Chemical Bonding NIOS Intext Questions With Answer

Chemical Bonding NIOS Intext Questions With Answer National Institute of Open Schooling | NIOS

Chemical Bonding NIOS Intext Questions With Answer

National Institute of Open Schooling (NIOS)
Sr. Secondary, Module-2
Lession-4: Chemical Bonding NIOS Intext Questions With Answer


1. Define electrovalent bond.

Answer: An electrovalent bond is formed when one or more electrons are transferred from one atom to another atom or atoms.

2. Show the formation of a nitrogen molecule from two nitrogen atoms in terms of Lewis theory.

Answer: :N⋮   ⋮N: → NN
According to Lewis's theory nitrogen is having a tripled bonded structure which is formed by the sharing of three electrons by the two nitrogen atoms to fulfil its octet.

3. What do you understand by a polar covalent bond? Give two examples.

Answer: In a covalent bond the shared pair of electrons is closer to the more electronegative atom. This leads to charge separation in the molecule and the bond becomes polar. HCl and H2O are the two examples of covalent bond.

4. What is a coordinate covalent bond ? How is it different from a covalent bond?

Answer: A coordinate covalent or dative bond is a type of bond in which the electrons pair are shared from the single atom. The difference between the coordinate covalent bond and a covalent bond is that a covalent bond is formed by the mutual sharing of electrons of similar or non-similar atoms while the coordinate covalent bond is formed by the sharing of pairs of electrons.


1. What are the basic postulates of VSEPR theory?

Answer: Valence Shell Electron Pair Repulsion theory was developed by Sidgwick and Powell which helps in determining the shapes of molecules. It is based on the fact that the shape of the molecule will be such that the repulsion will be minimum. The basic postulates of the valence shell electron pair repulsion theory is as follows-
i. The electron pairs (both bonding and non-bonding) around the central atom in a molecule arrange themselves in space in such a way that they minimize their mutual repulsion.
ii. The repulsion of a lone pair of electrons for another lone pair is greater than that between a bond pair and a lone pair which in turn is stronger than that between two bond pairs The order of repulsive force between different possibilities is as -
lone pair - lone pair > lone pair - bond pair > bond pair - bond pair

2. Predict the shape of methane ( CH4 ) on the basis of VSEPR theory.

Answer: In methane the central carbon atom would have four pairs of electrons in its valence shell. According to VSEPR theory these would be placed tetrahedrally around the carbon atom. Hence the methane molecule would have a tetrahedral shape.

3. It is a molecule the difference between the electro-negativity of two atom is 1.7. How much % will be ionic and covalent character?

Answer: The percentage ionic character is calculated by the following formula
% of ionic character = 16 (𝛘A − 𝛘B) + 3.5 (𝛘A − 𝛘B)2
Putting the value in the above equation we get-
% of ionic character = (16 x 1.7) + 3.5(1.7)2 = 37.315
Hence the percentage ionic character when electronegativity difference is 1.7 is 37.315%
The per centage of covalent character is 100 − per centage ionic character.
So the per centage covalent character will be-
100 − 37.315 = 62.685%
Therefore, per centage of ionic character = 37.315 and per centage of covalent character = 62.685.


1. What do you understand by the term, ‘hybridisation’?

Answer: Hybridisation is a concept which is quite useful in explaining the shapes of molecules. According to this two or more than two non equivalent orbitals with comparable energies and different shapes mix and give rise to an equal number of equivalent hybrid orbitals in an atom. The hybrid orbitals have identical energies and shapes.

2. How would you explain the shape of ammonia molecule on the basis of hybridisation?

Answer: In ammonia the 2s and three 2p orbitals hybridize to give four sp3 hybridized orbitals. Three of these overlap with the 1s orbitals of hydrogen and one remains nonbonding containing a lone pair. The sp3 hybridized orbitals are directed towards the corners of a regular tetrahedron. But due to the difference in the repulsion between lone pair - bond pair and bond pair - bond pair the ammonia molecule has a distorted tetrahedral shape which is some what like a trigonal pyramid.

3. Draw the canonical structures of CO3 −2 and SO2.

Cannonical structure of carbonate ion and sulphur dioxide ion


1. What is the basic difference between the valence bond and molecular orbital theories?

Answer: Valence bond theory visualises the bond formation to be localized whereas according to MOT it is delocalised.

2. Calculate the bond orders for Li2 and Be2 molecules using the molecular orbital diagrams given in Fig. 4.12.

Answer: Bond order = ½(nb – na) for Li2, Bond order = ½ [ 4 – 2] = ½ [2] = 1
for Be2,Bond order = ½ [ 4 – 4] = ½ [0] = 0

3. Predict the magnetic behaviour of O2.

Answer: MO configuration of O2 is σ2s2, σ*2s2, σ2pz2, π2px2 = π 2py2 π* 2px1 = π* 2py1
Due to two unpaired electrons O2, molecule is paramagnetic.

NIOS Sr. Secondary Chemistry Intext Questions of All Chapters with Answer