JEE Advanced · Chemistry · 23. Coordination Compounds
Among the following complexes \((K-P), \mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right](K)\),
\(\left[\right.\) Co \(\left.\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3(L), \mathrm{Na}_3\left[\mathrm{Co}(\mathrm{ox})_3\right](M)\),
\(\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2(\mathrm{~N}) ; \mathrm{K}_2\left[\mathrm{Pt}(\mathrm{CN})_4\right](\mathrm{O})\) and \(\left[\mathrm{Zn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]\left(\mathrm{NO}_3\right)_2(P)\)
the diamagnetic complexes are
- A \(K, L, M, N\)
- B \(K, M, O, P\)
- C \(L, M, O, P\)
- D \(L, M, N, O\)
Answer & Solution
Correct Answer
(C) \(L, M, O, P\)
Step-by-step Solution
Detailed explanation
For a diamagnetic complex, there should not be any unpaired electron in the valence shell of central metral. In \(\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]\), Fe (III) has \(d^5\)-configuration (odd electrons), hence it is paramagnetic. In [Co( \(\left.\left.\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3\), Co (III) has \(d^6\) configuration in a strong ligand field, hence all the electrons are paired and the complex is diamagnetic. In \(\mathrm{Na}_3\left[\mathrm{Co}(\mathrm{ox})_3\right]\), \(\mathrm{Co}\) (III) has \(d^6\)-configuration and oxalate being a chelating ligand, very strong ligand and all the six electrons remains paired in lower \(t_{2 g}\) level, diamagnetic. In \(\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2, \mathrm{Ni}\) (II) has \(3 d^8\)-configuration and \(\mathrm{H}_2 \mathrm{O}\) is a weak ligand hence

In \(\mathrm{K}_2\left[\mathrm{Pt}(\mathrm{CN})_4\right]\), Pt (II) has \(d^8\)-configuration and \(\mathrm{CN}^{-}\)is a strong ligand, hence all the eight electrons are spin-paired. Therefore, complex is diamagnetic.
In \(\left[\mathrm{Zn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]\left(\mathrm{NO}_3\right)_2, \mathrm{Zn}\) (II) has \({ }_3 \mathrm{~d}^{10}\) configuration spin paired, hence diamagnetic.

In \(\mathrm{K}_2\left[\mathrm{Pt}(\mathrm{CN})_4\right]\), Pt (II) has \(d^8\)-configuration and \(\mathrm{CN}^{-}\)is a strong ligand, hence all the eight electrons are spin-paired. Therefore, complex is diamagnetic.
In \(\left[\mathrm{Zn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]\left(\mathrm{NO}_3\right)_2, \mathrm{Zn}\) (II) has \({ }_3 \mathrm{~d}^{10}\) configuration spin paired, hence diamagnetic.
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