Match List I with List II: List I (Complexes) List II (Hybridisation) (A) $[\mathrm{Ni}(\mathrm{CO}{)}_{4}]$ I ${\mathrm{sp}}^{3}$ (B) ${[\mathrm{Cu}{({\mathrm{NH}}_{3})}_{4}]}^{2+}$ II ${\mathrm{dsp}}^{2}$ (C) ${[\mathrm{Fe}{({\mathrm{NH}}_{3})}_{6}]}^{2+}$ III ${\mathrm{sp}}^{3}{d}^{2}$ (D) ${[\mathrm{Fe}{({H}_{2}O)}_{6}]}^{2+}$ IV ${d}^{2}{\mathrm{sp}}^{3}$

JEE Main 2023 — Chemistry Inorganic Chemistry

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Match List I with List II:

List I (Complexes)		List II (Hybridisation)
(A)	$[\mathrm{Ni}(\mathrm{CO}{)}_{4}]$	I	${\mathrm{sp}}^{3}$
(B)	${[\mathrm{Cu}{({\mathrm{NH}}_{3})}_{4}]}^{2+}$	II	${\mathrm{dsp}}^{2}$
(C)	${[\mathrm{Fe}{({\mathrm{NH}}_{3})}_{6}]}^{2+}$	III	${\mathrm{sp}}^{3}{d}^{2}$
(D)	${[\mathrm{Fe}{({H}_{2}O)}_{6}]}^{2+}$	IV	${d}^{2}{\mathrm{sp}}^{3}$

Official previous-year question

Held on 30 Jan 2023 · Verified 6 Jul 2026.

Options

A
A – II, B – I, C – III, D – IV
B
A – I, B – II, C – III, D – IV
C
A – II, B – I, C – IV, D – III
D
A – I, B– II, C – IV, D – III

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Solution

$\mathrm{Ni}$ is in zero oxidation state in $[\mathrm{Ni}(\mathrm{CO}{)}_{4}]$ . So, the electronic configuration of $\mathrm{Ni}$ is $3{d}^{8}4{s}^{2}$ . As $\mathrm{CO}$ is a strong ligand, it pushes all the electrons in the 3d orbital, therefore the hybridisation of $[Ni(CO{)}_{4}]$ $_{ }$ is sp $^{3}$ and it has tetrahedral geometry. It is diamagnetic due to the absence of unpaired electrons.

${[\mathrm{Cu}{({\mathrm{NH}}_{3})}_{4}]}^{2+}$ ion, oxidation state of $\mathrm{Cu}$ is +2 and its valence shell electronic configuration is $3{d}^{9}$ . So, there would be a rearrangement of electrons in ${\mathrm{Cu}}^{2+}$ because of the ${\mathrm{NH}}_{3}$ strong field ligand . And the last electron in the d-orbital would be out waiting for the N's electrons to fill up first. So, the 4 electron pairs from $N$ would be in one 3d, one 4s, & two 4p orbitals and in the third place of 4p orbital, the e $^{-}$ from 3d would take place. So, you can say the hybridisation here would be ${\mathrm{dsp}}^{2}$ .

${[\mathrm{Fe}{({\mathrm{NH}}_{3})}_{6}]}^{2+}$ ion, oxidation state of $\mathrm{Fe}$ is +2 and its valence shell electronic configuration is $3{d}^{6}$ . ${\mathrm{NH}}_{3}$ is strong field ligand with complex. So, no pairing will be there. So, you can say the hybridisation here would be ${d}^{2}{\mathrm{sp}}^{3}$ .

${[\mathrm{Fe}{({H}_{2}O)}_{6}]}^{2+}$ ion, oxidation state of $\mathrm{Fe}$ is +2 and its valence shell electronic configuration is $3{d}^{6}$ . There are 4 unpaired electrons in 3d orbital. So, you can say the hybridisation here would be ${\mathrm{sp}}^{3}{d}^{2}$ .

So, correct order of matching is given in the below table:

List I (Complexes)	List II (Hybridisation)
$[\mathrm{Ni}(\mathrm{CO}{)}_{4}]$	${\mathrm{sp}}^{3}$
${[\mathrm{Cu}{({\mathrm{NH}}_{3})}_{4}]}^{2+}$	${\mathrm{dsp}}^{2}$
${[\mathrm{Fe}{({\mathrm{NH}}_{3})}_{6}]}^{2+}$	${d}^{2}{\mathrm{sp}}^{3}$
${[\mathrm{Fe}{({H}_{2}O)}_{6}]}^{2+}$	${\mathrm{sp}}^{3}{d}^{2}$

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Solution

So, correct order of matching is given in the below table:

List I (Complexes)	List II (Hybridisation)
$[\mathrm{Ni}(\mathrm{CO}{)}_{4}]$	${\mathrm{sp}}^{3}$
${[\mathrm{Cu}{({\mathrm{NH}}_{3})}_{4}]}^{2+}$	${\mathrm{dsp}}^{2}$
${[\mathrm{Fe}{({\mathrm{NH}}_{3})}_{6}]}^{2+}$	${d}^{2}{\mathrm{sp}}^{3}$
${[\mathrm{Fe}{({H}_{2}O)}_{6}]}^{2+}$	${\mathrm{sp}}^{3}{d}^{2}$

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

Physical Chemistry

Organic Chemistry

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