For octahedral Mn(II) and tetrahedral Ni(II) complexes, consider the following statements:
(I) both the complexes can be high spin.
(II) Ni(II) complex can very rarely be low spin.
(III) with strong field ligands, Mn(II) complexes can be low spin.
(IV)aqueous solution of Mn(II) ions is yellow in colour.
The correct statements are :

Solution

Chemistry ma’am · 3 steps

Compare splitting with pairing energy: octahedral can go either way, tetrahedral almost never pairs.

1. Electron counts. Mn²⁺ is 3d⁵ and Ni²⁺ is 3d⁸. A complex is high spin or low spin depending on whether the crystal field splitting Δ is smaller or larger than the pairing energy P.
2. Octahedral versus tetrahedral. In an octahedral field, Δₒ becomes large with strong field ligands such as CN⁻, so pairing can win. In a tetrahedral field, Δₜ ≈ 4/9Δₒ is almost always smaller than P, so tetrahedral complexes are nearly always high spin.

   Statement	Check	Verdict
   (I) both can be high spin	Weak field: octahedral [Mn(H₂O)₆]²⁺ has 5 unpaired electrons, and tetrahedral [NiCl₄]²⁻ has 2.	true
   (II) Ni(II) is very rarely low spin	Its geometry is tetrahedral, and Δₜ is too small to beat P.	true
   (III) strong field makes Mn(II) low spin	[Mn(CN)₆]⁴⁻ is t_(2g)⁵e_g⁰, with only 1 unpaired electron.	true
   (IV) aqueous Mn(II) is yellow	[Mn(H₂O)₆]²⁺ is very pale pink: every d–d transition of high-spin d⁵ must flip a spin, so absorption is weak.	false

3. So. Statements (I), (II) and (III) are correct and (IV) is wrong, which is option (d).

The idea to keep. Octahedral complexes go low spin only when Δₒ > P, which needs strong field ligands. Tetrahedral splitting is small, so treat tetrahedral complexes as high spin.

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