CBSE Class 12 Chemistry 2023 Delhi Set 1 Solved Paper

Question : 27

Total: 35

(a) On the basic of crystal field theory write the electronic configuration for d5 ion with a strong field ligand for which Δ0>P.
(b) [Ni(CO)4] has tetrahedral geometry while [Ni(CN)4]2− has square planar yet both exhibit diamagnetism. Explain.
[Atomic number : Ni=28 ]

Solution: 👈: Video Solution

(a) It is the magnitude difference in energy between the two sets of d-orbital i,e t2g and eg. electronic configuration of d5 if Δ0>P is t2g5 eg Because in a strong field ligand pairing of electrons takes place for eg [Ni(CN)4]2−
−CN is a strong field ligand

(b) Ni=28,3d84s2
Ni2+=3d84s0
[Ni(CN)4] is a strong field ligand.

=dsp2 hybridization showing square planar geometry
= All electrons are paired so it is diamagnetic in nature.
[Ni(CO)4]
Ni=28,3d84s2
Ni= valency is zero

=sp3 hybridization
→ showing tetrahedral geometry
→ All electrons are paired
so it is diamagnetic in nature.

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