Question

Which of the following has unpaired electron(s)?

• A. ${\text{O}}_2^{-}$
• B. ${\text{N}}_2^{2+}$
• C. ${\text{O}}_2^{2-}$
• D. $N_2$

Answer 1

Answer: (A)

The Molecular orbital electronic configurations are
$O_2^{-}\Rightarrow \sigma 1s^2<\stackrel{\ast }{\sigma }1s^2<\sigma 2s^2<\stackrel{\ast }{\sigma }2s^2<\sigma 2p_z^2<\pi 2p_x^2=\pi 2p_y^2<\stackrel{\ast }{\pi }2p_x^2=\stackrel{\ast }{\pi }2p_y^1$
(unpaired electron)
$N_2^{2+}\Rightarrow \sigma 1s^2<\stackrel{\ast }{\sigma }1s^2<\sigma 2s^2<\stackrel{\ast }{\sigma }2s^2<\pi 2p_x^2=\pi 2p_y^2<\sigma 2p_z^0$
$O_2^{2-}\Rightarrow \sigma 1s^2<\stackrel{\ast }{\sigma }{\delta }^2<\sigma 2s^2<\stackrel{\ast }{\sigma }2s^2<\sigma 2p_z^2<\pi 2p_x^2=\pi 2p_y^2<\stackrel{\ast }{\pi }2p_x^2=\stackrel{\ast }{\pi }2p_y^2$
$N_2\Rightarrow \sigma 1s^2<\stackrel{\ast }{\sigma }1s^2<\sigma 2s^2<\stackrel{\ast }{\sigma }2s^2<\pi 2p_x^2=\pi 2p_y^2<\sigma 2p_z^2$