid to evolve hydrogen chloride gas upon heating. It has a pungent smell.

Hydrogen bromide is also expected to be formed in the same way when a bromide is heated with concentrated sulphuric acid. However, the acid being a strong oxidising agent will oxidise HBr to bromine (Br₂) which gets evolved  as red vapours.

Please note HC1 (g) is quite stable and is not oxidised to chlorine with concentrated sulphuric acid.

Question 13.
Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions :

1. 2AgBr(s) + C₆H₆O₂(aq) →2Ag(s) + 2HBr(aq) + C₆H₄O₂(aq)
2. HCHO(l) + 2[Ag(NH₃)₂]⁺(aq) + 3OH^–(aq) → 2Ag(s) + HCOO^– (aq) + 4NH₃(aq) + 2H₂O(l)
3. HCHO(l) + 2Cu²⁺(aq) + 5OH^–(aq)→ Cu₂(O(s) + HCOO-(aq) + 3H₂O(l)
4. N₂H₄(l) + 2H₂O₂(l) →N₂(g) + 4H₂O(l)
5. Pb(s) + PbO₂(s) + 2H₂SO₄(aq)→ 2PbSO₄(s) + 2H₂O(l)

Answer:


AgBr         : reduced ; acts as oxidising agent
C₆H₆O₂    : oxidised ; acts as reducing agent.

Question 14.
Consider the reactions :
2S₂O^2-₃(aq) + I₂(s) → S₄O^2-₆(aq) + 2I^–(aq)
S₂O² (aq) + 2Br₂(l) + 5H₂O (l) → 2SO^2-₄(aq) + 4Br-(ag) + 10H +(aq)
Why does the same reductant the osulphate react differently with iodine and bromine ?
Answer:
I₂ oxidises thiosulphate ion to tetrathionate ion i.e., from O.S. of + 2 for S (in S₂O^2-₃ ) to S.O. of + 5/2 for S (in S₄O^2-₄ ion).
Br₂ oxidises thiosulphate ion to sulphate ion i.e., from O.S. of + 2 for S (in S₂O^2-₃) to O.S. of + 6 for S
(in SO^2-₄ ion)
This is because Br₂ is a stronger

Question 15.
Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Answer:
The halogens (X₂) have strong electron accepting tendency. They are therefore, powerful oxidising agents. The relative order of oxidising powers of halogens is :

Fluorine is the strongest oxidising agent (oxidant). It can be justified by the fact that it can liberate the other halogens from their respective compounds. For example,
2KC1 + F_{2  →}  2KF + Cl₂
2KBr + F₂ → 2KF + Br₂
2KI + F₂ → 2KF + I₂
Among halogen acids (HX), the HI is the strongest reducing agent or reductant because its bond dissociation enthalpy is the minimum (299 kJ mol^-1)

(kJ mol^-1)
The iodination of methane for examples, is of reversible nature because HI formed in the reaction being a very strong reducing agent converts iodomethane back to methane

However, the halogenation with other halogen acids is not of reversible nature. This justifies that HI is the strongest reducing agent in the halogen acids.

Question 16.
Why does the following reaction occur ?
XeO^4-₆ (aq) + 2F^–(aq) + 6H⁺(aq)→ XeO₃(g) + 3H₂O(l)
What conclusion about the compound Na₄XeO₆ (of which XeO^4-₆ is a part can be drawn from the reaction ?
Answer:

In the reaction, O.N. of Xe decreases from + 8 to + 6. Therefore, XeO₆ is reduced.O.N. of F increases from – 1 to 0. Therefore, F^– is oxidised. The redox reaction occurs because Na₂XeO₆ (of which XeOg“ is a part) is a stronger oxidising agent than F₂.

Question 17.
Consider the reactions :
(a) H₃PO₂ (aq) + 4AgNO₃ (aq) + 2H₂O (l)→H₃PO₄ (aq) + 4Ag (s) + 4HNO₃ (aq)
(b) H₃PO₂ (aq) + 2CuSQ₄ (aq) + 2H₂O (l)→ H₃PO₄ (aq) + 2Cu (s) + 2H₂SO₄ (aq)
(c) C₆H₅CHO (I) + 2[Ag(NH₃)₂⁺ (aq) + 3OH^– (aq)→ C₆H₅COO (aq) + 2Ag (s) + 4NH₃ (aq) + 2H₂O (l)
(d) C₆H₅CHO (l) + 2Cu²⁺(aq) + SOH^–(aq)→No change observed
What inference do you draw about the behaviour of Ag⁺ and Cu²⁺ ions from these reactions ?
Answer:
(a) Hypophosphorus acid (H₃PO₂) reduces Ag⁺ ions to Ag which gets precipitated
(b) Hypophosphorus acid (H₃PO₂) reduces Cu²⁺ ions to Cu which gets precipitated
(C) Benzaldehyde (C₆H₅CHO) reduces Ag+ions to Ag which gets precipitated
(d) Benzaldehyde (C₆H₅CHO) is not in a position to reduce Cu²⁺ ions to Cu.
This shows that Ag⁺ ions are stronger oxidising agent than Cu²⁺ ions in all the reactions. This is quite evident from their relative positions in the reactivity series.

Question 18.
Balance the following equations by ion-electron method

Answer:

Question 19.
Balance the following equations in basic medium by ion-electron method and oxidation number method and identify the oxidising agent and the reducing agent.
(a) p₄(s) + OH⁺(aq) —> PH₃(g) + H₂PO^–₂ (aq)
(b) N₂H₄(7) + CIO^–₃ (aq) —> NOfe) + C l^–(g)
(c) C1₂O₇(g) + H₂O₂(aq) —> CIO^–₂(aq) + O₂(g) + H^–
Answer:

Question 20.
Write four informations about the reaction :
(CN)₂ (g) + 2OH^– (aq)—> CN^– (aq) + CNO (aq) + H₂O (l).
Answer:

1. The reaction involves the decomposition of cyanogen (CN)₂ in the alkaline medium.
2. Both (CN)₂ i.e. cyanogen and CN^– i.e. cyanide ions are pseudo halogens in nature i.e. they behave like halogens in characteristics.
3. Cyanogen undergoes disproportionation in the reaction. It undergoes simultaneous increase and decrease in O.N.of the species involved.
4. It is an example of redox reaction.

Question 21.
The Mn³⁺ ion is unstable in solution and under goes disproportionation to give Mn²⁺, MnO₂ and H⁺ ion. Write balanced ionic equation for the reaction.
Answer:

Question 22.
Consider the elements : Cs, Ne, I, F

1. Identify the element that exhibits only -ve oxidation state.
2. Identify the element that exhibits only +ve oxidation state.
3. Identify the element that exhibits +ve and -ve oxidation states.
4. Identify the element that neither exhibits + ve and -ve oxidation states.

Answer:

1. F (fluorine) exhibits only -ve oxidation state (- 1) in its compounds because it is the most electronegative element.
2. Cs (cesium) exhibits only +ve oxidation state (+ 1) its compounds because it is the most electropositive element.
3. Iodine has seven electrons in the valance shell as well as vacant 5d orbital to which electrons from 5p and 5s orbitals can be shifted. Therefore, the elements exhibits – 1 oxidation state as well as variable positive oxidation states of + 1, +3, +5 and +7 in its compounds.
4. Ne (neon) neither exhibits +ve nor -ve oxidation states because it is a noble gas element with completely filled orbitals (Is² 2s²2p⁶). Moreover, it has no vacant d-orbitals.

Question 23.
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place.
(P.I.S.A. Based)
Answer:
Chlorine reacts with sulphur dioxide in the presence of water as follows
Cl₂ (aq) + SO₂ (aq) + H₂O (l) → Cl^– (aq) + SO₄^2- (aq)

Question 24.
From the periodic table, select three non-metals and three metals which can show disproportionation reactions.
Answer:
Non-metals and metals which show variable oxidation states can take part in the disproportionation reactions. For examples,
Non-metals : The non-metals showing disproportionation reactions states are : P₄, Cl₂ and S₈.

Question 25.
In the Ostwald process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.0 g of ammonia and reacting with 20.0 g of oxygen ?
Answer:



Question 26.
Using the standard electrode potentials given in the table, predict if the reaction between the following is feasible.
(a) Fe³⁺(aq) and I^–(aq)
(b) Ag⁺(aq) and Cu(s)
(c) Fe³⁺(aq) and Cu(s)
(d) Ag(s) and Fc³⁺(aq)
(f) Br₂(aq) and Fe²⁺(aq).
Answer:
A particular reaction can be feasible if e.m.f. of the cell based on the E° values is positive. Keeping this in mind, let us predict the feasibility of the reactions.

Question 27.
Predict the products of electrolysis of each of the following :

1. An aqueous solution of AgNO₃ using silver electrodes.
2. An aqueous solution of AgNO₃ using platinum electrodes.
3. A dilute solution of H₂SO₄ using platinum electrodes.
4. An aqueous solution of CuCl₂ using platinum electrodes.

Answer:
(1) An aqueous solution of AgNO₃ using platinum electrodes :     (P.I.S.A. Based)
Both AgNO₃ and water will ionise in aqueous solution

At cathode : Ag⁺ ions with less discharge potential are reduced in preference to H⁺ ions which will remain in solution. As a result, silver will be deposited at cathode.
Ag⁺ (aq) + e^– → Ag (deposited)
At anode : An equivalent amount of silver will be oxidised to Ag⁺ ions by releasing electrons.

As result of electrolysis, Ag from silver anode dissolves as Ag⁺(aq) ions while an equivalent amount of Ag⁺(aq) ions from the aqueous AgNO₃ solution get deposited on the cathode.

(2) An aqueous solution of AgNO₃ using platinum electrodes :

In the case, the platinum electrodes are the non-attackable electrodes. On passing current the following changes will occur at the electrodes.
At cathode : Ag⁺ ions will be reduced to Ag which will get deposited at the cathode.
At anode : Both NO^–₃ and OH^– ions will migrate. But OH^– ions with less discharge potential will be oxidised in preference to NO₃ ions which will remain in solution.
OH- (aq) → OH + e^–; 4OH → 2H₂O(l) + O₂ (g)
Thus, as a result of electrolysis, silver is deposited on the cathode while O₂ is evolved at the anode. The solution will be acidic due to the presence of HNO₃.

(3) A dilute solution of H₂SO₄ using platinum electrodes :
On passing current, both acid and water will ionise as follows :

At cathode :
H⁺ (aq) ions will migrate to the cathode and will be reduced to H₂.
H+ (aq) + e-→ H ; H + H→ H₂ (g)
Thus, H₂ (g) will evolve at cathode.
At anode : OH ions will be released in preference to SO^2-₄ ions because their discharge potential is less. They will be oxidised as follows :
OH^– (aq) → OH + e^– ; 4OH → 2H₂O(l) + O₂ (g)
Thus, O₂ (g) will be evolved at anode. The solution will be acidic and will contain H₂SO₄.

