A) \[{{[MnC{{l}_{4}}]}^{2-}}>{{[CoC{{l}_{4}}]}^{2-}}>{{[Fe{{(CN)}_{6}}]}^{4-}}\]
B) \[{{[MnC{{l}_{4}}]}^{2-}}>{{[Fe{{(CN)}_{6}}]}^{4-}}>{{[CoC{{l}_{4}}]}^{2-}}\]
C) \[{{[Fe{{(CN)}_{6}}]}^{4-}}>{{[MnC{{l}_{4}}]}^{2-}}>{{[CoC{{l}_{4}}]}^{2-}}\]
D) \[{{[Fe{{(CN)}_{6}}]}^{4-}}>{{[CoC{{l}_{4}}]}^{2-}}>{{[MnC{{l}_{4}}]}^{2-}}\]
Correct Answer: A
Solution :
As in Q. 124, number of unpaired electrons in \[{{[Fe{{(CN)}_{6}}]}^{4-}}\]is zero. Thus magnetic moment \[=\sqrt{n(n+2)}=0\,BM\] (\[n=\]unpaired electrons) \[n\]in\[{{[MnC{{l}_{4}}]}^{2-}}=5,\sqrt{35}\,BM\] \[n\]in\[{{[CoC{{l}_{4}}]}^{2-}}=3,\sqrt{15}\,BM\] (Atomic numbers\[Mn=25,\text{ }Fe=26,\text{ }Co=27\])
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