D-And F-Block Elements Ques 472
Question: The correct order of magnetic moments (spin only values in B.M.) among is (Atomic nos.: Mn = 25, Fe = 26, Co = 27)
Options:
A) $ {{[ Fe{{( CN )}_6} ]}^{4-}}>{{[ MnCl_4 ]}^{2}}^{-}>{{[ CoCl_4 ]}^{2-}} $
B) $ {{[MnCl_4]}^{2-}}>{{[Fe{{(CN)}_6}]}^{4-}}>{{[CoCl_4]}^{2-}} $
C) $ {{[MnCl_4]}^{2-}}>{{[Co{{(Cl)}_4}]}^{2-}}>{{[Fe{{(CN)}_6}]}^{4-}} $
D) $ {{[Fe{{(CN)}_6}]}^{4-}}>{{[Co{{(Cl)}_4}]}^{2-}}>{{[MnCl_4]}^{2-}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ {{[Fe{{(CN)}_6}]}^{4-}} $ no of unpaired electron = 0 no of unpaired electron = 5 no of unpaired electron = 3
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-}} $
