Match List I with List II
List I Complex List II $C F S E$ (stabilization) $(∆_{0})$
A. ${[C u {(N H_{3})}_{6}]}^{2 +}$ I. $- 0.6$
B. ${[T i {(H_{2} O)}_{6}]}^{3 +}$ II. $- 2.0$
C. ${[F e {(C N)}_{6}]}^{3 -}$ III. $- 1.2$
D. ${[N i F_{6}]}^{4 -}$ IV. $- 0.4$
Choose the correct answer from the options given below:

Question

Match List I with List II

	List I Complex		List II $C F S E$ (stabilization) $(∆_{0})$
A.	${[C u {(N H_{3})}_{6}]}^{2 +}$	I.	$- 0.6$
B.	${[T i {(H_{2} O)}_{6}]}^{3 +}$	II.	$- 2.0$
C.	${[F e {(C N)}_{6}]}^{3 -}$	III.	$- 1.2$
D.	${[N i F_{6}]}^{4 -}$	IV.	$- 0.4$

Choose the correct answer from the options given below:

TestHub · Accepted Answer

CFSE = ( - 0 . 4 nt 2 g + 0 . 6 ne g ) ∆ 0 nt 2 g = Number of electrons in t 2 g orbital ne g = number of electrons in eg orbital. Crystal Field Stabilisation Energy for the given complexes is as follows: (A) Cu NH 3 6 2 + Cu 2 + . . 3 d 6 , t 2 g 6 e g 3 C F S E = - 6 × 0 . 4 + 3 × 0 . 6 ∆ 0 = - 0 . 6 ∆ 0 (B) Ti H 2 O 6 3 + Ti 3 + : 3 d 1 , t 2 g 1 e g 0 C F S E = - 1 × 0 . 4 ∆ 0 = - 0 . 4 ∆ 0 (C) Fe CN 6 3 - Fe 3 + : 3 d 5 , t 2 g 5 e g 0 C F S E = - 5 × 0 . 4 ∆ 0 = - 2 . 0 ∆ 0 (D) NiF 6 4 - Ni 2 + : 3 d 8 , t 2 g 6 e g 2 C F S E = - 6 × 0 . 4 + 2 × 0 . 6 ∆ 0 = - 1 . 2 ∆ 0 So the correct option is B.

(A)	${[Cu {({NH}_{3})}_{6}]}^{2 +}$ ${Cu}^{2 + .} . 3 d^{6}, t_{2 g}^{6} e_{g}^{3}$	$C F S E$ $= (- 6 \times 0.4 + 3 \times 0.6) ∆_{0}$ $= - 0.6 ∆_{0}$
(B)	${[Ti {(H_{2} O)}_{6}]}^{3 +}$ ${Ti}^{3 +} : 3 d^{1}, t_{2 g}^{1} e_{g}^{0}$	$C F S E = - 1 \times 0.4 ∆_{0}$ $= - 0.4 ∆_{0}$
(C)	${[Fe {(CN)}_{6}]}^{3 -}$ ${Fe}^{3 +} : 3 d^{5}, t_{2 g}^{5} e_{g}^{0}$	$C F S E = - 5 \times 0.4 ∆_{0}$ $= - 2.0 ∆_{0}$
(D)	${[{NiF}_{6}]}^{4 -}$ ${Ni}^{2 +} : 3 d^{8}, t_{2 g}^{6} e_{g}^{2}$	$C F S E$ $= (- 6 \times 0.4 + 2 \times 0.6) ∆_{0}$ $= - 1.2 ∆_{0}$

Chemistry - Coordination Chemistry Question with Solution | TestHub

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