Question

The correct stability order for $N_2$ and its given ions is:

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

Answer 1

The correct option is A $N_2>N_2^{+}>N_2^{-}>N_2^{2-}$

Bond stability $\propto bondorder$

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

Order of filling MO is:

1. $N_2$:

valence electrons=$5+5=10$

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

BO=$1/2\times (8-2)=3$;

2. $N_2^{+}$:

valence electrons=$5+5-1=9$

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

BO=$1/2\times (7-2)=2.5$;

3.$N_2^{-}$:

valence electrons=$5+5+1=11$

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

BO=$1/2\times (8-3)=2.5$;

4. $N_2^{2-}$:

valence electrons=$5+5+2=12$

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

BO=$1/2\times (8-4)=2$;

bond strength or stability: $N_2$>$N_2^{+}$>$N_2^{-}$>$N_2^{2-}$

$N_2^{+}$

here $N_2^{+}$ and $N_2^{-}$ have same bond order but latter has one more electron in antibonding orbital which reduces its stability as compared to $N_2^{+}$.

Therefore, $A$ is correct.