Trigonal bipyramidal geometry is shown by:

To determine which compound exhibits trigonal bipyramidal geometry, we need to analyze the molecular geometry around the central xenon (Xe) atom using Valence Shell Electron Pair Repulsion (VSEPR) theory. Trigonal bipyramidal geometry occurs when the central atom has a steric number (SN) of 5 with five bonding domains and no lone pairs. The steric number is calculated as SN = (number of atoms bonded to central atom) + (number of lone pairs on central atom). We will evaluate each option step by step.

Option A: $$XeOF_2$$

First, calculate the total valence electrons: Xe has 8, O has 6, and two F atoms have $$7 \times 2 = 14$$, so total is $$8 + 6 + 14 = 28$$ electrons. In the Lewis structure, Xe is bonded to O via a double bond and to each F via a single bond. A double bond uses 4 electrons, and two single bonds use $$2 \times 2 = 4$$ electrons, so bonding electrons total $$4 + 4 = 8$$. Non-bonding electrons are $$28 - 8 = 20$$. Each F atom in a single bond has three lone pairs (6 electrons), so for two F atoms: $$2 \times 6 = 12$$ electrons. The O atom in a double bond has two lone pairs (4 electrons). So far, $$12 + 4 = 16$$ electrons are accounted for. The remaining electrons are $$20 - 16 = 4$$, which form two lone pairs on Xe. Thus, Xe has three bonding domains (one from the double bond to O and two from single bonds to F) and two lone pairs. Steric number SN = $$3 + 2 = 5$$. The electron geometry is trigonal bipyramidal, but with two lone pairs occupying equatorial positions, the molecular geometry is T-shaped, not trigonal bipyramidal.

Option B: $$XeO_3F_2$$

Total valence electrons: Xe has 8, three O atoms have $$6 \times 3 = 18$$, two F atoms have $$7 \times 2 = 14$$, so total is $$8 + 18 + 14 = 40$$ electrons. Xe is bonded to three O atoms via double bonds and to two F atoms via single bonds. Three double bonds use $$3 \times 4 = 12$$ electrons, and two single bonds use $$2 \times 2 = 4$$ electrons, so bonding electrons total $$12 + 4 = 16$$. Non-bonding electrons are $$40 - 16 = 24$$. Each O atom in a double bond has two lone pairs (4 electrons), so for three O atoms: $$3 \times 4 = 12$$ electrons. Each F atom in a single bond has three lone pairs (6 electrons), so for two F atoms: $$2 \times 6 = 12$$ electrons. Total non-bonding electrons accounted for: $$12 + 12 = 24$$, so no lone pairs on Xe. Thus, Xe has five bonding domains (three from double bonds to O and two from single bonds to F). Steric number SN = $$5 + 0 = 5$$. With five bonding domains and no lone pairs, the geometry is trigonal bipyramidal.

Option C: $$FXeOSO_2F$$

This compound is $$F-Xe-O-SO_2F$$, where the SO₂F group is sulfonyl fluoride. Total valence electrons: Xe has 8, F (attached to Xe) has 7, O (attached to Xe) has 6, S has 6, two O atoms in SO₂ have $$6 \times 2 = 12$$, F (attached to S) has 7, so total is $$8 + 7 + 6 + 6 + 12 + 7 = 46$$ electrons. Bonds: Xe-F single bond (2 electrons), Xe-O single bond (2 electrons), O-S single bond (2 electrons), S bonded to two O atoms via double bonds ($$2 \times 4 = 8$$ electrons), and S-F single bond (2 electrons). Total bonding electrons: $$2 + 2 + 2 + 8 + 2 = 16$$. Non-bonding electrons: $$46 - 16 = 30$$. Each F atom (both attached to Xe and S) in a single bond has three lone pairs (6 electrons), so $$2 \times 6 = 12$$ electrons. The O atom attached to Xe and S has two lone pairs (4 electrons). Each O atom in S=O has two lone pairs (4 electrons), so for two O atoms: $$2 \times 4 = 8$$ electrons. Total non-bonding electrons accounted for: $$12 + 4 + 8 = 24$$. The remaining electrons are $$30 - 24 = 6$$, which form three lone pairs on S (since S has four bonds and can expand its octet). For Xe, it is bonded to two atoms (F and O), so two bonding domains. Xenon has 8 valence electrons; it uses 2 electrons for the two bonds (one per bond), leaving 6 electrons, which form three lone pairs. Thus, Xe has two bonding domains and three lone pairs. Steric number SN = $$2 + 3 = 5$$. The electron geometry is trigonal bipyramidal, but with three lone pairs occupying equatorial positions, the molecular geometry is linear around Xe, not trigonal bipyramidal.

Option D: $$[XeF_5]^{2-}$$

Total valence electrons: Xe has 8, five F atoms have $$7 \times 5 = 35$$, and the 2- charge adds 2 electrons, so total is $$8 + 35 + 2 = 45$$ electrons. Xe is bonded to five F atoms via single bonds. Five single bonds use $$5 \times 2 = 10$$ electrons. Non-bonding electrons: $$45 - 10 = 35$$. Each F atom in a single bond has three lone pairs (6 electrons), so for five F atoms: $$5 \times 6 = 30$$ electrons. Remaining electrons: $$35 - 30 = 5$$. These 5 electrons are placed on Xe, forming two lone pairs (4 electrons) and one unpaired electron. However, an unpaired electron is not a domain in VSEPR; the two lone pairs count as two domains. Thus, Xe has five bonding domains and two lone pair domains. Steric number SN = $$5 + 2 = 7$$. The electron geometry is pentagonal bipyramidal, not trigonal bipyramidal.

Only option B, $$XeO_3F_2$$, has a steric number of 5 with five bonding domains and no lone pairs, resulting in trigonal bipyramidal geometry. Hence, the correct answer is Option B.