Question

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

• A. $N_2\to N_2^{+}$
• B. $O_2\to O_2^{+}$
• C. $O_2^{-}\to O_2^{2-}$
• D. $O_2^{-}\to O_2$

Answer 1

Answer: (D)

$O_2^{-}(17$ electrons $)$ :

$\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2=\pi 2p_y^2,{\pi }^{\ast }2p_x^2=$

${\pi }^{\ast }2p{,}_y^1{\sigma }^{\ast }2p_z$

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 $):-$

$\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2=\pi 2p_y^2,{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1$

So, Bond Order $=\frac{N_b-N_a}{2}=\frac{10-6}{2}=\frac{4}{2}=2$

No. of unpaired electrons $=2$