The hybridization involved in complex [Ni(CN)4]2− is (Atomic number of Ni = 28)
A
d2sp2
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B
d2sp3
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C
dsp2
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D
sp3
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Solution
The correct option is Cdsp2 In [Ni(CN)4]2−, the oxidation state of Ni is +2 . x−4=2 x=+2 Now, Ni2+→[Ar]3d84s0 Due to the presence of a strong field ligand (CN−) all unpaired electrons are paired up. Hybridisation of [Ni(CN)4]2− is dsp2