Co-Ordination Compounds Question 156
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-}} $
