JEE Chemistry 2025 Question with Solution

Given: We have to match the complexes in List–I with their hybridisations in List–II using valence bond theory.

Find: The correct correspondence among (I) $d^2sp^3$, (II) $sp^3$, (III) $sp^3d^2$, and (IV) $dsp^2$.

For (A) [CoF₆]³⁻:

• Oxidation state of Co is $+3$.
• F⁻ is a weak field ligand, so pairing does not occur.
• This forms an outer orbital octahedral complex. Therefore, hybridisation is $sp^3d^2$. So, (A) $\to$ (III).

For (B) [NiCl₄]²⁻:

• Oxidation state of Ni is $+2$.
• Cl⁻ is a weak field ligand, so pairing does not occur.
• The complex is tetrahedral. Therefore, hybridisation is $sp^3$. So, (B) $\to$ (II).

For (C) [Co(NH₃)₆]³⁺:

• Oxidation state of Co is $+3$.
• NH₃ acts as a strong ligand here, so pairing occurs.
• This gives an inner orbital octahedral complex. Therefore, hybridisation is $d^2sp^3$. So, (C) $\to$ (I).

For (D) [Ni(CN)₄]²⁻:

• Oxidation state of Ni is $+2$.
• CN⁻ is a strong field ligand, so pairing occurs.
• The complex becomes square planar. Therefore, hybridisation is $dsp^2$. So, (D) $\to$ (IV).

Thus the final matching is:

Therefore, the correct option is C.

Given: The complexes are [CoF₆]³⁻, [NiCl₄]²⁻, [Co(NH₃)₆]³⁺, and [Ni(CN)₄]²⁻.

Find: Match each complex with the correct hybridisation by examining ligand strength, electron pairing, and geometry.

Weak field ligands such as F⁻ and Cl⁻ generally do not cause pairing in the $d$ orbitals, so outer orbital complexes or tetrahedral arrangements are preferred. Strong field ligands such as NH₃ and CN⁻ can cause pairing, allowing inner orbital hybridisation.

Applying this:

1. [CoF₆]³⁻ is octahedral with weak ligand F⁻, so it is outer orbital: $sp^3d^2$.
2. [NiCl₄]²⁻ with weak ligand Cl⁻ is tetrahedral: $sp^3$.
3. [Co(NH₃)₆]³⁺ undergoes pairing and forms an inner orbital octahedral complex: $d^2sp^3$.
4. [Ni(CN)₄]²⁻ with strong ligand CN⁻ forms a square planar complex: $dsp^2$.

Hence the required match is (A)-(III), (B)-(II), (C)-(I), (D)-(IV), so the correct option is C.