QJEEM23Ch0732: JEE Main 2023 Chemistry - Class 12 JEEM23Ch0732 Solution: Match List I with List II .tg…

Question

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Match List I with List II

	LIST I Coordination Complex		LIST II Number of unpaired electrons
A.	${[Cr (CN)_{6}]}^{3 -}$	I.	0
B.	${[Fe {(H_{2} O)}_{6}]}^{2 +}$	II.	3
C.	${[Co {({NH}_{3})}_{6}]}^{3 +}$	III.	2
D.	${[Ni {({NH}_{3})}_{6}]}^{2 +}$	IV.	4

Choose the correct answer from the options given below:

Accepted Answer

For option (A)

\begin{aligned} {Cr}^{+ 3} : 3 d^{3} \\ {CN}^{-} \to SFL \end{aligned}

No. of unpaired electrons

=3

For option (B)

\begin{aligned} {Fe}^{+ 2} : 3 d^{6} \\ H_{2} O : WFL \end{aligned}

No. of unpaired electrons

=4

For option (C)

\begin{aligned} {Co}^{+ 3} : 3 d^{6} \\ {NH}_{3} : SFL \end{aligned}

No. of unpaired electrons

=0

For option (D)

\begin{aligned} {Ni}^{+ 2} : 3 d^{8} \\ {NH}_{3} : SFL \end{aligned}

No. of unpaired electrons

=2

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