- A$ - CHO$
- B$ > C = O$
- CBoth $(a)$ and $(b)$
- ✓None of these
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Statement I : $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ is a homoleptic complex whereas $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}$is a heteroleptic complex.
Statement II : Complex $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ has only one kind of ligands but $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}$has more than one kind of ligands.
In the light of the above statements, choose the correct answer from the options given below.
[Atomic Number: $\mathrm{Fe}=26, \mathrm{Mn}=25, \mathrm{Co}=27$ ]
| List-$I$ | List-$II$ |
| ($P$) $ t_{2 g}^6 e_g^0$ | ($1$)$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ |
| ($Q$) $t_{2 g}^3 e_g^2$ | ($2$) $\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ |
| ($R$) $e^2 t_2^3$ | ($3$)$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ |
| ($S$) $t_{2 g}^{+} e_g^2$ | ($4$)$\left[\mathrm{FeCl}_4\right]^{-}$ |
| $\left[\mathrm{CoCl}_4\right]^{2-}$ |
(atomic number : $\mathrm{Xe}=54, \mathrm{Ce}=58, \mathrm{Eu}=63$ )