CBSE Class 11

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 4 Important Extra Questions Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure Important Extra Questions Very Short Answer Type

Question 1.
What change in energy takes place when a molecule is formed from its atoms?
Answer:
There is a fall in energy.

Question 2.
Arrange the following in order of increasing bond strengths.
F₂, O₂, N₂, Cl₂
Answer:
F₂ < Cl₂ < O₂ < N₂

Question 3.
Name the shapes of the following molecules: CH₄, C₂H₂, CO₂.
Answer:
CH₄: Tetrahedral; C₂H₂: Cylindrical; CO₂: linear

Question 4.
Arrange the following in order of increasing strengths of hydrogen bonding O, F, S, Cl, N.
Answer:
Cl < S < N < O < F;

Question 5.
Identify the compound/compounds in the following in which S does not obey the Octet rule: SO₂, SF₂, SF₄, SF₆.
Answer:
SF₄, SF₆.

Question 6.
Name one compound each involving sp³, sp², sp hybridization.
Answer:
sp³: CH₄: sp²: C₂H₄: sp: C₂H₂

Question 7.
Which orbitals can overlap to form a cr-bond and which orbitals can do so to form a π bond?
Answer:
s-s, s-p, p-p form a bond, and only p-p form π bond.

Question 8.
Which electrons take part in bond formation?
Answer:
Valence electrons present in the outermost shell.

Question 9.
What are SI units of dipole moment?
Answer:
Coulomb meter (Cm).

Question 10.
Name the method generally used for the calculation of lattice energy or electron affinity.
Answer:
Borh-Haber cycle.

Question 11.
You are given the electronic configuration of five neutral atoms A, B, C, D, and E.
A – 1s², 2s² 2p⁶ 3s²;
B = 1s², 2s² 2p⁶, 3s¹
C = 1s², 2s² 2p¹;
D = 1s², 2s² 2p⁵,
E = 1s², 2s² 2p⁶.
Write the empirical formula for the substance containing
(i) A and D,
Answer:
AD₂

(ii) B and D
Answer:
BD

(iii) only D
Answer:
D₂

(iv) Only E.
Answer:
E.

Question 12.
Which type of forces holds the atoms together in an ionic compound?
Answer:
Electrostatic forces of attraction.

Question 13.
Choose the compounds containing ionic, covalent, and coordinate bonds out of the following: MgO, CH₄, CaCl₂, HCl,
NH⁺₄,O₃.
Answer:
IONIC: MgO, CaCl₂;
Covalent: CH₄, HCl;
Coordinate:NH⁺₄, O₃.

Question 14.
Write the Lewis structure of the polyatomic ions CN^–, SO₄^2-.
Answer:

Question 15.
Arrange the following in order of decreasing C-C bond length:
C₂H₆, C₂H₂, C₂H₄
Answer:
H₃C – CH₃ > H₂C = CH₂ > HC ≡ CH.

Question 16.
Arrange the following in order of decreasing boiling point.
H₂O, H₂S, H₂Se.
Answer:
H₂O > H₂S > H₂Se.

Question 17.
Which of the following has the maximum bond angle? Why?
H₂O, CO₂, NH₃ CH₄.
Answer:
CO₂. Because it is a linear molecule (bond angle is 180°).

Question 18.
Write down the resonance structures of nitrous oxide.
Answer:

Question 19.
Sodium metal vaporizes on heating and the vapors have diatomic molecules of sodium (Na₂). What type of bonding is present in these molecules?
Answer:
Covalent.

Question 20.
Predict the molecular geometries of (a) CIF₃ (b) ICI^–₄
Answer:
(a) T-shaped
(b) Square planar.

Question 21.
What kind of bond exists between two non-metallic elements?
Answer:
Covalent bonds because the difference of electronegativity between two non-metals is usually very small.

Question 22.
Which of the following bonds is most polar?
S – Cl, S – Br, Se – Cl, or Se – Br.
Answer:
Se – Cl.

Question 23.
Write the Lewis structure of PCl₃.
Answer:

Question 24.
Define valency.
Answer:
The number of electrons that an atom gains or loses or shares with other atoms to attain noble gas configuration is termed its valency.

Question 25.
Explain why O-O bond lengths in ozone molecule are equal?
Answer:
O-O bond lengths in ozone molecules are equal because of resonance between their structures.

Question 26.
MgCl₂ is linear, but SnCl₂ is angular. Why?
Answer:
MgCl₂ has sp hybridization, whereas SnCl₂ has sp² hybridization.

Question 27.
Why CCl₄ does not give white precipitates with silver nitrate?
Answer:
CCl₄ is a covalent molecule and does not give Cl^– ions in solution.

Question 28.
What is the approximate bond strength for an ionic bond?
Answer:
It is about 400 kJ mol^-1

Question 29.
What is the approximate bond strength for a covalent bond?
Answer:
It is about 100 kJ mol^-1.

Question 30.
Though chlorine has nearly the same electronegativity as nitrogen, yet there is no hydrogen bonding in HCl. Why?
Answer:
The size of the chlorine atom (3 orbits) is bigger than N (2 orbits).

Question 31.
Which will form a stronger ionic bond?
(a) Na and F
(b) Na and Cl.
Answer:
(a) Na and F.

Question 32.
Out of MgO and NaCl which has more lattice energy?
Answer:
MgO.

Question 33.
Draw Lewis structure of N^3- ion.
Answer:

Question 34.
On what factors the polarity of the bond depends?
Answer:
Difference of electronegativity of the two atoms.

Question 35.
Arrange NaCl, AlCl₃ MgCl₂ in increasing covalent character.
Answer:
NaCl < MgCl₂ < AlCl₃.

Question 36.
Arrange KCl, LiCl, NaCl in increasing ionic character.
Answer:
LiCl < NaCl < KCl.

Question 37.
BF₃ is non-polar. Why?
Answer:
The dipole moment of BF₃ is zero because the BF₃ molecule is symmetrical.

Question 38.
What type of bond is formed when atoms have
(i) zero difference of electronegativity
Answer:
Non-polar covalent

(ii) a little difference of electronegativity
Answer:
Polar covalent

(iii) high difference of electronegativity?
Answer:
Electrovalent.

Question 39.
What is the state of hybridization of C atoms in
(i) diamond
Answer:
It is sp³ in diamond

(ii) graphite?
Answer:
It is sp² in graphite.

Question 40.
What type of orbitals can overlap to form a covalent molecule?
Answer:
Half-filled atomic orbitals containing electrons with opposite spins.

Question 41.
Why glucose, fructose, sucrose, etc. are soluble in the water though they are covalent compounds?
Answer:
These compounds contain OH groups which can form H- bonds with water.

Question 42.
What is the shape of NH₄⁺ ion?
Answer:
Due to the sp³ hybridization of N in NH₄⁺, it is tetrahedral in shape.

Question 43.
Both FIF and H₂O are associated through hydrogen¬bonding, yet the boiling point of HF is lower than that of HzO. Why?
Answer:
H-bonds formed by H₂O are double in number than those formed by HF molecules.

Question 44.
Why ethyl alcohol is completely miscible with water?
Answer:
This is because the OH group present in ethyl alcohol forms H-bonds with water.

Question 45.
Benzene ring contains alternate single and double bonds, yet all the C-C bonds are of equal length. Why?
Answer:
This is due to resonance in benzene.

Question 46.
When is a molecule paramagnetic in nature?
Answer:
A molecule is paramagnetic when it contains half-filled molecular orbitals.

Question 47.
Arrange the following molecular species in increasing order of stability.
Answer:
N₂^2- < N^2- = N²⁺ < N₂

Question 48.
Explain on the basis of the molecular orbital diagram why O₂ should be paramagnetic?
Answer:
O₂ molecule contains one unpaired electron in each of one π2p_x and π2p_y orbitals.

Question 49.
Define antibonding molecular orbital.
Answer:
The molecular orbital formed by the subtractive effect of the electron waves of the combining atomic orbitals is called an antibonding molecular orbital.

Question 50.
What is the effect of the process C₂ → C₂⁺ + e^– on bond order of C₂? [A.IS.B. 2002]
Answer:

Chemical Bonding and Molecular Structure Important Extra Questions Short Answer Type

Question 1.
Which out of CH₃F and CH₃Cl has a higher dipole moment and why?
Answer:
The dipole moment of CH₃Cl is greater than that of CH₃F. The C-F bond length in CH₃F is smaller than the C-Cl bond length in CH₃Cl. The charge separation in the C-F bond is more than in the Cl-C bond- fluoride being more electronegative than chlorine. The bond length has a greater effect than the charge separation. Hence the dipole moment of CH₃C1 is greater than that of CH₃F.

Question 2.
Define the term chemical bond. What are its different types?
Answer:
The attractive forces which hold the constituent atoms in molecules or species in lattices etc. are called a chemical bond.

They are of the following types:

1. Electrovalent or ionic bond
2. Covalent bond
3. Coordinate or dative bond
4. metallic bond
5. hydrogen bond
6. van der Waals forces.

Question 3.
Why covalent bonds are called directional bonds whereas ionic bonds are called non-directional?
Answer:
A covalent bond is formed by the overlap of half-filled atomic orbitals which have definite directions. Hence covalent bond is directional. In ionic compounds, each ion is surrounded by a number of oppositely charged ions and hence there is no definite direction.

Question 4.
AlF₃ is a high melting solid whereas SiF₄ is a gas. Explain why?
Answer:
AlF₃ is an ionic solid due to the large difference in electronegativities of Al and F whereas SiF₄ is a covalent compound and hence there are only weak van der Waal’s forces among its molecules.

Question 5.
Using the VSEPR theory identifies the type of hybridization and draw the structure of OF₂ What are oxidation states of O and F?
Answer:
The electron dot structure of OF₂ is

Thus the central atom (O-atom) has 4 pairs of electrons (2 bond pairs and 2 lone pairs). Hence oxygen in OF₂ is sp³ hybridised and the molecule is v-shaped oxidation state of F = – 1, oxidation state of O = + 2.

Question 6.
Account for the following: The experimentally determined N-F bond length in NF₃ is greater than the sum of the single covalent radii of N and F.
Answer:
This is because both N and F are small and hence have high-electron density. So they repel the bond pairs thereby making the N-F bond length larger.

Question 7.
Explain why the bond angle of NH₃ is greater than that of NF₃ while the bond angle of PH₃ is less than that of. PF₃.
Answer:
Both NH₃ and NF₃ are pyramidal in shape with one lone pair on N. However as F has a higher electronegativity than H, the electron pair is attracted more towards F in NF₃, i.e., the bond pairs of electrons are away from N or in other words, the distance between bond pairs is more. Hence the repulsions between bond pairs in NF₃ are less than in NH₃. Thus the lone pair repels the bond pairs of NF₃ more than it does in NH₃ As a result the bond angles decrease to 102.4° whereas, in ammonia, it decreases to 107.5° only.

PH₃ and PF₃ are also pyramidal in shape with one lone pair of P. But PF₃ has a greater bond angle than PH₃ (opposite to NH₃ and NF₃). This is due to resonance in PF₃ leading to the partial double bond character as shown below

As a result, repulsions between P-F bonds are larger and hence the bond angle is large. There is no possibility for the formation of double bonds in PH₃.

Question 8.
Why HCl is polar whereas the Cl₂ molecule is non-polar?
Answer:
In Cl₂ both atoms have the same electronegativity. Hence the shared pair of electrons is attracted equally by both and remains exactly in the center. NO end acquires a negative or positive charge. In HCl, chlorine is more electronegative than H. Hence shared pair of electrons is more attracted towards chlorine, which, therefore acquires a negative charge while H acquires a positive charge.

Question 9.
Write the Lewis structure of the nitrate ion NO₂
Answer:
Total no. of valence electrons on N = 2s²2p³ = 5
Total no. of valence electrons on two oxygen atoms = 2s²2p⁴
= 2 × 6 = 12
add one electron for negative charge = 5 + 12 + 1 = 18.
The skeletal structure of NO₂^– is written as O N O^–

Draw a single bond (one shared electron pair) between the nitrogen atom and each of the two oxygen atoms completing the octets on oxygen atoms. This, however, does not complete the octet on Nitrogen if the remaining two electrons, constitute lone pair on it.

Hence we have to resort to multiple bonding between N and one of the oxygen atoms (in this ease a double bond). This leads to the following Lewis dot structures.

Question 10.
Draw the Lewis dot structure of CO₃ ion.
Answer:
Total no. of valence electrons of CO₃ = 4 + 3 × 6 = 22
total no. of valence electrons on CO₃ = 22 + 2 = 24
The skeletal structure of CO₃ is
Putting one shared pair of electrons between C and O and completing the Octets of oxygen, we have

In the above structure, an octet of oxygen is not complete. Hence multiple bonding is required between C and O.

Question 11.
Calculate the formal charge on each oxygen atom of O₃ molecules and write its structure with formal charges.
Answer:
Lewis structure of O₃ is

Formal charge on O₁ = 6 – 4 – \(\frac{1}{2}\)(4) = 0
Formal charge on 0(2) = 6 – 2 – \(\frac{1}{2}\) (6) = + 1
Formal charge on 0(3) = 6 – 6 – \(\frac{1}{2}\) (2) = – 1.

Hence we represent O3 along with formal charges as follows.

Question 12.
Describe briefly the limitations of the Octet rule.
Answer:

1. In some molecules like LiCl, BeCl₂, BF₃, AlCl₃, central atoms like Li, Be, B, Al have not completed their octets Li: Cl,
   
2. In some molecules, there are an odd number of electrons like
   
3. In some molecules like PF₅, SF₆, IF₇, the central atoms like P, S, I have 10,12,14 electrons respectively.
4. Some noble gases also combine with oxygen and F to form a no. of compounds like XeOF₂, XeF₂, XeF_6, etc.
5. Octet theory does not explain the shape of the molecules.
6. It does not explain the relative stability of the molecules being totally silent about the energy of a molecule.

Question 13.
Explain the structure of the CO₂ molecule on the basis of resonance.
Answer:
The experimentally determined C to O bond length in CO₂ is 115 pm. The lengths of a normal C to oxygen double bond (C = 0) and carbon, to oxygen triple bond (C ≡ O), are 121 pm and 110 pm respectively. The C-O bond lengths in CO₂ (115 pm) lie between the values of C =0 and C-O. Obviously, a single Lewis structure cannot depict this position and it becomes necessary to write more than one Lewis structure.

∴ CO₂ is best described as the resonance hybrid of the canonical or resonance forms I, II, and III

Question 14.
How would the bond lengths vary in the dicarbon species
C₂, C₂^–, C₂^2-?
Answer:


As bond lengths are inversely proportional to bond order
∴ Bond lengths will be in order.
C₂ > C₂^– > C₂^2-.

Question 15.
What is the effect of the following ionization process on the bond order of O₂?
O₂ → O₂⁺ + e^–
Answer:
O₂ → O₂ + e^–

Thus bond order of O₂ which is = 2
increases to 2.5 in O₂⁺.

Question 16.
Draw the Lewis dot diagram of nitric acid, sulphuric acid, phosphorus acid, and hypophosphorous acid.
Answer:


Question 17.
Out of sigma and pi bonds, which one is stronger and why?
Answer:
Sigma (a) bond is stronger This is because a bond is formed by head-on overlapping of atomic orbitals and such overlapping being on the internuclear axis is large, n bond is formed by sidewise overlapping which is small and so a Pi bond is weaker.

Question 18. Arrange the following in order of decreasing bond angles
(i) CH₄, NH₃ H₂O, BF₃, C₂H₂
Answer:
C₂H₂ (180°) > BF₃ (120°) > CH₄ (109°, 28′) > NH₃ (107°) > H₂O (104.5°)

(ii) NH₃ NH₂, NH₄
Answer:
NH₄⁺ > NH₃ > NH₂^– [because no. of lone pairs of electrons present on N are 0, 1, 2 respectively].

Question 19.
Predict which out of the following molecules will have a higher dipole moment and why? CS₂ and OCS.
Answer:

both are linear molecules, but bond moments in CS₂ cancel out so that the net dipole moment = O. But in OCS bond moment of C= O is not equal to that of C = S. Hence it has a net dipole moment. Thus dipole moment of OCS is higher.

Question 20.
Out of p-orbital and sp-hybrid orbital which has greater directional character and why?
Answer:
sp orbital has a greater directional character than p-orbital. This is because the p-orbital has equal-sized lobes with equal electron density in both the lobes whereas the sp-hybrid orbital has greater electron density on one side.

Question 21.
Interpret the non-linear shape of H2S and non-planar shape of PCl₃ using VSEPR theory.
Answer:
H₂S
No. of electron pairs around S = \(\frac{6+2}{2}=\frac{8}{2}\) = 4

Question 22.
Why mobility of H⁺ ions is greater in ice as compared to liquid water? [CBSE. PMT 2005]
Answer:
H⁺ ions (obtained from ionization of water H₂O) in liquid water get easily hydrated. As a result, the size of the hydrated H⁺ ions is much larger than in ice. Therefore the mobility of H⁺ ions in water is less than in ice.

