A) \[M{{n}^{2+}}\]
B) \[F{{e}^{2+}}\]
C) \[{{V}^{2+}}\]
D) \[C{{u}^{2+}}\]
Correct Answer: A
Solution :
Magnetic moment \[(\mu )=\sqrt{N(N+2)}\] where N = Number of unpaired electron. The electronic configuration of\[M{{n}^{2+}}\]is \[[Ar]\text{ }3{{d}^{5}}\]. It contains 5 unpaired electron. Hence, \[\mu =\sqrt{5(5+2)}\] \[\mu =\sqrt{35}=5.96\text{ }B.M\].
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