(4) An aqueous solution of CuCl₂ using platinum electrodes :

The electrolysis proceeds in the same manner as discussed in the case of AgNO₃ solution. Both CuCl₂ and H₂O will ionise as follows :

At cathode :  Cu²⁺ ions will be reduced in preference to H⁺ ions and copper will be deposited at cathode.
Cu²⁺ (aq) + 2e^– → Cu (deposited)
At anode : Cl^– ions will be discharged in preference to OH^– ions which will remain in solution.
Cl^–→Cl^–+ e^– ; Cl + Cl → Cl₂ (g) (evolved)
Thus, Cl₂ will evolve at anode.

Question 28.
Arrange the following metals in the order in which they displace each other from their salts. Al, Cu, Fe, Mg and Zn
Answer:
This is based upon the relative positions of these metals in the activity series. The metal placed lower in the series can displace the metals occupying a higher position present as its salt. Based upon this, the correct order is : Mg, Al, Zn, Fe, Cu.

Question 29.
Given the standard electrode potentials
K+/K = – 2.93 V, Ag+/Ag = 0.80V
Hg²+/Hg = 0.79 V ; Mg²⁺/Mg = – 2.37 V, Cr³⁺/Cr = – 0.74 Y
Arrange these metals in increasing order of their reducing power.
Answer:
If may be noted that lesser the E° value for an electrode, more will be reducing power. The increasing order of reducing power is :
Ag < Hg < Cr < Mg < K.

Question 30.
Depict the galvanic cell in which the reaction Zn(s) + 2Ag ⁺(aq)→ Zn²⁺(aq) + 2Ag(s) takes place. Further show :
(1) which electrode is negatively charged.
(2) the carriers of the current in the cell.
(3) individual reaction at each electrode.
Answer:
The galvanic cell for the reaction is

• Zinc electrode (anode) is negatively charged.
• Current flows from silver to zinc in outer circuit
• Anode : Zn (s) → Zn²⁺(aq) + 2e^–(oxidation)
  Cathode :   2Ag⁺ (ag) + 2e^– → 2Ag(s) (reduction)

We hope the NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions, help you. If you have any query regarding NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium.

Question 1.
A liquid is in equilibrium with its vapours in a sealed container at a fixed temperature. The volume of the container is suddenly increased,
(i) What is the initial effect of the change on the vapour pressure ?
(ii) How do the rates of evaporation and condensation change initially ?
(iii) What happens when equilibrium is restored finally and what will be the final vapour pressure ?
Answer:
(i) On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours are now distributed over a larger space.
(ii) On increasing the volume of the container, the rate of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.
(iii) Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.

Question 2.
What is Kc for the following reaction in a state of equilibrium :
2SO₂(g) + O₂(g) \(\rightleftharpoons \) 2SO₃(g) ?
Given : [SO₂] = 0.6 M ; [O₂]= 0.82 M ; and [SO₃] = 1.90 M
Answer:

Question 3.
At certain temperature and total pressure of 105Pa, iodine vapours contain 40t% by volume of iodine atoms in the equilibrium I₂ (g) \(\rightleftharpoons \) 2I (g). Calculate K_p for the equilibrium.
Answer:
According to available data :
Total pressure of equilibrium mixture = 10⁵ Pa
Partial pressure of iodine atoms (I) = \(\frac { 40 }{ 100 } \) × (10⁵ Pa) = 0.4 × 10⁵ Pa
Partial pressure iodine molecules (I₂) = \(\frac { 60 }{ 100 } \) × (10⁵ Pa) = 0.6 × 10⁵ Pa

Question 4.
Write the expression for the equilibrium constant for the following reactions
(i) 2NOCl(g)) \(\rightleftharpoons \) 2NO(g) + Cl₂(g)
(ii) 2CU(N0₃)₂(s) \(\rightleftharpoons \) 2CuO(s) + 4NO₂(g) + O₂(g)
(iii) CH₃COOC₂H₅(aq) + H₂O(l) \(\rightleftharpoons \) CH₃COOH(aq) + C₂H₅OH(aq)
(iv) Fe³⁺(aq) + 3OH^–(aq) \(\rightleftharpoons \) Fe(OH)₃(s)
(v) I₂(s) + 5F₂(g) \(\rightleftharpoons \) 2IF₅(l).
Answer:

Question 5.
Find the value of Kc for each of the following equilibria from the value of K_p:
(a) 2NOCl (g) \(\rightleftharpoons \) 2NO (g) + Cl₂ (g) ; K_p = 1.8 x 10^-2 atm at 500 K
(b) CaCO₃ (s) \(\rightleftharpoons \) CaO (s) + CO₂ (g) ; K_p = 167 atm at 1073 K.
Answer:
K_p and K_c are related to each other as K_p = K_c ( RT)^∆ng
The value of K_c can be calculated as follows :
(a) 2NOCl (g) \(\rightleftharpoons \) 2NO (g) + Cl₂ (g)
K_p = 1.8 × 10^-2 atm,
∆^ng = 3 – 2 = 1 ; R = 0.0821 litre atm K^-1 mol^-1 ; T = 500 K

(b) CaCO₃ (s) \(\rightleftharpoons \) CaO (s) + CO₂ (g)
K_p = 167 atm, ∆^ng = 1 – 0 = 1
R = 0.0821 liter atm K^-1 mol^-1 ; T = 1073 K

Question 6.
For the following equilibrium, K_c = 6.3 × 10¹⁴ at 1000 K
NO (g) + O₃ (g) \(\rightleftharpoons \) NO₂(g) + O₂ (g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc for the reverse reaction ?
Answer:
For the reverse reaction K_c = \(\frac { 1 }{ { K }_{ c } } \) = \(\frac { 1 }{ 6.3\times { 10 }^{ 14 } } \) = 1.59 × 10^-15

Question 7.
Explain why pure liquids and solids can be ignored while writing the value of the equilibrium constant.
Answer:
In writing the expression for the equilibrium constant, molar concentrations of the species involved in the reaction are taken into account.

We all know that the molar concentration of a substance implies number of moles per unit volume (moles/volume) or (mass/volume).

Since mass divided by volume represents density of the substance therefore, molar concentration of a substance is directly related to its density.
Molar concentration = \(\frac { No.ofmoles }{ Volume } \) α \(\frac { Mass }{ Volume } \) (α density)
Density as we know, is an intensive property and does not depend upon the mass of the substance.

Consequently, molar concentration of a pure substance (solid or liquid) has always the same value and it can be ignored while writing the value of the equilibrium constant.

However, for substances in gaseous state or in aqueous solution, the amount in a given volume can vary and their molar concentration does not remain constant and cannot be ignored while writing the expression for the equilibrium constant.

Question 8.
Reaction between nitrogen and oxygen takes place as follows :
2N₂ (g) + O₂ (g) \(\rightleftharpoons \) 2N₂O (g)
If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a reaction vessel of volume 10 L and allowed to form N₂0 at a temperature for which K_c = 2.0 × 10^-37, determine the composition of the equilibrium mixture.
Answer:
Let x moles of N₂ (g) take part in the reaction. According to the equation, \(\frac { x }{ 2 } \) moles of O₂ (g) will react to form x moles of N₂O (g). The molar concentration per litre of different species before the reaction and at the equilibrium point is :

The value of equilibrium constant (2.0 × 10^-37) is extremely small. This means that only small amounts of reactants have reacted. Therefore, x is extremely small and can be omitted as far as the reactants are concerned.

As x is extremely small, it can be neglected.
Thus, in the equilibrium mixture
Molar cone. of N₂ = 0.0482 mol L^-1
Molar cone. of O₂ = 0.0933 mol L^-1
Molar cone. of N₂O = 0.1 × x = 0.1 × 6.6 10^-20 mol L^-1
= 6.6 × 10^-21

Question 9.
Nitric oxide reacts with bromine and gives nitrosyl bromide as per reaction given below :
2NO (g) + Br₂ (g) \(\rightleftharpoons \) 2NOBr (g)
When 0.087 mole of NO and 0.0437 mole of Br₂ are mixed in a closed container at constant temperature, 0.0518 mole of NOBr is obtained at equilibrium. Determine the composition of the equilibrium mixture.
Answer:
The balanced chemical equation for the reaction is :
2NO (g) + Br₂ (g) \(\rightleftharpoons \) 2NOBr (g)
According to the equation, 2 moles of NO (g) react with 1 mole of Br₂ (g) to form 2 moles of NOBr (g). The composition of the equilibrium mixture can be calculated as follows :
No. of moles of NOBr (g) formed at equilibrium = 0.0518 mol (given)
No. of moles of NO (g) taking part in reaction = 0.0518 mol
No. of moles of NO (g) left at equilibrium = 0.087 – 0.0518 = 0.0352 mol
No. of moles of Br₂ (g) taking part in reaction = \(\frac { 1 }{ 2 } \) × 0.0518 = 0.0259 mol
No. moles of Br₂ (g) left at equilibrium = 0.0437 – 0.0259 = 0.0178 mol
The initial molar concentration and equilibrium molar concentration of different species may be represented as :

Question 10.
At 450 K, K_p = 2.0 × 10¹⁰ bar^-1 for the equilibrium reaction :
2SO₂ (g) = O₂ (g) \(\rightleftharpoons \) 2SO₃(g)
Answer:

Question 11.
A sample of HI (g) is placed in a flask at a pressure of 0.2 atm. At equilibrium partial pressure of HI (g) is 0.04 atm. What is K_p for the given equilibrium ?
2HI (g) \(\rightleftharpoons \) H₂(g) + I₂(g)
Answer:
pHI = 0.04 atm, pH₂ = 0.08 atm ; pI₂ = 0.08 atm
K_p = \(\frac { p{ H }_{ 2 }\times p{ I }_{ 2 } }{ { p }_{ HI }^{ 2 } } \) = \(\frac { (0.08atm)\times (0.08atm) }{ 0.04atm)\times (0.04atm) } \) = 4.0

Question 12.
A mixture of 1.57 mol of N₂, 1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant K_c for the reaction given as follows :
N₂(g) + 3H₂(g) \(\rightleftharpoons \) 2NH₃(g) is 1.7 × 10^-2.
Is this reaction at equilibrium ? If not, what is the direction of net reaction ?
Answer:
The reaction is : N₂(g) + 3H₂(g) \(\rightleftharpoons \) 2NH₃(g)

The equilibrium constant (Kc) for the reaction = 1.7 X 10^-2
As Q_c, ≠ K_c ; this means that the reaction is not in a state of equilibrium.
As Q_c > K_c ; the direction of the net reaction must be in the backward direction so that the molar concentration of NH₃ may decrease and equilibrium may be attained again.

