Bond order of 1.5 is shown by
- $\mathrm{O}_2^{+}$
- $\mathrm{O}_2^{-}$
- $\mathrm{O}_2^{2-}$
- $\mathrm{O}_2$
Solution
& \text{(a) } \text {MO configuration of } \mathrm{O}_2^{+}(8+8-1=15) \\
& =\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \stackrel{*}{\sigma} 2 s^2, \sigma 2 p_z^2, \\
& \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi 2 p_x^1 \approx \pi 2 p_y^0 \\
& \mathrm{BO}=\frac{N_b-N_a}{2}
\end{aligned}$
(where, $N_b=$ number of electrons in bonding molecular orbital
$N_a=$ number of electrons in anti-bonding molecular orbital
$\therefore \mathrm{BO}=\frac{10-5}{2}=2.5$
Similarly,
(b) $\mathrm{O}_2^{-}(8+8+1=17)$
so $\mathrm{BO}=\frac{N_b-N_a}{2}=\frac{10-7}{2}=1.5$
(c) $\mathrm{O}_2^{2-}(8+8+2=18)$
$\mathrm{BO}=\frac{N_b-N_a}{2}=\frac{10-8}{2}=1$
(d) $\mathrm{O}_2(8+8=16)$
$\mathrm{BO}=\frac{10-6}{2}=2$
Thus, $\mathrm{O}_2^{-}$shows the bond order 1.5.
Asked in: NEET 2012 (Screening)
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