Solutions-Part-1-Class-11-Science-Chemistry-Chapter-5-Chemical Bonding-Maharashtra Board

(b) CO₂

(a) covalent > hydrogen > vander waals’

(b) the shapes of molecules and ions.

(A ) C=O bond (Polar), CO₂ molecule (Non-polar)

(c) MOT

(d) H₂O

• Electron distribution in H₂O molecule

• Shape of H₂O molecule : V Shape
• H—O—H bond angle : 104⁰35’

Octet rule states that during the formation of chemical bond, atom loses, gains or shares electrons so that its outermost orbit (valence shell) contains eight electrons. Therefore the atom attains the nearest inert gas electronic configuration.

(a) Incomplete Octet: Some atoms can form stable compounds even when they possess fewer than eight valence electrons in their outermost shell.

Examples : Beryllium (Be), and Boron (B).

(b) Expanded Octet: Some molecules defy the octet rule by accommodating more than eight electrons around the central atom.

Examples :

Ionic bond : This bond is formed by a transfer of one or more electrons from one atom to another forming cation and anion.

Example : NaCl.

Covalent bond : A covalent bond is an attractive force formed by the mutual sharing of electrons between two atoms with similar or small differences in electronegativity. It is represented by a dash (—) between the two bonding atoms.

Example : Covalent bond formation in H₂ molecule

Co-ordinate covalent bond : This bond is formed by sharing a pair of electrons from one atom by two bonding atoms. It is represented by an arrow from a donor atom to an acceptor atom.

Example : SO₂

The concept of hybridisation was introduced by Pauling and Slater as a result of failure of valence bond theory to explain the following points.

(i) Valencies of certain elements : Valency of an element is equal to the number of unpaired electrons in the atoms. However, valency of certain elements cannot be explained on this basis.

For example :

• Beryllium has electronic configuration 1s², 2s². The expected valency is zero (as there is no unpaired electron) but the observed valency is 2 as in BeCl₂.
• Boron has electronic configuration 1s², 2s² 2p_x¹. The valency should be 1 but it is 3 as in BF₃.
• Carbon has electronic configuration 1s², 2s² 2p_x¹, 2p_y¹ The valency should be 2 but observed valency is 4 as in CH₄.

(ii) The shapes and geometry of certain molecules : The valence bond theory cannot explain shapes, geometry and bond angles in certain molecules.

For example :

• Tetrahedral shape of methane molecule.
• Bond angles in molecules like NH₃ (107°18') and H₂O (104°35’) molecules.
• But all these points can be explained only by the concept of hybridisation, which explains its need.

Geometry of methane molecule on the basis of sp³-hybridisation :

(i) Methane (CH₄) molecule has one carbon atom and four hydrogen atoms.

(ii) Electronic configuration of H (= 1) : 1s²

                                            C (=6) : 1s¹, 2s², 2p_x¹, 2p_y¹.

(iii) The central carbon atom undergoes sp³-hybridisation. Carbon is 1s¹, 2s², 2p_x¹, 2p_y¹ 2p_z⁰. In order to form four equivalent bonds with hydrogen the 2s and 2p orbitals undergo hybridization.

(iv) Tetravalency of carbon : In the ground state, carbon atom has only two unpaired electrons, hence, it should be divalent. But its tetravalency is explained by excitation of an electron from lower energy 2s—orbital to the higher energy vacant 2p_z-orbital giving four unpaired electrons.

(v) sp³-Hybridisation : In this, one s-orbital and three p-orbitals of C atom undergo sp³-hybridisation forming four equivalent sp³-hybridised orbitals directed towards four corners of tetrahedron with the angle 109°28’.

(vi) Formation of bonds : Each sp³-hybridised orbital of carbon atom overlaps coaxially with 1s-orbital of each hydrogen atom having unpaired electron with opposite spin forming four sp³-s sigma (σ) bonds.

• Therefore the geometry of methane molecule is tetrahedral.
• Each H—C—H bond angle is 109°28’.

• In NH₃ molecule, there are three bond pairs of 3 N—H bonds and one lone pair of electrons on N atom. 
• In H₂O molecule there are 2 O-H bond pairs and two lone pairs of electrons on oxygen atom.
• Due to greater repulsion between lone pair—lone pair and lone pair—bond pair in H₂O than in NH₃ molecule, the bond angle is reduced more in H₂O than in NH₃.
• Hence, the bond angle in NH₃ is 107° and in H₂O it is 104° 35’.