Question 13.
The equilibrium constant expression for a gas reaction is,

Write the balanced chemical equation corresponding to this expression.
Answer:
Balanced chemical equation for the reaction is
4 NO(g) + 6 H₂O (g) \(\rightleftharpoons \) 4NH₃ (g) + 5 O₂(g)

Question 14.
If 1 mole of H₂O and 1 mole of CO are taken in a 10 litre vessel and heated to 725 K, at equilibrium point 40 percent of water (by mass) reacts with carbon monoxide according to equation!
H₂O(g) + CO(g) \(\rightleftharpoons \) H₂(g) + CO₂(g)
Calculate the equilibrium constant for the reaction.
Answer:
Number of moles of water originally present = 1 mol
Percentage of water reacted =40%
Number of moles of water reacted = \(\frac { 1\times 40 }{ 100 } \) = 0.4 mol
Number of moles of water left = (1 – 0.4) = 0.6 mol
According to the equation, 0.4 mole of water will react with 0.4 mole of carbon monoxide to form 0.4 mole of hydrogen and 0.4 mole of carbon dioxide.
Thus, the molar cone./litre per litre of the reactants and products before the reaction and at the equilibrium point are as follows :

Question 15.
At 700 K, the equilibrium constant for the reaction H₂(g) + I₂(g) \(\rightleftharpoons \) 2HI(g) is 54.8. If 0.5 mol L^-1 of HI (g) is present at equilibrium at 700 K, what are the concentration of H₂(g) and I₂(g) ? Assume that we initially started with HI(g) and allowed it to reach equilibrium at 700 K.
Answer:

Question 16.
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ?
2ICl(g) \(\rightleftharpoons \) I₂(g) + Cl₂(g) ; K_c = 0.14
Answer:
Suppose at equilibrium, the molar concentration of both I₂(g) and Cl₂(g) is x mol L^-1

Question 17.
K_p = 0.04 atm at 898 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4 atm pressure and allowed to come to equilibrium.
C₂H₆(g) \(\rightleftharpoons \) C₂H₄(g) + H₂(g)
Answer:

Question 18.
The ester, ethyl acetate is formed by the reaction of ethanol and acetic acid and the equilibrium is represented as :
CH₃COOH(l) + C₂H₅OH(l) \(\rightleftharpoons \) CH₃COOC₂H₅(l) + H₂O(l)
(i) Write the concentration ratio (concentration quotient) Q for this reaction. Note that water is not in excess and is not a solvent in this reaction.
(ii) At 293 K, if one starts with 1.000 mol of acetic acid and 0.180 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.50 mol of ethanol and 1.000 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate
is found after some time. Has equilibrium been reached ?
Answer:

Question 19.
A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was reached, the concentration of PCl₅ was found to be 0.5 x 10^-1 mol L^-1. If K_c is 8.3 × 10^-3, what are the concentrations of PCl₃ and Cl₂ at the equilibrium ?
Answer:
Let the initial molar concentration of PCl5 per litre = x mol
Molar concentration of PCl₅ at equilibrium = 0.05 mol
∴ Moles of PCl₅5 decomposed = (x – 0.05) mol
Moles of PCl₃ formed = (x – 0.05) mol
Moles of Cl₂ formed = (x – 0.05) mol
The molar cone. /litre of reactants and products before the reaction and at the equilibrium point are:

Question 20.
One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by
carbon monoxide to give iron metal and CO₂
FeO(s) + CO(g) \(\rightleftharpoons \) Fe(s) + CO₂(g); K_p = 0.265 atm at 1050 K
What are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial pressures are :
Pco = 1.4 atm and Pco₂ = 0.80 atm ?
Answer:

Question 21.
Equilibrium constant Kc for the reaction, N₂ (g) + 3H₂ (g) \(\rightleftharpoons \) 2NH₃ (g) at 500 K is 0.061. At particular time, the analysis shows that the composition of the reaction mixture is : 3.0 mol L^-1 of N₂ ; 2.0 mol L^-1 of H₂ ; 0.50 mol L^-1 of NH₃. Is the reaction at equilibrium ? If not, in which direction does the reaction tend to proceed to reach the equilibrium ?
Answer:

Question 22.
Bromine monochloride (BrCl) decomposes into bromine and chlorine and reaches the equilibrium :
2BrCl (g) \(\rightleftharpoons \) Br₂ (g) + Cl₂ (g)
The value of K_c is 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3× 10^-3 mol L^-1, what is its molar concentration in the mixture at equilibrium ?
Answer:

Question 23.
At 1127 K and 1 atmosphere pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% by mass.
C(s) + CO₂(g) \(\rightleftharpoons \) 2CO(g)
Calculate K_c for the reaction at the above temperature.
Answer:

Question 24.

Answer:

Question 25.
Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume ?
(i) PCl₅(g) \(\rightleftharpoons \) PCl₃(g) + Cl₂
(ii) CaO(s) + CO₂(g) \(\rightleftharpoons \) CaCO₃(s)
3Fe(s) + 4H₂O(g) \(\rightleftharpoons \) Fe₃O₄(s) + 4H₂(g)
Answer:
(i) Pressure will increase in the forward reaction and number of moles of products will increase.
(ii) Pressure will increase in backward reaction and number of moles of products will decrease.
(iii) The change in pressure will have no effect on the equilibrium constant and there will be no change in the number of moles.

Question 26.
Which of the following reactions will get affected by increase in pressure ? Also mention whether the change will cause the reaction to go to the right or left direction.

Answer:
Only those reactions will be affected by increasing the pressure in which the number of moles of the gaseous reactants and products are different (np -/ nr) (gaseous). With the exception of the reaction (1) ; all the remaining five reactions will get affected by increasing the pressure. In general,
• The reaction will gq to the left if n_p > n_r.
• The reaction will go to the right if n_p < n_r.
Keeping this in mind,
(i) Increase in pressure will not affect equilibrium because n_p = n_r =3
(ii) Increase in pressure will favour backward reaction because n_p (2) > n_r (1)
(iii) Increase in pressure will favour backward reaction because n_p (10) > n_r (9)
(iv) Increase in pressure will favour forward reaction because n_p(1) < n_r (2)
(v) Increase in pressure will favour backward reaction because n_p (2) > n_r ( 1)
(vi) Increase in pressure will favour backward reaction because n_p (1) > n_r (0).

Question 27.
The equilibrium constant for the following reaction is 1.6 x 10⁵ at 1024 K.
H₂(g) + Br₂(g) \(\rightleftharpoons \) 2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.
Answer:

Question 28.
Hydrogen gas is obtained from the natural gas by partial oxidation with steam as per following endothermic reaction :
CH₄ (g) + H₂O (g) \(\rightleftharpoons \) CO (g) + 3H₂ (g)
Write the expression for K_p for the above reaction
How will the value of K_p and composition of equilibrium mixture be affected by :
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst ?
Answer:
The expression for K_p for the reaction is :

(i) By increasing the pressure, the no. of moles per unit volume will increase. In order to decrease the same, the equilibrium gets shifted to the left or in the backward direction. As a result, more of reactants will be formed and the value of K_p will decrease.
(ii) If the temperature is increased, according to Le Chatelier’s principle, the forward reaction will be favoured as it is endothermic. Therefore, the equilibrium gets shifted to the right and the value of K_p will increase.
(iii) The addition of catalyst will not change the equilibrium since it influences both the forward and the backward reactions to the same extent. But it will be attained more quickly.

Question 29.
What is the effect of :
(i) addition of H₂
(ii) addition of CH₃OH
(iii) removal of CO
(iv) removal of CH₃OH
On the equilibrium 2H₂ (g) + CO (g) \(\rightleftharpoons \) CH₃OH ?
Answer:
(i) Equilibrium will be shifted in the forward direction.
(ii) Equilibrium will be shifted in the backward direction.
(iii) Equilibrium will be shifted in the backward direction.
(iv) Equilibrium will be shifted in the forward direction.

Question 30.
At 473 K, the equilibrium constant Kc for the decomposition of phosphorus pentachloride (PCl₅) is 8.3 × 10^-3. If decomposition proceeds as :
PCl₅ (g) \(\rightleftharpoons \) PCl₃ (g) + Cl₂ (g) ; ∆H = + 124.0 kJ mol^-1
(a) Write an expression for Kc for the reaction.
(b) What is the value of Kc for the reverse reaction at the same temperature ?
(c) What would be the effect on Kc if
Answer:

Question 31.
Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction.
CO(g) + H₂O(g) \(\rightleftharpoons \) CO₂(g) + H₂(g)
If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam so that pco = PH₂O = 4.0 bar, what will be the partial pressure of H₂ at equilibrium ? K_p = 0.1 at 400°C.
Answer:

Question 32.
Predict which of the following will have appreciable concentration of reactants and products :
(a) Cl₂ (g) \(\rightleftharpoons \) 2Cl (g) ; K_c = 5 ×10^-39
(b) Cl₂ (g) + 2NO (g) \(\rightleftharpoons \) 2NOCl (g) ; K_c = 3.7 × 10⁸
(c) Cl₂ (g) + 2NO₂ (g) \(\rightleftharpoons \) 2NO₂Cl (g) ; K_c = 1.8.
Answer:
Following conclusions can be drawn from the values of Kc.
(a) Since the value of K_c is very small, this means that the molar concentration of the products is very small as compared to that of the reactants.
(b) Since the value of K_c is quite large, this means that the molar concentration of the products is very large as compared to that of the reactants.
(c) Since the value of K_c is 1.8, this means that both the products and reactants have appreciable concentration.

Question 33.
The value of K_c for the reaction 3O₂(g) \(\rightleftharpoons \) 2O₃(g) is 2.0 × 1(10^-50 at 25°C. If equilibrium concentration of O₂ in air at 25°C is 1.6 × 10^-2, what is the concentration of O₂ ?
Answer:

Question 34.
The reaction CO(g) + 3H₂(g) \(\rightleftharpoons \) CH₄(g) + H₂O(g)
is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.
Answer:

Question .35.
What is meant by conjugate acid-base pair ? Find the conjugate acid/base for the following species :
HNO₂,CN^–,HClO₄,OH^–,CO₃^2-,S^2-.
Answer:
An acid-base pair which differs by a proton only (HA \(\rightleftharpoons \) N A^– + H⁺ ) is known as conjugate acid-base pair. Conjugate acid : HCN, H₂O, HCO₃^–, HS^–.
Conjugate base : NO₂^–, ClO₄^–, O₂ .

Question 36.
Which of the following are Lewis Acids ?
H₂O, BF₃, H⁺ and NH₄⁺
Answer:
BF₃, H⁺ , NH₄⁺ ions are Lewis acids.

Question 37.
What will be the conjugate bases for the Bronsted acids ? HF, H₂SO₄ and H₂CO₃ ?
Answer:
Conjugate bases : F^–, HSO₄^– , HCO₃^–

Question 38.
Write the conjugate acids for the following Bronsted acids.
NH₂^–, NH₃ and HCOO^–
Answer:
NH₃, NH₄⁺ and H₂CO₃.

Question 39.
The species H₂O, HCO₃^–, HSO₄^– and NH₃ can act both as Bronsted acid and base. For each case, give the corresponding conjugate acid and base.
Answer:

Question 40.
Classify the following species into Lewis acids and Lewis bases and show how these can act as Lewis acid/Lewis base ?
(a) OH^– ions
(b) F^–
(c) H⁺
(d) BCl₃
Answer:
(a) OH^– ions can donate an electron pair and act as Lewis base.
(b) F^– ions can donate and electron pair and act as Lewis base.
(c) H⁺ ions can accept an electron pair and act as Lewis acid.
(d) BCl₃ can accept an electron pair since Boron atom is electron deficient. It is a Lewis acid.

