| I. has five completely filled anti-bonding molecular orbitals |
| II. is diamagnetic |
| III. has bond order one |
| IV. is isoelectronic with neon |
A) II and III
B) I, II and IV
C) I, III and IV
D) I and IV
Correct Answer: A
Solution :
Peroxide ion is\[O_{2}^{2-}.\] \[O_{2}^{2-}(18)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\sigma 2p_{z}^{2}\] \[\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{2}\approx \overset{*}{\mathop{\pi }}\,2p_{y}^{2}\] Bond order\[=\frac{{{N}_{b}}-{{N}_{a}}}{2}\] \[=\frac{10-8}{2}=1\] It contains four completely filled anti bonding 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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