JEE Chemistry 2025 Question with Solution

Given: Match the species in LIST-I with their hybridisation in LIST-II.

Find: The correct correspondence among the four entries.

Determine the hybridisation of the central atom in each molecule/ion by counting the number of sigma bonds and lone pairs.

For $PF_5$:

So A $\to$ II.

For $SF_6$:

So B $\to$ III.

For $Ni(CO)_4$:

In presence of ligand field, $Ni(0)$ uses $sp^3$ hybridisation. So C $\to$ IV.

For $[PtCl_4]^{2-}$:

In presence of ligand field, $Pt^{2+}$ uses $dsp^2$ hybridisation. So D $\to$ I.

Therefore, the correct match is A-II, B-III, C-IV, D-I. Hence, the correct option is A.

Given:

• A = $PF_5$
• B = $SF_6$
• C = $Ni(CO)_4$
• D = $[PtCl_4]^{2-}$

Find: The correct hybridisation match.

1. $PF_5$ has trigonal bipyramidal geometry. The central phosphorus is bonded to five fluorine atoms, so its hybridisation is $sp^3d$, which corresponds to II.
2. $SF_6$ has octahedral geometry. The central sulfur is bonded to six fluorine atoms, so its hybridisation is $sp^3d^2$, which corresponds to III.
3. $Ni(CO)_4$ is tetrahedral. Nickel forms bonds with four CO ligands, so the hybridisation is $sp^3$, which corresponds to IV.
4. $[PtCl_4]^{2-}$ is square planar. Platinum forms four bonds in a square planar arrangement, so the hybridisation is $dsp^2$, which corresponds to I.

Thus the final matching is:

• A $\to$ II
• B $\to$ III
• C $\to$ IV
• D $\to$ I

Therefore, the correct option is A.