Question 41.
The concentration of hydrogen ions in a sample of soft drink is 3.8 × 10^-3 M. What is its pH value ?
Answer:
pH = – log [H⁺] = – log (3.8 × 10^-3)
= (log 3 – log 3.8) = 3 – 0.5798 = 2.4202

Question 42.
The pH of soft drink is 3.76. Calculate the concentration of hydrogen ions in it.
Answer:
pH = – log [H⁺] or log [H⁺] = -pH = – 3.76 = \(\bar { 4 } \).24
[H⁺] = Antilog (\(\bar { 4 } \).24) = 1.74 x 10^-4M.

Question 43.
The ionisation constants of HF, HCOOH and HCN at 298 K are 6.8 x 10^-4, 1-8 × 10^-4 and 4.8 × 10^-9 respectively. Calculate the ionisation constant of the corresponding congugate bases.
Answer:

Question 44.
The ionisation constant of phenol is 1.0 x 10^-10. What is the concentration of phenate ion in 0.05 M solution of phenol and pH of solution ? What will be the degree of ionisation if the solution is also 0.01 M in sodium phenate ?
Answer:

Question 45.
The first dissociation constant H2S is 9.1 × 10^-8. Calculate the concentration of HS^– ions in its 0.1 M solution and how much will this concentration be affected if the solution is 0.1 M in HCl also. If the second dissociation constant of H₂S is 1.2 × 10^-13, calculate the concentration of S^2- ions under both the conditions.
Answer:

Question 46.
The ionization constant of acetic acid is 1.74 × 10^-5. Calculate the degree of dissociation of acetic acid in 0.05 M solution. Calculate the concentration of acetate ions in solution and the pH of the solution.
Answer:

Question 47.
It has been found that the pH of 0.01 M solution of organic acid is 4.15. Calculate the concentration of the anion, ionization constant of the acid and the pK_a.
Answer:

Question 48.
Assuming complete dissociation, calculate the pH of the following solutions :
(a) 0.003 M HCl
(b) 0.005 M NaOH
(c) 0.002 M HBr
(d) 0.002 M KOH.
Answer:

Question 49.
Calculate the pH of the following solutions :
(a) 2 g of TlOH dissolved in water to give 2 litre of solution.
(b) 0.3 g of Ca(OH)₂ dissolved in water to give 500 mL of solution.
(c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
(d) 1 mL of 13.6 M HCl diluted with water to give 1 litre of solution.
Answer:

Question 50.
The degree of ionization of 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and pKa of bromoacetic acid.
Answer:

Question 51.
The pH of 0-005 M codeine (C₁₈H₂₁NO₃) solution is 9.95. Calculate its ionizationconstant and pK_b.
Answer:

Question 52.
What is the pH of 0.001 M aniline solution ? The ionization constant of aniline is 4.27 × 10^-10. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Answer:

Question 53.
Calculate the degree of ionization of 0.05 M acetic acid if its pK_a value is 4.74. How is degree of dissociation affected when the solution contains (a) 0.01 M HCl (b) 0.1 M HCl ?
Answer:

Question 54.
The ionization constant of dimethylamine is 5.4 × 10^-4. Calculate the degree of ionization of its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.01 M NaOH ?
Answer:

Question 55.
Calculate the hydrogen ion concentration in the following biological fluids whose pH values are as follows :
(a) Human muscles fluid 6.83
(b) Human stomach fluid 1.2.
(b) Human blood 7.38
(d) Human saliva 6.4
Answer:
(a) [H⁺] of human muscles fluid
pH = 6.83 or – log [H⁺] = 6.83 or log [H⁺] = – 6.83 = \(\bar { 7 } \).17
[H⁺] = Antilog [H⁺] = Antilog [\(\bar { 7 } \).17] = 1.479 ×10^-7 M

(b) [H⁺ ] of human stomach fluid
pH = 1.2 or – log [H⁺] = 1.2 or log [H⁺] = – 1.2
[H⁺] = Antilog (-1-2) = Antilog (2-8) = 6.309 × 10^-2 M

(c) [H⁺ ] of human blood
pH = 7.38 or – log [H⁺] = 7.38 or log [H⁺] = – 7.38
[H⁺] = Antilog (-7.38) = Antilog (\(\bar { 8 } \).62) = 4.168 × 10^-8 M

(d) [H⁺] of human saliva
pH = 6.4 or – log [H⁺] = 6.4 or log [H⁺] = – 6.4
[H⁺] = Antilog (-6.4) = Antilog (\(\bar { 7 } \).6) = 3.98 × 10^-7 M.

Question 56.
The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each case.
Answer:
(a) [H⁺] of milk
pH = 6.8 or – log [H⁺] = 6.8 or log [H⁺] = – 6.8
[H⁺] = Antilog (-6.80) = Antilog (7.20) = 1.58 × 10^-7 M

(b) [H⁺] of black coffee
pH = 5.0 or – log [H⁺] = 5.0 or log [H⁺] = – 5.0
[H⁺] = Antilog (-5.0) = Antilog (\(\bar { 5 } \)) = 10^-5 M

(c) [H⁺] of tomato juice
pH = 4.2 or – log [H⁺] = 4.2 or log [H⁺] = – 4.2
[H⁺] = Antilog (-4.2) = Antilog (\(\bar { 5 } \).80) = 6.31 × 10^-5 M

(d) [H⁺] of lemon juice
pH = 2.2 or – log [H⁺]= 2.2 or log [H⁺] = – 2.2
[H⁺] = Antilog (-2.2) = Antilog (\(\bar { 3 } \).80) = 6.310 × 10^-3 M

(e) [H⁺] of egg white
pH = 7.8 or – log [H⁺] = 7.8 or log [H+] = – 7.8
[H⁺] = Antilog (-7.8) = Antilog (\(\bar { 8 } \).20) = 1.58 × 10^-8 M

Question 57.
0.561 g KOH is dissolved in water to give 200 mL of solution at 298K. Calculate the concentration of potassium, hydrogen and hydroxyl ions. What is the pH of the solution ?
Answer:

Question 58.
The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution. (Atomic mass of Sr = 87.6)
Answer:

Question 59.
The ionisation constant of propionic acid is 1.32 × 10^-5. Calculate the degree of ionisation of acid in its 0.05 M solution and also its pH. What will be its degree of ionisation if the solution is 0.01 M in HCl also ?
Answer:

Question 60.
The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate ionisation constant of the acid and also its degree of dissociation in the solution.
Answer:

Question 61.
The ionisation constant of nitrous acid is 4.5 × 10^-4. Calculate the pH value of 0.04 M NaNO₂ solution and also its degree of hydrolysis.
Answer:

Question 62.
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionisation constant of pyridine.
Answer:

Question 63.
Predict if the solutions of the following salts are neutral, acidic or basic :
NaCl, KBr, NaCN, NH₄NO₃, NaNO₂, and KF
Answer:
Neutral :NaCl, KBr
Basic : NaCN, NaNO₂, KF
Acidic :NH₄NO₃

Question 64.
The ionisation constant of chloroacetic acid is 1.35 × 10^-3. What will be the pH of 0.1 M acid solution and of its 0.1 M sodium salt solution ?
Answer:

Question 65.
The ionic product of water at 310 K is 2.7 × 10^-14. What is the pH value of neutral water at this temperature ?
Answer:
Since water is neutral ; [H₃O⁺] = [OH^–]
We know that, [H₃O⁺] × [OH^–] = K_w = 2.7 × 10^-14
∴ [H₃O⁺]² = 2.7 × 10^-14
or [H₃O⁺] = (2.7 × 10^-14)^1/2 = 1.643 × 10^-7
pH of water = – log [H₃O⁺] = – log (1.643 × 10^-7)
= (-) (-7) + log (1.643) = (7 – log 1.643) = 7 – 0.2156 = 6.78.

Question 66.
Calculate the pH of the resultant mixtures
(a) 10 mL. of 0.2 M Ca(OH)₂ + 25 mL of 0.1 M HCl
(b) 10 mL of 0.01 M H₂SO₄ + 10 mL of 0.01 M Ca(OH)₂
(c) 10 mL of 0.1 M H₂SO₄ + 10 mL of 0.1 M KOH.
Answer:

Question 67.
Determine the solubilites of silver chromate, barium chromate and ferric hydroxide at 289 K from their solubility product constants. Determine also the molarities of the individual ions.
Given : K_spAg₂CrO₄) = 1.1 × 10^-12
K_sp(BaCrO₄) = 1.2 × 10^-10
K_sp[Fe(OH)₃] = 1.0 × 10^-38
Answer:

Question 68.
The solubility product constants of Ag₂CrO₄ and AgBr are 1.1 × 10^-12 and 5.0 × 10^-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Answer:

Question 69.
Equal volumes of 0.002 M solutions of sodium iodate and copper chlorate are mixed together. Will it lead to the precipitation of copper iodate ? (For copper iodate K_sp = 7.4 × 10^-8).
Answer:

Question 70.
The ionization constant of benzoic acid is 6.46 × 10^-5 and K_sp for silver benzoate is 2.5 × 10^-13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water ?
Answer:

Question 71.
What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide ? (For iron sulphide, K_sp = 6.3 × 10^-18).
Answer:

Question 72.
What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K. For calcium sulphate K_sp = 9.1 × 10^-6?
Answer:

Question 73.
The concentration of sulphide ion in 0.1 M HCl solution saturated with hydrogen sulphide is 1.0 × 10^-19 M. If 10 mL of this solution is added to 5 mL of 0.04 M solution of FeSO₄, MnCl₂, ZnCl₂ and CaCl₂, in which solutions precipitation will take place ? Given K_sp for FeS = 6.3 × 10^-18, MnS = 2.5 × 10^-13, ZnS = 1.6 × 10^-24 and CdS = 8.0 × 10^-27.
Answer:

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NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics.

Question 1.
Choose the correct answer.
A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.
Answer:
(ii) whose value is independent of path

Question 2.
For the process to occur under adiabatic conditions, the correct condition is :
(i) ∆T = 0
(ii) ∆p = 0
(iii) q = 0
(iv) w = 0
Answer:
(iii) q = 0

Question 3.
The enthalpies of all elements in their standard states are :
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element
Answer:
(ii) zero

Question 4.

Answer:

Question 5.
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are – 890.3 kJ mol^-1, -393.5 kJ mol^-1 and – 285.8 kJ mol^-1 respectively. Enthalpy of formation of CH₄ (g) will be
(i) -74.8 kJ mol^-1 (ii) -52.27 kJ mol^-1
(iii) + 74.8 kJ mol^-1 (iv) + 52.26 kJ mol^-1
Answer:

Question 6.
A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(iv) possible at any temperature
Answer:
(iv) possible at any temperature

Question 7.
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process ?
Answer:
Heat absorbed by the system, q = 701 J
Work done by the system = – 304 J
Change in internal energy (∆U) = q + w = 701 – 394 = 307 J.

