question_answer

The pair having the same magnetic moment is [at. no. \[Cr=24,\]\[Mn=25,\]\[Fe=26\] and \[Co=27\]]

A) \[[Cr\,{{({{H}_{2}}O)}_{6}}^{2+}\text{and  }\!\![\!\!\text{ }Fe\,{{\text{(}{{H}_{2}}O)}_{6}}{{]}^{2+}}\]

B) \[{{[Mn\,{{({{H}_{2}}O\text{)}}_{6}}\text{ }\!\!]\!\!\text{ }}^{2+}}\text{and  }\!\![\!\!\text{ }Fe\,{{\text{(}{{H}_{2}}O)}_{6}}{{]}^{2+}}\]

C) \[{{[CoC{{l}_{4}}]}^{2-}}\text{and}\,{{[Fe\,{{({{H}_{2}}O)}_{6}}]}^{2+}}\]

D) \[{{[Cr\,{{({{H}_{2}}O)}_{6}}]}^{2+}}\text{and}\,{{\text{ }\!\![\!\!\text{ }CoC{{l}_{4}}]}^{2-}}\]

Correct Answer: A

Solution :

[a] The complexes that has same number of unpaired electrons will have same magnetic moment. The number of unpaired electrons in complexes are as follows:

Complex	Electronic configuration of metal ion	Number of unpaired electrons (n)
\[{{[Cr\,{{({{H}_{2}}O)}_{6}}]}^{2+}}\]	\[C{{r}^{2+}};[Ar]\,3{{d}^{4}}\]
\[{{[Fe\,{{({{H}_{2}}O)}_{6}}]}^{2+}}\]	\[F{{e}^{2+}};[Ar]\,3{{d}^{6}}\]
\[{{[Mn\,{{({{H}_{2}}O)}_{6}}]}^{2+}}\]	\[M{{n}^{2+}};[Ar]\,3{{d}^{5}}\]
\[{{[CoCl]}^{2-}}\]	\[C{{o}^{2+}};[Ar]3{{d}^{7}}\]

Thus, \[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}},\,\,{{[Fe\,{{({{H}_{2}}O)}_{6}}]}^{2+}}\] have same magnetic moment.