A) \[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}},{{[CoC{{l}_{4}}]}^{2-}}\]
B) \[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}},{{[Fe{{({{H}_{2}}O)}_{6}}]}^{2+}}\]
C) \[{{[Mn{{({{H}_{2}}O)}_{6}}]}^{2+}},{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}}\]
D) \[{{[CoC{{l}_{4}}]}^{2-}},{{[Fe{{({{H}_{2}}O)}_{6}}]}^{2+}}\]
Correct Answer: B
Solution :
[b] Same magnetic moment = Same number of unpaired electrons. \[\mu =\sqrt{n(n+2)}\] where n = number of unpaired electrons \[C{{o}^{2+}}=3{{d}^{7}},3\] unpaired electrons; \[C{{r}^{2+}}=3{{d}^{4}},4\] unpaired electrons; \[M{{n}^{2+}}=3{{d}^{5}},5\] unpaired electrons; \[F{{e}^{2+}}=3{{d}^{6}},4\] unpaired electrons.
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