Question 8.
The reaction of cyanamide, NH₂CN(s) with oxygen was affected in a bomb calorimeter and ∆U was found to be – 742.7 kJ mol^-1 of cyanamide at 298 K. Calculate the enthalpy change for the reaction at 298 K.
NH₂CN(y) + 3/2O₂(g) → N₂fe) + CO₂(g) + H₂O(l)
Answer:
∆U = – 742.7 kJ mol^-1 ; ∆^ng = 2 – \(\frac { 3 }{ 2 } \) = + \(\frac { 1 }{ 2 } \) mol.
R = 8.314 × 10^-3 kJ K^-1 mol^-1 ; T = 298 K
According to the relation, ∆H = ∆U + ∆^ng RT
AH = (- 742.7 kJ) + (\(\frac { 1 }{ 2 } \) mol) × (8.314 × 10^-3 kJ K^-1 mol^-1) × (298 K)
= -742.7 kJ + 1.239 kJ = -741.5 kJ.

Question 9.
Calculate the number of kJ necessary to raise the temperature of 60 g of aluminium from 35 to 55°C. Molar heat capacity of Al is 24 J mol^-1 K^-1.
Answer:
No. of moles of Al (m) = \(\frac { 60g }{ 27gmo{ l }^{ -1 } } \) = 2.22 mol
Molar heat capacity (C) = 24 J mol^-1 K^-1
Rise in temperature (∆T) = 55 – 35 = 20°C = 20 K
Heat evolved (q) = C × m × T = (24 J mol^-1 K^-1) × (2.22 mol) × (20 K)
= 1065.6 J = 1.067 kJ.

Question 10.
Calculate the enthalpy change on freezing of 1.0 mole of water at – 10.0°C to ice at – 10.0°C. ∆_fusH = 6.03 kJ mol^-1 at 0°C; C_p[H₂O(l)] = 75.3 J mol^-1 K^-1 ; C_p[H₂O(s)l = 36.8 J mol^-1 K^-1.
Answer:
Total change in enthalpy (AH) for the freezing process may be calculated as :
∆H = (1 male of water at 10°C → 1 mole of water at 0°C) + (1 mole of water at °C → 1 mole of ice at 0°C)
+ (1 mole of ice at °C → 1 mole of ice at – 10°C)
= C_p[H₂O(l)] × ∆T + ∆H_freezmg + C_p [H₂O(s)] × ∆T.
= (75.3 Jk^-1 mol^-1) (0 – 10 K) + (-6.03 kJ mol^-1) + (36.8Jk^-1 mol^-1) × (-10 K)
= (-753 J mol^-1) – (6.03 kJ mol^-1) – (368 J mol^-1)
= (-0.753 kJ mol^-1) – (6.03 kJ mol^-1) – (0.368 kJ mol^-1)
= – 7.151 kJ mol^-1

Question 11.
Enthalpy of combustion of carbon to carbon dioxide is – 393.5 kJ mol^-1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and oxygen gas.
Answer:
The combustion equation is :
C(s) + O₂(g) → CO₂(g) ; ∆_cH = – 393.5 kJ mol^-1 (44 g)
(44g)
Heat released in the formation of 44g of CO₂ = 393.5 kJ
Heat released in the formation of 35.2 g of CO₂ = \(\frac { (393.5kJ)\times (35.2g) }{ (44g) } \) = 314.8 kJ.

Question 12.
Calculate the enthalpy of the reaction :
N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)
Given that;
∆_fH CO(g) = – 110 kJ mol^-1 ;
∆_fH CO₂(g) = – 393 kJ mol^-1
∆_fH. N₂O(g) = 81 kJ mol^-1 ;
∆_fH N₂O₄(g) = – 9.7 kJ mol^-1.
Answer:
Enthalpy of reaction (∆_rH) = [81 + 3(- 393)] – [9.7 + 3(- 110)]
= [81 – 1179] – [9.7 – 330] = – 778 kJ mol^-1.

Question 13.

Answer:

Question 14.

Answer:

Question 15.
Calculate the enthalpy change for the process
CCl₄ (g) → C (g) + 4 Cl (g) and calculate bond enthalpy of C—Cl in CC14 (g)
Given : ∆_vap H° (CCl₄) = 30.5 kJ mol^-1 ; ∆_fH°(CCl₄) = – 135.5 kJ mol^-1
∆_aH° (C) = 715.0 kJ mol^-1 where ∆_a H° is enthalpy of atomisation
∆_aH° (Cl₂) = 242 kJ mol^-1.
Answer:

Question 16.
For an isolated system ∆U = 0 ; what will be ∆S ?
Answer:
Change in internal energy (∆U) for an isolated system is zero because it does not exchange any energy with the surroundings. But entropy tends to increase in case of spontaneous reaction. Therefore, ∆S > 0 or positive.

Question 17.
For a reaction at 298 K
2 A + B → C
∆H = 400 kJ mol^-1 and ∆S = 0.2 kJ K’1 mol^-1.
At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range ?
Answer:
According to Gibbs-Helmholtz equation :
∆G = ∆H – T∆S
For ∆G = 0; ∆H = T∆S or T : = \(\frac { \triangle H }{ \triangle s } \)

T = \(\frac { (400kJmo{ l }^{ -1 }) }{ (0.2kJ{ K }^{ -1 }mo{ l }^{ -1 }) } \)
Thus, reaction will be in a state of equilibrium at 2000 K and will be spontaneous above this temperature.

Question 18.
For the reaction ; 2Cl (g) → Cl₂(g) ; what will be the signs of ∆H and ∆S ?
Answer:
∆H : negative (-ve) because energy is released in bond formation
∆S : negative (-ve) because entropy decreases when atoms combine to form molecules.

Question 19.
For a reaction ; 2A (g) + B (g) → 2D(g)

Calculate ∆U₂₉₈ for the reaction and predict whether the reaction is spontaneous or not.
Answer:
Let the mass of H₂ in the mixture = 20 g
The mass of O₂ in the mixture will be = 80 g

Question 20.

Answer:

Question 21.

Answer:

Question 22.

Answer:

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NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter Gases and Liquids

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter Solids, Liquids and Gases.

Question 1.
What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200 dm³ at 30° C ?
Answer:
From the given data :
P₁ = 1 bar P₂ = ?
V₁ = 500 dm³ V₂ = 200 dm³
P₁V₁ = V₁P₂ or P₂ = \(\frac { { P }_{ 1 }{ V }_{ 1 } }{ { V }_{ 2 } } \)

Question 2.
A vessel of 120 mL capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°C. What would be its pressure ?
Answer:
From the given data :
P₁ = 1.2 bar P₂ = ?
V₁ = 120 mL V₂ = 180 mL
Since temperature is constant, Boyle’s Law can be applied.
P₁V₁ = V₁P₂ or P₂ = \(\frac { { P }_{ 1 }{ V }_{ 1 } }{ { V }_{ 2 } } \) ; P₂ = \(\frac { (1.2bar)\times (120mL) }{ 180mL } \) = 0.8 bar

Question 3.
Using the equation of state PV = nRT show that at a given temperature, the density of the gas is proportional to the gas pressure p.
Answer:

Question 4.
At 0°C, the density of a gaseous oxide at 2 bar is the same as that of nitrogen at 5 bar. What is the molecular mass of the oxide ?
Answer:

Question 5.
Pressure of lg of an ideal gas A at 27°C is found to be 2 bar. When 2g of another ideal gas B is introduced in the same flask at same temperature, the pressure becomes 3 bar. Find the relationship between their molecular masses.
Answer:
Let MA and MB be the molar masses of the two gases A and B. According to available data :

Question 6.
The drain cleaner, ‘Drainex’ contains small bits of aluminium which react with caustic soda to produce hydrogen. What volume of hydrogen at 20°C and one bar will be released when 0.15 g of aluminium reacts ?
Answer:
Step I. Calculation of the volume of hydrogen released under N. T.P. conditions.
The chemical equation for the reaction is :
2Al + 2NaOH + 2H₂O → 2NaAl0₂ + 3H₂
2×27=54 g Sod. meta aluminate 3×22400 mL
54g of Al at N.T.P. release H₂ gas = 3 × 22400 mL
0.15g of Al at N.T.P. release H₂ gas = \(\frac { (3\times 22400mL)\quad \times \quad (0.15g) }{ (54g) } \) =186.7mL.
Step II. Calculation of volume of hydrogen released at 20°C and 1 bar pressure.

Question 7.
What will be the pressure exerted (in pascal) by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm³ flask at 27°C ?
Answer:

Question 8.
What will be the pressure of the gaseous mixture when 0.5 L of H₂ at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at 27°C ?
Answer:
Step I. Calculation of partial pressure of H2 in 1 L vessel.
V₁, = 0-5 L V₂ = 1.0 L
P₁ = 0.8 bar P₂ = ?
According to Boyle’s Law, P₁V₁ = P₂V₂
P₂ = \(\frac { { P }_{ 1 }{ V }_{ 1 } }{ { V }_{ 2 } } \) = \(\frac { (0.8bar) \times  (0.5L) }{ (1.0L) } \) = 0.4bar
Step II. Calculation of partial pressure of O₂ in 1 L vessel.
V₁ = 2.0 L V₂ = 1.0L
P₁ = 0.7 bar P₂ = ?
According to Boyle’s Law, P₁V₁ = P₂V₂
P₂ = \(\frac { { P }_{ 1 }{ V }_{ 1 } }{ { V }_{ 2 } } \) = \(\frac { (0.7bar) \times  (2.0L) }{ (1.0L) } \) =1.4 bar
Step III. Calculation of total pressure of gaseous mixture.
P = P₁ +P₂ = (0.4 + 1.4 ) bar = 1.8 bar

Question 9.
The density of a gas is found to be 5.46 g/dm³ at 27°C and under 2 bar pressure. What will be its density at STP ?
Answer:

Question 10.
34.05 mL of phosphorus vapours weigh 0-0625 g at 546°C and 1.0 bar pressure. What is the molar mass of phosphorus ?
Answer:
According to ideal gas equation,
PV = n RT ; PV = \(\frac { WRT }{ M } \) or M = \(\frac { WRT }{ PV } \)
According to available data :
Mass of phosphorus vapours (W) = 0.0625 g
Volume of vapours (V) = 34.05 mL = 34.05 × 10³ L
Pressure of vapours (P) = 1.0 bar.
Gas constant (R) = 0.083 bar L K^-1 mol^-1
Temperature (T) = 546 + 273 = 819 K

Question 11.
A student forgot to add the reaction mixture to the round bottomed flask at 27°C but put it on the flame. After a lapse of time, he realised his mistake. By using a pyrometer, he found that the temperature of the flask was 477°C. What fraction of air would have been expelled out ?
Answer:
Since the student was working in the laboratory, there is no change in pressure. Thus, Charles’ Law is applicable.
According to given data : V₁ = V L (say) V₂ = ?
T₁ = 27+273 =300 K ; T₂ = 477 + 273 = 750K
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } \) = \(\frac { { V }_{ 2 } }{ { T }_{ 2 } } \) or V₂ = \(\frac { { V }_{ 1 }{ T }_{ 2 } }{ { T }_{ 1 } } \) = \frac { (VL)\times (750K) }{ 300K } = 2.5 VL
Thus, volume of air expelled = 2.5 V – V = 1.5 V 1.5V
Fraction of air expelled = \(\frac { 1.5V }{ 2.5V } \) = \(\frac { 3 }{ 5 } \) .

