Read the passage carefully and answer.

The crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. Ligands are treated as point charges in case of anions or dipoles in case of neutral molecules. The five d orbitals in an isolated gaseous metal atom/ion have same energy, i.e., they are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ion. However, when this negative field is due to ligands (either anions or the negative ends of dipolar molecules like $NH_3$ and $H_2O$) in a complex, it becomes asymmetrical and the degeneracy of the d orbitals is lifted. It results in splitting of the d orbitals. The pattern of splitting depends upon the nature of the crystal field.

Which of the following pair exists as high spin, tetrahedral complexs?

$[NiCl)_4]^{2-}$ and $[MnBг_4]^{2-}$

The correct answer is Option (3) → $[NiCl)_4]^{2-}$ and $[MnBг_4]^{2-}$

Step 1: Spot the geometry

• If ligand is strong field (like CN⁻, CO) → think square planar (especially d⁸ metals like Ni²⁺)

• Otherwise → assume tetrahedral

• Step 2: Remember this golden rule

 All tetrahedral complexes are HIGH SPIN
(no need to calculate splitting)

Step 3: Identify weak vs strong ligands

Strong field (eliminate for high spin tetrahedral):

• CN⁻, CO, NH₃ (moderate but often causes pairing)

 Weak field (keep these):

• Cl⁻, Br⁻, F⁻, I⁻, SCN⁻ (usually weak)

• Step 4: Special trap (VERY IMPORTANT)

 Ni²⁺ (d⁸) + strong ligand (like CN⁻)
= square planar (low spin)

 Apply to the question

• Remove anything with CN⁻ 

• Keep Cl⁻, Br⁻ 

• Both become tetrahedral → automatically high spin

 Final: [NiCl₄]²⁻ and [MnBr₄]²⁻

 Ultra-short memory trick:

“Weak ligand → tetrahedral → high spin (always)”