CBSE 2023 · Region 5 · Set 3 · Q27 · 3 marks
(a)On the basis of crystal field theory, write the electronic configuration for $\displaystyle \mathrm{d}^{4}$ with a strong field ligand for which $\displaystyle \Delta_{0}>\mathrm{P}$.(b)A solution of $\displaystyle \left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ is green but a solution of $\displaystyle \left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ is colourless. Explain. [Atomic number : Ni = $\displaystyle 28$]
(a)
On the basis of crystal field theory, write the electronic configuration for $\displaystyle \mathrm{d}^{4}$ with a strong field ligand for which $\displaystyle \Delta_{0}>\mathrm{P}$.
(b)
A solution of $\displaystyle \left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ is green but a solution of $\displaystyle \left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ is colourless. Explain. [Atomic number : Ni = $\displaystyle 28$]
Marking-scheme solution
(a) \(\displaystyle t_{2 g}^{4} e_{g}^{0}\)
In \(\displaystyle \left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) d-d transition is possible due to unpaired electrons as \(\displaystyle \mathrm{H}_{2} \mathrm{O}\) is a weak field ligand.
In \(\displaystyle \left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) d-d transition is not possible due to no unpaired electrons as CO is a strong field ligand.
Coordination CompoundsBonding in Coordination CompoundsUnderstandshort_answerhard
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CBSE Class 12 Chemistry past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.