KCET · Chemistry · Chemical Bonding and Molecular Structure
Peroxide ion .......
- A (ii) and (iii)
- B (i), (ii) and (iv)
- C (i), (ii) and (iii)
- D (i) and (iv)
Answer & Solution
Correct Answer
(A) (ii) and (iii)
Step-by-step Solution
Detailed explanation
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.
\[
\begin{aligned}
&\mathrm{O}_{2}^{2-}(18)=\sigma 1 \mathrm{~s}^{2}, \sigma * 1 \mathrm{~s}^{2}, \sigma 2 \mathrm{~s}^{2}, \sigma * 2 \mathrm{~s}^{2}, \sigma 2 \mathrm{p}_{\mathrm{z}}^{2}, \\
&\pi 2 \mathrm{p}_{\mathrm{x}}^{2} \approx \pi 2 \mathrm{p}_{\mathrm{y}}^{2}, \pi^{*} 2 \mathrm{p}_{\mathrm{x}}^{2} \approx \pi * 2 \mathrm{p}_{\mathrm{y}}^{2} \\
&\text { Bond order }=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{10-8}{2}=1
\end{aligned}
\]
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.
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