Question 23.
What is the hybrid state of BeCl₂? What will be the change in the hybrid state of BeCl₂ in the solid-state? [CBSE PMT 2005]
Answer:
In the vapor state at high temperature, BeCl₂ exists as a linear molecule. Cl – Be – Cl due to sp hybridization of Be.

In the solid-state, it has a polymeric structure as follows:

It is possible only due to the sp³ hybrid state of Be. Two half-filled hybrid orbitals will form normal covalent bonds with two Cl atoms. The other two Cl atoms are coordinated to Be atom by donating electron pairs into the empty hybrid orbitals of Be.

Question 24.
In both water and diethyl ether, the central atom viz. O- atom has the same hybridization. Then why do they have different bond angles? Which one has a greater bond angle?
Answer:
Both water and diethyl ether have the central atom O in an sp³ hybrid state with two lone pairs of electrons on O.

But due to the greater repulsions between two ethyl (C₂H₅) groups in diethyl ether than between two H-atoms in H₂O result in a greater bond angle (110°) in diethyl ether than 104.5° in that of water (H₂O).

Question 25.
BCl₃ is planar but anhydrous AlCl₃ is tetrahedral. Explain.
Answer:
B atom in BCl₃ is sp² hybridized due to which the shape of BCl₃ is triangular planar. Anhydrous AlCl₃ exists as a dimer with the formula Al₂Cl₆.

Thus, each atom of A1 in Al₂Cl₆ is surrounded by four bond pairs and hence it assumes a tetrahedral structure.

Question 26.
(a) How bond energy varies from N₂^– to N₂⁺ and why?
Answer:
Both N₂^– and N₂⁺ is, almost the same [though N₂^– is slightly less stable and hence has less bond energy than N₂⁺ due to the presence of a greater number of electrons in the antibonding molecular orbitals]

(b) On the basis of molecular orbital theory, what is a similarity between
(i) F₂, O₂^––
Answer:
Both F₂ and O₂^–– have the same bond order = 1 and so the same bond length

(ii) CO, N₂, NO⁺? [CBSE PMT 2004]
Answer:
All of them have the same bond order and bond length.

Question 27.
Phosphorus pentachloride dissociates in the vapor phase as
PCl₅ ⇌ PCl₄⁺ + Cl^–
The hybridization of P in PCl₅ is sp³d. What is it in PCl₄⁺?
Answer:
Phosphorus is sp³ hybridized in PCl₄ since it is isostructural with NH₄ which also has sp³ hybridization of Nitrogen.

Question 28.
Explain giving reasons whether BH₄ and H₃O⁺ will have the same/different geometry?
Answer:
The central atom in both the ions is surrounded by the same number of pairs of electrons, that is, 4. Hence they have the same tetrahedral geometry.

Question 29.
State with reasons, which is more polar CO₂ or N₂O?
Answer:
N₂O is more polar than CO₂ which is a linear molecule and thus symmetrical. Its net dipole moment is zero.

N₂O is linear but unsymmetrical. It is a resonance hybrid of the following canonical structures:

It has a net dipole moment of 0.116 D.

Question 30.
Out of peroxide ion (O₂) and superoxide ion (O₂ ) which has larger bond length and why?
Answer:
The bond order of O₂^– is 1.5 while that of O₂^2- is 1.0.
The lesser the bond order, the greater is the bond length as the bond order is inversely proportional to bond length. ( Hence O₂^2- has a larger bond length than O₂^2-.

Chemical Bonding and Molecular Structure Important Extra Questions Long Answer Type

Question 1.
Explain the formation of the following molecules according to the orbital concept, F₂, HF, O₂, H₂O, N₂, NH₃ molecules.
Answer:
1. Formation of F₂ molecule. Atomic number (Z) of fluorine is 9 and its orbital electronic configuration is 1s² 2s² 2p²_x, 2p²_y, 2p¹_z. Thus, a fluorine atom has one half-filled atomic orbital. Therefore, two atoms of fluorine combine to form the fluorine molecule as a result of the combination for their half-filled atomic orbitals shown in Fig. The two atoms get linked by a single covalent bond.

Formation of F₂ molecule

2. Formation of HF molecule. Fluorine atom, as stated above, has one half-filled atomic orbital. Hydrogen atom (Z = 1) has only one electron in Is orbital. Thus, the hydrogen fluoride (HF) molecule. is formed as a result of the combination (or overlap) of the half-filled orbitals belonging to the participating atoms.

Formation of HF molecule

3. Formation of O₂ molecules. The atomic number (Z) of oxygen is 8 and its orbital electronic configuration is 1s² 2s² 2p²_x 2p¹_x 2p¹_z. This means that an oxygen atom has two half-filled orbitals with one electron each. Two such atoms will combine to form a molecule of oxygen as a result of the overlap of the half-filled orbitals with opposite spins of electrons.

Formation of O₂ molecule
Thus, the two atoms of oxygen are bonded to each other by two covalent bonds or double bonds (O = O).

4. Formation of H₂O molecule. In the formation of the H₂O molecule, the two half-filled orbitals of the oxygen atom combine with the half-filled orbitals (1s) of the hydrogen atoms. Thus, the oxygen atom gets linked to the two hydrogen atoms by single covalent bonds as shown in

Formation of H₂O molecule

5. Formation of N₂ molecule. The atomic number of nitrogen is 7 and its orbital electronic configuration is 1s² 2s² 2p¹_x 2p¹_y 2p¹ _z. This shows that the nitrogen atom has three half-filled atomic orbitals. Two such atoms combine as a result of the overlap of the three half-filled orbitals and a triple bond gets formed (N = N).

Formation of N₂ molecule

6. Formation of NH₃ molecule. In the formation of ammonia (NH₃) molecule, three half-filled orbitals present in the valence shell of nitrogen atom combine with 1s orbital of three hydrogen atoms with one electron each. As a result, the nitrogen atom completes its octet and a molecule of NH₃ is formed in which the nitrogen atom is linked to three hydrogen atoms by covalent bonds.

Formation of NH₃ molecule

Question 2.
What is a hydrogen bond, what are its causes, and give the conditions for hydrogen bonding? What is the strength of hydrogen bonding? Describe the two types of hydrogen bonding.
Answer:
When hydrogen is connected to small highly electronegative atoms such as F, O, and N in such cases hydrogen forms an electrostatic weak bond with an electronegative atom of the second molecule, this type of bond binds the hydrogen atom of one molecule and the electronegative atom of the 2nd molecule is called as hydrogen bond. It is a weak bond and it is denoted by dotted lines e.g., in HF, hydrogen forms a weak bond with the electronegative F atom of the 2nd molecule neighboring HF.

So it means hydrogen is acting as a bridge between two molecules by one covalent bond and the other by a hydrogen bond. Due to this hydrogen bonding, HF will not exist as a single molecule but it will exist as an associated molecule (HF)_n. So hydrogen bond may be defined as a weak electrostatic bond that binds the hydrogen atom of one molecule and electronegative bond atoms (F, O, N) of the second neighboring molecule.

Cause of hydrogen bonding: When a hydrogen atom is bonded to an electronegative atom (say F, O, N) through a covalent bond, due to electronegativity difference, the electronegative atom attracts the shared pair of electrons towards its side with a great force as a result of which the shared pair of electrons will be displaced toward electronegative atom and away from a hydrogen atom.

Due to which hydrogen atom will acquire a slightly negative charge and if another molecule is brought nearer to it in such a way that electronegative atom of the second molecule faces hydrogen atom of the 1st molecule, due to opposite charges present on the atoms, an electrostatic bond will be formed between the hydrogen atom of one molecule and electronegative atom of 2nd molecule and this is called as hydrogen bond.

Conditions for hydrogen bonding. The following two necessary conditions for hydrogen bonding are:

1. Hydrogen atom should be connected to highly electronegative atom say F, O, or N.
2. The electronegative atom of which the hydrogen atom is connected should be the same in size.

The smaller the size of the electronegative atom greater will be the attraction of that atom for shared pair of electrons and hence that pair will be displaced more nearer to that atom and hence that atom will develop greater negative charge and the hydrogen atom will develop a greater positive charge and hence hydrogen atom of this molecule will easily attract negative atom of the IInd molecule and hence a hydrogen bond will be easily formed.

As both these conditions are satisfied only by F, O, N atoms so only three atoms show hydrogen bond.

Strength of Hydrogen Bond: A hydrogen bond is a very weak bond. It is weaker than an ionic or a covalent bond. Its strength range from 13 kJ mol-1 to 42 kJ mol^-1. The strength of the hydrogen bond for some of the molecules is the order H-F H (40 kJ mol^-1) > O-H…… O (28 kJ mol^-1) > H-N…… H (13 kJ mol^-1) whereas the strength of a covalent bond is quite high. For example, the bond strength of the H-H bond in H₂ is 433 kJ mol^-1.

Types of H-bonding
There are two types of hydrogen-bonds

1. Intermolecular hydrogen bond. It is formed between two different molecules of the same or different compounds. For example H-bond in case of HF, alcohol, or water.
2. Intramolecular Hydrogen bond. It is formed when a hydrogen atom is in between the two highly electronegative (F, O, N) atoms present within the same molecule. For example, in o-nitrophenol, hydrogen is in between the two oxygen atoms.
   

Question 3.
Compare the properties of electrovalent and covalent compounds.
Answer:
The main points of difference between electrovalent and covalent compounds are summed up in the table below:

Electrovalent/Ionic compounds	Covalent compounds
1. They are formed by sharing of electrons between two atoms.	1. They are formed by sharing of electrons between two atoms.
2. These compounds may be solids, liquids, or gases.	2. These compounds may be solids, liquids, or gases.
3. They are made up of molecules held together by weak van der Waals forces	3. They are made up of molecules held together by weak van der Waals forces of attraction.
4. They have generally high M and B points	4.            They have generally low melting and boiling points.
5. They are generally soluble in polar solvents like water and insoluble in organic solvents.	5. Covalent compounds are generally soluble in non-polar solvents such as benzene and insoluble in polar solvents.
6. Ionic compounds conduct electricity in the molten or dissolved state.	6. Covalent compounds are generally bad conductors of electricity.
7. Ionic compounds are non-directional and do not show isomerism.	7. Covalent compounds are rigid and directional. They show isomerism.
8. These compounds undergo ionic reactions which are very fast, almost instantaneous.	8. These compounds undergo molecular reactions which are very slow.

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 3 Important Extra Questions Classification of Elements and Periodicity in Properties

Classification of Elements and Periodicity in Properties Important Extra Questions Very Short Answer Type

Question 1.
An element is present in the third period of the p-block. It has 5 electrons in its outermost shell. Predict its group. How many unpaired electrons does it have?
Answer:
It belongs to the 15th group (P). It has 3 unpaired electrons.

Question 2.
An element X with Z = 112 has been recently discovered. Predict its electronic configuration and suggest the group in which it is present.
Answer:
[Rn] 5f¹⁴ 6d¹⁰ 7s². It belongs to the 12th group.

Question 3.
The electronic configuration of an element is 1s² 2s² 2p⁶ 3s² 3p⁵. Name the period and the group to which it belongs?
Answer:
Third-period Group 17.

Question 4.
Arrange Cl, Cl^–, Cl⁺ ion in order of increasing size.
Answer:
CP < Cl < CP.

Question 5.
Arrange the following in increasing order of size.
N^3-, Na⁺, F^–, O^2-, Mg²⁺
Answer:
Mg²⁺< Na⁺< F^– < O^2- < N^3-

Question 6.
Give the formula of one species positively charged and one negatively charged that will be isoelectronic with Ne.
Answer:
Na⁺, F^–.

Question 7.
Argon has atomic number 18 and belongs to the 3rd period and 18th group. Predict the group and period for the element having atomic number 19.
Answer:
Group I, Period 4th.

Question 8.
Which of these belong to
(i) the same period and
(ii) the same group.

Answer:
A and B belong to the same group (18th group)
B and C belong to the same period (II period).

Question 9.
Among the elements of the 3rd period, Na to Al pick out the element.
(i) With the highest first ionization enthalpy.
Answer:
Argon

(ii) With the largest atomic radius.
Answer:
Sodium

(iii) That is the most reactive non-metal.
Answer:
Chlorine

(iv) That is the most reactive metal.
Answer:
Sodium.

Question 10.
Arrange the following ions in the order of increasing size:
Be²⁺, Cl^– , S^2-, Na⁺, Mg²⁺, Br.
Answer:
Be²⁺ < Mg²⁺ < Na⁺ < Cl^– < S^2- < Br^–

Question 11.
Arrange the following elements in the increasing order of metallic character: B, Al, Mg, K.
Answer:
B < Al < Mg < K

Question 12.
Among the elements, Li, K, Ca, S and Kr which one is expected to have the lowest first ionization enthalpy and which one the highest first ionization enthalpy?
Answer:
K has the lowest IE₁
Kr has the highest IE₁

Question 13.
Predict the position of the element in the periodic table satisfying the electronic configuration (n – 1)d¹ ns² for n = 4.
Answer:
(n – 1 )d¹ ns² for n = 4 becomes 3d¹ 4s². It lies in the 4th period and, in the 3rd group.,

Question 14.
Predict the formulae of the stable binary compounds that would be formed by the following pairs of elements:
(a) Silicon and Oxygen
Answer:
SiO₂

(b) Aluminium and Bromine
Answer:
AlBr₃

(c) Calcium and Iodine
Answer:
CaI₂

(d) Element 114 and Fluorine
Answer:
Uuq F₄

(e) Element 120 and Oxygen.
Answer
(Ubn) O

Question 15.
Consider the elements N, P, O, and S and arrange them in order of
(a) increasing first ionization enthalpy
Answer:
S < O < P < N

(b) increasing negative electron gain enthalpy
Answer:
N < P < O < S

(c) increasing non-metallic character.
Answer:
P < N < S < O

Question 16.
Among the elements B, Al, C, and Si
(a) Which has the highest IE?
Answer:
C (Carbon)

(b) Which has the highest negative electron gain enthalpy?
Answer:
C (Carbon)

(c) Which has the largest atomic radius?
Answer:
Al

(d) Which has the most metallic character?
Answer:
Al

Question 47.
Arrange the following ions in order of decreasing ionic radii: Li²⁺, He⁺, Be³⁺.
Answer:
He⁺, Li²⁺, Be³⁺ are all isoelectronic ions
∴ He⁺ > Li²⁺ > Be³⁺.

Question 18.
Name the group of elements classified as s-, p-, d-blocks.
Answer:

• s-block consists up of groups 1 and 2. Alkali and alkaline earth metals.
• p-block consists up of groups 13-18.
• d-block consists up of groups 3-12, called transition elements.

Question 19.
Al atom loses electrons successively to form Al⁺, Al²⁺, Al³⁺ ions. Which step will have the highest ionization enthalpy?
Answer:

will have the highest IE.

Question 20.
In terms of electronic configuration, what the elements of a given period and a given group have in common?
Answer:
In a given period, the no. of shells is equal and in a given group, the no. of electrons in the valence shell is the same.

Question 21.
Which of the following elements has the most positive electron gain enthalpy? Fluorine, Nitrogen, Neon.
Answer:
Neon.

Question 22.
Why electron affinities of Be and Mg are positive?
Answer:
They have fully s-orbitals and the additional electron cannot be placed in the much higher energy p-orbitals of the valence shell.

Question 23.
What would be the IUPAC name and symbol for the element with atomic number 120?
Answer:
Unbinillium and its symbol are Ubn.

Question 24.
Why ionization enthalpy of Nitrogen is greater than that of oxygen?
Answer:
Nitrogen has. exactly half-filled orbitals, due to which it is difficult to remove an electron from the valence shell of N atoms.

Question 25.
Give four examples of species that are isoelectronic with Ca²⁺.
Answer:
Ar, K⁺, S^2-, P^3- are isoelectronic with Ca2+.

Question 26.
Comment on the statement, “The electron gain enthalpy of halogens decrease in the order F > Cl > Br > I.”
Answer:
The statement is wrong. The actual order is I > Br > F > Cl.

Question 27.
Arrange the following in order of increasing radii I, I⁺, I^–.
Answer:
I⁺ < I < I^–

Question 28.
Which of the following pairs would have a larger size
(i) K or K⁺
Answer:
K

(ii) Br or Br
Answer:
Br

(iii) O^2- or F^–
Answer:
O^2-

(iv) Na⁺ or K⁺
Answer:
K⁺

Question 29.
Select from each group the one with the smallest radius.
(i)O, O^–, O^2-
Answer:
O

(ii) K⁺, Sr²⁺, Ar
Answer:
Sr²⁺

(iii) Si, P, Cl
Answer:
Cl

Question 30.
Which of the following has the largest and smallest size? Mg, Mg²⁺, Al, Al³⁺.
Answer:
Mg has the largest and Al³⁺ the smallest size.

Question 31.
What is the recommended, IUPAC name and symbol of the element with atomic number 108?
Answer:
Element with At. No. = 108 is Uno. Its recommended name is Unniloctium and its official IUPAC .name is Hassnium.

