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 :
Spin only magnetic moment,\[{{\mu }_{s}}=\sqrt{n(n+2)}\] where, \[n=\]number of unpaired electrons. Number of unpaired electrons in\[C{{r}^{2+}}([Ar]\,3{{d}^{4}})\] is 4. in \[C{{o}^{2+}}\]\[([Ar]\,3{{d}^{7}})\] is 3, in \[F{{e}^{2+}}\]\[([Ar]\,3{{d}^{6}})\] is 4, in \[M{{n}^{2+}}\]\[([Ar]\,3{{d}^{5}})\] is 5. As the number of unpaired electrons in \[C{{r}^{2+}}\] and \[F{{e}^{2+}}\] are same hence \[{{[Cr\,{{({{H}_{2}}O)}_{6}}]}^{2+}}\] and \[{{[Fe\,{{({{H}_{2}}O)}_{6}}]}^{2+}}\] will have same magnetic moment.
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