Chapter 4: Chemical Bonding and Molecular Structure

A covalent bond is formed when 2 atoms combine with each other by sharing their valence electrons. The shared pairs of electrons present between the bonded atoms are called bond pairs. Each and every electron cannot participate in bonding. The pairs of electrons which do not participate in bonding are called lone pairs.

E.g. a) Ethane has 7 bond pairs but zero lone pair.

b) Water has 2 bond pairs and 2 lone pairs on O- atom.

Assuming 2 H- atoms X and Y with nuclei N_X and N_Y and electrons e_X and e_Y, respectively.

When X and Y are far from each other, there is no interaction between them. As soon as they come closer, the attractive force and repulsive force become active.

The repulsive forces are:

(i) Between electrons of both the atoms, i.e., e_X and e_Y.

(ii) Between nuclei of both the atoms, i.e., N_X and N_Y.

The attractive forces are:

(i) Between the electron and nucleus of the same atom i.e. N_X – e_X and N_Y -e_Y.

(ii) Between the electron of one atom and the nucleus of the other atom, i.e., N_X–e_Y and N_Y–e_X.

The repulsive force pushes the 2 atoms apart, whereas the attractive force tends to bring them together.

Repulsive forces:

Attractive forces:

The values of repulsive forces are less than that of attractive forces. Thus, 2 atoms approach each other. There is a decrease in potential energy. In the end, a stage is reached when the repulsive forces balance the attractive forces, and the system achieves the minimum energy, which leads to the formation of H2 molecule.

The condition that is required for the linear combination of atomic orbitals to form molecular orbitals are as follows:

(i) The joining of atomic orbitals must have approximately the same energy. This implies in a homo-nuclear molecule, the 1s-orbital of one atom can join with the 1s- orbital of another atom but cannot join with the 2s-orbital.

(ii) The joining atomic orbitals must have legitimate orientations to ensure the maximum overlap.

(iii) The overlapping must be to a large extent.

Electronic configuration of Be:

1s² 2s²

The molecular orbital electronic configuration of Be₂ is:

Thus, bond order of Be₂: 0.5(N_b – N_a).

N_b: No. of electrons in the bonding orbitals

N_a: No. of electrons in the anti-bonding orbitals

Therefore, bond order of Be₂ = 0.5(4 – 4) = 0

Zero value of bond order indicates that the given molecule is unstable. Thus, Be₂ doesn’t exist.

O₂ contain 16 electrons i.e. 8 electrons from each O- atom.

• The electronic configuration of O₂ is:

As 1s- orbital of each O- atom does not involve in the bonding,

No. of bonding electrons = N_b = 8

No. of anti-bonding electrons = Na = 4

Now,

Bond order = 0.5(8 – 4) = 2

• Electronic configuration of
  \(\begin{array}{l}O_{2}^{+}\end{array} \)
  is:

No. of bonding electrons = N_b = 8

No. of anti-bonding electrons = Na = 3

Now,

Bond order = 0.5(8 – 3) = 2.5

As, bond dissociation energy

Hence, the higher the bond order, the higher stability.

The arrangement according to decreasing order of stability is given as:

• Electronic configuration of
  \(\begin{array}{l}O_{2}^{-}\end{array} \)
  (Superoxide) is:

No. of bonding electrons = N_b = 8

No. of anti-bonding electrons = Na = 5

Now,

Bond order = 0.5(8 – 5) = 1.5

• Electronic configuration of
  \(\begin{array}{l}O_{2}^{2-}\end{array} \)
  (peroxide) is:

No. of bonding electrons = N_b = 8

No. of anti-bonding electrons = Na = 6

Now,

Bond order = 0.5(8 – 6) = 1

Generally, molecular orbital is represented by the ‘wave function’.

Positive (+) sign in representing a molecular orbital indicates positive wave function.

Negative (-) sign in representing a molecular orbital indicates negative wave function.

The electronic configurations of the outer orbital of phosphorus in excited state and in ground state are given below:

Ground State:

Excited State:

Phosphorus atom has sp³d hybridization. These orbitals are filled due to the donation of electron pairs by 5 Cl- atoms: PCl₅

The 5 sp³d hybrid orbitals present here are directed towards 5 corners of trigonal bipyramidal. Thus, the geometry of PCl₅ is given below:

PCl₅ contains 5 P- Cl sigma bonds. Out of which, 3 P-Cl bonds lie in only 1 plane, and they are making

Out of 2 remaining P-Cl bonds, one bond lie above the equatorial plane, and one bond lie below the equatorial bond. And they are making

Equatorial bond pairs repel axial bond pairs to a large extent. So, equatorial bonds are slightly shorter than axial bonds.

“H-bond is defined as an attractive force acting between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule”.

As there is a difference in the electronegativities between the atoms, the bond pair electronegative atom and hydrogen atom is drifted away from H- atom. Therefore, the hydrogen atom gets electropositive w.r.t. the other atom and procures a positive charge.

The value of H- bond is minimum in gaseous state and maximum in the solid state.

Two types of hydrogen bonds are there:

(a) Intramolecular hydrogen bond, e.g., o- nitrophenol

(b) Intermolecular hydrogen bond, e.g., HF, H₂O etc.

H- bonds are stronger than Van der Waals forces as H- bonds are regarded as an extreme form of the dipole-dipole interaction.

Bond Order: It is defined as 0.5 times the difference between the “number of electrons present in bonding orbitals and number of electrons present in anti-bonding orbitals” of a molecule.

Bond Order = 0.5(N_b – N_a);

Na: No. of anti-bonding electrons

Nb: No. of bonding electrons

O₂ contain 16 electrons, i.e., 8 electrons from each O- atom.

• Electronic configuration of O₂ is:

No. of bonding electrons = N_b = 8

No. of anti-bonding electrons = N_a = 4

Now,

Bond order = 0.5(8 – 4) = 2

• Electronic configuration of
  \(\begin{array}{l}O_{2}^{-}\end{array} \)
  (superoxide) is:

No. of bonding electrons = N_b = 8

No. of anti-bonding electrons = N_a = 5

Now,

Bond order = 0.5(8 – 5) = 1.5

• Electronic configuration of
  \(\begin{array}{l}O_{2}^{+}\end{array} \)
  is:

No. of bonding electrons = N_b = 8

No. of anti-bonding electrons = N_a = 3

Now,

Bond order = 0.5(8 – 3) = 2.5

• Electronic configuration of N₂ is:

No. of bonding electrons = N_b = 10

No. of anti-bonding electrons = Na = 4

Now,

Bond order = 0.5(10 – 4) = 3