A) \[{{\left[ Cr{{({{H}_{2}}O)}_{6}} \right]}^{+2}},\,\,{{\left[ CoC{{l}_{4}} \right]}^{-2}}\]
B) \[{{\left[ Cr{{({{H}_{2}}O)}_{6}} \right]}^{+2}},\,\,{{\left[ Fe{{({{H}_{2}}O)}_{6}} \right]}^{+2}}\]
C) \[{{\left[ Cr{{({{H}_{2}}O)}_{6}} \right]}^{+2}},\,\,{{\left[ Fe{{({{H}_{2}}O)}_{6}} \right]}^{+2}}\]
D) \[{{\left[ CoC{{l}_{4}} \right]}^{-2}},\,\,{{\left[ Fe{{({{H}_{2}}O)}_{6}} \right]}^{+2}}\]
Correct Answer: B
Solution :
Same magnetic moment = same no. of unpaired electron \[\mu =\sqrt{n(n+2)}\] \[where,\,\,n=No.\,\,of\,\,unpaired\,{{e}^{-}}\] \[\,C{{O}^{+2}}=3{{d}^{7}},3\,\,unpaired\,\,{{e}^{-}}\] \[M{{n}^{+2}}=3{{d}^{5}},5\,\,unpaired\,\,{{e}^{-}}\] \[~C{{r}^{+2}}=3{{d}^{4}},\,\,4\,\,unpaired\,\,{{e}^{-}}\] \[F{{e}^{+2}}\,\,=\,\,3{{d}^{6}},4\,\,unpaired\,\,{{e}^{\text{-}}}\]
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