• Sigma (σ) bond is formed by coaxial overlapping while pi (π) bond is formed by sidewise or colateral overlapping of atomic orbitals of bonding atoms.
• Hence the extent of overlapping in a sigma (σ) bond is greater than in pi (π) bond.
• The bond enthalpy of a sigma bond is higher than pi bond.

Therefore sigma bond is stronger than pi bond.

In HF molecule, fluorine being highly electronegative, there is a partial shift of electron bond pair towards F atom, making it partially negative and H atom partially positive. Hence HF is a polar molecule.

Carbon has electronic configuration ₆C 1s² 2s² 2p² since it has four valence electrons it is tetravalent in nature.

• Types of hybridization : sp³
• Geometry : Pyramidal
• Bond angle : 107°18’

In F—Be—F molecule, there is sp-hybridisation hence it is linear. But in H—O—H molecule, there is sp³-hybridisation and there are two lone pairs of electrons on oxygen, hence it is angular.

BF₃ molecule has sp²-hybridisation hence it is planar while NH₃ molecule has sp³-hybridisation and N has a lone pair of electrons on oxygen, hence it is pyramidal.

In case of acetylene molecule :

(a) Number of covalent bonds = 5

(b) Number of sigma bonds = 3, Number of pi bonds = 2

(c) Type of hybridisation = sp.

• It is the amount of energy (or enthalpy charge) required to break one mole of a particular bond present between two atoms in the gaseous state.
• Bond enthalpy of a diatomic molecule is equal to its dissociation energy.
• Bond enthalpy is a measure of strength of the bond between two atoms.
• Larger the bond dissociation energy stronger is the bond.

• In case of polyatomic molecules, the bond enthalpy is the average of the sum of successive bond dissociation energies.

Example :

H₂O(g) →  H(g) + OH(g) ΔH₁ = 502 kJ mol^-1

OH(g) → H(g) + O(g) ΔH₂ = 427 kJ mol^-1

Hence average bond enthalpy of O—H bond is

ΔH = (502 + 427)/2  = 464.5 kJ mol^-1

• Bond length implies the equilibrium distance between the nuclei of two covalently bonded atoms in a molecule.

• Bond lengths are measured by X-ray and Electron diffraction techniques.

• Each atom of the bonded pair contributes to the bond length. Bond length depends upon the size of atoms and multiplicity of bonds.
• It increases with increase in size of atom and decreases with increase in multiplicity of bond.
• It is generally measured in picometre (pm) or in Angstrom unit (Å)

(a) As unsaturation increases, bond length decreases.

(b) Since ethyne has a triple bond (-C ≡ C-), the least bond length and the highest bond enthalpy, it is the most stable compound.

(c) As bond length decreases, bond enthalpy increases.

(d) As bond length decreases, bond enthalpy increases, bond strength increases and stability of the compound increases.

The stability of ionic compound depends upon :

(i) size of cation and anion

(ii) charge on ions

(iii) lattice energy.

Decreasing order of lattice energies :

There are six S-F covalent bonds, hence there are six bond pairs and 12 electrons around S in SF₆.

A covalent bond is formed by co-axial over-lapping of atomic orbitals like s-s, p-p and s-p. Hence it is directional.

The interacting forces present during the formation of a molecule are :

• Attractive forces between electrons of atom A and nucleus of atom B.
• Repulsive forces between electrons of A and B as well as repulsive forces between nuclei of A and B.

Pi (π) bond is formed by colateral or sidewise overlap of p-orbitals of two bonding atoms.

Following steps are involved in hybridisation :

(i) Formation of an excited state.

(ii) Mixing and recasting of atomic orbitals.

(iii) Orientation of hybrid orbitals in space.

(iv) Conditions for hybridisation.

Bond order = \(\frac{N_b-N_a}{2}\)

where N_b and N_a are number of electrons present in bonding and antibonding molecular orbitals respectively.

Since O₂ has two unpaired electrons in two antibonding molecular orbitals (π*2p'_y and π *2p'_z), it is paramagnetic in nature.

Formal charge : The formal charge of an atom in a polyatomic molecule or an ion is defined as the difference between the number of valence electrons of the atom in an isolated or a free state and the number of electrons assigned to that atom in the Lewis structure.

Significance of formal charge :

• It helps us in assigning bonds when several structures for the given compound are
• The best Lewis structure is chosen such that the formal charge is as close to zero as possible.
• The structure having the lowest formal charge has the lowest energy and hence the highest stability.