The shape of $$XeOF_4$$ by VSEPR theory is:

To determine the shape of the molecule $$XeOF_4$$ using VSEPR theory, we first recall that VSEPR theory predicts molecular geometry based on the repulsion between electron pairs around the central atom. The central atom here is xenon (Xe).

We start by calculating the total number of valence electrons in $$XeOF_4$$. Xenon (Xe) is in group 18 and has 8 valence electrons. Oxygen (O) is in group 16 and has 6 valence electrons. Each fluorine (F) is in group 17 and has 7 valence electrons, and there are four fluorine atoms. So, total valence electrons = valence electrons from Xe + valence electrons from O + valence electrons from four F atoms = $$8 + 6 + (4 \times 7) = 8 + 6 + 28 = 42$$.

Next, we draw the Lewis structure. Xenon is bonded to one oxygen atom and four fluorine atoms, forming five bonds. To minimize formal charges and achieve stability, xenon has one lone pair. Each fluorine atom forms one single bond with xenon and has three lone pairs to complete its octet. Oxygen forms one single bond with xenon and has three lone pairs. Now, let's verify the electron count: the lone pair on xenon accounts for 2 electrons. The five bonds (one Xe-O and four Xe-F) each consist of 2 electrons, so they account for $$5 \times 2 = 10$$ electrons. Each fluorine has three lone pairs (6 electrons), so four fluorines account for $$4 \times 6 = 24$$ electrons. Oxygen has three lone pairs, accounting for 6 electrons. Total electrons = lone pair on Xe + bonding electrons + lone pairs on F + lone pairs on O = $$2 + 10 + 24 + 6 = 42$$, which matches the total valence electrons. Formal charges: xenon has a formal charge of +1 (calculated as valence electrons - lone pair electrons - half bonding electrons = $$8 - 2 - \frac{10}{2} = 8 - 2 - 5 = +1$$), oxygen has a formal charge of -1 ($$6 - 6 - \frac{2}{2} = 6 - 6 - 1 = -1$$), and each fluorine has formal charge 0 ($$7 - 6 - \frac{2}{2} = 7 - 6 - 1 = 0$$), summing to zero for the molecule.

In VSEPR theory, we consider electron domains (bonding pairs and lone pairs) around the central atom. Xenon has five bonding pairs (one to O and four to F) and one lone pair, giving a total of six electron domains. Six electron domains arrange themselves in an octahedral geometry to minimize repulsion, as this geometry maximizes the angles between domains.

The molecular geometry is determined by the positions of the atoms, ignoring lone pairs. In octahedral electron geometry, when one position is occupied by a lone pair, the five bonding pairs occupy the remaining positions, resulting in a square pyramidal molecular geometry. This is because the lone pair occupies one apex, and the five atoms form a square base with xenon slightly above the plane.

Comparing with the options: trigonal bipyramidal (option A) has five domains with no lone pairs, pentagonal planar (option C) is not standard for six domains, and octahedral (option D) would require all six positions occupied by atoms. Hence, the correct geometry is square pyramidal.

Hence, the correct answer is Option B.