Match the LIST-I with LIST-II.

Choose the correct answer from the options given below :

The hybridisation of a central atom is obtained from its steric number (total number of $$\sigma$$-bonds + lone pairs).
For transition-metal complexes we also look at the observed geometry.

Case A: $$PF_5$$
Phosphorus forms five $$\sigma$$-bonds with fluorine and has no lone pair.
Hence steric number $$=5$$  $$\Rightarrow$$  hybridisation $$= sp^3d$$ (trigonal bipyramidal).
Therefore $$PF_5$$ matches List-II entry II.

Case B: $$SF_6$$
Sulphur forms six $$\sigma$$-bonds with fluorine and has no lone pair.
Steric number $$=6$$  $$\Rightarrow$$  hybridisation $$= sp^3d^2$$ (octahedral).
Therefore $$SF_6$$ matches List-II entry III.

Case C: $$Ni(CO)_4$$
Nickel in $$Ni(CO)_4$$ is in the zero oxidation state and the complex is tetrahedral.
A tetrahedral arrangement uses four equivalent hybrid orbitals: $$sp^3$$.
Therefore $$Ni(CO)_4$$ matches List-II entry IV.

Case D: $$[PtCl_4]^{2-}$$
The ion is square-planar; Pt(II) uses one $$d$$, one $$s$$ and two $$p$$ orbitals forming $$dsp^2$$ hybrids.
Therefore $$[PtCl_4]^{2-}$$ matches List-II entry I.

Collecting all results:
A → II,  B → III,  C → IV,  D → I.

The correct option is Option A.