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.,You need to login to perform this action.
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