Question

Which of the following species has the shortest bond length?

• A. $N_2^{+}$
• B. $N_2$
• C. $N_2^{-}$
• D. $N_2^{2-}$

Answer 1

The correct option is B $N_2$

Bond order = $1/2\times (N_B-N_{AB})$

Order of filling MO is:

1. $N_2^{+}$: valence electrons=$5+5-1=9$

${\left(\sigma 2s\right)}^2{(\sigma \ast 2s)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^3$

BO = $1/2\times (5-0)=2.5$;

2.$N_2$: valence electrons=$5+5=10$

${\left(\sigma 2s\right)}^2{(\sigma \ast 2s)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4$

BO=$1/2\times (6-0)=3$;

3. $N_2^{-}$: valence electrons=$5+5+1=11$

${\left(\sigma 2s\right)}^2{(\sigma \ast 2s)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4{\left({\pi }^{\ast }2p\right)}^1$

BO=$1/2\times (6-1)=2.5$;

4. $N_2^{2-}$: valence electrons=$5+5+2=12$

${\left(\sigma 2s\right)}^2{(\sigma \ast 2s)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4{\left({\pi }^{\ast }2p\right)}^2$

BO=$1/2\times (6-2)=2$;

Since bond length $\propto \frac{1}{bondorder}$

Shortest bond length is of $N_2$

Hence, $B$ is correct.