Question

In which change, bond order $\&$ number of unpaired electron $(s),$ both are increased?

• A. $N _{2} \rightarrow N _{2}^{+}$
• B. $O _{2} \rightarrow O _{2}^{+}$
• C. $O _{2}^{-} \rightarrow O _{2}^{2-}$
• D. $O _{2}^{-} \rightarrow O _{2}$

Answer 1

Answer: (D)

$O _{2}^{-}(17$ electrons $)$ :

$\sigma 1 s ^{2}, \sigma^{*} 1 s ^{2}, \sigma 2 s ^{2}, \sigma^{*} 2 s ^{2}, \sigma 2 p _{z}^{2}, \pi 2 p _{ x }^{2}=\pi 2 p _{ y }^{2}, \pi^{*} 2 p _{ x }^{2}=$

$\pi^{*} 2 p,_{y}^{1} \sigma^{*} 2 p_{z}$

Bond Order $=\frac{N_{6}-N_{a}}{2}=\frac{10-7}{2}=\frac{3}{2}$

So, bond order $=1.5$

No. of unpaired electron $=1$

$O _{2}(16$ electrons $):-$

$\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2}, \pi^{*} 2 p_{x}^{1}=\pi^{*} 2 p_{y}^{1}$

So, Bond Order $=\frac{N_{b}-N_{a}}{2}=\frac{10-6}{2}=\frac{4}{2}=2$

No. of unpaired electrons $=2$