A) \[O_{2}^{-}<NO<C_{2}^{2-}<He_{2}^{+}\]
B) \[NO<C_{2}^{2-}<O_{2}^{-}He_{2}^{+}\]
C) \[C_{2}^{2-}<He_{2}^{+}<NO<O_{2}^{-}\]
D) \[He_{2}^{+}<O_{2}^{-}<NO<C_{2}^{2-}\]
Correct Answer: D
Solution :
Key Idea: Bond order\[=\frac{{{N}_{b}}-{{N}_{a}}}{2}\] \[He_{2}^{+}:{{(\sigma 1s)}^{2}}{{({{\sigma }^{*}}1s)}^{1}}\] Bond order\[=\frac{2-1}{2}=\frac{1}{2}\] \[C_{2}^{2-}:\] \[KK{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{1}}{{(\pi 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}\] Bond order \[=\frac{8-2}{2}=3\] \[O_{2}^{-}:KK{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}{{(\pi 2{{p}_{x}})}^{2}}\] \[{{(\pi 2{{p}_{y}})}^{2}}{{({{\pi }^{*}}2{{p}_{x}})}^{2}}{{({{\pi }^{*}}2{{p}_{y}})}^{1}}\] Bond order\[=\frac{8-5}{2}=1\frac{1}{2}\] \[NO:KK{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}{{(\pi 2{{p}_{x}})}^{2}}\] \[{{(\pi 2{{p}_{y}})}^{2}}{{({{\pi }^{*}}2{{p}_{x}})}^{1}}\] Bond order\[=\frac{8-3}{2}=2\frac{1}{2}\] Hence, the order of increasing bond order is as:\[\underset{(\frac{1}{2})}{\mathop{He_{2}^{+}}}\,<\underset{(1\frac{1}{2})}{\mathop{O_{2}^{-}}}\,<\underset{(2\frac{1}{2})}{\mathop{NO}}\,<\underset{(3)}{\mathop{C_{2}^{2-}}}\,\]
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