| (i) has five completely filled antibonding molecular orbitals |
| (ii) is diamagnetic |
| (iii) has bond order one |
| (iv) is isoelectronic with neon |
A) (ii) and (in)
B) (i), (ii) and (iv)
C) (i), (ii) and (iii)
D) (i) and (iv)
Correct Answer: A
Solution :
Peroxide ion is \[O_{2}^{2-}\] \[O_{2}^{2-}(18)=\sigma 1{{s}^{2}},\,\,\,\,\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\,\,\,\,\sigma 2{{s}^{2}},\,\,\,\,\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\,\,\,\,\sigma 2p_{z}^{2}\], \[\pi 2p_{x}^{2}\approx 2p_{y}^{2},\,\overset{*}{\mathop{\pi }}\,2p_{x}^{2}\approx \overset{*}{\mathop{\pi }}\,2p_{y}^{2}\] Bond order \[=\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{10-8}{2}=1\] It contains four completely filled antibonding molecular orbitals. Since, all the electrons are paired, \[O_{2}^{2-}\] is diamagnetic. Peroxide ion is isoelectronic with argon, not with neon.
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