Question

$N_2$ and $O_2$ are converted into monoanions, $N_2^{-}$ and $O_2^{-}$ respectively. Which of the following statement(s) is/are correct?

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

Answer 1

Answer: (A), (C), (D)

In $N_2^{-}$, $N-N$ bond becomes weaker than in $N_2$
$N_2^{-}$ becomes paramagnetic
In $O_2^{-}$, $O-O$ bond order decreases than in $O_2$

For $O_2^{-}$

Total valence electrons=13

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

Bond Order = $\frac{8-5}{2}=1.5$

For $N_2^{-}$

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

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

Bond Order = $\frac{8-3}{2}=2.5$

For $N_2$,

valence electrons=$5+5=10$

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

Bond Order = $\frac{8-2}{2}=3$

For $O_2$

Total valence electrons=12

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

Bond Order = $\frac{8-4}{2}=2$

1) Since, bond order of $N_2^{-}$ is less than $N_2$, so the bond weakens.
2) The bond order of $O_2^{-}$ is less than the bond order of $O_2$.
3) As while making monoanions, we are filling the electron in antibonding orbital which decreases the bond order and making $N_2^{-}$ as paramagnetic.