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 :
\[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}}=C{{r}^{2+}}=3{{d}^{4}}=\] four unpaired electrons \[{{[CoC{{l}_{4}}]}^{2-}}=C{{o}^{2+}}=3{{d}^{7}}=3\] unpaired electrons \[{{[Fe{{({{H}_{2}}O)}_{6}}]}^{2+}}=F{{e}^{2+}}=3{{d}^{6}}=4\] unpaired electrons \[{{[Mn{{({{H}_{2}}O)}_{6}}]}^{2+}}=M{{n}^{2+}}=3{{d}^{6}}=5\] unpaired electrons So, \[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}}\] and \[{{[Fe{{({{H}_{2}}O)}_{6}}]}^{2+}}\] has same magnetic moment.
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