question_answer

The pair having the same magnetic moment is:- \[[\text{At}.\text{No}.:\text{Cr}=\text{24},\text{Mn}=\text{25},\text{Fe}=\text{26},\text{Co}=\text{27}]\] [JEE Main Solved Paper-2016 ]

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

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

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

D) \[{{[Mn{{({{H}_{2}}O)}_{6}}]}^{2+}}\text{and}\,Cr{{({{H}_{2}}O)}_{6}}{{]}^{2+}}\]

Correct Answer: C

Solution :

                In option (1) : \[{{[CoC{{l}_{4}}]}^{2-}},C{{o}^{2+}}(3{{d}^{7}})\] with W.F.L., \[\And {{[Fe{{({{H}_{2}}O)}_{6}}]}^{2+}},F{{e}^{2+}}(3{{d}^{6}})\]with W.F.L., In option (2) : \[[Cr{{(Cr{{({{H}_{2}}O)}_{6}}]}^{2+}},C{{r}^{2+}}(3{{d}^{4}})\]with W.F.L., \[{{[CoC{{l}_{4}}]}^{2-}}C{{o}^{2+}}(3{{d}^{7}})\]with W.F.L., In option (3) : \[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}},C{{r}^{2+}}(3{{d}^{4}})\]with W.F.L., \[\And {{[Fe{{({{H}_{2}}O)}_{6}}]}^{2+}},F{{e}^{2+}}(3{{d}^{6}})\] with W.F.L., Here both complexes have same unpaired electrons i.e. = 4 In option (4) : \[{{[Mn{{({{H}_{2}}O)}_{6}}]}^{2+}},M{{n}^{2+}}(3{{d}^{5}})\] with \[\And {{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}},C{{r}^{2+}}(3{{d}^{4}})\]with W.F.L.,