[Co(NH$_3$)$_6$]$^{3+}$:} In this complex, cobalt is in the $+3$ oxidation state with an electronic configuration of [Ar]3$d^6$. Since ammonia (NH$_3$) is a weak field ligand, it causes a small splitting of d-orbitals in the octahedral field, leading to a spin paired (low-spin) configuration. Therefore, [Co(NH$_3$)$_6$]$^{3+}$ is a spin paired complex.
[CoF$_6$]$^{3-}$: In this complex, cobalt is also in the $+3$ oxidation state with an electronic configuration of [Ar]3$d^6$. Fluoride (F$^-$) is a weak field ligand, causing less splitting of d-orbitals in the octahedral field, resulting in a spin free (high-spin) complex configuration. Therefore, [CoF$_6$]$^{3-}$ is a spin free complex.
Conclusion: [Co(NH$_3$)$_6$]$^{3+}$ is a spin paired complex and [CoF$_6$]$^{3-}$ is a spin free complex.
Assertion (A): [Cr(H_2O)_6]Cl_2 and [Fe(H_2O)_6]Cl_2 are examples of homoleptic complexes. Reason (R): All the ligands attached to the metal are the same.
Low spin tetrahedral complexes are not known.
Co2+ is easily oxidized to Co3+ in the presence of a strong ligand (At. No. of Co = 27).
Why is a solution of \(\text{Ni(H}_2\text{O})_6^{2+}\) green while a solution of \(\text{Ni(CN)}_4^{2-}\) is colourless? (At. No. of Ni = 28)
The color of a coordination complex depends on the d–d transitions in the visible region of the spectrum, which are influenced by the ligand field strength.
- In \(\text{Ni(H}_2\text{O})_6^{2+}\), Ni\(^{2+}\) has an electronic configuration of \([Ar]3d^8\). Water (\(\text{H}_2\text{O}\)) is a weak field ligand, causing a small crystal field splitting (\(\Delta\)), allowing d–d transitions in the visible region, which results in a green color.
- In \(\text{Ni(CN)}_4^{2-}\), cyanide (\(\text{CN}^-\)) is a strong field ligand,