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] \[{{[Cr\,({{H}_{2}}{{O}_{6}})]}^{2+}},{{[Fe{{({{H}_{2}}O)}_{6}}]}^{2+}}\] \[\mu =\sqrt{n(n+2)}\] No, of unpaired electron (n) are: \[Cr{{2}^{2+}}\,(3{{d}^{4}}),n=4\] \[F{{e}^{2+}}(3{{d}^{6}})\,,n=4\] \[C{{o}^{2+}}(3{{d}^{7}})\,,n=3\] \[M{{n}^{2+}}(3{{d}^{5}})\,,n=5\] Thus, \[C{{r}^{2+}}\] and\[F{{e}^{2+}}\]have same number of unpaired electrons. Hence in [b] option both species will have same magnetic moment.
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