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Q. When $O_2$ is converted into $O_2^+$

KCETKCET 2011Chemical Bonding and Molecular Structure

Solution:

(i) $O _{2}-(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{\star} 2 s\right)^{2}$

$\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\pi * 2 p_{x}\right)^{1}$

$\left(\pi^{\star} 2 p_{y}\right)^{1}$

Bond order $=\frac{N_{b}-N_{a}}{2}$

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

$=2$

$O _{2}$ molecule having 2 unpaired electron.

(ii) $O _{2}^{+}-(\sigma 1 s)^{2}\left(\sigma^{\star} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{\star} 2 s\right)^{2}$

$\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\pi^{\star} 2 p_{x}\right)^{1}$

$\left(\pi^{*} 2 p_{y}\right)^{0}$

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

$=2.5$

$O _{2}^{+}$ ion having only 1 unpaired electron.

Hence, when $O _{2}$ is converted into $O _{2}^{+}$ paramagnetic character decrease and the bond order increases.