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 :

Easy

Solution

$\begin{array}{lll} {[{CrF}_{6}]}^{4 -} & n = 4 & (3 d^{4}) \\ {[{MnF}_{6}]}^{4 -} & n = 5 & (3 d^{5}) \\ {[Cr {(CN)}_{6}]}^{4 -} & n = 2 & (3 d^{4}) \\ {[Mn {(CN)}_{6}]}^{4 -} & n = 1 & (3 d^{5}) \end{array}$
n = 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