Question 12.
Calculate the temperature of 4.0 moles of a gas occupying 5 dm³ at 3.32 bar (R = 0.083 bar dm³ K^-1 mol^-1).
Answer:
According to idea gas equation, PV = n RT or T = [/latex] = \(\frac { PV }{ nR } \)
According to available data :
No. of moles of the gas (n) = 4.0 moles
Volume of the gas (V) = 5 dm³
Pressure of the gas (P) = 3.32 bar
Gas constant (R) = 0.083 bar dm³ K^-1 mol^-1

Question 13.
Calculate the total number of electrons present in 1.4 g of nitrogen gas.
Answer:
Molecular mass of N₂ = 28g
28g of N₂ represent molecules = 6.022 × 10²³
1.4 g of N₂ represent molecules = 6.022 × 10²³ × \(\frac { 1.4g }{ 28g } \) = 3.011 × 10²²
Atomic number of nitrogen (N) = 7
1 molecule of N₂ has electrones = 7 × 2 = 14
3.011 × 10²² molecules of N₂ have electrons = 4 × 3.011 × 10²² = 4.215 × 10²³ electrons

Question 14.
How much time would it take to distribute one Avogadro number of wheat grains if 10¹⁰ grains are distributed each second ?
Answer:
Time taken to distribute 10¹⁰ grains = 1 s

Question 15.
Calculate the total pressure of a mixture of 8 g of oxygen and 4 g of hydrogen confined in a vessel of 1 dm3 at 27°C (R = 0-083 bar dm³K^-1mol^-1)
Answer:

Question 16.
Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m and mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m^-3 and R = 0.083 bar dm³K^-1mol^-1).
Answer:
Step I Calculation of the mass of displaced air
Radius of balloon (r) = 10 m
Volume of balloon (V) = \(\frac { 4 }{ 3 } \) πr³ = \(\frac { 4 }{ 3 } \) × \(\frac { 22 }{ 7 } \) × (10 m)³ = 4190.5 m³
Mass of the displaced air = Volume of air (balloon) × Density of air
= (4190.5 m³) × (1.2 kg m^-3) = 5028.6 kg
Step II Calculation of the mass of the filled balloon.

= 279.37 × 10³ mol.
Mass of He present = Moles of He × Molar mass of He
= (279.37 × 10³ mol) × (4 g mol^-1)
= 1117.48 × 10³g = 1117.48 kg
Mass of filled balloon = 100 + 1117.48 = 1217.48 kg
Step III. Calcu lation of the pay load
Pay load = Mass of displaced air – Mass of filled balloon
= 5028.6 – 1217.48 =3811.12 kg.

Question 17.
Calculate the volume occupied by 8-8 g of Co₂ at 31.1 °C and 1 bar pressure (R = 0.083 bar LK^-1mol^-1).
Answer:

Question 18.
2.9 g of a gas at 95°C occupied the same volume as 0.184 g of hydrogen at 17°C and at the same pressure. What is the molar mass of the gas ?
Answer:

Question 19.
A mixture of hydrogen and oxygen at one bar pressure contains 20% by weight of hydrogen. Calculate the partial pressure of hydrogen.
Answer:
Let the mass of H₂ in the mixture = 20 g
The mass of O₂ in the mixture will be = 80 g

Question 20.
What would be SI units of a quantity pV²T²/n ?
Answer:

Question 21.
In term of Charles’ Law explain why -273 °C is the lowest possible temperature ?
Answer:
The temperature -273°C (or 0 K) is known as absolute zero temperature. Below this temperature, a substance cannot exist as a gas and changes to the liquid state. This means that the Charles’ Law can be applied only upto a temperature – 273°C since a substance fails to exist as gas below this temperature.

Question 22.
Critical temperature for carbon dioxide and methane are 31.1°C and -81.9°C respectively. Which of these has stronger intermolecular forces and why ?
Answer:
The values of critical temperatures suggest that the attractive forces in the molecules of carbon dioxide are more. In fact, both are non-polar gases but the van der Waals forces of attraction in carbon dioxide molecules are more because of greater molecular size.

Question 23.
Explain the physical significance of van der Waals parameters.
Answer:
Significance and Units of the van der Waals’ constants
The significance and the units of the constant ‘a’ and ‘b’ used in the van der Waals equation are discussed as van der Waals’ constant ‘a’ The constant ‘a’ is related to the magnitude of the attractive forces among the molecules in a particular gas. Thus, greater the value of ‘a’ more will be the attractive forces. It may be noted that ‘a’ is higher for gases like NH₃, HCl and SO₂ which can be easily liquefied than for gases like He, H₂ and N₂ which are liquefied with diffieculty. Actually the former are of polar nature and the attractive forces in their molecules are higher than the latter gases which are of non-polar nature. The units of the constant ‘a’ can be calculated as follows :

Units of ‘a’ are related to the units in which pressure ‘p’ and volume ‘V’ are expressed.
• Pressure is in atmospheres and volume in litres, a = atm. L² mol^-2
• Pressure is in atmospheres and volume in m³, a = atm m⁶ mol^-2
• Pressure is in Nm”2 and volume in m3, a = Nm^-2 m⁶ mol^-2
Van der Waals’ constant ‘b’. The constant ‘b’ represents the co-volume or excluded volume which is effective volume of the molecules in a gas. It is, in fact, four times the volume occupied by the gas molecules. The units of ‘h’ can be calculated as follows :
Units of ‘b’ may be L mol^-1 or m³ mol^-1.

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter Solids

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 5(A) States of Matter Solids.

Question 1.
Define the term ‘amorphous’. Give a few examples of amorphous solids.
Answer:
Amorphous or amorphous solids imply those solids in which the constituent particles have short range order. These have irregular shapes and are isotropic in nature. Apart from that they do not have sharp melting points. A few examples of amorphous solids are : glass, rubber, plastic, celluose etc.

Question 2.
What makes glass different from a solid such as quartz ? Under what conditions could quartz be converted into glass ?
Answer:
Glass is a super cooled liquid and an amorphous substance. Quartz is the crystalline form of silica (SiO₂) ‘n which tetrahedral units SiO₄ are linked with each other in such a way that the oxygen atom of one tetrahedron is shared with another Si atom. Quartz cap be converted into glass by melting it and cooling the melt very rapidly. In glass, SiO₄ tetrahedra are joined in a random manner.

Question 3.
Classify each of the following solids as ionic, metallic, molecular, net work (covalent) or amorphous:
(a) Tetra phosphorus decoxide (P₄O₁₀)
(b) Graphite
(c) Brass
(d) Ammonium phosphate (NH₄)₃PO₄
(e) SiC
(f) Rb
(g)I₂
(h) LiBr
(i) P₄
(j) Si (k) Plastic.
Answer:
(a) Molecular solid
(b) Covalent (Net-work) solid
(c) Metallic solid
(d) Ionic solid
(e) Covalent solid (Network)
(f) Metallic solid
(g) Molecular solid
(h) Ionic solid
(i) Molecular solid
(j) Covalent solid
(k) Amorphous solid

Question 4.
(a)What is meant by the term coordination number ?
(b)What is the co-ordination number of atoms
(i) in a cubic close packed structure
(ii) in a body centred cubic structure ?
Answer:
(a)The co-ordination number of a constituent particle (atom, ion or molecule) is the number of nearest neighbours in its contact in the crystal lattice. „
(b)(i) Twelve, (ii) Eight

Question 5.
How can you determine the atomic mass of an unknown metal if you know its density and dimensions of its unit cell ? Explain your answer.
Answer:

We have learnt that different types of the cubic unit cells differ in their packing fractions i.e. with respect to the the space occupied by the constituting particles. These are expected to differ in their relative densities also. Let us derive a general relation between the edge length and the density in a unit cell.
Consider a unit cell with edge length = a pm = a × 10^-10 cm.
(∵ Because density is normally expressed in g cm^-3)
Volume of the unit cell = (a × 10^-10 cm)³ = a³ × 10^-30 cm³
Density of unit cell =\(\frac { Mass\quad of\quad unit\quad cell }{ Volume\quad of\quad unit\quad cell } \) ….(i)
Now, mass of unit cell = No. of atoms in a unit cell × Mass of each atom
Z = \(\frac { Atomic\quad mass\quad (M) }{ Avogadro’s\quad number(No) } \) ….(ii)
By substituting the value of mass of unit cell in equation (i)

The value of Z is different for different types of cubic unit cells.
It may be noted that the density of any crystalline solid is the same as that of its unit cell.

• While deriving the above relationship, the edge length is expressed in picometre (pm) units. In case, it is expressed in any other units, please convert the same into pm. There is no need to mention its units in the calculations.

Question 6.
(a) Stability of a crystal is reflected in the magnitude of the melting point. Comment.
(b) Collect the melting point of (i) Ice (ii) ethyl alcohol (iii) diethyl ether (iv) methane from a data book. What can you say about intermolecular forces between the molecules ?
Answer:
(a) The stability of a crystal depends upon the magnitude of force of interaction in the constituting particles. Greater the force of attraction present, more will be the stability of the crystal. For example, ionic solids such as NaCl, KC1 etc. have very high melting and boiling points while the molecular crystals such as naphthalene, iodine etc. have low values of melting and boiling points.

(b) The melting points of different substances are :
(i) Ice = 273 K
(ii) Ethyl alcohol = 155.7 K
(iii) Diethyl ether = 156.8 K
(iv) Methane = 90.5 K
The intermolecular forces in molecules of ice and ethyl alcohol are mainly hydrogen bonding. The magnitude is more in ice than in alcohol as is evident from the values of melting points.
The forces in the molecules of diethyl ether are dipolar forces while in methane, these are mainly the van der Waal’s forces which is quite evident from its melting point (least among the compounds listed).

