The hybridization involved in… - NEET UG 2015 Chemistry PYQ

NEET UG 2015 — Chemistry Inorganic Chemistry

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The hybridization involved in complex ${[Ni{(CN)}_{4}]}^{2-}$ is: (Atomic number of Ni = 28)

Official previous-year question

Held on 30 Apr 2015 · Verified 9 Jul 2026.

Options

A
$ds{p}^{2}$
B
$s{p}^{3}$
C
${d}^{2}s{p}^{2}$
D
${d}^{2}s{p}^{3}$

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Solution

${[Ni{(CN)}_{4}]}^{2-}$

Oxidation state of Ni is +2

$x-4=2$

$x=+2$

Now, ${\mathrm{Ni}}^{2+}=[\mathrm{Ar}]3{d}^{8}4s^{\circ}$

NEET 2015 Chemistry, Coordination Compounds — solution figure 1

therefore, ${\mathrm{CN}}^{-}$ is a strong field ligand. Hence, all unpaired electrons are paired up.

NEET 2015 Chemistry, Coordination Compounds — solution figure 2

therefore, hybridisation of ${[\mathrm{Ni}(\mathrm{CN}{)}_{4}]}^{2-}$ is ${\mathrm{dsp}}^{2}$

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Solution

${[Ni{(CN)}_{4}]}^{2-}$

Oxidation state of Ni is +2

$x-4=2$

$x=+2$

Now, ${\mathrm{Ni}}^{2+}=[\mathrm{Ar}]3{d}^{8}4s^{\circ}$

NEET 2015 Chemistry, Coordination Compounds — solution figure 1

therefore, ${\mathrm{CN}}^{-}$ is a strong field ligand. Hence, all unpaired electrons are paired up.

NEET 2015 Chemistry, Coordination Compounds — solution figure 2

therefore, hybridisation of ${[\mathrm{Ni}(\mathrm{CN}{)}_{4}]}^{2-}$ is ${\mathrm{dsp}}^{2}$

Options

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Solution

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Options

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Solution

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NEET UG Inorganic Chemistry in other years

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Physical Chemistry

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