Match List I with List II List I Complex List II $CFSE$ $({\Delta }_{0})$ A. ${[Cu{(N{H}_{3})}_{6}]}^{2+}$ I. $-0.6$ B. ${[Ti{({H}_{2}O)}_{6}]}^{3+}$ II. $-2.0$ C. ${[Fe{(CN)}_{6}]}^{3-}$ III. $-1.2$ D. ${[Ni{F}_{6}]}^{4-}$ IV. $-0.4$ Choose the correct answer from the options given below:

$\mathrm{CFSE}=(-0.4{\mathrm{nt}}_{2g}+0.6{\mathrm{ne}}_{g}){\Delta }_{0}$ ${\mathrm{nt}}_{2g}=$ Number of electrons in ${t}_{2g}$ orbital ${\mathrm{ne}}_{g}=$number of electrons in $\mathrm{eg}$ orbital. Crystal Field Stabilisation Energy for the given complexes is as follows: (A) ${[\mathrm{Cu}{({\mathrm{NH}}_{3})}_{6}]}^{2+}$ ${\mathrm{Cu}}^{2+.}.3{d}^{6},{t}_{2g}^{6}{e}_{g}^{3}$ $CFSE$ $=(-6\times 0.4+3\times 0.6){\Delta }_{0}$ $=-0.6{\Delta }_{0}$ (B) ${[\mathrm{Ti}{({H}_{2}O)}_{6}]}^{3+}$ ${\mathrm{Ti}}^{3+}:3{d}^{1},{t}_{2g}^{1}{e}_{g}^{0}$ $CFSE=-1\times 0.4{\Delta }_{0}$ $=-0.4{\Delta }_{0}$ (C) ${[\mathrm{Fe}{(\mathrm{CN})}_{6}]}^{3-}$ ${\mathrm{Fe}}^{3+}:3{d}^{5},{t}_{2g}^{5}{e}_{g}^{0}$ $CFSE=-5\times 0.4{\Delta }_{0}$ $=-2.0{\Delta }_{0}$ (D) ${[{\mathrm{NiF}}_{6}]}^{4-}$ ${\mathrm{Ni}}^{2+}:3{d}^{8},{t}_{2g}^{6}{e}_{g}^{2}$ $CFSE$ $=(-6\times 0.4+2\times 0.6){\Delta }_{0}$ $=-1.2{\Delta }_{0}$ So the correct option is B.

JEE Main 2023 — Chemistry Inorganic Chemistry

2023mcqmedium

Match List I with List II

	List I Complex		List II $CFSE$ $({\Delta }_{0})$
A.	${[Cu{(N{H}_{3})}_{6}]}^{2+}$	I.	$-0.6$
B.	${[Ti{({H}_{2}O)}_{6}]}^{3+}$	II.	$-2.0$
C.	${[Fe{(CN)}_{6}]}^{3-}$	III.	$-1.2$
D.	${[Ni{F}_{6}]}^{4-}$	IV.	$-0.4$

Choose the correct answer from the options given below:

Official previous-year question

Held on 12 Apr 2023 · Verified 6 Jul 2026.

Options

A
A(III), B(IV), C(I), D(II)
B
A(I), B(IV), C(II), D(III)
C
A(I), B(II), C(IV), D(III)
D
A(II), B(III), C(I), D(IV)

Did you get this right?

Solution

$\mathrm{CFSE}=(-0.4{\mathrm{nt}}_{2g}+0.6{\mathrm{ne}}_{g}){\Delta }_{0}$

${\mathrm{nt}}_{2g}=$ Number of electrons in ${t}_{2g}$ orbital

${\mathrm{ne}}_{g}=$ number of electrons in $\mathrm{eg}$ orbital.

Crystal Field Stabilisation Energy for the given complexes is as follows:

(A)	${[\mathrm{Cu}{({\mathrm{NH}}_{3})}_{6}]}^{2+}$ ${\mathrm{Cu}}^{2+.}.3{d}^{6},{t}_{2g}^{6}{e}_{g}^{3}$	$CFSE$ $=(-6\times 0.4+3\times 0.6){\Delta }_{0}$ $=-0.6{\Delta }_{0}$
(B)	${[\mathrm{Ti}{({H}_{2}O)}_{6}]}^{3+}$ ${\mathrm{Ti}}^{3+}:3{d}^{1},{t}_{2g}^{1}{e}_{g}^{0}$	$CFSE=-1\times 0.4{\Delta }_{0}$ $=-0.4{\Delta }_{0}$
(C)	${[\mathrm{Fe}{(\mathrm{CN})}_{6}]}^{3-}$ ${\mathrm{Fe}}^{3+}:3{d}^{5},{t}_{2g}^{5}{e}_{g}^{0}$	$CFSE=-5\times 0.4{\Delta }_{0}$ $=-2.0{\Delta }_{0}$
(D)	${[{\mathrm{NiF}}_{6}]}^{4-}$ ${\mathrm{Ni}}^{2+}:3{d}^{8},{t}_{2g}^{6}{e}_{g}^{2}$	$CFSE$ $=(-6\times 0.4+2\times 0.6){\Delta }_{0}$ $=-1.2{\Delta }_{0}$

So the correct option is B.

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Solution

$\mathrm{CFSE}=(-0.4{\mathrm{nt}}_{2g}+0.6{\mathrm{ne}}_{g}){\Delta }_{0}$

${\mathrm{nt}}_{2g}=$ Number of electrons in ${t}_{2g}$ orbital

${\mathrm{ne}}_{g}=$ number of electrons in $\mathrm{eg}$ orbital.

Crystal Field Stabilisation Energy for the given complexes is as follows:

(A)	${[\mathrm{Cu}{({\mathrm{NH}}_{3})}_{6}]}^{2+}$ ${\mathrm{Cu}}^{2+.}.3{d}^{6},{t}_{2g}^{6}{e}_{g}^{3}$	$CFSE$ $=(-6\times 0.4+3\times 0.6){\Delta }_{0}$ $=-0.6{\Delta }_{0}$
(B)	${[\mathrm{Ti}{({H}_{2}O)}_{6}]}^{3+}$ ${\mathrm{Ti}}^{3+}:3{d}^{1},{t}_{2g}^{1}{e}_{g}^{0}$	$CFSE=-1\times 0.4{\Delta }_{0}$ $=-0.4{\Delta }_{0}$
(C)	${[\mathrm{Fe}{(\mathrm{CN})}_{6}]}^{3-}$ ${\mathrm{Fe}}^{3+}:3{d}^{5},{t}_{2g}^{5}{e}_{g}^{0}$	$CFSE=-5\times 0.4{\Delta }_{0}$ $=-2.0{\Delta }_{0}$
(D)	${[{\mathrm{NiF}}_{6}]}^{4-}$ ${\mathrm{Ni}}^{2+}:3{d}^{8},{t}_{2g}^{6}{e}_{g}^{2}$	$CFSE$ $=(-6\times 0.4+2\times 0.6){\Delta }_{0}$ $=-1.2{\Delta }_{0}$

So the correct option is B.

JEE Main 2023 — Chemistry Inorganic Chemistry

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JEE Main 2023 — Chemistry Inorganic Chemistry

Options

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Solution

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JEE Main Inorganic Chemistry in other years

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Other Chemistry topics for JEE Main

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Organic Chemistry

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