1. Check which molecules involve $sp^3d$ hybridization

For $sp^3d$ hybridization, the central atom must have 5 electron domains.

• A: $\mathrm{XeF_4}$
  Central Xe has 4 bond pairs + 2 lone pairs $= 6$ electron domains
  Hybridization $= sp^3d^2$
  So, not $sp^3d$.

• B: $\mathrm{XeF_2}$
  Central Xe has 2 bond pairs + 3 lone pairs $= 5$ electron domains
  Hybridization $= sp^3d$
  So, this satisfies condition (a).

• C: $\mathrm{PF_5}$
  Central P has 5 bond pairs + 0 lone pairs $= 5$ electron domains
  Hybridization $= sp^3d$
  So, this satisfies condition (a).

• D: $\mathrm{SF_4}$
  Central S has 4 bond pairs + 1 lone pair $= 5$ electron domains
  Hybridization $= sp^3d$
  So, this satisfies condition (a).

2. Check for different bond lengths

In a trigonal bipyramidal electron geometry, axial and equatorial bonds are different.

• $\mathrm{XeF_2}$: linear molecule; both Xe–F bonds are equivalent, so bond lengths are same.
• $\mathrm{PF_5}$: trigonal bipyramidal; axial and equatorial P–F bonds are different, so bond lengths differ.
• $\mathrm{SF_4}$: seesaw shape; axial and equatorial S–F bonds are different, so bond lengths differ.

Thus, condition (b) is satisfied by $\mathrm{PF_5}$ and $\mathrm{SF_4}$.

3. Check for lone pair on the central atom

• $\mathrm{XeF_2}$: yes, 3 lone pairs on Xe.
• $\mathrm{PF_5}$: no lone pair on P.
• $\mathrm{SF_4}$: yes, 1 lone pair on S.

Thus, condition (c) is satisfied by $\mathrm{XeF_2}$ and $\mathrm{SF_4}$.

4. Find the molecule satisfying all three conditions simultaneously

We need:

• $sp^3d$ hybridization,
• different bond lengths,
• lone pair on central atom.

Checking each:

• $\mathrm{XeF_4}$: fails (a)
• $\mathrm{XeF_2}$: fails (b)
• $\mathrm{PF_5}$: fails (c)
• $\mathrm{SF_4}$: satisfies all (a), (b), and (c)

Therefore, the correct answer is:

$\boxed{\mathrm{SF_4}}$