Number of complexes from the following with even number of unpaired "d" electrons is ________ [V(H₂ O)₆]³⁺,[Cr(H₂ O)₆]²⁺,[Fe(H₂ O)₆]³⁺,[Ni(H₂ O)₆]³⁺,[Cu(H₂ O)₆]²⁺ [Given atomic numbers: V=23, Cr=24, Fe=26, Ni=28 Cu=29]

Solution

Chemistry ma’am · 3 steps

Find each metal's d count, fill the octahedral levels high-spin since water is weak, then count.

1. Oxidation state first. Remove the 4s electrons first, then 3d. This gives V³⁺ d², Cr²⁺ d⁴, Fe³⁺ d⁵, Ni³⁺ d⁷ and Cu²⁺ d⁹.
2. Water is weak field. H₂O gives a small splitting Δₒ, so electrons spread out into e_g before pairing in t_(2g). All five complexes are high spin.

   Complex	d count	Filling	Unpaired	Even?
   [V(H₂O)₆]³⁺	d²	t_(2g)²	2	yes
   [Cr(H₂O)₆]²⁺	d⁴	t_(2g)³ e_g¹	4	yes
   [Fe(H₂O)₆]³⁺	d⁵	t_(2g)³ e_g²	5	no
   [Ni(H₂O)₆]³⁺	d⁷	t_(2g)⁵ e_g²	3	no
   [Cu(H₂O)₆]²⁺	d⁹	t_(2g)⁶ e_g³	1	no

3. So. Only the vanadium and chromium complexes have an even number of unpaired d electrons, so the count is 2, option (c).

The idea to keep. For octahedral H₂O complexes, think high spin: unpaired electrons are n for d¹ to d⁵ and 10 - n for d⁶ to d¹⁰.

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