Find the shape of $[XeF_5]^-$

We use VSEPR theory.

• Xenon has $8$ valence electrons.
• Because of the negative charge, add $1$ electron.
• Total electrons on central Xe to count electron domains: $8+1=9$

Xe forms $5$ bonds with $5$ fluorine atoms.

In VSEPR steric number calculation: $\text{Steric number} = \frac{V + M - C + A}{2}$ where for $[XeF_5]^-$,

• $V=8$ (Xe valence electrons)
• $M=5$ (monovalent atoms attached)
• $C=1$ (negative charge)

So, $\text{SN} = \frac{8+5+1}{2} = 7$

Thus, there are $7$ electron pairs around Xe:

• $5$ bond pairs
• $2$ lone pairs

Electron pair geometry for steric number $7$ is pentagonal bipyramidal.

In pentagonal bipyramidal arrangement, lone pairs prefer the two axial positions to minimize repulsion, leaving the five F atoms in one plane.

Therefore, the molecular shape is: $[XeF_5]^- : \text{pentagonal planar}$

Find the shape of $XeO_3F_2$

Count electron domains around Xe.

Xenon is bonded to:

• $3$ oxygen atoms by double bonds
• $2$ fluorine atoms by single bonds

In VSEPR, each multiple bond counts as one electron domain.

So total bonded domains on Xe: $3 + 2 = 5$

Now calculate steric number: $\text{SN} = \frac{V + M}{2}$ Here effectively Xe has five sigma-bond domains and no lone pair on Xe in this structure.

Hence Xe has:

• $5$ bond pairs
• $0$ lone pairs

Therefore, the geometry is: $XeO_3F_2 : \text{trigonal bipyramidal}$

More specifically, the more electronegative F atoms prefer axial/equatorial arrangement depending on repulsions, but overall shape remains trigonal bipyramidal.

Match with options

• $[XeF_5]^-$ : pentagonal planar
• $XeO_3F_2$ : trigonal bipyramidal

This matches: Option A

Comparison with stored correct answer

Stored correct answer = A

My derived answer = A

Therefore, they agree.