Question 7.
How will you distinguish between the following pairs of terms :
(a) Hexagonal close packing and cubic close packing
(b) Crystal lattice and unit cell
(c) Tetrahedral void and octahedral void.
Answer:
(a) In hexagonal close packing (hep), the spheres of the third layer are vertically above the spheres of the first layer
(ABABAB type). On the other hand, in cubic close packing (ccp), the spheres of the fourth layer are present
above the spheres of the first layer (ABCABC type).
(b) Crystal Lattices and Unit Ceils
We have studied that in the crystalline solids, the constituent particles (atoms, ions or molecules) have a regular orderly arrangement throughout and it gets repeated again and again resulting in a definite pattern. These points are generally shown as dots (.) and the sites which are occupied by these points are known as lattice sites or crystal lattices. Crystal lattices may be either two dimensional or three dimensional in nature.
Two dimensional crystal lattices : A two dimensional lattice is planar and is represented in terms of two basic vectors (a and b) and the angle (L) between them. There are five different types of two dimensional lattices :

Three dimensional crystal lattices : A three dimensional lattice is a further extension of the two dimensional lattice. It depicts the actual shape as well as size of the constituent particles in the crystal.
It is therefore, called space lattice or cyrstal lattice. It may be defined as :
the regular three dimensional arrangement of the points in a crystal.
Please note that
• Each point in the lattice is known as lattice point or lattice site.
• Each point in the lattice represents the constituting particle. These particles may be atoms, ions or molecules depending upon the nature of the crystalline solid.
• Lattice points are joined by straight lines in order to depict the geometry of the solid. A typical space lattice has been shown in the fig. 5.37.

Unit Cell
If we carefully examine the space lattice given in the figure, we can find a group of lattice points (shown as shaded) which get repeated in different directions to form the complete lattice. This smallest repeating pattern is called unit cell which may be defined as :
the smallest but complete unit in the space lattice which when repeated over and again in the three dimensions, generates the crystal of the given substance.
In order to illustrate a unit cell, let us consider a block made of bricks of the same size. Each brick represents the unit cell while the block is similar to the space or crystal lattice. Thus, a unit cell is the fundamental building block of the space lattice.
It may be noted that crystals differ from each other with respect to the shape and size of the unit cells. A unit cell is completely characterised in terms of edge lengths and the angles between the edges.

(i) Edges or edge lengths. The edges a, b and c represent the dimensions of the unit cell in space along the three axes. The edges may or may not be mutually perpendicular.

(ii) Angles between the edges. There are three angles between the edges. These are denoted as \(\alpha \) (between b and c), \(\beta \) (between a and c) and \(\Upsilon \) (between a and b). Thus, a unit cell may be characterised by six parameters as shown in the Fig. 5.38. The various types
of crystal systems differ with respect to edge lengths as well as angles between the edges.

(c) Tetrahedral and Octahedral Voids or Interstitial Sites
In the discussion about the close packing of spheres in space, we have seen that even in the closest packed structure when the spheres (or particles) touch each other, certain space in between the particles is left unoccupied. For example, in both hep and ccp, 26% of the space is left unoccupied. In fact, the spheres can touch each other only at certain points. The space which is left in between the closest packed arrangement is known as void, hole or interstice. Two types of voids or interstitial sites are formed in close packed structures. These are tetrahedral and octahedral voids.

Tetrahedral Voids.
We have seen that a triangular void is formed in hexagonal two dimensional close packing . A tetrahedral void is formed when the triangular void made by three spheres of a particular layer and touching each other has a contact with one sphere either in the upper layer or the layer below it. Thus, a tetrahedral void is formed when four spheres placed at the comers of a tetrahedron touch each other as shown in the Figure 5.54.

It is interesting to note that the actual shape of the void is triangular but the arrangement of the spheres around the void is tetrahedral.
In a close packed structure, each sphere is in contact with three spheres in the layer above it and also three spheres in the layer below it. Thus, there are two tetrahedral voids associated with each sphere (or constituting particle). The radius of a tetrahedral void in a closest packed structure is 22.5% of the sphere which is involved in the arrangement. Thus,

Octahedral Voids
An octahedral void or site is formed when three spheres arranged at the corners of an equilateral triangle are placed over another set of spheres pointing in the opposite directions as shown in the Figure 5.56.
Thus, octahedral void is the empty space created at the centre of six spheres which are. placed octahedrally.

It has been found that there is only one octahedral site for each sphere. The radius of an octahedral void in a closest packed structure is 0.414 of the sphere which is involved in the arrangement.

Question 8.
How many lattice points are there in one unit cell of each of the following lattices
(a) face centred cubic (b) face centred tetragonal (c) body centred cubic ?
Answer:
(a) In face centred cubic arrangement,
Lattice points located at the corners of the cube = 8
Lattice points located at the centre of each face = 6
Total no. of lattice points = 8 + 6 = 14

(b) In face centred tetragonal, the number of lattice points is also the same i.e., 8 + 6 = 14.
(In both the cases, .particles per unit cell = 8 × \(\frac { 1 }{ 8 } \) + 6 × \(\frac { 1 }{ 2 } \) – 4)

(c) In body centred cubic arrangement,
Lattice points located at the corners of the cube = 8
Lattice points located in the centre of the body = 1
Total no. of lattice points =8+1=9
(the total number of particles per unit cell = 8 × \(\frac { 1 }{ 8 } \) + 1 = 2)

Question 9.
Explain :
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.
Answer:
(i) Basis of similarities. The basis of similarities between the metallic and ionic crystals are the presence of strong electrostatic forces of attraction. These are present among the ions in the ionic crystals and among the kernels and valence electrons in the metallic crystals. That is why both metals and ionic compounds are good conductors of electricity and have high melting points.
Basis of differences. The basis of differences is the absence of mobility of ions in the ionic crystals while the same is present in the valence electrons and kernels in case of metallic crystals. As a consequence, the ionic compounds conduct electricity only in the molten state while the metals can do so even in the solid state.
(ii) The ionic solids are hard and brittle because of strong electrostatic forces- of attraction which are present in the oppositely charged ions.
(iii) The ionic solids are hard because of the presence of strong inter ionic forces of attraction in the oppositely charged ions. These ions are arranged in three dimensional space. The ionic solids are brittle because the ionic bond is non-directional.

Question 10.
Calculate the efficiency of packing in case of metal crystal for :
(i) Simple cubic
(ii) Body centred cubic
(iii) Face centred cubic (with the assumption that the atoms are touching each other).
Answer:
(i) Simple Cubic Unit Cell
We have calculated the number of atoms or spheres (since an atom or particle may be regarded as the sphere) in different types of cubic unit cells. Now, let us calculate the relation between the edge length (a) and the radius (r) of the spheres present in these unit cells in case of pure elements i.e. elements composed of the same type of atoms.

It may be noted that :
(a) Distance between the centres of the spheres present on the corners of the edges of the cube is called edge length (a).
(b) Distance between the centres of the two nearest spheres is called nearest neighbour
distance (d) and is equal to r + r = 2r. (Here r is the radius of the spheres or the atomic radius.) .
We know that in a simple cubic unit cell, there is one atom (or one sphere) per unit cell. If r is the radius of the sphere, volume occupied by one sphere present in the unit cell = \(\frac { 4 }{ 3 } \) πr³.
Since the spheres at the corners touch each other ; Edge length (a) = r + r = 2r = d

(ii) Body Centred Cubic Unit Cell
We know that a body-centred cubic unit cell has 2 spheres (atoms) per unit cell. If r is the radius of the sphere Volume of one sphere = 4/3 nr3
Volume of two spheres present in the unit cell = \(\frac { 4 }{ 3 } \) πr³ × 2 = \(\frac { 8 }{ 3 } \) πr^{3</sup .
From the Fig. 5.62, it is evident that the sphere present in the centre of the cube touches a corner sphere so that the distance in their centres = 2r}

Length of the body diagonal (BD) in the cube = 4r.
By considering the triangle BCD and by applying Pythagorean theorem,
BD² = CD² + BC²
But BC² = AB² + AC²
∴ BD² = CD² + AB² + AC²
edge length = a, then
(4r)² = a² + a² + a²
16r² = 3a² or a = (16/3 r²)^1/2
a = \(\frac { 4r }{ \sqrt { 3 } } \) or r = \(\frac { \sqrt { 3 } a }{ 4 } \)
or d = 2r = \(\frac { 2\times \sqrt { 3 } a }{ 4 } \) = \(\frac { \sqrt { 3 } a }{ 2 } \)

(iii)Face centred cubic unit cell
We know that a face centred cubic unit cell (fcc) contains four spheres (or atoms) per unit cell.
The spheres at the corners are touching the sphere or atom present in the centre. However, they can not touch each other.
Figure 5.61, Faced centred cubic unit cell.

Volume occupied by one sphere of radius r = \(\frac { 4 }{ 3 } \) πr³
∴ Volume occupied by four spheres present in the unit cell = \(\frac { 4 }{ 3 } \) πr³ × 4 = \(\frac { 16 }{ 3 } \) πr³
Distance in their centres = r + r = 2r
Length of the face diagonal = r + 2r + r = 4r.
Now, according to Pythagorean theorem,
the square of the hypotenuse of a right triangle is equal to sum of the squares of the other two sides. If a is the edge length of the unit cell, then
a² + a² = (4r)² or 2a² = 16r²
a² = \(\frac { 16 }{ 2 } \) r² = 8r² or a = (8r²)^1/2 = \(\sqrt { 8 } \)r
a = 2r\(\sqrt { 2 } \) or r = \(\frac { a }{ 2\sqrt { 2 } } \)
d = 2r = \(\frac { 2a }{ 2\sqrt { 2 } } \) = \(\frac { a }{ \sqrt { 2 } } \) (d is the nearest neighbour distance.)

Question 11.
Silver crystallises in a face centred cubic lattice with all the atoms at the lattice points. The length of the edge of the unit cell as determined by X-ray diffraction studies is found to be 4.077 × 10^-8 cm. The density of silver is 10.5 g cm^-3. Calculate the atomic mass of silver.
Answer:

Question 12.
A cubic solid is made of two elements P and Q. Atoms Q are at the corners of the cube and P at the body centre. What is the formula of the compound ? What is the co-ordination number of P and Q ?
Answer:
Contribution by atoms Q present at the eight comers of the cube = \(\frac { 1 }{ 8 } \) × 8 = 1
Contribution by atom P present at the body centre = 1
Thus, P and Q are present in the ratio 1:1.
∴ Formula of the compound is PQ.
Co-ordination number of atoms P and Q = 8.

