A) both\[H_{2}^{+}\]and\[H_{2}^{-}\]are equally stable
B) both\[H_{2}^{+}\]and\[H_{2}^{-}\]do not exist
C) \[H_{2}^{-}\] is more stable than\[H_{2}^{+}\]
D) \[H_{2}^{+}\]is more stable than\[H_{2}^{-}\]
Correct Answer: D
Solution :
\[H_{2}^{+}:\sigma 1{{s}^{1}},\,\,\overset{*}{\mathop{\sigma }}\,1{{s}^{0}}\] Bond order\[=\frac{1}{2}\] \[H_{2}^{-}:\sigma 1{{s}^{2}},\,\,\overset{*}{\mathop{\sigma }}\,1{{s}^{1}}\] Bond order\[=\frac{2-1}{2}=\frac{1}{2}\] The bond order of\[H_{2}^{+}\]and\[H_{2}^{-}\]are same but\[H_{2}^{-}\]is more stable than \[H_{2}^{-}\]. It is due to the presence of one electron in the antibonding molecular orbital in\[H_{2}^{-}\].
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