Number of Complexes with even number of electrons in t_(2 g) orbitals is -
[Fe(H₂ O)₆]²⁺,[Co(H₂ O)₆]²⁺,[Co(H₂ O)₆]³⁺,[Cu(H₂ O)₆]²⁺,[Cr(H₂ O)₆]²⁺

Solution

Chemistry ma’am · 4 steps

Get the d count of each ion, fill the octahedral levels, then check only the t_(2g) number.

1. Find the d count. Work out the oxidation state of each metal and its d-electron count. Then fill the octahedral levels: t_(2g) is lower in energy, e_g is higher.
2. Weak-field ligand. H₂O is a weak-field ligand, so Δₒ is small and most of these ions are high spin. Electrons enter e_g singly before they pair up in t_(2g).

   Complex	d count	Configuration	t_(2g) electrons
   [Fe(H₂O)₆]²⁺	Fe²⁺, d⁶	t_(2g)⁴ e_g²	4, even
   [Co(H₂O)₆]²⁺	Co²⁺, d⁷	t_(2g)⁵ e_g²	5, odd
   [Co(H₂O)₆]³⁺	Co³⁺, d⁶	t_(2g)⁶ e_g⁰ (low spin)	6, even
   [Cu(H₂O)₆]²⁺	Cu²⁺, d⁹	t_(2g)⁶ e_g³	6, even
   [Cr(H₂O)₆]²⁺	Cr²⁺, d⁴	t_(2g)³ e_g¹	3, odd

3. Two careful rows. Co³⁺ has a high charge, so even water pairs its electrons. Either way its t_(2g) count is even: 6 if low spin, 4 if high spin. For Cu²⁺ (d⁹), t_(2g) is full with 6, even though the total d count is odd.
4. So. [Fe(H₂O)₆]²⁺, [Co(H₂O)₆]³⁺ and [Cu(H₂O)₆]²⁺ have an even number of t_(2g) electrons, so the answer is 3, option (a).

The idea to keep. Fill t_(2g) and e_g properly before counting. For d⁴ to d⁷, high or low spin changes the t_(2g) count, so decide the ligand field strength first.

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