Ncert Exercises & Solutions

4.40 What is meant by the term bond order? Calculate the bond order of : N2, O2, O2+ and O2–.

Bond order is defined as one half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule.
If N

_{a}

is equal to the number of electrons in an anti-bonding orbital, then N

_{b}

is equal to the number of electrons in a bonding orbital.
Bond order =

\frac{1}{2} (N_{b} - N_{a}) .

If N

_{b}

> N

_{a}

, then the molecule is said be stable. However, if N

_{b}

≤ N

_{a}

, then the molecule is considered to be unstable.
Bond order of N

_{2}

can be calculated from its electronic configuration as:

{[σ (1 s)]}^{2} {[σ^{*} (1 s)]}^{2} {[σ (2 s)]}^{2} {[σ^{*} (2 s)]}^{2} {[π (2 p_{x})]}^{2} {[π (2 p_{y})]}^{2} {[π (2 p_{y})]}^{2} {[σ (2 p_{z})]}^{2}

Number of bonding electrons, N

_{b}

= 10
Number of anti-bonding electrons,N

_{a}

= 4

Bond order of nitrogen molecule

= \frac{1}{2} (10 - 4) .

= 3

There are 16 electrons in a molecule of dioxygen, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:

{[σ - (1 s)]}^{2} {[σ^{*} (1 s)]}^{2} {[σ (2 s)]}^{2} {[σ^{*} (2 s)]}^{2} {[σ (1 p_{z})]}^{2} {[π (2 p_{x})]}^{2} {[π (2 p_{y})]}^{2} {[π^{*} (2 p_{x})]}^{1} {[π^{*} (2 p_{y})]}^{1}

Since the 1s orbital of each oxygen atom is not involved in boding, the number of bonding electrons = 8 = N

_{b}

and the number of anti-bonding orbitals = 4 = N

_{a}

Bond order =

\frac{1}{2} (N_{b} - N_{a}) .

\frac{1}{2} (8 - 4)

= 2

Hence, the bond order of oxygen molecule is 2.

Similarly, the electronic configuration of

O_{2}^{+}

can be written as:

K K {[σ (2 s)]}^{2} {[σ^{*} (2 s)]}^{2} {[σ (2 p_{z})]}^{2} {[π (2 p_{x})]}^{2} {[π (2 p_{y})]}^{2} {[π^{*} (2 p_{x})]}^{1}

_{b}

= 8
N

_{a}

= 3
Bond order of

O_{2}^{+} = \frac{1}{2} (8 - 3)

= 2.5

Thus, the bond order of

O_{2}^{+}

is 2.5.

The electronic configuration of

O_{2}^{-}

ion will be:

K K {[σ (2 s)]}^{2} {[σ^{*} (2 s)]}^{2} {[σ (2 p_{z})]}^{2} {[π (2 p_{x})]}^{2} {[π (2 p_{y})]}^{2} {[π^{*} (2 p_{x})]}^{2} {[π^{*} (2 p_{y})]}^{1}

_{b}

= 8
N

_{a}

= 5
Bond order of

O_{2}^{-} = \frac{1}{2} (8 - 5)

= 1.5

Thus, the bond order of

O_{2}^{-}

ion is 1.5.

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