Question 32.
What is the general outer shell electronic configuration of p-block elements?
Answer:
ns² np^1-6 where n = 2 to 6.

Question 33.
What is the general outer shell electronic configuration of s-block elements?
Answer:
ns1 ~ 2, where n = 2 to 7.

Question 34.
Give the general outer shell electronic configuration of d-block elements?
Answer:
(n – 1 )d^1-10 ns^0-2 where n = 4 to 7.

Question 35.
What is the general outer shell electronic configuration of f-block elements?
Answer:
(n -2)f^0-14 (n – l)d^0-2 ns² where n = 6 – 7.

Question 36.
Which are the three elements among the first three transition series (1st, 2nd, and 3rd) which generally do not show v that properties shown by other members of the series.
Answer:
Zn, Cd, and Hg (Zinc, cadmium, and mercury).

Question 37.
Which is the shortest period in the modern periodic table? Name the elements in it.
Answer:
First period. Hydrogen (H) and helium (He).

Question 38.
What is the number of groups in the d-block?
Answer:
Group numbers 3, 4, 5, 6, 7, 8, 9, 10,11, 12 belong to the d- block

Question 39.
Give the electronic configuration of the first and the last element of the I transition series.
Answer:
Sc (21) = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰4s² .
Zn (30) = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰ 4s².

Question 40.
Give the names and symbols of two metalloids.
Answer:
Arsenic (As) and antimony (sb).

Question 41.
What do you mean by Vander Waals’ radius?
Answer:
It is defined as one-half the distance between the nuclei of two identical non-bonded isolated atoms or two adjacent identical atoms belonging to two neighboring molecules of an element in the solid state.

Question 42.
Arrange the following ions in order of decreasing ionic radii: Li²⁺, He⁺, Be³⁺.
Answer:
He⁺> Li²⁺> Be³⁺.

Question 43.
Arrange the following in decreasing, order of their van der Waals radii: Cl, H, O, N.
Answer:
Cl > N > O > H. .

Question 44.
What are super heavy elements?
Answer:
Elements with Z > 100 which have high densities are called superheavy elements.

Question 45.
Which transition element has the maximum oxidation state?
Answer:
Osmium (Os) shows an oxidation state of + 8.

Question 46.
Why do the elements of the 2nd period show anomalous properties than the other members of their respective groups?
Answer:

1. small size
2. large charge/radius ratio
3. high electronegativity
4. Absence of d-orbitals.

Question 47.
Which group elements in the periodic table show the maximum electronegativity in their compounds?
Answer:
Halogens belonging to the 17th group.

Question 48.
For each of the following pairs, predict which one has lower first ionization enthalpy?
(i) N or O
Answer:
O

(ii) Na or Na+
Answer:
Na

(iii) Be⁺ or Mg²⁺
Answer:
Be⁺

(iv) I or I^–
Answer:
I^–

Question 49.
Which of the following electronic configurations of the elements do you expect to have a higher value of ionization enthalpy?
(i) 1s² 2s² 2p³ or 1s² 2s² 2p⁴
Answer:
1s² 2s² 2p³

(ii) 1s² or 1s² 2s² 2p⁶
Answer:
1s²

(iii) 3s² or 3s² 3p¹
Answer:
3s²

Question 50.
Arrange the following elements in order of decreasing electron gain enthalpy: B, C, N O.
Answer:
N > B > C > O.

Classification of Elements and Periodicity in Properties Important Extra Questions Short Answer Type

Question 1.
Do elements with high I.E. have high E.A.?
Answer:
Normally is true that the elements with haying high value of I.E. have a high value of E affinity. But however, there are marked exceptions. It is seen that elements, with stable electronic configurations, have very high values of I-Energies as it is difficult to remove electrons as is the case with 15th and 18th group elements but in such case electron cannot be added easily so that is why elements of 15th group have almost zero E.A. and elements of 18th group have got zero E.A. whereas their Ionisation energy values are very high.

Question 2.
What is a periodic classification of elements?
Answer:
By periodic classification of the elements we mean the arrangement of the elements in such a way that the elements with similar physical and chemical properties are grouped together and for this various scientists made contributions but however the contributions made by Mendeleev are of great significance and he gave a periodic table which called as Mendeleev’s Periodic ‘Table which was older and replaced by the long form of the periodic table.

Question 3.
Distinguish between s and p block elements.
Answer:
They can be distinguished as follows: s block elements:

1. They have got the general configuration of the valence shell, ns^1-2.
2. They are all metals.
3. Their compounds are mostly ionic.
4. They are generally strong reducing agents.
5. They mostly impart characteristic color to the flame.
6. They have low ionization energies.
7. They show fixed oxidation states,

p block elements:

1. The valence shell electronic configuration of p block elements in ns² p^1-6.
2. They are mostly non-metals.
3. Their compounds are mostly covalent.
4. They are generally strong oxidizing agents.
5.  Mostly they do not impart color to the flame.
6. They have got a comparatively higher value of I.E.
7. They show variable oxidation states. ,

Question 4.
Explain why ionization enthalpies decrease down a group of the Periodic Table.
Answer:
The decrease in ionization enthalpies down any group is because of the following factors:

1. There is an increase in the number of the main energy shells
2. moving from one element to another.
3. There is also an increase in the magnitude of the screening effect due to the gradual increase in the number of inner electrons.

Question 5.
Why does the first ionization enthalpy increase as we go . from left to right across a given period of the Periodic Table?
Answer:
The value of ionization enthalpy increases with the increase in atomic number across the period.

This is due to the fact that in moving across the period from left to right.,

1. Nuclear charge increases regularly by one unit.
2. The progressive addition of electrons occurs at the same level.
3. Atomic size decreases.

This is due to the gradual increase in nuclear charge and a simultaneous decrease in atomic size the electrons are more and more tightly bound to the nucleus. This results in a gradual increase in ionization energy across the period.

Question 6.
How do atomic radii vary across a period with an atomic number in the periodic table? Explain.
Answer:
Variation of Atomic radii across a period: Atomic radii decrease with the increase in the atomic number in a period. For example, atomic radii decrease from lithium to fluorine in the second period.

In moving from left to right across the period, the nuclear charge increases progressively by one unit but the additional electron goes” to the same principal shell. As a result, the electron cloud is pulled closer to the nucleus by increased effective nuclear charge. This causes a decrease in atomic size.

Variation of the atomic radius with an atomic number across the second period

Question 7.
In each of the following pairs, which species has a larger size? Explain.
(i) Kor K⁺
Answer:
K is larger in size than K⁺.
The electronic configurations of K and K⁺ are:
K: 1s², 2s² 2p⁶, 3s² 3p⁶, 4s¹
K⁺: 1s², 2s² 2p⁶, 3s² 3p⁶.

K⁺ is formed from K when the latter loses its 4s electrons.
K → K⁺ + e^–

As seen from the electronic configurations of K and K⁺, the K⁺ ion has 18 electrons and the K atom has 19 electrons but the nuclear charge in both these species is the same (+ 19). Since the K⁺ ion has 1 electron less than the K atom, the forces of attraction between the nucleus and the electrons are more strong in the K⁺ ion than in the K atom. This results in more inward pulling of electrons towards the nucleus in K+ ion and hence its size decreases.

(ii) Br or Br
Answer:
Br has formed .by the gain of one electron by Br atom. In Br, the nuclear charge is the same as that in the Br atom but the number of electrons has increased. Since the same nuclear charge how acts on the increased number of electrons, the effective nuclear charge per electron decreases in Br”. The electron cloud is held less tightly by the nucleus. This causes an increase in size.

Question 8.
Explain why chlorine has a higher electron affinity than Fluorine? :
Answer:
Cl has a higher electron affinity than F. This is due to the small size of fluorine. As a result of its very small size the inter-electronic repulsions in relatively compact 2p-subshell of a chlorine atom. On comparing chloride ion with fluoride ion we find that electron density per unit volume in fluoride ion (F^–) is more than in chloride (Cl^–) ion. This means that the coming election in the fluorine atom finds less attraction than in the chlorine atom. Consequently, the electron affinity of chlorine is higher than that of fluorine.

Question 9.
Two elements C and D have atomic numbers 36 and 58 respectively. On the basis of electronic configuration predict the following:
(i) The group, period, and block to which each element belongs,
Answer:
The electronic configuration of elements C and D are:
C (At. no 36) = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰, 4s² 4p⁶.
D (At. no. 58) = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰, 4s² 4p⁶, 5s² 4d¹⁰ 5p⁶,6s² 5d¹ 4f¹.

The element C belongs to the 18th group, lies in the 4th period, and belongs to the p-block of elements. The element D belongs to lanthanides, lies in the 6th period, and belongs to/-block of elements.

(ii) Are they representative elements?
Answer:
The element C is a noble gas, and D is the inner transition element. They are not representative elements.

Question 10.
Account for the fact that the fourth period has 18 and not 8 elements.
Answer:
The first element of fourth period is potassium (Z = 19) having electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. Thus it starts with the filling of 4s orbital., After the filling of 4s orbital, instead of 4p orbitals, there start the filling 3d orbitals. This is in keeping with the fact that 3d orbitals have fewer energies than 4p orbitals.

Thus 10 elements are built up by the filling of 3d orbitals. After the filling of 3d orbitals, 4p orbitals are filled up and that process is completed at krypton [Kr, Z = 36]. Hence the fourth period consists up of 18 and not 8 elements.

Question 11.
Explain the term ‘valency’ of an element. How does it vary in a period and in a group in the periodic table?
Answer:
The chemical properties of elements depend upon the number of electrons in the outermost shell of an atom. These electrons are called valence electrons and thus determines the valency of the atom (or element).

In representative elements, the valency is generally equal to either n or (8 – n), where n is no. of. valence electrons in the atom.

In a period valence electrons increase from 0 to 8 on moving from left to light. The valency of an element w.r.t. H and Cl increase from 1 to 4 and then decrease to zero. However w.r.t. oxygen, valency increases from 1 to 7 and then becomes zero in noble gases.

In a group, the number of valence electrons remains the same, and therefore all elements in a group exhibit the same valency, e.g., all elements of group I have valency one and those of group 2 have valency two. However, the transition elements exhibit variable valency.

Question 12.
The elements, Z = 107 and Z = 109 have been made recently, element Z = 108 has not yet been made. Indicate the groups in which you will place the above elements.
Answer:
The element with Z = 107, will be placed in the 7th Group, an element with Z = 108 in group 8th and the element with Z = 109 will be placed in group 9th of the periodic table. All three elements will belong to the d-block of the periodic table. (The valence shell electronic configuration will be 7s² 5f¹⁴ 6d⁵ to 7s²5f¹⁴ 6d⁷).

Question 13.
Lanthanides and actinides are placed in separate rows at the bottom of the periodic table. Explain the reason for this arrangement.
Answer:
Lanthanides and actinides are placed in separate rows at the bottom of the periodic table

• to save space
• to keep elements? with similar properties in a single column and
• because their antepenultimate 4/and 5/orbitals are filled respectively.

Question 14.
Which of the following species will have the largest and the smallest size? Mg, Mg²⁺, Al, Al³⁺.
Answer:
Atomic radii decrease from left to right across a period. Cations are always smaller than their parent atoms. Among the isoelectronic species, the one with a larger positive nuclear charge will have a smaller ionic radius.

Hence the largest species is Mg, the smallest one is Al³⁺

Question 15.
The first ionization enthalpy (IE) of the third-period elements Na, Mg, and Si are respectively 496, 737, and 786 kJ mol^-1. Predict whether the first IE value of Al will be more close to 575 or 760 kJ mol^-1 Justify your answer.
Answer:
The first ionization enthalpy (IE) of Al will be more close to 575 kJ mol^-1. The value of Al should be lower than that of Mg because of the effective shielding of 3p electrons from the nucleus by 3s electrons.

Question 16.
Which of the following will have the most negative electron gain enthalpy and which the least?
P, S, Cl, F. Explain your answer.
Answer:
Electron gain enthalpy generally becomes more negative across a period as we move from left to right. Within a group, electron gain enthalpy becomes less negative down a group. However, adding to electron to 2p orbital leads to greater repulsion than adding an electron to the larger 3p orbital. Hence the element with the most negative electron gain enthalpy is chlorine; the one with the least negative electron gain enthalpy is phosphorus.

Question 17.
Predict the formulae Of compounds which right be formed by the following pairs of elements
(a) Silicon and bromine
Answer:
Silicon (Si) is a group 14 element with a valence of 4; bromine belongs to the group 17 halogen family with a valence of one. Hence the formula of the compound will be SiBr₄.

(b) aluminum and sulfur.
Answer:
Aluminum belongs to group 13 with a valence of 3, sulfur belongs to group 16 with a valence of 2. Hence the formula of the compound would be Al₂S₃.

Question 18.
Show by a chemical reaction with water that Na₂O is a basic oxide and C₁₂O₇ is an acidic oxide.
Answer:
Na₂O reacts with water to form a strong base whereas C₁₂O₇ forms a strong acid.
Na₂O + H₂O → 2NaOH (strong base)
C₁₂O₇ + H₂O → 2HClO₄ (strong acid) ,
Their basic or acidic nature can be qualitatively tested with litmus paper.

Question 19.
Describe the nature of oxides formed by the extreme left elements of the s-blocks and elements on the extreme right Of the p- block.
Answer:
Elements on two extremes of a period easily combine with oxygen to form oxides. The normal oxide formed by the element in the extreme left is the most basic (e.g., Na₂O) whereas that formed by the element on the extreme right is the most acidic (e.g., C₁₂O₇). Oxides of elements in the center are amphoteric (e.g., Al₂O₃ As₂O₃) or neutral (e.g., CO, NO. N₂O).

Question 20.
What are super heavy elements?
Answer:
Elements that have Z > 100 and have high densities are called superheavy elements.

Question 21.
Calculate the energy required to convert all atoms of magnesium to magnesium ions present in 24 mg of magnesium vapors. IE₁ and IE₂ of Mg are 737.76 and 1450.73 kJ mol-1 respectively.
Answer:
Mg (g) + IE₁ → Mg+ (g) + e^– (g)
IE₁ = 737.76 kJ mol^-1

Mg+ (g) + IE₂ → Mg²⁺ + e^– (g)
IE₂ = 1450.73 kJ mol^-1

Total amount of energy needed to convert Mg (g) into ,
Mg²⁺ ion = IE₁ + IE₂

Now 24 mg of Mg = \(\frac{24}{1000}\) g = \(\frac{24}{1000 \times 24}\) mol [1 Mole of Mg = 24 g]
= 10^-3 mole.

Question 22.
The IE₁ and IE₂ of Mg. (g) are 740 and 1450 kJ mol^-1. Calculate the percentage of Mg⁺ (g) and Mg²⁺ (g) if 1 g of Mg (g)
absorbs 50 kJ of energy.
Answer:
No. of moies of Mg vapours present in I g = \(\frac{1}{24}\) = 0.0147

Energy absorbed to convèrt Mg (g) to Mg⁺ (g) = 0.0417 × 740
=30.83 kJ

Energy left unused = 50 – 30.83
= 19.17 kJ
Now 19.17 kJ will be used to e Mg⁺ (g) to Mg²⁺ (g) .
∴ No. of moles of Mg+ (g) converted to Mg2+ (g) = \(\frac{19.17}{1450}\) = 0.132

%age of Mg⁺ (g) = \(\frac{0.0285}{0.0417}\) × 100 = 68.35
and %age of Mg²⁺ (g) = 100 – 68.35 = 31.65

Question 23.
Which of the elements Na, Mg, Si, P would have the greatest difference between the first and the second ionization enthalpies. Briefly explain your answer
Answer:
Among Na, Mg, Si, P, Na is an alkali metal. It has only one electron in the valence shell. Therefore, its IE₁ is low: However, after the removal of the first electron, it acquires a Neon gas configuration i.e., Na+ (1s², 2s² 2p⁶). Therefore its IE₂ is expected to be very high. Consequently, the difference in IE₁ and IE₂ comes to be greatest in the case of Na.

Question 24.
The IE₂ of Mg is higher than that of Na. On the other hand, the IE₂ of Na is much higher than that of Mg. Explain,
Answer:
The first electron in both cases has to be removed from the 3s orbital, but the nuclear charge of Na is less than that of Mg. After the removal of the first electron from Na, the electronic configuration of Na⁺ is 1s², 2s² 2p⁶, i.e., that of noble gas which is very stable and the removal of the 2nd electron is very difficult. In the case of Mg after the removal of the first electron, the electronic configuration of Mg⁺ is 1s², 2s² 2p⁶ 3s. The 2nd electron to be removed is again from 3s orbital which is easier.

