A) \[O_{2}^{2+}>O_{2}^{+}>O_{2}^{-}\]
B) \[O_{2}^{2+}<O_{2}^{-}<O_{2}^{+}\]
C) \[O_{2}^{+}>O_{2}^{-}<O_{2}^{2+}\]
D) \[O_{2}^{-}<O_{2}^{+}>O_{2}^{2+}\]
Correct Answer: A
Solution :
Bond order of \[O_{2}^{2+},O_{2}^{+}\] and \[O_{2}^{-}\] are \[O_{2}^{2+}=\sigma 1{{s}^{2}}\overset{*}{\mathop{\sigma }}\,1{{s}^{2}}\sigma 2{{s}^{2}}\overset{*}{\mathop{\sigma }}\,2{{s}^{2}}\sigma 2p_{z}^{2}\] \[(\pi 2p_{x}^{2}=\pi 2p_{y}^{2})\] \[(\overset{*}{\mathop{\pi }}\,2p{{x}^{0}}=\overset{*}{\mathop{\pi }}\,2p{{y}^{0}})\] Bond order \[=\frac{10-4}{2}=\frac{6}{2}=3\] \[O_{2}^{+}=\sigma 1{{s}^{2}}\overset{*}{\mathop{\sigma }}\,1{{s}^{2}}\sigma 2{{s}^{2}}\overset{*}{\mathop{\sigma }}\,2{{s}^{2}}\sigma 2p_{z}^{2}\] \[(\pi 2p_{x}^{2}=\pi 2p_{y}^{2})(\overset{*}{\mathop{\pi }}\,2p_{x}^{2}=\overset{*}{\mathop{\pi }}\,2p_{y}^{1})\] Bond order = \[\frac{10-7}{2}=\frac{3}{2}=1.5\] So, the correct bond order sequence is \[O_{2}^{2+}>O_{2}^{+}>O_{2}^{-}\]
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