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Passage / Comprehension

Column - I	Column - II	Column-III
(a) $[Pt (NH_{3})_{4} Cl_{2}] Br_{2}$	(i) Inner orbital complex	(P)Total possible isomers = 3
(b) $[Co (en) (NH_{3})_{2} Cl_{2}] Cl$	(ii) Low spin complex	(Q)Total possible isomers = 6
(c) $K_{2} [Ni (CN)_{4}]$	(iii) Outer orbital complex	(R) Total possible isomers = 8
(d) $K_{4} [Ni (SCN)_{6}]$	(iv) High spin complex	(S) Total possible isomers = 10

Select the CORRECT combination is :

Options:

Answer:

Solution:

Complex (d) is $K_{4} [N i (SCN)_{6}]$ , where the central metal ion is $N i^{2 +}$ with a $d^{8}$ electronic configuration.

Since it has a coordination number of 6, a $d^{8}$ system cannot vacate inner $d$ -orbitals and must use outer $4 d$ orbitals, making it an outer orbital complex (iii).

$SC N^{-}$ is an ambidentate ligand that can bind through either N or S atoms, resulting in linkage isomerism with a total of 10 possible isomers (S).

Therefore, the combination d-iii-s is perfectly correct.

Chemistry - Coordination Chemistry Question with Solution | TestHub

Options:

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