Question 25.
The electron gain enthalpy of chlorine is – 349 kJ mol^-1 How much energy in kJ is released when 3.55 g of chlorine is converted completely into Cl^– ion in the gaseous state.
Answer:
According to the definition of electron gain enthalpy
Cl(g) + e^– (g) → Cl^– (g) + 349 kJ mol^-1

The energy released when 1 mole (= 35.5 g) of chlorine atoms change completely with Cl^– (g) = 349 kJ

Question 26.
The amount of energy released when 1 × 10¹⁰ atoms of chlorine in vapor state are converted to Cl^– ions according to the equation.
Cl (g) + e^– → Cl^– (g) is 57.86 × 10^-10 J
Calculate the electron gain enthalpy of the chlorine atom in terms of kJ mol^-1 and eV pet atom.
Answer:
The electron gain enthalpy of chlorine

Question 27.
What is electronegativity? How does it vary along a period and within a group?
Answer:
Electronegativity is defined as the tendency/power/urge of an atom in a molecule to attract the shared pair of electrons v towards itself.

• Variation across a period: Electronegativity generally increases from left to right across a period.
• Variation within a group: It generally decreases in going from top to bottom within a group.

Question 28.
Electronic configuration of the four elements are given below:
Arrange these elements in increasing order of their metallic character. Give reasons for your answer.
(i) [Ar]4s²
Answer:
[Ar]4s² is Calcium metal with At. no. = 20.

(ii) [Ar]3d¹⁰ 4s²
Answer:
[Ar]3d¹⁰ 4s² is Zinc metal with At. no. = 30.

(iii) [Ar]3d¹⁰ 4s² 4p⁶ 5s²
Answer:
[Ar]3d¹⁰ 4s² 4p⁶ 5s² is Strontium metal with At. no. = 38.

(iv) [Arl 3d¹⁰ 4s² 4p⁶ 5s¹
Answer:
[Ar] 3d¹⁰ 4s² 4p⁶, 5s¹ is*Rubidium metal with At. no. = 37.

Alkali metals are the most metallic, followed by alkaline earth metals and transition metals. Among alkali metals – Rubidium (37) is the most metallic. Among alkaline earth metals (Ca, Sr) Sr (Strontium) is more metallic than Calcium (Ca) as the metallic character increases from top to bottom in a group. Zinc – the transition metal is the least metallic. Thus metallic character increases from
Zn < Ca < Sr < Rb or (ii) < (i) < (iii) < (iv)

Question 29.
lthough F is more electronegative than chlorine but the electron gain enthalpy of Cl is more negative than that of F. Why?
Answer:
L. Pauling gave the highest value of 4.0 to F due to its highest electronegative character. Electronegativity decreases from top to bottom in a group and therefore Cl is less electronegative than F. Electron gain enthalpy of F.is less negative than Cl because of the compact size of its atom (2 orbits) as compared to Cl (3 orbits) the mutual electronic repulsion in F is more than that of Cl.

Question 30.
The formulation of F (g) from F (g) is exothermic whereas that of O^2- (g) from O (g) is endothermic. Explain.
Answer:
F (g) + e^– (g) → F (g); ΔH = Negative
Energy is released when an extra electron from outside is added to a neutral isolated gaseous atom of an element.
To convert, O (g) to O^2- (g) two steps are required
(i) O (g) + e^– (g) → O^– (g); ΔH₁ = – 141 kJ mol^-1
(ii) O^– (g) + e^– (g) → O^2- (g); ΔH₂ = + 780 kJ mol^-1

Hence the over all processes endothermic (+ 780 – 141 = + 639 kJ mol^-1) whereas F (g) to F^– (g) is exothermic.

Classification of Elements and Periodicity in Properties Important Extra Questions Long Answer Type

Question 1.
Explain the important general characteristics of groups in the modem periodic table in brief.
Answer:
The elements of a group show the following important similar characteristics.
(0 Electronic configuration. All elements in a particular group have similar outer electronic configuration e.g., all elements of group I’, i.e., alkali metals have ns1 configuration in their valency shell. Similarly, group 2 elements (alkaline Earths) haye ns² outer configuration and halogens (group 17) have ns² np⁵ configuration (where n is the outermost shell).

(it) Valency. The valency of an element depends upon the number of electrons in the outermost shell. So elements of a group show the same valency, e.g., elements of group 1 show + 1 valency and group 2 show + 2 valencies i.e. valency i.e., NaCl > MgCl₂ etc.

(iii) Chemical properties. The chemical properties of the elements are related to the number of electrons in the outermost shell of their atoms. Hence all elements belonging to the same group show similar chemical properties. But the degree of reactivity varies gradually from top to bottom in a group. For example, in group 1 all the elements are highly reactive metals but the degree of reactivity increases from Li to Cs. Similarly, elements of group 17, i.e., halogens: F, Cl, Br, I are all non-metals and they’re- reactivity goes on decreasing from top to bottom.

Question 2.
Explain the electronic configuration in periods in the periodic table. „
Answer:
Each successive period in the periodic table is associated with the filling Up of the next higher principal energy level (n – 1, n – 2, etc.). It can be readily seen that the number of elements in each period is twice the number of atomic orbitals available in the energy level that is being filled. The first period starts with the filling of the lowest level (1s) and has thus the two elements – hydrogen (1s¹) and helium (1s²) when the first shell (K) is completed. The second period starts with lithium and the third electron enters the 2s orbital.

The next element, beryllium has four electrons and has the electronic configuration 1s² 2s². Starting from the next element boron, the 2p orbitals are filled with electrons when the L shell is completed’ at neon (2s² 2p⁶). Thus there are 8 elements in the second period. The third period (n = 3) being at sodium, and the added electron enters a 3s orbital. Successive filling of 3s and 3p orbitals give rise to the third period of 8 elements from sodium to argon.

The fourth period (n = 4) starts at potassium with the filling up of 4p of 4s orbital. Before the 4p orbital is filled, the filling up of 3d orbitals becomes energetically favorable and we come across the so-called 3d transition series of elements. The fourth period ends at krypton with the filling up of the 4p orbitals. Altogether we have 18 elements in the fourth period. The fifth period (n = 5) beginning with rubidium is similar to the fourth period and contains the 4d transition series starting at yttrium (Z = 39).

This period ends at xenon with the filling up of the 5p orbitals. The sixth period (n = 6) contains 32 elements and successive electrons enter 6s, 4/, 5d, and 6p orbitals, in that order. Filling up of the 4/ orbitals being with cerium, (Z = 58) and ends at lutetium (Z = 71) to give the 4/-inner transition series which is called the lanthanide series. The seventh period (n = 7) is similar to the sixth period with the successive filling up of the 7s, 5f, 6d, and 7p orbitals and includes most of the man-made radioactive elements.

This period will end at the element with atomic number 118 which would belong to the noble gas family. Filling up of the 5f orbitals after actinium (Z = 89) gives the 5f-inner transition series known as the actinide series. The 4f and 5f transition series of elements are placed separately in the periodic table to maintain its structure and to preserve the principle of classification by keeping elements with similar properties in a single column.

Question 3.
Explain the variation of valence in the periodic table.
Answer:
Variation of valence in a group as well as across a period in the periodic table occurs as follows:
1. In a group: All elements in a group show the same valency. For example, all alkali metals (group 1) show a valency of 1+. Alkaline earth metals (group 2) show a valency of 2+.

However, the heavier elements of p-block elements (except noble gases) show two valences: one equal to the number of valence electrons or 8-No. of valence electron# and the other two less. For example, thallium (Tl) belongs to group 13. It shows valence of 3+ and 1+.

Lead (Pb) belongs to group 14. If shows valance of 4+ and 2+.
Antimony (Sb) and Bismuth (Bi) belong to group 15. They show valence of 5+ and 3+ being more stable.

This happens due to the non-participation of tie two s-electrons present in the valence shell of these elements. This non-participation of one pair of s-electrons in bonding is called the inert-pair effect.

2. In a period: The number of the valence electrons increases – in going from left to right in a period of the periodic table. Therefore the valency of the elements in a period first increases, and then decreases.

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 2 Structure of Atom Chemistry. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 2 Important Extra Questions Structure of Atom Chemistry

Structure of Atom Chemistry Important Extra Questions Very Short Answer Type

Question 1.
How many total electrons are present in nitrate ion?
Answer:
No. of electrons in NO₃^– ion
= No. of electrons on N + No. of electrons on 3 oxygen atom + one ē
= 7 + 3 × 8 + 1 = 32 electrons.

Question 2.
The nucleus of the atom of an element does not contain a neutron. Name the element and what does its nucleus consists of.
Answer:
The nucleus of hydrogen. It contains only one proton.

Question 3.
What are nucleons?
Answer:
The neutrons and protons present in the nucleus of an atom are collectively called nucleons.

Question 4.
Write electronic configurations of Chromium (At. Np. = 24).
Answer:
Cr = 24 = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁵, 4s¹.

Question 5.
Which of the following has the smallest de-Broglie wavelength? O₂, H₂, a proton, an electron.
Answer:
According to the de-Broglie equation λ = \(\frac{h}{m \times v}\) for same value of velocity λ ∝\(\frac{1}{m}\)
∴ O₂ molecule has shortest wavelength.

Question 6.
How many unpaired electrons are there is a carbon atom in the ground state?
Answer:
C = 6 = 1s², 2s², 2p¹_x 2p¹_y. There are only two unpaired electrons.

Question 7.
What type of spectrum is obtained when light emitted from the discharge tube containing hydrogen gas is analyzed?
Answer:
Emission line spectrum.

Question 8.
What is the maximum number of electrons in an atom having n = 3, l = 1 and s = + \(\frac{1}{2}\)?
Answer:
Three electrons (one each in 3p_x’ 3p_y’, 3p_z’ ).

Question 9.
Name the spectral line in the spectrum of H-atom obtained when an electron jumps from n = 4 to n = 2.
Answer:
Balmer Series.

Question 10.
Give some examples of electromagnetic radiation.
Answer:
Y-rays, X-rays, UV-rays, visible rays, radio waves, etc.

Question 11.
State two properties of electromagnetic radiation.
Answer:
Electromagnetic radiation shows the phenomenon of:

1. Interference,
2. Diffraction.

Question 12.
What is meant by the quantization of electron energy?
Answer:
It means that an electron in an atom has a certain, specific, discrete amount of energy.

Question 13.
What does a principal quantity number denote?
Answer:
It denotes a specific stationary state.

Question 14.
Why Bohr’s orbits are also called ‘energy levels’?
Answer:
Because they are associated with a certain definite amount of energy.

Question 15.
How many spherical nodal surfaces are there in a 3s- sub-shell?
Answer:
Two.

Question 16.
Out of 6s and 4f orbitals, which has higher energy and why?
Answer:
4forbital has higher energy, ((n + l) value of 4f is 7 while that of 6s is 6). The higher the (n + l) value of an orbital higher is the energy.

Question 17.
List the value/values of quantum numbers n and l for 4f electrons.
Answer:
n = 4, l = 3.

Question 18.
Out of 4s and 3d orbitals, which will have higher energy and why?
Answer:
3d orbital has higher energy as it has a higher value of (n + l).

Question 19.
Which of the following orbitals are not possible?
1p, 2p, 2d, 3f, 4f?
Answer:
1p, 2d, 3f is not possible.

Question 20.
Which orbital does not have directional characteristics?
Answer:
s-orbital.

Question 21.
An electron is in 3p-orbital. What are the permitted values of n, l, and m?
Answer:
n = 3, l = 1, m = – 1, 0, + 1.

Question 22.
Write designation of an orbital having n = 5,1 = 3.
Answer:
5f -orbital.

Question 23.
Consider the electronic configuration 1s° 2s° 2p° 3s¹.
Name the element having this configuration. Is it in an excited state or ground state?
Answer:
It is the configuration of the H-atom. It is in an excited state.

Question 24.
Which quantum number determines the
(i) size,
Answer:
Principal

(ii) orientation,
Answer:
Magnetic

(in) the shape of orbital?
Answer:
Azimuthal quantum number.

Question 25.
Which energy level does not have a p-orbital?
Answer:
First energy level (i.e., n – 1, K-shell).

Question 26.
Name an element that has only one d-electron.
Answer:
Scandium (atomic no. = 21).

Question 27.
Given an isotone of C-13 atom.
Answer:
Isotones are the atoms of different elements which have the same number of neutrons. N-14 is an isotone of a C-13 atom.

Question 28.
Which of the following orbitals does not make sense?
5s, 4f, 3p, 2d.
Answer:
2d orbital does not exist and thus makes no sense.

Question 29.
Name the famous experiment which showed for the first time that an atom has a nucleus.
Answer:
Rutherford experiment of scattering of a-particles.

Question 30.
Write the value of four quantum numbers for the valence electron of the sodium atom.
Answer:
Sodium atom has 11 electrons and its valence electron is 3s¹ (as configuration is 1s², 2s²p⁶, 3s¹)
∴ The value are: n = 3, l = 0, m = 0, s = + \(\frac{1}{2}\)

Question 31.
What is the relationship between velocity, wavelength, and frequency of radiation?
Answer:
These three characteristics of wave motion are related to each other as frequency = \(\frac{\text { Velocity }}{\text { Wavelength }}\)
In terms of symbols v = \(\frac{c}{λ}\).

Question 32.
How wave number (\(\vec{v}\) ) and wave length (λ) are related?
Answer:
\(\vec{v}\) = \(\frac{1}{λ}\) . Wave number is the reciprocal of wavelength.

Question 33.
Which series of hydrogen spectrum lies in the visible spectrum?
Answer:
Balmer series.

Question 34.
How do you distinguish the two electrons present in the same orbital?
Answer:
By their spin quantum no. s which has + \(\frac{1}{2}\) and – \(\frac{1}{2}\) values.

Question 35.
Name the principle which establishes that two electrons cannot have the same values for all the 4 quantum numbers.
Answer:
Pauli exclusion principle.

Question 36.
Write down the electronic configuration of Cu (= 29) in the ground state.
Answer:
Cu = 29 = 1s², 2s² 2p⁶, 3s² 3p⁶, 3d¹⁰ 4s¹.

Question 37.
What is the lowest value of n that allows g orbitals to exists?
Answer:
n = 5.

Question 38.
Why do an atom M and its ion M²⁺ have the same mass?
Answer:
Both have the same no. of neutrons and protons which are responsible for the mass of an atom.

Question 39.
The nucleus of an atom has 6 protons and 8 neutrons. What are its atomic number and mass number? What is this element1?
Answer:
At. No. = 6; Mass No. = 6 + 8 = 14; Element is Carbon.

Question 40.
What is the number of electrons having l = 0 in an atom with an atomic number 29?
Answer:
7. [1s², 2s², 3s², 4s¹, i.e., 2 + 2 + 1 = 7]

Question 41.
Express s, p, d, f for a shell in increasing order of energy.
Answer:
s < p < d < f. .

Question 42.
For n = 5; what are the possible values of l?
Answer:
When n = 5; l = 0,1, 2, 3, 4.

Question 43.
For l = 3, what are the possible values of m?
Answer:
m = – 3, – 2, – 1, 0, + 1, + 2, + 3.

Question 44.
The ionization potential of an atom is 13.6 V. How much energy is required to ionize it?
Answer:
13.6 eV.

Question 45.
The threshold wavelength for a metal surface is λ₀. How is it related to the work function of the metal?
Answer:
W₀ = \(\frac{h c}{\lambda_{0}}\)

Question 46.
Write the energy E of a photon in terms of frequency.
Answer:
E = hv where h is called Planck’ constant.

Question 47.
How much energy is required for the removal of the only electron present in the hydrogen atom?
Answer:
ΔE = E_∞ – E₁ = 0 – (- 1312 kJ mol^-1) – 1312 kJ mol^-1.

Question 48.
Which quantum number determines the
(a) shape
Answer:
Azimuthal

(b) orientation and
Answer:
Magnetic

(c) size of the orbital?
Answer:
Principal.

Question 49.
Write down the values of n, l, m, s of the last electron in potassium (Z = 19)?
Answer:
The last electron in potassium is present in 4s.
Its n = 4; l = 0; m = 0, s = + \(\frac{1}{2}\) or – \(\frac{1}{2}\).

Question 50.
What is the sequence of energies of 3s, 3p, and 3d orbitals in
(i) H-atom
Answer:
H-atom: 3s = 3p = 3d

(ii) a multi-electron atom?
Answer:
3s < 3p < 3d.

Question 51.
Name the cations which do not have any electrons.
Answer:
H⁺, He²⁺.

Question 52.
How many quantum numbers are required to specify an orbital? Name them.
Answer:
Three quantum numbers.
These are

1. Principal quantum no. (n)
2. Azimuthal quantum No. (l)
3. Magnetic quantum no. (m).

Question 53.
What is observed when an opaque object is placed in the path of cathode rays?
Answer:
A shadow of an opaque object is observed on the wall opposite to cathode.

Question 54.
What happens when a very light paddle wheel is placed. in the path of cathode rays?
Answer:
It begins to undergo a rotatory motion.

Question 55.
What name was given to the particles which constitute cathode rays?
Answer:
Electron.

Question 56.
Arrange the following orbitals in the order in which electrons may be normally expected to fill them.
3s, 2p, 3p, 2s, 3d, 4s.
Answer:
According to the Aufbau principle, the given orbitals will be filled in the order: 2s, 2p, 3s, 3p, 4s, 3d.

