The shape / structure of [XeF₅] – and XeO₃F₂, respectively, are

Solution

Chemistry ma’am · 4 steps

Count sigma bonds and lone pairs on Xe for each species; the lone pairs decide the final shape.

1. Electron pairs on Xe. Xenon brings 8 valence electrons. Each F takes one Xe electron for a single bond. Each O takes two Xe electrons for a double bond. A negative charge adds one electron. The electrons left over pair up as lone pairs.
2. Fluoride ion. In [XeF₅]⁻, Xe has 8 + 1 = 9 electrons. Five go into the five Xe–F bonds, and the other 4 make 2 lone pairs. That gives 5 + 2 = 7 electron pairs, so the geometry is pentagonal bipyramidal (sp³d³). The two lone pairs sit opposite each other in the axial positions, which leaves the five F atoms in one plane: pentagonal planar.
3. Oxyfluoride. In XeO₃F₂, the three Xe=O bonds use 6 of Xe's electrons and the two Xe–F bonds use the other 2. No lone pairs are left, so Xe has 5 sigma bonds (sp³d). The shape is trigonal bipyramidal.

   Species	Sigma bonds	Lone pairs on Xe	Shape
   [XeF₅]⁻	5	2 (axial)	pentagonal planar
   XeO₃F₂	5	0	trigonal bipyramidal

4. So. The pair, in order, is pentagonal planar and trigonal bipyramidal. That is option (a).

The idea to keep. Find the steric number first: sigma bonds plus lone pairs. A double-bonded O uses two Xe electrons but counts as only one sigma bond. Then let the lone pairs take the positions that keep them farthest apart.

Answer from the NTA final answer key. Solution by PrepWiser.