A) \[{{[Fe{{(CN)}_{6}}]}^{4-}}>{{[CoC{{l}_{4}}]}^{2-}}>{{[MnC{{l}_{4}}]}^{2-}}\]
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) \[{{[MnC{{l}_{4}}]}^{2-}}>{{[CoC{{l}_{4}}]}^{2-}}>{{[Fe{{(CN)}_{6}}]}^{4-}}\] (Atomic no. of Mn = 25, Fe = 26, Co = 27)
Correct Answer: D
Solution :
[d] \[\therefore \]Number of unpaired electrons=5 \[\therefore \]Number of unpaired electrons=3 \[\therefore \]Number of unpaired electrons=0 As we know that the greater the number of unpaired electrons, greater the magnitude of magnetic moment, Hence the correct order will be. \[{{[Mnc{{l}_{4}}]}^{2-}}>{{[CoC{{l}_{4}}]}^{2-}}>{{[Fe{{(CN)}_{6}}]}^{4-}}\]You need to login to perform this action.
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