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CBSE Class 12 Chemistry 2023 Delhi Set 1 Solved Paper

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Question : 27 of 35
Marks: +1, -0
(a) On the basic of crystal field theory write the electronic configuration for d5d^5 ion with a strong field ligand for which Δ0>P\Delta_0 > P.
(b) [Ni(CO)4][\mathrm{Ni}(\mathrm{CO})_4] has tetrahedral geometry while [Ni(CN)4]2[\mathrm{Ni}(\mathrm{CN})_4]^{2-} has square planar yet both exhibit diamagnetism. Explain.
[Atomic number : Ni=28\mathrm{Ni}=28 ]
(a) It is the magnitude difference in energy between the two sets of d-orbital i,e t2gt_{2g} and eg. electronic configuration of d5d^5 if Δ0>P\Delta_0 > P is t2g5t_{2g}^5 eg Because in a strong field ligand pairing of electrons takes place for eg [Ni(CN)4]2[\mathrm{Ni}(\mathrm{CN})_4]^{2-}
CN-\mathrm{CN} is a strong field ligand
(b) Ni=28,3d84s2\mathrm{Ni}=28,3 d^8 4 s^2
Ni2+=3d84s0\mathrm{Ni}^{2+}=3 d^8 4 s^0
[Ni(CN)4][\mathrm{Ni}(\mathrm{CN})_4] is a strong field ligand.
=dsp2=dsp^2 hybridization showing square planar geometry
== All electrons are paired so it is diamagnetic in nature.
[Ni(CO)4][\mathrm{Ni}(\mathrm{CO})_4]
Ni=28,3d84s2\mathrm{Ni}=28,3 d^8 4 s^2
Ni=\mathrm{Ni}= valency is zero
=sp3=sp^3 hybridization
\rightarrow showing tetrahedral geometry
\rightarrow All electrons are paired
so it is diamagnetic in nature.
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