Which among the following molecules is (a) involved in $$sp^3d$$ hybridization, (b) has different bond lengths and (c) has lone pair of electrons on the central atom ?

We have to identify the molecule that simultaneously satisfies three conditions:
(a) central atom undergoes $$sp^3d$$ hybridization,
(b) the molecule possesses more than one type of bond length, and
(c) the central atom contains at least one lone pair of electrons.

• Central atom $$P$$: valence electrons = $$5$$, all used for $$5$$ $$P-F$$ bonds.
• Hybridization: $$sp^3d$$ (trigonal-bipyramidal).
• Lone pairs on $$P$$: $$0$$, so condition (c) is not met.
Hence $$PF_5$$ is rejected.

• Central atom $$Xe$$: valence electrons = $$8$$, uses $$4$$ for bonds and retains $$2$$ lone pairs.
• Hybridization: $$sp^3d^2$$ (octahedral arrangement, square-planar shape).
• Geometry is symmetric; all $$Xe-F$$ bonds are equivalent in length.
Condition (a) fails (hybridization not $$sp^3d$$), so $$XeF_4$$ is rejected.

• Central atom $$S$$: valence electrons = $$6$$.
  - $$4$$ electrons form $$4$$ $$S-F$$ sigma bonds.
  - $$2$$ electrons remain as one lone pair.
• Steric number $$= 5$$ ⇒ hybridization $$sp^3d$$ leading to a trigonal-bipyramidal electron geometry.
• Molecular shape: seesaw (because one equatorial position is occupied by the lone pair).
• In a trigonal-bipyramidal framework, axial bonds (aligned vertically) are longer than equatorial bonds (lying in the trigonal plane) due to greater repulsion along the axial-equatorial angle $$90^{\circ}$$.
Therefore $$SF_4$$ possesses two distinct $$S-F$$ bond lengths.
Conditions (a), (b) and (c) are all satisfied.

• Central atom $$Xe$$: valence electrons = $$8$$.
  - $$2$$ electrons pairs form $$2$$ $$Xe-F$$ bonds.
  - $$3$$ lone pairs remain.
• Steric number $$= 5$$ ⇒ hybridization $$sp^3d$$.
• All three lone pairs occupy equatorial positions, giving a linear molecule F-Xe-F.
• Both $$Xe-F$$ bonds occupy identical axial positions; thus the two bond lengths are equal.
Condition (b) fails, so $$XeF_2$$ is rejected.

Only $$SF_4$$ fulfils all three requirements.

Hence the correct option is Option C.