Q7 / 125 · BITSAT 2011 · Chemistry
The correct bond order in the following species is:
- A$\mathrm{O}_{2}^{2+}<\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{+}$
- B$\mathrm{O}_{2}^{+}<\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{2+}$
- Correct answerC$\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{+}<\mathrm{O}_{2}^{2+}$
- D$\mathrm{O}_{2}^{2+}<\mathrm{O}_{2}^{+}<\mathrm{O}_{2}^{-}$
Solution
$\mathrm{O}_{2}^{+}$ ion - Total number of electrons
$(16-1)=15$
E.C.:
$$
\begin{array}{r}
\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{2} \\
\pi 2 p_{x}^{2}=\pi 2 p_{y}^{2} \pi^{*} 2 p_{x}^{1}=\pi^{*} 2 p_{y} \\
\text { Bond order }=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{10-5}{2}=\frac{5}{2}=2 \frac{1}{2}
\end{array}
$$
$\mathrm{O}_{2}^{-}$ (Super oxide ion): Total number of
electrons $(16+1)=17$
E.C.:
$$
\begin{array}{l}
\text { \sigma1s }^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{2} \\
\pi 2 p_{x}^{2}=\pi 2 p_{y}^{2} \pi^{*} 2 p_{x}^{2}=\pi^{*} 2 p_{y}^{1} \\
\text { Bond order }=\frac{\left(\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)}{2}=\frac{10-7}{2}=\frac{3}{2}=1 \frac{1}{2}
\end{array}
$$
$\mathrm{O}_{2}^{2+}$ ion: Total number of electrons $=(16-2)=14$
E.C.
$$
\begin{array}{l}
\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2} \sigma 2 p_{z}^{2} \\
\text { Bond order }=\frac{\left(N_{b}-N_{a}\right)}{2}=\frac{10-4}{2}=\frac{6}{2}=3
\end{array}
$$
So bond order: $\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{+}<\mathrm{O}_{2}^{2+}$
Hence, the correct answer is option: 'C'.