Question

What do you meant by term bond order? Calculate the bond order of $N_2,C_2,H_2,N_2$.

Answer 1

Bond ion the define as one half the difference between the number of electrons present in the bonding and the anti-bonding orbitals.
Bond order (B.O) $=1/2(N_b-N_a)$

where, $N_b$ and $N_a$ are number of electron in bonding and anti bonding molecular orbitals respectively. Integral bond order values of $1,2$ or $3$ correspond ti single, double or triple bonds respectively.

A positive order (i.e., $N_b>N_a$) means molecule while a negative (i.e., $N_b<N_a$) or zero (i.e., $N_b=N_a$) bond order means unstable molecules

(a) Bond order of $N_2$
Molecular electronic configuration of $N_2$

$\left[\sigma \right(1s\left){]}^2\right[\sigma \ast \left(1s\right){]}^2\left[\sigma \right(2s\left){]}^2\right[\sigma \ast \left(2s\right){]}^2\left[\pi \right(2p_x\left){]}^2\right[\sigma \ast \left(2s\right){]}^2\left[\pi \right(2p_y\left){]}^2\right[\sigma \ast \left(2s\right){]}^2\left[\pi \right(2p_z){]}^2$

Number of bonding electrons $=10$
Number of anti-bonding electrons $=4$

Bond order of nitrogen molecule $=1/2(10-4)=3$

(b) Bond order of $C_2$
Total number of electrons in $C_2$ molecule is $6+6=12$
Molecular orbital electronic configuration of $C_2$
$\left[\sigma \right(1s\left){]}^2\right[\sigma \ast \left(1s\right){]}^2\left[\sigma \right(2s\left){]}^2\right[\sigma \ast \left(2s\right){]}^2\left[\pi \right(2p_x){]}^2=[\pi \left(2p_y\right){]}^2$
Bond order (B.O) $=1/2(N_b-N_a)=1/2(8-4)=2$

The bond order is $2$ and so carbon molecule will have a double bond.

(c) Bond order of $H_2^{-}$
Total number of electrons $=2+1=3$
Molecular orbital electrons is configuration
$\left[\sigma \right(1s\left){]}^2\right[\sigma \ast \left(1s\right){]}^1$

Bond order $=1/2(2-1)=\frac{1}{2}$

(d) Bond order $N_2^{-}$

Total number of electron $=14+1=15$
Molecular orbital electronic configuration

$[\sigma 1s{]}^2,[\sigma 2s{]}^2,[\sigma \ast 2s{]}^2,[\pi 2p_x{]}^2$ $=[\pi 2p_y{]}^2,[\sigma 2p_z{]}^2,[\pi 2p_x{]}^2,[\pi 2p_y{]}^2$

Electrons in bonding orbitals $:10$

Electrons in anti-bonding orbitals $:5$

Bond order of $N_2^{-}=182(10-5)=2.5$