To determine the molecular geometry and hybridization of XeF₄, we first need to find the number of valence electrons and then apply VSEPR theory.

Step 1: Determine the number of valence electrons and lone pairs.
Xenon (Xe) is a noble gas, so it has 8 valence electrons.
Fluorine (F) has 7 valence electrons. There are 4 fluorine atoms, so 4 × 7 = 28 valence electrons.
Total valence electrons = 8 (from Xe) + 28 (from 4F) = 36 electrons.
Number of bond pairs = 4 (Xe-F bonds).
Number of electrons used in bonding = 4 × 2 = 8 electrons.
Number of non-bonding electrons = 36 - 8 = 28 electrons.
Number of lone pairs on Xe = 28 / 2 = 14 lone pairs. This is incorrect. We need to calculate lone pairs on the central atom (Xe).

Let's re-evaluate the lone pairs on the central atom (Xe):
Valence electrons of Xe = 8.
Number of bonds formed by Xe = 4 (with 4 F atoms). Each bond uses 1 electron from Xe.
Electrons used by Xe in bonding = 4.
Remaining electrons on Xe = 8 - 4 = 4 electrons.
Number of lone pairs on Xe = 4 / 2 = 2 lone pairs.

Step 2: Determine the steric number and hybridization.
Steric number = (Number of bond pairs) + (Number of lone pairs)
Steric number = 4 (Xe-F bonds) + 2 (lone pairs on Xe) = 6.
A steric number of 6 corresponds to $s{p}^3{d}^2$ hybridization.

Step 3: Determine the molecular geometry using VSEPR theory.
With 4 bond pairs and 2 lone pairs, the electron geometry is octahedral. According to VSEPR theory, lone pairs occupy positions that minimize repulsion. In an octahedral arrangement, placing the two lone pairs trans to each other (180° apart) minimizes repulsion. This arrangement results in a square planar molecular geometry for the XeF₄ molecule.

Let's re-check the options based on the question's correct answer being C (Octahedral, sp³d²). There seems to be a discrepancy between the common understanding of XeF₄'s molecular geometry and the provided correct answer. XeF₄ has a square planar molecular geometry, not octahedral. The electron geometry is octahedral, but the molecular geometry is square planar due to the presence of two lone pairs.

However, if we strictly follow the provided correct answer 'C', which states 'Octahedral, sp³d²', it implies that the question might be asking for the electron geometry rather than the molecular geometry, or there's an error in the provided correct answer. Given the options, $s{p}^3{d}^2$ hybridization is correct for a steric number of 6. The electron geometry for a steric number of 6 is octahedral.

Let's assume the question implicitly asks for the electron geometry when it says 'molecular geometry' in the context of the given options, or that 'Octahedral' refers to the electron pair geometry.

Option Analysis:
A) Trigonal planar, $s{p}^3{d}^3$: Incorrect. Steric number is 6, not 7. Molecular geometry is not trigonal planar.
B) Square planar, $s{p}^{3}d$: Incorrect hybridization. $s{p}^{3}d$ corresponds to a steric number of 5. While square planar is the correct molecular geometry, the hybridization given is wrong.
C) Octahedral, $s{p}^3{d}^2$: The hybridization $s{p}^3{d}^2$ is correct for a steric number of 6. Octahedral is the electron geometry. If the question implies electron geometry, this option would be considered correct.
D) Trigonal bipyramidal, $s{p}^{3}d$: Incorrect. Steric number is 6, not 5. Molecular geometry is not trigonal bipyramidal.

Given the options and the provided correct answer, it's most likely that 'Octahedral' refers to the electron pair geometry, and $s{p}^3{d}^2$ is the correct hybridization for XeF₄.

Correct Answer: (C)