Co Ordination Compounds Question 275

Question: The correct order of magnetic moments (spin values in B.M.) among is:

Options:

A) $ {{[Fe{{(CN)}_6}]}^{4-}}>{{[CoCl_4]}^{2-}}>{{[MnCl_4]}^{2-}} $

B) $ {{[MnCl_4]}^{2-}}>{{[Fe{{(CN)}_6}]}^{4-}}>{{[CoCl_4]}^{2-}} $

C) $ {{[Fe{{(CN)}_6}]}^{4-}}>{{[MnCl_4]}^{2-}}>{{[CoCl_4]}^{2-}} $

D) $ {{[MnCl_4]}^{2-}}>{{[CoCl_4]}^{2-}}>{{[Fe{{(CN)}_6}]}^{4-}} $ (Atomic no. of Mn = 25, Fe = 26, Co = 27)

Show Answer

Answer:

Correct Answer: D

Solution:

$ \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.

$ {{[Mncl_4]}^{2-}}>{{[CoCl_4]}^{2-}}>{{[Fe{{(CN)}_6}]}^{4-}} $



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