A) (ii) and (iii)
B) (i), (ii) and (iv)
C) (i), (ii) and (iii)
D) (i) and (iv)
Correct Answer: A
Solution :
Peroxide ion is \[O_{2}^{2-}\] \[O_{2}^{2-}(18)=\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2p_{z}^{2},\] \[\pi 2p_{x}^{2}\approx 2p_{y}^{2},{{\pi }^{*}}2p_{x}^{2}\approx {{\pi }^{*}}2p_{y}^{2}\] 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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