Identify the homoleptic complexes with odd number of d electrons in the central metal :
(A) [FeO₄]²⁻
(B) [Fe(CN)₆]³⁻
(C) [Fe(CN)₅ NO]²⁻
(D) [CoCl₄]²⁻
(E) [Co(H₂ O)₃ F₃]
Choose the correct answer from the options given below :

Solution

Chemistry ma’am · 3 steps

Two filters, one after the other: first keep only one-ligand-type complexes, then count the d electrons.

1. Homoleptic first. A homoleptic complex has only one kind of ligand around the metal. (C) has both CN⁻ and NO, and (E) has both H₂O and F⁻, so both are heteroleptic and drop out.
2. Count d electrons. Find the oxidation state of the metal from the charge on the complex. Then remove electrons from Fe (3d⁶ 4s²) or Co (3d⁷ 4s²), taking the 4s electrons out first.

   Complex	Ligands	Metal oxidation state	d electrons
   (A) [FeO₄]²⁻	only O²⁻, homoleptic	x - 8 = -2, so +6	3d², even
   (B) [Fe(CN)₆]³⁻	only CN⁻, homoleptic	x - 6 = -3, so +3	3d⁵, odd
   (C) [Fe(CN)₅NO]²⁻	CN⁻ and NO⁺, heteroleptic	+2	3d⁶, even
   (D) [CoCl₄]²⁻	only Cl⁻, homoleptic	x - 4 = -2, so +2	3d⁷, odd
   (E) [Co(H₂O)₃F₃]	H₂O and F⁻, heteroleptic	x - 3 = 0, so +3	3d⁶, even

3. So. Only (B) with d⁵ and (D) with d⁷ are homoleptic and have an odd number of d electrons. The answer is (c).

The idea to keep. For d-electron questions, get the metal's oxidation state from the overall charge, then subtract from the neutral atom's count. Remember that 4s electrons leave before 3d.

Answer from the NTA final answer key. Solution by PrepWiser.