To determine the geometry of the ions ${\mathrm{XeF}}_3^{+}$ and ${\mathrm{XeF}}_5^{+}$, we need to find the steric number (number of bond pairs + lone pairs) around the central Xe atom and apply VSEPR theory.

Step 1: Determine the geometry of ${\mathrm{XeF}}_3^{+}$

Xenon (Xe) is a noble gas with 8 valence electrons. In ${\mathrm{XeF}}_3^{+}$, one electron is removed, so Xe has 7 valence electrons. Three fluorine atoms form single bonds with Xe.

• Number of bond pairs = 3
• Number of lone pairs = (7 - 3) / 2 = 4 / 2 = 2
• Steric number = 3 (bond pairs) + 2 (lone pairs) = 5

For a steric number of 5, the electron geometry is trigonal bipyramidal. With 3 bond pairs and 2 lone pairs, the lone pairs occupy equatorial positions to minimize repulsion. This results in a bent T-shaped molecular geometry.

Step 2: Determine the geometry of ${\mathrm{XeF}}_5^{+}$

In ${\mathrm{XeF}}_5^{+}$, Xe has 7 valence electrons (8 - 1 charge). Five fluorine atoms form single bonds with Xe.

• Number of bond pairs = 5
• Number of lone pairs = (7 - 5) / 2 = 2 / 2 = 1
• Steric number = 5 (bond pairs) + 1 (lone pair) = 6

For a steric number of 6, the electron geometry is octahedral. With 5 bond pairs and 1 lone pair, the lone pair occupies one position, leading to a square pyramidal molecular geometry.

Step 3: Option Analysis

• A) Square pyramidal, T-shaped: Incorrect. ${\mathrm{XeF}}_3^{+}$ is bent T-shaped, not T-shaped (which implies 3 bond pairs, 2 lone pairs, but with 90 degree angles, not 180 degrees).
• B) Bent T-shaped, square pyramidal: Correct. This matches our derivations for ${\mathrm{XeF}}_3^{+}$ and ${\mathrm{XeF}}_5^{+}$.
• C) See-saw, square pyramidal: Incorrect. See-saw geometry corresponds to 4 bond pairs and 1 lone pair (steric number 5).
• D) Square pyramidal, see-saw: Incorrect.

Correct Answer: (B)