Question

$N_2$ and $O_2$ are converted into monoanions, $N_2^{-}$ and $O_2^{-}$ respectively. Which of the following statements is wrong?

• A. In $N_2^{-},N-N$ bond weakens
• B. In $O_2^{-},O-O$ bond order increases
• C. In $O_2^{-},O-O$ bond order decreases
• D. $N_2^{-}$becomes paramagnetic

Answer 1

Answer: (B)

In $O_2^{-},O-O$ bond order increases

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

Order of filling MO is:

1. $N_2$:

valence electrons=$5+5=10$

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

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

2. $N_2^{-}$:

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

3.$O_2$:

valence electrons=$6+6=12$

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

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

4. $O_2^{-}$:

valence electrons=$6+6+1=13$

${\left(\sigma 2s\right)}^2{(\sigma \ast 2s)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4{\left({\pi }^{\ast }2p\right)}^3$

BO=$1/2\times (6-3)=1.5$;

Bond strength $\propto bondorder$

bond strength/order: $N_2$>$N_2^{-}$ and $O_2$>$O_2^{-}$

Therefore, $B$ is correct.