Question 57.
Which fundamental property of an atom is not understood if we assume that an atom consists of a nucleus containing protons only and an extranuclear part containing an equal number of electrons?
Answer:
The mass number of atoms and stability of the nucleus cannot be explained.

Question 58.
The following ions are isoelectronic: F^–, Mg²⁺, O^2-. Write the common electronic configuration.
Answer:
Each given ion contains 10 electrons. The common configuration is that of the Neon atom, i.e., 1s², 2s² p⁶.

Question 59.
What is the atomic number of an element whose mass number is 23 and contains 12 neutrons in its nucleus? What is the symbol of an element?
Answer:
Atomic number = No. of protons in the nucleus
= Mass no. – No. of neutrons
= 23 – 12 = 11
The element is sodium and its symbol is Na.

Question 60.
An atom has 2 K, 8 L, and 5 M electrons. Write the electronic configuration of the atom. How many unpaired electrons are there in the atom?
Answer:
Electronic configuration of atom is: 1s², 2s²p⁶, 3s², 3p¹_x, 3p¹_y, 3p¹_z. It has three unpaired electrons.

Question 61.
Write the various possible quantum numbers for unpaired electron of Aluminium atom?
Answer:
Al = 13 = 1s², 2s² 2p⁶ 3s² 3p¹_x
n = 3; l = 1; m = – 1, 0, + 1; s = \(\frac{1}{2}\) or – \(\frac{1}{2}\) .

Question 62.
Give the values of quantum numbers for the electron with the highest energy in sodium atom.
Answer:
n = 3; l = 0; m = 0; s = + \(\frac{1}{2}\) or – \(\frac{1}{2}\) .

Question 63.
What do you observe in the spectrum of NaCl?
Answer:
Two yellow lines with a wavelength of 5890 Å and 5896 Å

Question 64.
What do you mean by saying that the energy of the electron is quantized?
Answer:
This means that the electrons in an atom have only definite values of energy.

Question 65.
Why are Bohr’s orbits called stationary states?
Answer:
This is because the energies of the orbits in which the electrons revolve are fixed.

Question 66.
What is the difference between a quantum and a photon?
Answer:
The smallest packet of energy of any radiation is called a quantum whereas that of light is called a photon.

Question 67.
Which quantum number does not follow from the solution of the Schrodinger wave equation?
Answer:
Spin quantum number.

Question 68.
How many orbitals will be possible in a g-subshell?
Answer:
For g-subsheil l = 4; m = 2l + 1; – 4, – 3, – 2, – 1, 0, + 1, + 2, + 3, + 4; 9 orbitals.

Structure of Atom Chemistry Important Extra Questions Short Answer Type

Question 1.
Enumerate the important characteristics of anode-rays (or positive rays). How this study led to the discovery of proton?
Answer:

1. The mass of positive particles which constitute these rays depend upon the nature of the gas in the tube.
2. The charge/mass (e/m) ratio of anode-rays is not constant but depends upon the nature of gas in the tube. The value of e/m is greatest for the lightest gas, hydrogen the electric charge on a lightest positively charged particle from hydrogen gas was found to be exactly equal in magnitude but opposite in sign to that of the electron. This lightest positively charged particle from hydrogen gas was named the proton. The mass of a proton is almost 1836 times that of the electron.

Question 2.
What are anode-rays? Illustrate their formation by a diagram.
Answer:
Anode-rays. If a perforated cathode is used in the discharge tube experiment, it is found that a certain type of radiation also travels from anode to cathode. These are called anode rays or positive rays.

Production of anode rays

Question 3.
Describe the important properties of cathode-rays. What is concluded about the nature of these rays?
Answer:
The cathode rays possess the following properties:

1. Travel in straight lines perpendicular to the surface of the cathode.
2. Consists of material particles.
3. Have got the heating effect.
4. Consists of negatively charged particles.
5. Produce X-rays when they strike against hard metals like copper, tungsten, platinum, etc.
6. Produce fluorescence when they strike glass or certain other materials like zinc sulfide.
7. Penetrate through thin aluminum foils and other metals.
8. Affect the photographic plates.

Question 4.
What are the main features of Rutherford’s model of an atom?
Answer:
The main features of this model are:

1. Atom is spherical and consists of two parts: Nucleus and extra-nuclear part.
2. The entire mass and entire positive charge are concentrated in a very small region at the center known as the nucleus.
3. The space surrounding the nucleus known as the extra-nuclear part is negatively charged so an atom as a whole is neutral.
4. Most of the extra-nuclear part is empty.
5. The electrons are not stationary but are revolving around the nucleus at very high speeds like planets revolving around the Sun.

Question 5.
What is meant by the dual nature of radiation?
Answer:
The fact that light energy is carried in terms of packets of energy (i.e., photons) as suggested by Planck’s theory means that light has a particle character. At the same time, the fact light has a wave character. These experimental facts led Einstein to suggest that light has a dual character, i.e., it behaves both like a wave and like a particle.

Question 6.
Describe the drawback to Rutherford’s model of the atom.
Answer:
The main drawback is that it could not explain the stability of an atom. Maxwell has shown that when electric charge is subjected to acceleration, it emits energy in the form of radiations. In Rutherford’s model of the atom, electrons are orbiting the nucleus and hence the direction of their velocity is constantly changing, i.e., electrons are accelerating.

This will cause the electrons will have lesser and lesser energy and will get closer and closer to the nucleus until at last, it spirals into the nucleus and thus does not provide a stable model of the atom.

Question 7.
What is the value of
(i) charge to mass ratio (e/m) of electrons,
Answer:
J. Thomson determined the value of e/m for electron by the study of deflection of electron beam under the simultaneous influence of electric and magnetic field perpendicular to each other, the e/m value is 1.76 × 10⁸ coulomb per gram of electrons.

(ii) charge of electrons,
Answer:
The charge of electrons was measured by Millikan in 1909 by his famous ‘oil drop’ experiment. It was found to be 1.60 × 10^-19 coulombs.

(iii) mass of an electron?
Answer:
The mass of electrons is 9.1 × 10^-28 g.

Question 8.
How is it concluded that electrons are a universal constituent of all matter?
Answer:
The charge/mass (e/m) ratio for the particles in the cathode rays (i.e., electron) is found to be the same irrespective of the nature of the cathode or the nature of the gas taken in the discharge tube. This shows that electrons are universal constituents of all matter.

Question 9.
Distinguish between an Emission spectrum and an Absorption spectrum.
Answer:
The important differences between the emission and absorption spectra are given in the following table:

Question 10.
What are the shortcomings of Bohr’s atomic model?
Answer:

1. It couldn’t explain the spectra of multi-electron atoms.
2. It fails to explain the splitting of spectral lines when subjected to the electrostatic or magnetic fields (Stark or Zeeman’s effect).
3. It does not account for the fine splitting of spectral lines.
4. It affords a two-dimensional picture of the revolution of electrons while actually electron revolves around the nucleus in three dimensions.
5. It does not account for the shapes of molecules.

According to it, this is possible to determine simultaneously both the position and momentum of the electron accurately. But this is contrary to Heisenberg’s Uncertainty Principle.

Question 11.
Account for the stability of the atom with the help of Bohr’s theory.
Answer:
According to Bohr’s theory, an electron revolves around the nucleus only in a definite orbit and cannot lose energy continuously. It can lose energy only if it jumps from a higher orbit to a lower orbit but this is possible only if the electron has already acquired a higher energy level by absorbing a certain amount of energy. If no lower level is available, the electron cannot lose energy at all, i.e., an atom does not collapse. In other words, it is quite stable.

Question 12.
What are the main achievements of Bohr’s theory of the atom?
Answer:
The main achievements of Bohr’s theory of atom are:

1. It can explain the stability of the atom.
2. It successfully explains the line spectrum of hydrogen.
3. It explains the line spectra of single-electron ions like He⁺ and Li²⁺.

Question 13.
Write a short note on de-Broglie relation (or de-Borglie) equation.
Answer:
A moving material particle, like an electron, proton, etc. having mass m and velocity v is associated with wavelength X related by:
λ = \(\frac{h}{m v}=\frac{h}{p}\)
where p stands for the momentum of the particle and h is Planck’s constant. This relation is known as a de-Broglie equation.

Question 14.
Do atomic orbitals have sharp boundaries? Explain why or why not?
Or
Why don’t we draw a boundary surface diagram within the probability of finding the electron is 100%?
Answer:
No, atomic orbitals do not have sharp boundaries because the probability of finding the electron even at large distances may become very small, but not equal to zero.

Question 15.
What is the difference between the angular momentum of an electron present in 3p and that present in 4p?
Answer:
No difference, because angular momentum is given by = \(\frac{h}{2 \pi} \sqrt{l(l+1)}\), i.e., it depends only on the value of l and not the principal quantum no. n.

Question 16.
Why splitting of spectral lines takes place when the source giving the spectrum is placed in a magnetic field?
Answer:
In the presence of a magnetic field, the orbitals present in a subshell (which were degenerate) take up different orientations.

Question 17.
How many electrons in Sulphur (Z = 16) can have n + l = 3?
Answer:
16S = 1s², 2s² 2p⁶ 3s² 3p⁴

• For 1s², n + l = l+ 0 = l
• For 1s², n + l = 2 + 0 = 2
• For 2p⁶, n + 1 = 2 + 1= 3
• For 3s², n + l = 3 + 0 = 3
• For 3p⁴, n + l = 3 + l = 4.

Thus n + l = 3 for 2p6 and 3s2 electrons, i.e. for 8 electrons.

Question 18.
Why Pauli Exclusion principle is called the exclusion principle?
Answer:
This is because according to the principle, if one electron in an atom has some particular values for the four quantum numbers, then all the other electrons in that atom are excluded from having the same set of values.

Question 19.
Why Hund’s rule is called the rule of maximum multiplicity?
Answer:
This is because out of the various possible electronic configurations, only that Configuration is correct for which the total spin value is maximum.

Question 20.
How many orbitals are present in the M-shell?
Answer:

Structure of Atom Chemistry Important Extra Questions Long Answer Type

Question 1.
Describe the shape of s – and p – orbitals What do you mean by node or nodal surface?
Answer:
Shapes of Orbitals:
s-orbitals: These are spherically symmetrical and non-directional. Shapes of 1s and 2s orbitals are shown in Fig. The effective volume of 2s orbital is larger than Is orbital. Another important feature of 2s orbital is that there is a spherical shell within 2s (region without dots) where the probability of finding the electron is zero. This is called a node or a nodal surface. There are (n – 1) nodes in an s-orbital (where n is the energy level).

p-orbitals: There are three p-orbitals designated as p_x’, p_y’ or p_z’ which are oriented along the three mutually perpendicular axis x, y, and 2. Each, orbital consists of two lobes symmetrical about the particular axis and has a dumbbell shape. The two lobes are separated by a nodal plane.

Shapes of three 2p orbitals

The two lobes of each orbital are separated by a plane having x zero electron density. This plane is known as a nodal plane.

Question 2.
How does the Schrodinger wave equation help to understand the probability of finding the electron near the nucleus? What do you mean by an orbital?
Answer:
Probability Picture of Electrons:
Schrodinger incorporated the requirements of the uncertainty principle and de Broglie’s concept of matter waves and proposed a mathematical equation to describe the behavior of an electron in an atom. The equation was known as the Schrodinger wave equation.

The Schrodinger wave equation is

where x, y, and z are three space coordinates,
m is the mass of the electron,
h is Planck’s, constant

E is the total energy and V is the potential energy of the electron, φ (Greek letter psi) is the amplitude of the wave, called wave function, \(\frac{\partial^{2} \psi}{\partial x^{2}}\) refers to the second derivative of φ with respect to x only and so on.

The solution of this equation gave the mathematical expression which gives information about the various energy states and other measurable properties such as the radiation frequencies emitted or absorbed for the hydrogen atom. The solutions of the Schrodinger wave equation are called wave functions and are denoted by the symbol φ.

The physical significance of wave function: In the physical sense φ gives the amplitude of the wave associated with the electron. We know that in the case of light waves, the square of the amplitude, of the wave at a point is proportional to the intensity of light. Extending the same concept of electron wave motion, the square of the wave function, φ² may be taken as the intensity of electrons at any point.

In other words, φ² determines the probability density. Thus, φ² has been called the probability density and φ the probability amplitude. Thus, the solutions of the Schrodinger wave equation replace the discrete energy levels or orbits proposed by Bohr and led to the concept of the most probable regions in space in terms of φ2².

A large value of φ² means a high probability of finding the electron at that place and a small value of φ² means low probability. If φ² is almost zero at a particular point, it means that the probability of finding the electron at that point is negligible. Therefore, the wave mechanics approach gives meaningful wave functions which describe the position and energy levels of electrons in an atom.

Concept of Orbital: An orbital is a region in space around the nucleus where the probability of finding the electrons is maximum.

Structure of Atom Chemistry Important Extra Questions Numerical Problems

Question 1.
How many nucleons are present in an atom Nobelium, No? How many electrons are present in the atom? How many nucleons may be considered neutrons?
Answer:
Nucleons = 254, electrons = 102 and neutrons 254 – 102 = 152.

Question 2.
Complete the following table:


Answer:

Question 3.
Complete the following table:

Answer:

Question 4.
Find the number of protons, electrons an-d neutrons in
(a) ₁₃²⁷A³⁺
Answer:
p = 13, ē = 10, n = 14

(b) ₈¹⁵O^2-
Answer:
p = 8, ē = 10, n = 7.

Question 5.
Name the element whose atomic nucleus does not contain any neutrons.
Answer:
The nucleus of the hydrogen atom does not contain any neutron.

Question 6.
Calculate the total no. of electrons present in one mole of methane.
Answer:
1 Molecule of methane (CH₄) contains electrons = 6 + 4 = 10
1 Mole, i.e., 6.022 × 10²³ molecules will contain electrons = 6.022 × 10²⁴

Question 7.
Find (a) the total number and
(b) the total mass of neutrons in 7 mg of 14C
(Assume that the mass of a neutron = 1.675 × 10^-27 kg).
Answer:
1 mol of 14C = 6,022 × 10²³ atoms
i. e. 14 g of 14C = 6.022 × 10²³ atoms

∴ 14 g of it = 6.022 × 10²³ × 0.008 neutrons [∵ one atom ofuC contains = 14 – 6 = 8 neutrons]
and now 0.007 g of it will have
= 6.022 × 10²³ × 8 × 0.007/13
= 2.409 × 10²¹ neutrons .

(b) Mass of 1 neutrons = 1.675 × 10^-27 kg
∴ Mass of 2.409 × 10²¹ neutrons = 4.035 × 10^-6 kg.

Question 8.
A particular radio station broadcasts at a frequency of 1120 kHz (kilohertz). Another radio station broadcasts at a frequency of 98.7 MHz (megahertz). What are the wavelengths of the radiations from each station?
Answer:

Question 9.
Calculate the wave number of radiations having a frequency of 4 × 10¹⁴ Hz.
Answer:
v = 4 × 10¹⁴ Hz = 4 × 10¹⁴ sec^-1
Wave lemgth v̅ = \(\frac{1}{\lambda}=\frac{v}{c}\)
= \(\frac{4 \times 10^{14} \sec ^{-1}}{3 \times 10^{8} \mathrm{msec}^{-1}}\)
= 1.33 × 10⁶ m^-1
= 1.33 × 10⁴ cm^-1.

Question 10.
A photon of wave length 4 × 10^-7 m strikes on metal surface, the work function of the metal being 2.13 eV.
Calculate
(i) the energy of the photon (eV)
Answer:
Energy of the photon = E = hv

(ii) the kinetic energy of the emission and
Answer:
Kinetic energy of emission = \(\frac{1}{2}\) mv² = hv – hv₀
= 3.10 – 2.13 = 0.97 eV

(iii) the velocity of the photoelectron [leV = 1.602 × 10^-19 J].
Answer:

Question 11.
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom. Calculate the Ionisation energy of solution in kJ mol^-1
Answer:

Question 12.
The wavelength of the first line in the Balmer Series is 656 nm. Calculate the wavelength of the second line and limiting line in the Balmer series.
Answer:
According to Rydberg’s formula

For the Balmer series n₁ = 2 and for the first line n₂ = 3

Divide (i) by (ii)

Question 13.
Calculate the wavelength of the spherical line in the Lyman series corresponding to n₂ = 3.
Answer:
For the Lyman series n₁ = 1
∴ v̅ = R\(\left[\frac{1}{1^{2}}-\frac{1}{3^{2}}\right]\)
= 109577 × \(\frac{8}{9}\) = 97490.7 m^-1
λ = \(\frac{1}{v}=\frac{1}{97490.7}\) cm^-1
= 102.6 × 10^-7cm
= 102.6 nm

Question 14.
Calculate the velocity of electrons in the first Bohr orbit of the hydrogen atom. Given that Bohr’s radius = 0,529 A. Planck’s constant h = 6.626 × 10^-34 Js mass of electron = 9.11 × 10^-31 kg and 1 J = 1 kg m² s^-1.
Answer:

Question 15.
The electron energy in a hydrogen atom is given by EH = (- 2.18 × 10^-18)/n² J. Calculate the energy required to, remove the electron completely from n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Answer:

Question 16.
Give the values of the quantum numbers for the electron with the highest energy in the sodium atom.
Answer:
Electronic configuration of riNa = 1s², 2s²2p⁶ 3s¹
The electron with the highest energy is 3s¹ for which n = 3 and l = 0

Question 17.
Find the number of unpaired electrons in Fe²⁺ and Fe³⁺. At.No. of Fe = 26.
Answer:
Electronic configuration of 26Fe = 1s², 2s²2p⁶, 3s² 3p⁶ 3d⁶ 4s²
Fe²⁺ = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁶ It has 4 unpaired electrons
Fe³⁺ = 1s², 1s² 2p⁶, 3s² 3p⁶ 3d⁵ It has 5 unpaired electrons.

