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 :
Same magnetic moment = same number of unpaired electrons\[=\sqrt{n(n+2)},\]where n represents 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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