To calculate the Effective Atomic Number (EAN), use the following equation: \[ EAN = \text{Atomic number of the metal} - \text{Oxidation state} + 2 \times \text{Coordination number} \] Considering the complex \( [Co(NH_3)_6]^{3+} \):
- The atomic number of cobalt is 27.
- The oxidation state of cobalt in the complex is +3.
- The coordination number, representing the number of ligands bound to cobalt, is 6.
Applying these values to the formula: \[ EAN = 27 - 3 + 2 \times 6 = 36 \]
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,