Question 18.
What atoms are indicated by the following electronic cofigurations?
(i) 1s², 2s²2p¹,
Answer:
Total no. of electrons = 2 + 2 + 1 = 5
∴ At. no. of element is 5.
Hence the element is Boron (B).

(ii) [Ar] 4s² 3d¹
Answer:
Total no. of electrons in [Ar] 4s² 3d¹ = 18 + 2 + 1 = 21
∴ At. no. of element is 21.
Hence the element is Scandium (Sc).

Question 19.
Give the electronic configuration of N^-3, K⁺, P^-3, O^2-.
Answer:

• N = 7 = 1s²,2s² 2p³
• N^3- = 10e = 1s², 2s² 2p⁶
• K = 19 = 1s², 2s² 2p⁶, 3s² 3p⁶ 4s¹
• K+ = 18e = 1s², 2s² 2p⁶, 3s² 3p⁶
• P = 15 = 1s², 2s² 2p⁶, 3s² 3p³
• p^3- = 18e^– = 1s², 2s² 2p⁶, 3s² 3p⁶
• O = 8 = 1s², 2s² 2p⁴
• O^2- = 10e^– = 1s², 2s² 2p⁶

Question 20.
Which out of Cu²⁺, Fe²⁺, Cr³⁺ has the highest paramagnetism and why?
Answer:
Paramagnetism is a property of unpaired electrons. More the no. of unpaired electrons, more the paramagnetism.
Cu²⁺ = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁹
∴ It has only one unpaired electron.
Fe²⁺ = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁶

∴ It has 4 unpaired electrons.
Cr³⁺ = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d³

∴ It has 3 unpaired electrons.
Hence Fe²⁺ with 4 unpaired electrons is having the highest paramagnetism.

Question 24.
Find the velocity (in ms^-1) of the electron in the first Bohr orbit of radius a0. Also, find the de Broglie wavelength in m. Find the orbital angular momentum of 2p orbital of hydrogen in units of \(\frac{h}{2 \pi}\)
Answer:
For H-like particle, velocity in the nth orbit
V_n = 2.188 × 10⁶ × \(\frac{Z}{n}\) ms^-1

For H-atom, Z = 1; For 1st orbit n = 1
∴ v = 2.188 × 10⁶ ms^-1

∴ de Broglie wavelength λ = \(\frac{h}{mv}\)
= \(\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 2.188 \times 10^{6}}\) m
= 3.33 × 10^-10 m

∴ Orbital angular momentum = \(\sqrt{l(l+1)} \frac{h}{2 \pi}\)
For 2p orbital, l = 1
∴ Orbital angular momentum = \(\sqrt{l(l+1)} \frac{h}{2 \pi}\) = –\(\sqrt{2} \frac{h}{2 \pi}\) .

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 1 Some Basic Concepts of Chemistry. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 1 Important Extra Questions Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry Important Extra Questions Very Short Answer Type

Question 1.
What is the number of significant figures in 1050 × 10⁴?
Answer:
Four.

Question 2.
What is the SI unit of length?
Answer:
Metre.

Question 3.
What is the SI unit of density?
Answer:
kg m^-3.

Question 4.
What is the law called which deals with the ratios of volumes of the gaseous reactants and products?
Answer:
Gay Lussac’s law of combining volumes.

Question 5.
What is one a.m.u. or one u?
Answer:
1 a.m.u. or 1 u = \(\frac{1}{12}\)th of the mass of an atom C-12.

Question 6.
Which isotope of carbon is used for getting relative atomic masses?
Answer:
Carbon-12 (C-12).

Question 7.
Write down the empirical formula of benzene.
Answer:
CH.

Question 8.
The mass of human DNA molecules is 1 FG. What is its mass in kg?
Answer:
10^-18 kg

Question 9.
Vanadium metal is added to steel to impart strength. The density of vanadium is 5.96 g/cm³. Express this in the SI unit.
Answer:
5960 kg/m³.

Question 10.
What physical quantities are represented by the following units and what are their most common names?
(i) kg ms²
Answer:
Force (Newton)

(ii) kg m²s².
Answer:
Work (joule)

Question 11.
If in the reaction HgO(s) → Hg(l) + \(\frac{1}{2}\)O₂(g) 100.0 g of HgO on heating in a closed tube gives 92.6 g of Hg, what is the weight of oxygen formed?
Answer:
Weight of oxygen formed = 100 – 92.6 = 7.4 g.

Question 12.
Name three compounds formed by dinitrogen and dioxygen.
Answer:
1. N₂O (Nitrous Oxide)
2. NO (Nitric Oxide)
3. N₂O₅ (Nitrogen pentoxide).

Question 13.
If 700 mL of H₂ at STP contains x molecules of it, how many molecules of O₂ are present in 700 mL of it at the same temperature and pressure?
Answer:
x molecules.

Question 14.
nitrogen combines with dihydrogen according to the reaction.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
What is the ratio in their volumes under similar conditions of temperature and pressure?
Answer:
The ratio in their volumes is 1: 3: 2.

Question 15.
What is the relationship between molecular weight and vapor density of a gas?
Answer:
Molecular weight = 2 × Vapour density.,

Question 16.
What is the mass in gms of 11.2 L of N₂ at STP?
Answer:
22.4 L of N₂ at STP weighs = 28.0 g.
11.2 L of N₂ at STP weighs = \(\frac{28}{22.4}\) × 11.2 = 14.0 gm.

Question 17.
What is the mass of one molecule of sodium chloride?
Answer:
6.023 × 10²³ molecules of NaCl weighs = 58.5 g

I molecule of NaCI weighs = \(\frac{58.5}{6.023 \times 10^{23}}\) = 9.71 × 10^-23 g.

Question 18.
How many total electrons are present in 1.4 g of nitrogen gas?
Answer:
1.4 g of N₂ = \(\frac{14}{28}\) mol = 0.05 and 0.05 × 6.02 × 10²³ molecules
= 3.01 × 10²¹ molecules
= 3.01 × 10²¹ × 14 electrons [ ∵ 1 Molecules of 2 = 14 ē]
= 4.214 × 10²² electrons.

Question 19.
Why atomic masses are the average values?
Answer:
Most of the elements exist in different isotopes. Hence their average values are taken.

Question 20.
What is the percentage composition of Ca in CaCO₃?
Answer:
% of Ca = \(\frac{40 \times 100}{40+12+3 \times 16}\) = 40.0%

Question 21.
What is the mass of 2 moles of CO₂?
Answer:
1 mole of CO₂ = 1 × 12 + 2 × 16 = 44 g.
2 mole of CO₂ = 2 × 44.0 = 88.0 g.

Question 22.
Express the number 45000 in exponential notation to show
(i) two significant figures
Answer:
4.5 × 10⁴

(ii) four significant
Answer:
4.500 × 10⁴.

Question 23.
How many moles of sulphuric acid are present in 1 dm3 of 0.5 M solution?
Answer:
0.5 mole.

Question 24.
Give one example each of a molecule in which empirical formula and molecular formula are
(i) different
(ii) same.
Answer:

Question 25.
The radius of an atom is 10-10 m. What will be its value in micro metre?
Answer:
10^-10 m = \(\frac{10^{-10}}{10^{-6}}\) = 10^-4 micrometer.

Question 26.
Define the limiting reagent.
Answer:
The reacting substance gets used up first in the reactions.

Question 27.
What is meant by 1 gram atom of iron?
Answer:
1 g atom of iron means the atomic mass of iron expressed in g i.e., 56 g.

Question 28.
What is the number of oxygen atoms in one mole of CuSO₄.5H₂O?
Answer:
It is moles of oxygen atoms = 9 × 6.02 × 10²³ = 5.418 × 10²⁴ atoms.

Question 29.
What is the SI unit of velocity?
Answer:
ms-1.

Question 30.
A worker reads the resistance of the wire as 20.01 ohm. What may be the actual range of resistance?
Answer:
20.01 ± 0.01 ohm, i.e., 20.00 to 20.02 ohm.

Question 31.
What is AZT? To which use is it put?
Answer:
Azidothymidine used for AIDS victims.

Question 32.
Write down the empirical formula of acetic acid.
Answer:
CH₂O.

Question 33.
What is the advantage of using Molality over Molarity?
Answer:
Molality does not depend upon temperature.

Question 34.
Balance the equation
C₃Hg(g) + O₂(g) → CO₂(g) + H₂O(1)
Answer:
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l)

Question 35.
Balance the equation
Fe(s) + O₂(g) → Fe₂O₃(s)
Answer:
4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s)

Question 36.
Balance the equation
P₄(s) + O₂(g) → P₄O₁₀(s)
Answer:
P₄(s) + 5 O₂(g) → P₄O₁₀(s)

Question 37.
What is the SI scale of temperature?
Answer:
Kelvin (K).

Question 38.
What is the relation between Empirical formula and molecular formula?
Answer:
Molecular formula = n × Empirical formula
where n = \(\frac{\text { Molar mass }}{\text { empirical formula mass }}\)

Question 39.
What is the mass of NH₃ in 0.5 mol of it?
Answer:
0.5 mol of NH₃ weights 8.5 g.

Question 40.
What is the mass of 3.01 × 10²² g of SO₂?
Answer:
3.2 g.

Question 41.
How many atoms of silver are present in a 0.5 g atom of silver? At the mass of silver = 108.
Answer:
3.01 × 10²³ atoms.

Question 42.
Which of the two is heavier – 2 g of iron or 2 g atom of iron.
Answer:
2 g atom of iron.

Question 43.
How many molecules of CO₂ are present in 1.12 L of it at STP?
Answer:
3.011 × 10²² molecules.

Question 44.
What is the weight of H₂SO₄ in 0.01 M solution of it?
Answer:
0.98 g.

Question 45.
What volume of CO₂ at STP can be obtained on the decomposition of 100.0 g of CaCO₃?
Answer:
22.4 L.

Question 46.
How many molecules of cane sugar (C₁₂H₂₂O₁₁) are present in 34.20 grams of it.
Answer:
6.022 × 10²² molecules.

Question 47.
How many molecules of water are present in a drop of it having a mass of 0.05 g?
Answer:
1.673 × 10²¹ molecules.

Question 48.
Calculate the mass of 0.1 moles of KNO₃.
[At. wt. of K = 39, N = 14, O = 16]
Answer:
1 Mole of KNO-, = 1 × 39 + 1 × 14 + 3 × 16 = 101 g
∴ 0.1 mole of it = 101 × 0.1 = 10.1 g.

Question 49.
What is the molarity of a solution of oxalic acid containing 0.63 g of it in 250 cm³ of the solution?
Answer:
Molarity = 0.02 M.

Question 50.
A solution of NaCl has been prepared by dissolving 5.85 g of it 1 L of water. What is its molality?
Answer:
0.1 m.

Some Basic Concepts of Chemistry Important Extra Questions Short Answer Type

Question 1.
Define Mole. What is its numerical value?
Answer:
A mole is the amount of a substance that contains as many entities (atoms, molecules, or other particles) as there are atoms in exactly 0.012 kg or 12 g of the carbon-12 isotope.

Its numerical value is 6.023 × 10²³.

Question 2.
Define molarity. Is it affected by a change in temperature?
Answer:
The molarity of a solution is defined as the number of moles of the solute present per liter of the solution. It is represented by the symbol M. Its value changes with the change in temperature.

Question 3.
What do you mean by Precision and accuracy?
Answer:
Precision and accuracy: The term precision refers to the closeness of the set of values obtained from identical measurements of a quantity.

Accuracy refers to the closeness of a single measurement to its true value.

Question 4.
Distinguish between fundamental and the derived units.
Answer:
Fundamental units: Fundamental units are those units by which other physical units can be derived. These are mass (M), Length (L), time (T), temperature (°).

Derived units: The units which are obtained by the combination of the fundamental units are called derived units.

Question 5.
Define molality and write its temperature dependence.
Answer:
Molality is defined as the number of g moles of the solute dissolved per kilogram of the solvent.
Molality (m) = \(\frac{\text { Mole of solute }}{\text { Mass of the solvent in } \mathrm{kg}}\)

The molality of the solution does not depend upon the temperature.

Question 6.
Distinguish between an atom and a molecule.
Answer:
Atom: An atom is the smallest particle of an element that takes part in a chemical reaction. It may or may not be capable of independent existence.

Molecule: It is the smallest particle of a substance (element or compound) that is capable of independent existence

Question 7.
Derive the SI unit of Joule (J) in terms of fundamental units.
Answer:
Joule is the SI unit of work or energy

Question 8.
Define the SI unit of energy.
Answer:
Energy = Force × distance through which point of application of force moves.
S.l. unit of energy = SI unit of force × SI unit of distance.
= 1 Newton × 1 metre
= 1 Nm

But 1 N = 1 kg ms^-2
∴ SI unit of energy = 1 kg ms^-2 × 1 m = 1 kg m²s^-2

The unit of energy is joule (J)
Thus 1 Joule (J) = 1 kg m² s^-2.

Question 9.
One volume of a gaseous compound requires 2 volumes of O₂ for combustion and gives 2 volumes of CO₂ and 1 volume of N₂2. Determine the molecular formula of the compound.
Answer:
X + 2O₂ → 2 CO₂ + N₂ (According to problem)
Since the product has 2 atoms of C (in 2 CO₂) and 2 atoms of N (in N₂) these must have come from X. Hence X is C₂N₂.

Question 10.
What is a chemical equation? What are its essential features?
Answer:
The shorthand representation of a chemical change in terms of symbols and formulae is called a chemical equation.

Essential features:

• It should represent a true chemical change.
• It should be balanced.
• It should be molecular.

Question 11.
Write down the significance of a chemical equation.
Answer:
Significance of a chemical equation:

1. It tells about the reactants taking part in the chemical reaction and products formed as a result of chemical change.
2. It tells about the relative number of atoms or- molecules of reactants and products.
3. It tells about the relative weights of reactants and products

Question 12.
How is mole related to
(a) number of atoms/molecules
Answer:
A mole contains 6.022 × 10²³ atoms/molecules.

(b) mass of the substance?
Answer:
A mole of a substance denotes the molecular mass of that substance.

Question 13.
How is mole related to the volume of a gas?
Answer:
A mole of a gas occupies 22.4 L of the gas measured at STP, i. e., 0° C and 1 atmospheric pressure.

Question 14.
Why atomic masses are the average values?
Answer:
Most of the elements exist in different isotopic forms. Chlorine has 2 isotopes with mass numbers 35 and 37 existing in the ratio of 3: 1. Hence the average value is taken.

Question 15.
How many molecules approximately do you expect to be present in a small crystal of sugar which weighs 10 mg?
Answer:
10 mg sugar (C₁₂H₂₂O₁₁) = 0.01 g = \(\frac{0.01}{342}\) mol
= 2.92 × 10^-5 mole
= 2.92 × 10^-5 × 6.02 × 10²³ molecules
= 1.76 × 10¹⁹ molecules.

Question 16.
Two containers of equal capacity A₁ and A₂ contain 10 g of oxygen (O₂) and ozone (O₃) respectively. Which of the two will have greater no. of O-atoms and which will give greater no. of molecules?
Answer:
10 g of O₂ = \(\frac{10}{32}\) mol = \(\frac{10}{32}\) × 6.02 × 10²³ molecules
= 1.88 × 10²³ molecules
= 3.76 × 10²³ atoms.

10 g of O₂ = \(\frac{10}{48}\) mole = \(\frac{10}{48}\) × 6.02 × 10²³ molecules
= 1.254 × 10²³ molecules
= 3.76 × 10²³ atoms

Thus both A₁ and A₂ contain the same no. of atoms, but A₁ contains more numbers of molecules.

Question 17.
What do you mean by the term ‘Formality’? To what type of compounds it is applied?
Answer:
Ionic compounds like Na⁺Cl^– are not molecular. To express the concentration of their solution the term formality is used in place of molarity. Formality is the number of formula weights present in one liter of a solution.

