1. Recall the shape corresponding to square pyramidal geometry

A molecule is square pyramidal when the electron-pair arrangement is octahedral with one position occupied by a lone pair, i.e. type:

$AX_5E$

This gives a molecular shape of square pyramidal.

2. Check each molecule

(i) $\mathrm{BrF}_5$

• Central atom: $\mathrm{Br}$
• Valence electrons on Br = 7
• It forms 5 bonds with F, leaving 1 lone pair.

So steric number = 6:

$AX_5E$

Hence, shape = square pyramidal.

(ii) $\mathrm{PCl}_5$

• Central atom: $\mathrm{P}$
• Valence electrons on P = 5
• It forms 5 bonds and has no lone pair.

So steric number = 5:

$AX_5$

Hence, shape = trigonal bipyramidal, not square pyramidal.

(iii) $\mathrm{SbF}_5$

• Central atom: $\mathrm{Sb}$
• Valence electrons on Sb = 5
• It forms 5 bonds and has no lone pair.

So:

$AX_5$

Hence, shape = trigonal bipyramidal, not square pyramidal.

(iv) $\mathrm{XeOF}_4$

• Central atom: Xe
• Xe has 8 valence electrons.
• Around Xe: 4 Xe–F bonds + 1 Xe=O bond + 1 lone pair.

In VSEPR, the double bond to O counts as one electron domain. So total electron domains = 6:

$AX_5E$

Hence, electron-pair geometry = octahedral and molecular shape = square pyramidal.

3. Evaluate options

• A: $\mathrm{BrF}_5$ and $\mathrm{PCl}_5$
  $\mathrm{BrF}_5$ is square pyramidal, but $\mathrm{PCl}_5$ is not. ❌

• B: $\mathrm{SbF}_5$ and $\mathrm{PCl}_5$
  Neither is square pyramidal. ❌

• C: $\mathrm{BrF}_5$ and $\mathrm{XeOF}_4$
  Both are square pyramidal. ✅

• D: $\mathrm{SbF}_5$ and $\mathrm{XeOF}_4$
  $\mathrm{SbF}_5$ is not square pyramidal. ❌

4. Final answer

The molecules having square pyramidal geometry are:

$\boxed{\mathrm{BrF}_5 \text{ and } \mathrm{XeOF}_4}$

So, the correct option is:

$\boxed{C}$