CBSE 2023 · Region 5 · Set 1 · Q27 · 3 marks
(a)On the basis of crystal field theory write the electronic configuration for $\displaystyle \mathrm{d}^{5}$ ion with a strong field ligand for which $\displaystyle \Delta_{0}>\mathrm{P}$.(b)$\displaystyle \left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ has tetrahedral geometry while $\displaystyle \left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ has square planar yet both exhibit dimagnetism. Explain. [Atomic number : Ni = $\displaystyle 28$]
(a)
On the basis of crystal field theory write the electronic configuration for $\displaystyle \mathrm{d}^{5}$ ion with a strong field ligand for which $\displaystyle \Delta_{0}>\mathrm{P}$.
(b)
$\displaystyle \left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ has tetrahedral geometry while $\displaystyle \left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ has square planar yet both exhibit dimagnetism. Explain. [Atomic number : Ni = $\displaystyle 28$]
Marking-scheme solution
(a) \(\displaystyle \mathrm{t}^{5}{ }_{\mathrm{g}} \mathrm{e}_{\mathrm{g}}^{0}\)
(b) \(\displaystyle \left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) has \(\displaystyle \mathrm{sp}^{3}\) hybridisation
\(\displaystyle \left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) has \(\displaystyle \mathrm{dsp}^{2}\) hybridisation
In both, all electrons are paired.
(or explain using V.B. theory).
Coordination CompoundsBonding in Coordination CompoundsApplyshort_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.