Question 18.
What is meant by a ‘standard solution’?
Answer:
A standard solution is one whose molarity or normality is known. For example 0.1 M NaOH.

Question 19.
Assuming the density of water to be 1 g/cm³, calculate the volume occupied by one molecule of water.
Answer:
1 Mole of H₂O = 18 g = 18 cm³[∵ density of H₂O = 1 g/cm³]
= 6.022 × 10²³ molecules of H₂O
1 Molecule will have a volume
= \(\frac{18}{6.022 \times 10^{23}}\) cm- = 2.989 × 10^-23 cm³.

Question 20.
Define Avogadro’s number. What is its equal to?
Answer:
Avogadro’s number may be defined as the number or number of molecules present in one gram molecule of the substance.
Avogadro’s No. = 6.022 × 10²³.

Some Basic Concepts of Chemistry Important Extra Questions Long Answer Type

Question 1.
State the law of Multiple Proportions. Explain with two examples.
Answer:
The Law of Multiple Proportions states:
“When two elements combine to form two or more than two chemical compounds than the weights of one of elements which combine with a fixed weight of the other, bear a simple ratio to one another.

Examples:
1. Compound of Carbon and Oxygen: C and O combine to form two compounds CO and CO₂.
In CO₂ 12 parts of wt. of C combined with 16 parts by wt. O.
In CO₂ 12 parts of wt. of C combined with 32 parts by wt. of O.
If the weight of C is fixed at 12 parts by wt. then the ratio in the weights of oxygen which combine with the fixed wt. of C (= 12) is 16: 32 or 1: 2.
Thus the weight of oxygen bears a simple ratio of 1: 2 to each other.

2. Compounds of Sulphate (S) and Oxygen (O):
S forms two oxides with O, viz., SO₂ and SO₃
In SO₂, 32 parts of wt. of S combine with 32 parts by wt. of O.
In SO₃, 32 parts of wt. of S combine with 48 parts by wt. of O.
If the wt. of S is fixed at 32 parts, then’ the ratio in the weights of oxygen which combine with the fixed wt. of S is 32: 48 or 2: 3.

Thus the weights of oxygen bear a simple ratio of 2: 3 to each other.

Question 2.
State the law of Constant Composition. Illustrate with two examples.
Answer:
Law of Constant Composition of Definite Proportions states: “A chemical compound is always found, to be made up of the same elements combined together in the same fixed proportion by weight”.

Examples:

1. CO₂ may be prepared in the laboratory as follows:

In all the above examples, CO₂ is made up of the same elements i. e., Carbon (C) and Oxygen (O) combined together in the same fixed proportion by weight of 12: 32 or 3: 8 by weight.

Question 3.
Define empirical formula and molecular formula. How will you establish a relationship between the two? Give examples.
Answer:
The empirical formula of a compound expresses the simplest whole-number ratio of the atoms of the various elements present in one molecule of the compound.

For example, the empirical formula of benzene is CH and that of glucose is CH₂O. This suggests that in the molecule of benzene one atom of Carbon (C) is present for every atom of Hydrogen (H). Similarly in the molecule of glucose (CH₂O), for every one atom of C, there are two atoms of H and one atom of O present in its molecule. Thus, the empirical formula of a compound represents only the atomic ratio of various elements present in its molecule.

The molecular formula of a compound represents the true formula of its molecule. It expresses the actual number of atoms of various elements present in one molecule of a compound. For example, the molecular formula of benzene is C₆H₆ and that of glucose is C₆H₁₂O₆. This suggests that in one molecule of benzene, six atoms of C and 6 atoms of H are present. Similarly, one molecule of glucose (C₆H₁₂O₆) actually contains 6 atoms of C, 12 atoms of H, and 6 atoms of O.

Relation between the empirical and molecular formula
Molecular formula = n × Empirical formula where n is an integer such as 1, 2, 3…
When n = 1; Molecular formula = Empirical formula
When n = 2; Molecular formula = 2 × Empirical formula.
The value of n can be obtained from the relation.
n = \(\frac{\text { Molecular mass }}{\text { Empirical formula mass }}\)

The molecular mass of a volatile substance can be determined by Victor Meyer’s method or by employing the relation.
Molecular mass = 2 × vapour density .

Empirical formula mass can however be obtained from its empirical formula simply by adding the atomic masses of the various atoms present in it.

Thus the empirical formula mass of glucose CH20
= 1 × 12 + 2 × 1 + 1 × 16 = 30.0 u.

Some Basic Concepts of Chemistry Important Extra Questions Numerical Problems

Question 1.
In the commercial manufacture of nitric acid, how many moles of NO2 produce 7.33 mol HN03 in the reaction
3 NO₂(g) + H₂O(1) → 2HNO₃(aq) + NO(g).
Answer:
2 mols of HNO₃ are produced by 3 mols of NO₂
7.33 mol HNO₃ are produced by \(\frac{3 \times 7.33}{2}\) mol of NO₂
= 10.995 mols.

Question 2.
A sample of NaNO₃ weighing 0.83 g is placed in a 50,0 mL volumetric flask. The flask is then filled with water upon the etched mark. What is the molarity of the solution?
Answer:
Molar mass of NaNO₃ = 23 + 14 + 3 × 16 = 85 g mol^-1

Molarity = \(\frac{\text { Number of moles of solute }}{\text { Volume of solution in } L}\)
= \(\frac{0.83 \times 1000}{85 \times 50}\)
= 0.196 M.

Question 3.
Potassium bromide’ KBr contains 32.9% by mass of potassium. If 6.40 g of bromine reacts with 3.60 g of potassium, calculate the number of moles of potassium that combine with bromine to form KBr.
Answer:
Mass percentage of K = 32.9
∴ Mass percentage of Br = 100 – 32.9 = 67.1
Now 67.1 g of bromine combines with 32.9 g of potassium

6.40 g of bromine combines with = \(\frac{32.9}{67.1}\) × 6.40 = 3.14 g of K
Potassium (K) which remains unreacted = 3.60 – 3.14 = 0.46 g

Thus, Br₂ is the limiting reagent.
Mass of K = \(\frac{\text { Mass of } \mathrm{K}}{\text { Atomic mass of } \mathrm{K}}=\frac{3.14 \mathrm{~g}}{39 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.0802 mol.

Question 4.
Calculate the molarity of water in a sample of pure water.
Answer:
1 L of pure water = 1000 cm³
= 1000 g (assuming density = 1 g cm^-3)

No. of moles of H₂O in pure water = \(\frac{1000}{18.0}\) = 55.55 L^-1

Question 5.
How many molecules are there in 10.0 liters of a gas at a pressure of 75 cm at 27°C?
Answer:
PV = nRT

∴ Number of molecules present in 0.4 mol
= 6.023 × 10²³ × 0.4 = 2.409 × 10²³ molecules.

Question 6.
Two acids H₂SO₄ and H₃PO₄ are neutralized separately by the same amount of an alkali when sulfate and dihydrogen orthophosphate are formed respectively. Find the ratio of the masses of H₂SO₄ and H₃PO₄ (P = 31).
Answer:
1 g equivalent of alkali (NaOH) will neutralize 1 g Eqn. of H₂SO₄ and 1 g eq. of H₃PO₄. For the given neutralization reaction.
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
H₃PO₄ + NaOH = Na H₂PO₄ + H₂O

Eq. wt. of H₂SO₄ = \(\frac{\text { Mol. wt. }}{2}=\frac{98}{2}\) = 49;
Eq wt. of H₃PO₄ .
\(\frac{\text { Mol. wt. }}{1}\) = 98

Hence the ratio of masses of H₂SO₄ and HPO₄ = 49: 98 = 1: 2.

Question 7.
What weight of iodine is liberated from a solution, of potassium iodide when 1 liter of Cl₂ gas at 10° C and 750 mm pressure is passed through it?
Answer:
The chemical equation is
2KI + Cl₂ → 2KCl + I₂
22.4 L   2 × 127
at STP = 254 g

(V₁) volume of Cl₂ gas=1 L
V₂ = vol. of gas at
STP =?

(P₁) Pressure = \(\frac{750}{760}\)atm
P₂ = 1 atm

(T₁) Temperature = 10°C + 273 = 283 K
T₂ = 273 K

∴ V₂ = \(\frac{P_{1} V_{1}}{T_{1}^{-}} \times \frac{T_{2}}{P_{2}}=\frac{750 \times 1 \times 273}{760 \times 283 \times 1}\)
= 0.952 L

∴ Vol of Cl₂ g passed at STP = 0.952 L
Now 22.4 L of Cl₂ produces at STP = 254 g of I₂

0.952 L of Cl₂ at STP produces = \(\frac{254}{22.4}\) × 0.952 g of I₂
= 10.78 g of I₂.

Question 8.
A crystalline salt on being rendered anhydrous loses 45.6% of its weight. The percentage composition of the anhydrous salt is Aluminium = 10.50%, Potassium = 15.1% Sulphur = 24.96%, Oxygen = 49.92%. Find the simplest formula of the anhydrous and crystalline salt.
Answer:
Step 1: To calculate the empirical formula of the anhydrous salt


The empirical formula of the anhydrous salt is KAlS2O8

Step 2: Empirical formula mass of anhydrosis salt
= 39 + 27 + 2 × 32 + 8 × 16 = 258 u

Step 3: To calculate the empirical formula mass of the hydrated salt.
Loss of water due to dehydration = 45.6%

∴ The empirical formula mass of anhydrous salt (assuming empirical formula mass of the hydrated sample to be 100) = 100 – 45.6 = 54.4 u

When empirical formula mass.of anhydrous salt is 258 u, that of 100 hydrated is = \(\frac{100}{54.4}\) × 258 = 473.3 u

Step 4: To calculate the no. of water molecules
Total loss in wt. due to dehydration = 473.3 – 258 = 215.3
No. of water molecules = \(\frac{215.3}{18}\) = 11.96 ≈ 12

Step 5: Empirical formula of the hydrated salt = KAlS2O8.12H2O

Question 9.
An organic compound containing C, H, and O gave the following percentage composition: C = 40.687%.
H = 5.085% O = 54.228%
The vapor density of the compound is 59. Calculate the molecular formula of the compound.
Answer:
Calculation of the empirical formula of the compound

∴ The empirical formula of the compound is C2H3O2
Molecular massof compound = 2 × V.D. = 2 × 59 = 118
n = \(\frac{\text { Molecular Mass }}{\text { Empirical formula mass }}\)
= \(\frac{118}{2 \times 12+3 \times 1+2 \times 16}=\frac{118}{59}\) = 2

∴ Molecular formula of the compound
= n × empirical formula = 2 × C₂H₃O₂
= C₄H₆O₄.

Question 10.
Calculate the number of molecules present in 350 cm³ of NH₃ gas at 273 K and 2-atmosphere pressure.
Answer:
Calculate the volume of the gas at STP.
V₁ = 350 cm³
V₂ = ?
T₁ = 273 K
T₂ = 273 K
P₁ = 2 atm.
P₂ = 1 atm

V₂ = \(\frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}}=\frac{2 \times 350 \times 273}{273 \times 1}\) = 700 cm³

22,400 cm³ of NH₃ at STP contains
= 6.023 × 10²³ molecules of NH₃

700 cm3 of NH₃ at STP will contain
= 1.882 × 10²² molecules.

Question 11.
Calculate the percentage of water of crystallization in the sample of blue vitriol (CuS04.5H20).
Answer:
Mol. wt. of CuSO₄.5H₂O) = 63.5 + 32 + 4 × 16 + 5 × 18 = 249.5
No. of parts bv wt. of H₂O = 5 × 18 = 90
∴ % of H₂O = \(\frac{90}{249.5}\) × 100 = 36.07

Question 12.
Calculate the weight of iron which will be converted into its oxide (Fe₃O₄) by the action of 18 g of steam on it.
Answer:
The chemical equation representing the reaction is

Now 72 g of steam [H₂O(g)] react with 168 g of Iron

∴ 18 g of steam will’reacts with = \(\frac{168}{72}\) × 18 = 42.0 g of iron
Thus, the weight of iron required = 42.0 g

Question 13.
To account for the atomic mass of nitrogen as 14.0067, what should be the ratio of 15N and 14N atoms in natural nitrogen?
[At. mass of 14N = 14.00307 u and 15N = 15.001 u]
Answer:
Let the % of 14N = x
% of 15N = 100 – x
According to the definition of average atomic mass

Question 14.
2.16 g of copper metal, when treated with nitric acid followed by ignition of the nitrate, gave 2.70 g of copper oxide. In another experiment, 1.15 g of copper oxide upon reduction with hydrogen gave 0.92 g of copper. Show that the above data illustrates the law of definite proportions.
Answer:
% of Cu in copper oxide in 1st case = \(\frac{2.16 \times 100}{2.70}\) = 80
% of oxygen = 20%
% of Cu in copper oxide in 2nd case = \(\frac{0.92 \times 100}{1.15}\) = 80
% of oxygen = 20%

Thus, the percentage of copper in copper oxide from both experiments is the same. Hence the above data illustrates the law of definite proportions.

Question 15.
The mass of precious stones is expressed in terms of ‘carat’. Given that 1 carat = 3.168 grains and 1 gram = 15.4 grains, calculate the total mass of a ring in grams and kilograms which contains 0.500 carat diamond and 7.00 gram gold.
Answer:
The unit conversion factors to be used will be

Question 16.
“The star of India” sapphire weighs 563 carats If one carat is equal to 200 mg, what is the weight of the gemstone in grams?
Answer:
The weight of the gemstone in grams will be
= 563 carats × \(\frac{200 \mathrm{mg}}{1 \text { carat }}=\frac{1 \mathrm{~g}}{100 \mathrm{mg}}\)
= 112.6 g

Question 17.
When 4.2 g of NaHCO₃ is added to a solution of acetic acid (CH₃COOH) weighing 10.0 g, it is observed that 2.2 g of CO₂ is released into the atmosphere. The residue left behind is found to weigh 12.0 g. Show that these observations are in agreement with the law of conservation of mass.
Answer:

Total mass of the reactants = 10.0 + 4.2 = 14.2 g
Total mass of the products = 12.0 + 2.2 = 14.2 g

Thus, during the above chemical reactions, the total mass of the reactants is equal to the total mass of the products, i.e., the matter is neither gained nor lost.

This is in keeping with the law of conservation of mass.

Question 18.
Calculate the volume of 0.05 M KMnO₄ solution required to oxidize completely 2.70 g of oxalic acid in acidic solution.
Answer:
The equation representing the chemical change is
2KMnO₄ + 3 H₂SO₄ → K₂SO₄ + 2 MnSO₄ + 3 H₂O = 5[0]
H₂C₂O₄ + [O] → 2 CO₂ + H₂O × 5

Step I. To calculate the no. of moles of KMn04 required to completely oxidize 2.70 g of’H₂C₂O₄ in acidic medium:
Molar mass of H₂C₂O₄ = 2 × 1 +2 × 12 + 4 × 16
= 90.0 g mol^-1

∴ No. of moles of H₂C₂O₄ contained in 2.70 g of it = \(\frac{2.70}{90.0}\) = 0.03

5 Moles of H₂C₂O₄ are oxidized by 2 moles of KMnO₄
∴ 0.03 moles of it oxidized by = \(\frac{2}{5}\) × 0.03
= 0.012 mole of KMnO₄

Step II. To calculate the volume of 0.05 M KMnO₄ solution
Now 0.05 mole of KMnO₄ are contained in 1000 cm³ of solution
∴ 0.012 mole of KMnO₄ will be contained in
\(\frac{1000}{0.05}\) × 0.012 = 240 cm³ of solution

Thus, the required volume of 0.05 M KMnO₄ solution = 240 cm³.

Question 19.
4 g carbon was heated with 8 g of sulfur. How much carbon disulfide (CS₂) will be formed when the reaction is complete? What will be its percentage purity?
Answer:

Obviously sulphur will be limiting reactant.

8 g of sulphur will produce CS₂ = \(\frac{76}{64}\) × 8 = 9.5 g

Amount of carbon reacted = \(\frac{12}{64}\) × 8 = 1.5g

Amount of carbon left = 4 – 1.5 = 2.5 g
Total weight of the products = 9.5 + 2.5 = 12 g

% purity of CS₂ in the product = \(\frac{9.5}{12}\) × 100 = 79.2.

Question 20.
A gas mixture of 3.0 liters of propane and butane on complete combustion at 25°C produced 10 liters of CO₂. Find out the composition of the gas mixture.
Answer:
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
C₄H₁₀ + 6 \(\frac{1}{2}\)O₂ → 4 CO₂ + 5 H₂O

Suppose volume of propane (C₃H₈) = xL
∴ Volume of butane (C₄H₁₀) = (3 – x)L
1 L of C₃H₈ gives 3 L of CO₂ and 1 L of C₄H₁₀ gives 4 L of CO₂
∴ CO₂ produced = 3x + 4(3 – x) = 12 – x (Given)
12 – x = 10
Hence x = 2 L
∴ Volume of propane = 2 L
and volume of butane = 3 – 2 = 1 L

Here we are providing Class 11 Biology Important Extra Questions and Answers Chapter 21 Neural Control and Coordination. Important Questions for Class 11 Biology are the best resource for students which helps in Class 11 board exams.

