Q12 / 180 · KCET 2010 · Chemistry
Peroxide ion .......
- Correct answerA(ii) and (iii)
- B(i), (ii) and (iv)
- C(i), (ii) and (iii)
- D(i) and (iv)
Solution
Peroxide ion is $\mathrm{O}_{2}^{2-}$
$$
\begin{aligned}
&\mathrm{O}_{2}^{2-}(18)=\sigma 1 \mathrm{~s}^{2}, \sigma * 1 \mathrm{~s}^{2}, \sigma 2 \mathrm{~s}^{2}, \sigma * 2 \mathrm{~s}^{2}, \sigma 2 \mathrm{p}_{\mathrm{z}}^{2}, \\
&\pi 2 \mathrm{p}_{\mathrm{x}}^{2} \approx \pi 2 \mathrm{p}_{\mathrm{y}}^{2}, \pi^{*} 2 \mathrm{p}_{\mathrm{x}}^{2} \approx \pi * 2 \mathrm{p}_{\mathrm{y}}^{2} \\
&\text { Bond order }=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{10-8}{2}=1
\end{aligned}
$$
It contains four completely filled antibonding molecular orbitals. Since, all the electrons are paired, $\mathrm{O}_{2}^{2-}$ is diamagnetic.
Peroxide ion is isoelectronic with argon, not with neon.
Hence, the correct answer is option: 'A'.