Questions

Match List-I (complex ions) with List-II (number of unpaired electrons) and select the correct answer using the codes given below lists:
List-I
(Complex ions)
List-II
(Number of Unpaired Electron
A) ${[{CrF}_{6}]}^{4 -}$ 1) One
B) ${[{MnF}_{6}]}^{4 -}$ 2) Two
C) ${[Cr {(CN)}_{6}]}^{4 -}$ 3) Three
D) ${[Mn {(CN)}_{6}]}^{4 -}$ 4) Four
5) Five
Code :

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detailed solution

Correct option is C

CrF64−n=43d4MnF64−n=53d5CrCN64−n=23d4MnCN64−n=13d5n = number of unpaired electrons and that can be calculated by CFT.

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List-I (Complex ions)		List-II (Number of Unpaired Electron
A)	${[{CrF}_{6}]}^{4 -}$	1)	One
B)	${[{MnF}_{6}]}^{4 -}$	2)	Two
C)	${[Cr {(CN)}_{6}]}^{4 -}$	3)	Three
D)	${[Mn {(CN)}_{6}]}^{4 -}$	4)	Four
		5)	Five

Match List-I (complex ions) with List-II (number of unpaired electrons) and select the correct answer using the codes given below lists:List-I(Complex ions) List-II(Number of Unpaired ElectronA) CrF64−1) OneB) MnF64−2) TwoC) CrCN64−3) ThreeD) MnCN64−4) Four 5) FiveCode :

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