Question 13.
Niobium crystallises in body centred cubic structure. If the density is 8.55 g cm^-3, calculate atomic radius of niobium given that atomic mass of niobium is 93 g mol^-1.
Answer:

Question 14.
If the radius of octahedral void is r and the radius of the atom in close packing is R, derive the relation between r and R.
Answer:
A tetrahedral void has been shown in the Figure 1.54. Here three spheres form the triangle base while the fourth sphere lies at the top. The shaded sphere occupies the tetrahedral void. Let the radii of void and sphere be r and R respectively. Let the edge length of the cube be a.
From the right hand triangle ABC,
AB² = AC² + BC²
Here AB is the face diagonal while AC and BC are the edge lengths (a). Therefore,
AB² = a² + a²
or AB = (2a²) ^1/2= a\(\sqrt { 2 } \)

Question 15.
Copper crystallises into a fee lattice with edge length 3.61 × 10^-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm^-3.
Answer:

Question 16.
Analysis shows that nickel oxide has formula Ni_0.98 O_1.00. What fraction of nickel exists as Ni²⁺ and as Ni³⁺ ions ?
Answer:
The ratio of Ni and O atoms in pure nickel oxide (NiO) = 1 : 1
Let x be the no. of Ni (II) atoms replaced by Ni (III) atoms in the oxide.
∴ No. of Ni (II) atoms present = (0.98 – x)
Since the oxide is neutral in nature,
Charge on Ni atoms = Charge on oxygen atoms
2(0.98 -x) + 3x =2
1.96 – 2x + 3x =2
x = 2 – 1.96 = 0.04
% of Ni (III) atoms in nickel oxide \(\frac { No.of\quad Ni(III)atoms }{ Total\quad no.Ni\quad atoms } \) × 100 = \(\frac { 0.04 }{ 0.98 } \) = 4.01%
% of Ni (II) atoms in nickel oxide = 100 – 4.01 = 95.99%

Question 17.
What are semi-conductors ? Describe the two main types of semiconductors and contrast their conduction mechanisms.
Answer:
Semi-conductors are the substances whose conductivity lies in between those of conductors and insulators. The two main types of semi-conductors are n-type and p-type.
Applications of n-type and p-type semi-conductors
Germanium and silicon are the elements which belong to group 14 and have four valence electrons in their atoms. These are both metalloids and also semi-conductors. A large variety of solid state materials have been prepared by the combination of elements belonging to group 13 and 15 or group 12 and 16. These have also four valence electrons. A few examples of compounds resulting from 13-15 combination are : InSb, A1P and GaAs. Similarly, compounds resulting from 12-16 combination are CdS, CdSe, HgTe. (Cd and Hg are the transition metals present in group 12). These have resulted in many sophisticated semi-conductors which have brought about a revolution in this field.
If a single crystal is doped with indium at one end and with arsenic at the other end, then one part is a p-type semi¬conductor while the other is «-type semi-conductor. In the middle, the region where the two sides meet represents n-p junction These junctions are used in making microcompressor devices and integrated circuits in the solid state electronics.
Some important applications of n- and p-type semi-conductors are listed.

1. Rectifiers : The role of rectifiers is to allow current from an outside source to flow in one direction and thus, converts an alternate current (AC) into direct current. For this, a circuit made from four diodes is needed. A diode is basically a transistor consisting of two zones ; one p-type and the other n-type along with a n-p junction in between.

2. Photovoltaic cells : A photovoltaic cell is a device which can convert light energy into electrical energy. In order to achieve this, a n-p junction is irridiated with photons of light having energy more than band gap. As a result, some n-p bonds break and generate electrons and positive holes. These electrons then get promoted from valence band to the conduction band. These extra electrons present in the conduction band make the n-type region more negative. At the same time in the p-type region, the electrons get trapped by some positive holes. By connecting the two regions in the external circuit, the electrons are in a position to flow from n-type region to the p-type region. The current flows in the opposite direction i.e., from p-type region to the n-type region. In this manner, photo-voltaic cells function. As pointed earlier, they work at the expense of light energy into electrical energy.

3. Transistors : These are basically single crystals of silicon which have been doped to give three zones i.e., p-n-p and n-p-n. These transistors have wide range of applications. They work as amplifiers and oscillators in radios, televisions and also in computers. In addition, these are used as photo-transistors, tunnel diodes, solar cells etc. Now-a-days, it has become possible to manufacture computer chips which are equivalent to many thousand single crystal transistors.

Question 18.
Non-stoichiometric cuprous oxide (Cu₂O) can be prepared in the laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semi-conductor ?
Answer:
The ratio less than 2 : 1 in Cu₂0 shows that some cuprous (Cu⁺) ions have been replaced by cupric (Cusup>2+) 10ns. In order to maintain the electrical neutrality, every two Cusup>+ ions will be replaced by one Cusup>2+ ion which results in creating cation vacancies leading to positive holes. Since the conduction is due to positive holes, it is a p-type semiconductor.

Question 19.
Ferric oxide crystallises in a hexagonal close packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of ferric oxide.
Answer:
There is one octahedral hole for each atom in hexagonal close packed arrangement.
If the number of oxide ions (O^2-) per unit cell is 1, then the number of Fe³⁺ ions = \(\frac { 2 }{ 3 } \) × octahedral holes = \(\frac { 2 }{ 3 } \) ×1 = 2/3. Thus, the formula of the compound = Fe_2/3O₁ or Fe₂C₃.

Question 21.
Gold (atomic radius = 0.14411m) crystallises in a face centred unit cell. What is the length of the side of the unit cell ?
Answer:
For a face centred cubic unit cell (fcc)
Edge length (a) = 2\(\sqrt { 2r } \) = 2 × 1.4142 × 0.144 mm = 0.407 nm

Question 22.
In terms of Band Theory, what is the difference (i) between a conductor and an insulator (ii) between a conductor and semi-conductor ?
Answer:
Explanation for Conductors, Insulators and Semi-conductors
The’variation in the electrical conductivity of the solids can be explained with the help of the band theory. In metals, the energy gap between the partially filled valence band and unoccupied conduction band is negligible. Rather they overlap and electron flow readily takes place. In semi-conductors, there is small energy gap between valence band and conduction band. However, some electrons may jump to the conduction band and these semi-conductors can exhibit a little electrical conductivity.

For example, in case of silicon and germanium, these are 111 kJ mol^-1 and 63 kJ mol^-1 respectively and the electron jump is feasible. The conductivity of the semi-conductors increases with the rise in temperature because more electrons are now in a position to jump from the valence band to the conduction band. Such type of semi-conductors are known as intrinsic semiconductors However, the conductivity of these semi-conductors is so low that it is hardly useful practically. In insulators, the energy gaps are very large and the no electron jump is feasible from the valence band to the conduction band. The energy gaps also called forbidden zones. The insulators therefore, do not conduct electricity. For example, in case of diamond the energy gap is of the order of 5 IT kJ mol^-1. Therefore, diamond is an insulator. From the above discussion, we conclude that the conductivity in the solids is primarily linked with the energy difference between the valence and conduction bands.

Question 23.
Explain the following terms with suitable examples.
(i) Schottky defect, (ii) Frenkel defect, (iii) Interstitials, (iv) F-centres.
Answer:
(i) Schottky defect. The defect was noticed by German scientist Schottky in 1930. This arises because certain ions are missing from the crystal lattice and vacancies or holes are created at their respective positions. Since a crystal is electrically neutral, the number of such missing cations (A⁺) and anions (B^–) must be the same. The crystal lattice of an ideal crystal and that sufferring from Schottky defect has been shown below. Only one cation and one anion have been indicated as missing.

These are only the planar representation of the crystal lattice and not the three dimensional or space lattice.

(ii) Frenkel Defect. This defect in the ionic crystals was discovered by a Russian scientist Frenkel in 1926. It results when certain ions leave their normal sites and occupy positions elsewhere in the crystal lattice. Holes are created at their respective positions. Since cations are smaller in size as compared to anions normally these are involved in Frenkel defect.
Frenkel defect is depicted in the Fig. 5.67 in which only one cation has been shown to be dislocated from its position.

(iii)Interstitial defect. This defect is noticed when certain constituent particles (atoms or molecules) occupy the interstitial sites in the crystal lattice.. As a result, the number of particles per unit volume increases and so is the density of the solid. These defects are very common in the interstitial compounds of transtion metals. For details, consult unit 8 (d and fblock elements).

Question 24.
Aluminium crystallises in a cubic close packed structure. Its metallic radius is 125 pm.
(a) What is the length of the side of the unit cell ?
(b) How many unit cells are there in 1-00 cm3 of aluminium ?
Answer:
Step I. Calculation of length of side of the unit cell
Forf.c.c. unit cell, a = 2\(\sqrt { 2r } \) = 2\(\sqrt { 2 } \)(125pm) = 2 × 1.4142 × (125 pm) = 354 pm.
Step II. Calculation of no. of unit cells in 1.00 cm³ of aluminium.
Volume of one unit cell = (354 pm)³ = (354 x 10^-10 cm)³ = 44174155 × 10^-30 cc.
No. of unit cells in 1.00 cc of Al metal = \(\frac { 1(cc) }{ 44174155\times { 10 }^{ -30 }(cc) } \) = 2.26 x 10²².

Question 25.
If NaCl is doped with 10^-3 mol % of SrCl₂, what is the concentration of cation vacancies ?
Answer:

Question 26.
Explain the following with suitable examples
(a) Ferromagnetism (b) Piezoelectric effect
(c) Paramagnetism (d) Ferromagnetism
(e) 12-16 and 13-15 compounds.
Answer:
(a)Ferromagnetic solids (Ferromagnetism). A few solids like iron, cobalt, nickel, gadolinium and Cr02 are attracted very strongly by magnetic fields. These are known as ferromagnetic solids. Apart from that, they can be even permanently magnetised or become permanent magnet. Actually, in the solid state, the metal ions of these substances are grouped together into small regions known as domains. Each domain behaves like a small magnet. When a ferromagnetic solid is placed in a magnetic field, all the domains get oriented in the direction of the magnetic field (Fig. 5.76 (a)). As a result, a strong magnetic effect is produced. The effect persists even when the magnetic field is removed. Thus, a ferromagnetic substance becomes a permanent magnet. CrO₂ is used in magnetic tapes for audio frequency.

(b) Piezoelectric effect.  It may be noted that all the three types of solids (ferromagnetic, anti-ferromagnetic and ferrimagnetic) listed above may become paramagnetic upon heating upto a certain temperature known as Curie temperature. The change in the magnetic character is because of the reallignment of the magnetic moments or domains in one direction upon heating.For example,
(i) Ferricmagnetic solid Fe₃O₄ becomes paramagnetic when heated to 850 K.
(ii) Antiferromagnetic solid V₂O₃ becomes paramagnetic when heated to 750 K.

(c) Paramagnetic solids (Paramagnetism). These are the solids attracted by a magnet and this property is known as paramagnetism. Actually, the atoms of the elements present have certain unpaired electrons. Their spins or magnetic moments may lead to magnetic character. However, the electron spins may mutually cancel due to random orientation under normal conditions and the substances may not have any magnetic character. Under the influence of an external magnetic field or magnet, the spins of certain electrons may allign to produce temporary magnetism i.e. they may be attracted by the magnet. However, they may lose their orientation and also magnetic character once the magnet is removed. This means that their magnetic character is temporary and is present as long as external magnetic field is present. Many transition metals such as Co, Ni, Fe, Cu, etc. and their ions are paramagnetic. Similarly, dioxygen (O₂) and nitric oxide (NO) have also unpaired electrons and they exhibit paramagnetism.

(d).Ferrimagnetic solids (Ferrimagnetism). In certain ferromagnetic solids, the magnetic moment of the domains allign in parallel and anti-parallel directions in unequal numbers as shown in the Fig. 5.76(c). As a result, they-‘have certain resultant magnetic moment or magnetic character which is of permanent nature. However, ferrimagnetic solids are less magnetic than ferromagnetic solids. For example, magnetic oxide of iron (Fe₃O₄) and ferrites with general formula MFe₂O₄ (where M = Mg, Cu, Zn etc.) are ferrimagnetic in nature.

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