Class 11 Biology Chapter 21 Important Extra Questions Neural Control and Coordination

Neural Control and Coordination Important Extra Questions Very Short Answer Type

Question 1.
What are the major divisions of the forebrain?
Answer:
Cerebrum, Thalamus, Hypothalamus.

Question 2.
Which parts of the central nervous system constitute the grey matter?
Answer:
Areas that contain cell bodies of the neurons.

Question 3.
Name the major lobes of the cerebral hemisphere?
Answer:
Frontal, parietal, temporal and occipital.

Question 4.
What is the function of cerebral spinal fluid?
Answer:
It maintains a constant pressure inside the cranium.

Question 5.
What is the junction between two neurons known as?
Answer:
Synapse.

Question 6.
What is the polarized state of the nerve membrane?
Answer:
It is the state of the nerve membrane when its inner side is electronegative to its outer side.

Question 7.
Give two examples of unconditioned reflexes.
Answer:
(i) Salivation on tasting food.
(ii) Constriction of the pupil on the illumination of the eye.

Question 8.
Name the types of cells present in the retina.
Answer:
Rods, cones, bipolar neurons, ganglion cells, supporting cells.

Question 9.
Where is iodopsin present in the eye?
Answer:
In the cone cells of the retina

Question 10.
Where are taste buds located?
Answer:
In the mucous membrane over the papillae on the tongue.

Question 11.
Name the main parts of the human brain.
Answer:
Cerebrum, cerebellum, medulla, thalamus, and hypothalamus.

Question 12.
How many cranial nerves and spinal nerves do we possess?
Answer:
12 pairs of cranial nerves and 31 pairs of spinal nerves.

Question 13.
Which part of the brain controls posture and equilibrium?
Answer:
Cerebellum.

Question 14.
What is a polarised membrane?
Answer:
It is electrically positive outside and negative inside.

Question 15.
Compare rods and cones.
Answer:
Rods work in dim light and dark. Cones work in bright light.

Question 16.
What is the blind spot?
Answer:
The point of the retina from where the optic nerve starts and receptor cells are absent.

Neural Control and Coordination Important Extra Questions Short Answer Type

Question 1.
What are receptors?
Answer:
Receptors are peripheral nerve endings in the skin or special sense organs. They collect information from the external or internal environment of the body; transform them into electrical potential changes, which then pass along the afferent neurons to CNS.

Question 2.
Why does vitamin A deficiency produce night blindness?
Answer:
Vitamin ‘A’ is the constituent of rhodopsin, a pigment present in the photoreceptor cells of the eye. Rhodopsin breaks up into opsin and rode to visualize things in bright and dim light. There is constant consumption of vitamin A in rod cells. Deficiency of vitamin A causes impairment of synthesis of rhodopsin leading to night blindness, i.e., inability to see in the dark.

Question 3.
Why does the nerve impulse flow more rapidly in myelinated nerve fibers than in the non-myelinated fibers?
Answer:
Due to the following reasons nerve impulse flows more rapidly in myelinated nerve fibers:

1. Myelin sheath provides insulation to the nerve fibers from electrical disturbances between the neighboring fibers.
2. Myelin sheath is impermeable to free ions present in the extracellular fluid. So, it prevents the exchange of ions between the extracellular fluid and the interior of the myelinated axon.
3. The myelin sheath is absent at the Nodes of Ranvier, so, action potential jumps from one Node of Ranvier to the next. Thus, the nerve impulse flows in the form of leaps or jumps. This is known as the saltatory conduction of impulse.
4. It is more rapid than the smooth flow of impulse.

Question 4.
What is a synapse?
Answer:
It is the junction between axon terminals of a neuron and dendrites or the cell body of another neuron. There is a narrow fluid-filled space, called Synaptic Cleft separating axon terminals and dendrites at the synaptic junction. So, the two-neurons forming synapse does not form actual continuity between the neurons.

Structure of Synapse

Question 5.
Draw a labeled diagram of a section of the retina to illustrate its structure.
Answer:

Diagrammatic representation of the sectional view of the retina.

Question 6.
What functions does the hypothalamus serve in coordinating the various activities of the body?
Answer:

1. It contains nerve centers for temperature regulation, hunger, thirst, and emotional reactions.
2. It secretes neurohormones, which control the secretion of anterior pituitary hormones.
3. It synthesizes the posterior pituitary hormones and controls their release into the blood.

Question 7.
What is a nerve fiber? How is it classified according to myelin sheath?
Answer:
A nerve fiber is a long axon or dendrite of a neuron. According to the presence and absence of myelin sheath around the fibers.

These are classified as:

1. Myelinated nerve fiber (i.e., presence of myelin sheath) and
2. Non-myelinated nerve fiber (i.e., absence of myelin sheath).

Question 8.
Explain Motor-end plate.
Answer:
A Motor-end plate is a specialized structure formed by the muscle fiber at the point where the axon terminal is applied to it. The axon of the motor neuron is divided into branches near the muscle fibers. Each branch loses its myelin sheath near its termination and ends in an expanded foot-like form which is supplied closely to a muscle fiber.

There is no actual continuity between the neuron and muscle fiber. The membranes of the two are separated from each other by a narrow cleft-like fluid-filled space.

Question 9.
What are the biological functions of Dorsal and Ventral spinal nerve roots fibers?
Answer:
Dorsal spinal nerve root fibers bring impulses from the peripheral tissues, giving rise to sensations like touch, temperature, and pain, or to involuntary spontaneous activities called Reflexes.

Ventral spinal nerve root fibers: Some of the root fibers go to skeletal muscle fibers directly to stimulate or inhibit their activities; many others go to autonomic ganglia and end in them.

Question 10.
Our rods and cones evenly distributed over the entire surface of the retina? Why or not? At which point on the retina is a point-to-point image formed?
Answer:
The retina is composed of several layers of cells. First, there are the photoreceptor cells, the rods, and cones, partially embedded in the microvilli of pigmented epithelium cells of the choroids. The rod cells are present on the periphery of the retina in the human eye. The total number of rod cells has been estimated to be between 110 – 125 million. They contain a visual pigment called Rhodopsin.

The cone cells are shorter, thicker, and conical in shape. Cone cells are responsible for the perception of different colors. The total number of cone cells is 6.36 – 6.8 million. Cones are abundant on the rear wall and fovea centralis of the retina.

The point-to-point image is formed on the blind spot. From it, the optic nerve and blood vessels exit the retina and join the diencephalon of the brain.

Question 11.
Blindspot in the eye is devoid of the ability of vision. Why is it so?
Answer:
It is devoid of rods and cone cells. It is unstable to light rays.

Question 12.
If a strong odor is smelled continuously for some time, the sensation of that weakens. Justify.
Answer:
When a person continuously inhales the fumes in the air of a strong-smelling substance the sense of smell progressively and rapidly declines and finally disappears. This is because the olfactory cells get fatigued rapidly due to overstimulation. This is called olfactory adaptation, which develops from various changes in the olfactory epithelium and olfactory centers of the brain.

Question 13.
Which part of the nervous system participates in the maintenance of balance and co-ordinate body movements?
Answer:
The cerebellum process all the data and co-ordinates muscle movement in conjunction with the cortex and sends signals to the muscles to adjust.

Question 14.
What is a reflex action? What units of the nervous systems are involved with a typical vertebrate reflex arc?
Answer:
It is a spontaneous, automatic, mechanical, nerve-mediated response evoked at the unconscious level by the stimulation of any specific receptor without exercising the will of an organism.

There, are more than 200 reflexes “wired” into our nervous system all following the sequence from stimulus to reflex along the specific neural pathway that makes up the reflex arc. The simplest reflex arc involves some specific receptor, afferent sensory neuron towards an aggregation of nervous tissue which may be ganglion or the spinal cord.

Question 15.
Which nerve tract connects the right and left hemispheres of the cerebrum? Into what four lobes in each hemisphere divided?
Answer:
A longitudinal fissure splits the brain into two halves, the left, and right cerebral hemispheres. Other grooves divide the surface of each cerebral hemisphere into four lobes. The frontal lobe, temporal lobe at the front, parietal lobe, and occipital lobe at the back.

Question 16.
What is the primary function of neuroglia cells? What special structure is produced by Schwann cells?
Answer:
The neuroglia cells perform many house-keeping functions, provide nutritional support to the neurons and consume waste products. They also insulate, separating each neuron from the others.

Schwann cells, a type of neuroglial, wrap around the axon with concentric layers of the insulating plasma membrane.

Question 17.
How does a wave of depolarization spread along with a nerve fiber?
Answer:
Nerve cells have polarized membranes, having an electrical potential difference across the membrane. The trigger zone for a particular neuron is the place on the membrane where voltage-gated channels are clustered most densely. When stimulated opening of voltage-gated Na⁺ ion channel brings Na⁺ ions into the cell, a temporary, very localized, but rapid inflow of Na⁺ ions into the cells occurs, wiping out the local electrical potential difference in the immediate vicinity. This is called depolarization.

When the site of stimulation has less charge difference than the membrane surface surrounding it, this potential difference establishes a small, localized current in the immediate vicinity, which influences the nearby Na⁺ channels to open and depolarizing these cells.

The depolarization thus spreads, producing a local current, which induces passive Na⁺ channels to open and so to depolarize the near % site. In this way, initial depolarization passes outward over the membrane and spreads out in all directions along with the nerve fiber, from dendron to axon.

Question 18.
What is a synapse? How does the nerve cell across the synapse?
Answer:
A nerve signal travels from neuron to neuron all around the body. These associations are called Synapse.

There are mainly two types of synapses:

1. Electrical and
2. Chemical depending upon the nature of the transfer of information across the synapse.

In electrical synapses, cells are separated by a gap, the synaptic cleft, of only 0.2 mm. So that an action potential arriving at the presynaptic side of the cleft can sufficiently depolarize the postsynaptic membrane to directly trigger its action potential.

Chemical synapses are the common type of synapse consists of a bulbous expansion of a nerve terminal called a synaptic knob. The cytoplasm of the synaptic knob contains numerous tiny round sacs synaptic vesicles. Each vesicle contains a neurotransmitter substance responsible for the transmission of nerve impulses across the synapse.

Question 19.
What is the action potential of a neuron? Do all neurons possess the same action potential?
Answer:
Depolarisation is caused by rapid change in membrane permeability and a corresponding shift in the balance of ions. If the shift of ions and consequent shift in electrical charges is sufficient, it will trigger a wave of transient membrane depolarisation known as nerve impulse or Action potential. Different neurons possess different densities of Na⁺ ion channels, different neurons exhibit different action potentials. However, for anyone neuron, the action potential is always the same.

Question 20.
Why is the mode of conduction of electrical impulse along the myelinated neuron is advantageous to a non-myelinated neuron? What is this type of conduction called?
Answer:
The myelinated nerve fibers carry impulses nearly 20 times faster than the non-myelinated nerve fibers. These avoid the dissipation of impulses into adjacent fibers. The myelin sheath serves as a highly insulating layer that prevents the flow of ions between the fluid external to the myelin sheath and within the axon.

In non-myelinated fiber, ionic charges and depolarization are repeated over the membrane along the length of the fiber and action potential flow over the entire length of the fiber. But in myelinated fibers, ionic changes and depolarization are repeated only at the nodes. Thus the impulse is more rapid in myelinated fibers and requires less energy. This jumping of depolarization from node to node is called saltatory conduction of nerve impulse.

Saltatory conduction

Question 21.
(a) Make a clearly labeled diagram of the inner ear of a human being.
Answer:

Diagrams showing the inner ear

(b) Describe how each of the following is achieved in us
(i) hearing
(ii) balance.
Answer:

Neural Control and Coordination Important Extra Questions Long Answer Type

Question 1.
(a) Describe the reflex arc with a diagram.
Answer:
The neurons forming the pathway taken by the nerve impulses in the reflex action form the Reflex Arc.
The reflex arc consists of

1. Receptor,
2. An afferent neuron or sensory neurons from receptor to CN system,
3. The efferent neuron of motor neurons from CN system to specific muscle fibers or gland cells,
4. a number of connectors or intermediate neurons conducting impulses form the afferent to the efferent neurons.

Reflex Arc

When a specific stimulus is applied to a specific group of receptors, it stimulates the receptor to initiate a nerve impulse along the afferent neurons. This impulse travels along with the afferent connector and efferent neurons to reach an effector-muscle or gland for that reflex. Thus the flow of impulse can only be in a single direction in a reflex arc, i.e.,

Stimulus → receptor → afferent neuron → CN system efferent neuron ← (connector neuron)

(b) Distinguish between conditioned reflex and unconditioned reflex.
Answer:
Differences between conditioned reflex and unconditioned reflex:

Question 2.
(a) Give an account of spinal nerves in man.
Answer:
There are 31 pairs of a spinal nerve in man. From each segment of the spinal cord, there are two spinal nerves. Each spinal nerve is a mixed nerve, containing both sensory and motor nerve fibers. It runs between the spinal cord and peripheral tissue. The two roots, i.e., motor or ventral and sensory or dorsal connect the spinal nerve to the spinal cord.

The DORSAL ROOT carries sensory or afferent fiber and has a dorsal root ganglion at its middle. The VENTRAL ROOT contains a motor or efferent nerve fibers. The dorsal root fibers bring impulses from the peripheral tissue and give rise to sensations like touch, temperature, and pain.

The ventral nerve root fibers pass impulses to muscles and glands in the peripheral tissues. The spinal nerve has been named according to its relation with the vertebral column.
These are

1. Eight pairs of cervical
2. 12 pairs of thoracic
3. 5 pairs of lumber,
4. 5 pairs of sacral and
5. a pair of coccygeal or caudal.

(b) What biological functions are served by the skeletal system?
Answer:

1. The skeletal system forms the rigid structural framework of the body and supports the weight of the body along with its limbs.
2. It affords protection to the internal organs against mechanical injury by forming cage-like compartments, e.g., skull.
3. It serves as a storage depot for calcium and phosphate, which are released for a number of functions of the body.
4. It participates in movement and locomotion.

The spinal nerve in man

Question 3.
Distinguish between:
(a) Afferent neurons and efferent neurons.
Answer:
Afferent neurons and efferent neurons:
Afferent neurons: These conduct sensory impulses from the receptors present in the peripheral organs and tissues towards the central nervous system. Their bodies are called afferent neurons.

Efferent neurons: These conduct motor impulses from the central nervous system to the peripheral organs and tissues serving as effectors. Their cell bodies are called efferent neurons.

(b) Rods and cones
Answer:
Rods: Rod cells are rod-like, elongated cells, bearing long, thin cylinders, containing a visual pigment called Rhodopsin. Rod cells are present on the periphery of the retina in the human eye. These cells do not form color vision.

Cones: Cone cells are shorter, thicker, and conical in shape. These are highly sensitive to bright light and colors. They contain a violet color pigment called rhodopsin. Cone cells are responsible for the perception of different colors. Cones are abundant on the rear wall and fovea centralis of the retina.

(c) Resting membrane potential and action potential.
Answer:
Resting membrane potential: The surface of the axon carries a positive charge relative to its interior and this electrical potential difference across the plasma membrane is called resting membrane potential.

Action potential: The shift of ions and consequents shift in electrical charges is sufficient enough; it will trigger a wave of transient membrane depolarization known as nerve impulse or Action potential.

(d) Impulse conduction in myelinated nerve fiber and unmyelinated nerve fiber.
Answer:
Impulse conduction in myelinated nerve fiber: The myelinated fibers carry impulses nearly 20 times faster than the non-myelinated nerve fibers. These avoid dissipation of impulse into adjusting fibers. The myelin sheath serves as a highly insulating layer that prevents the flow of ions. Impulses are rapid.

Non-myelinated nerve fiber: Ionic changes and depolarization are repeated over the membrane all along with the fiber. Impulse requires less energy and does not need to run all along with the fiber.

(e) Aqueous humor and vitreous humor.
Answer:
Aqueous humor: The chamber between the cornea and lens is filled with a clear watery fluid, the aqueous humor.

Vitreous humor; The chamber behind the lens is filled with a semisolid gelatinous material the vitreous humor.

(f) Blindspot and yellow spot.
Answer:
Blindspot: It s a small insensitive light area of about 0.5 cm. in diameter. It is devoid of rod and cone cells. It is unable to receive light rays.
Yellow spot: A tiny circular area, about 6 mm in diameter in the retina is a yellow spot. Here the vision is sharpest. It has rod and cone cells.

(g) Cranial nerve and spinal nerves.
Answer:
Cranial nerve: There are 12 pairs of cranial nerves, 10 originate from the brain stem, but all pass through the foramina of the skull. Cranial nerves contain only sensory fibers. The remainder contains both sensory and motor fibers.

Spinal Nerve: They arise from the cord. 31 pairs of segmental spinal nerves arise from the cord. They contain both receptor neurons and effectors neurons.