Question

The d-electron configuration of \(\mathrm{[Ru(en)_3]Cl_2}\) and \(\mathrm{[Fe(H_2O)_6]Cl_2}\), respectively are :

• A. \(t_{2g}^4 e_g^2\) and \(t_{2g}^6 e_g^0\)
• B. \(t_{2g}^6 e_g^0\) and \(t_{2g}^6 e_g^0\)
• C. \(t_{2g}^6 e_g^0\) and \(t_{2g}^4 e_g^2\)
• D. \(t_{2g}^4 e_g^2\) and \(t_{2g}^4 e_g^2\)

Hint:

Both metals are d⁶ (+2). Ru (4d series) with the strong-field ligand en is low spin; Fe²⁺ with weak-field H₂O is high spin.

Answer 1

The Correct Option is C

Oxidation state is +2 in both complexes.
Ru (Z = 44): [Kr] 4d⁷5s¹, so Ru²⁺ is d⁶. Fe (Z = 26): 3d⁶4s², so Fe²⁺ is d⁶.
[Ru(en)₃]²⁺: en is a strong-field ligand and 4d metals have a large \(\Delta_o\) (greater than the pairing energy P), so all six electrons pair up in the lower set: \(t_{2g}^6 e_g^0\) (low spin, diamagnetic).
[Fe(H₂O)₆]²⁺: H₂O is a weak-field ligand with a 3d metal, \(\Delta_o \lt P\), so electrons occupy e_g before pairing: \(t_{2g}^4 e_g^2\) (high spin, 4 unpaired electrons).
Answer: (3) \(t_{2g}^6 e_g^0\) and \(t_{2g}^4 e_g^2\)