A) \[{{H}_{2}}\]
B) \[H_{2}^{-}\]
C) \[H_{2}^{+}\]
D) \[He_{2}^{+}\]
Correct Answer: A
Solution :
Diamagnetic species have all electrons paired, whereas paramagnetic species have unpaired electrons. \[{{H}_{2}}=\sigma 1{{s}^{2}}\] No. of unpaired electron\[=0\Rightarrow \]diamagnetic \[H_{2}^{-}=\sigma 1s,\,\,\sigma *1{{s}^{1}}\] No. of unpaired electron\[=1\Rightarrow \]paramagnetic \[H_{2}^{-}=\sigma 1{{s}^{1}}\] No. of unpaired electron\[=1\Rightarrow \]paramagnetic \[He_{2}^{+}=\sigma 1{{s}^{2}},\,\,\sigma *1{{s}^{1}}\] No. of unpaired electron\[=1\Rightarrow \]paramagnetic
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