Question

The shape/structure of ${\left[{\mathrm{XeF}}_5\right]}^{–}$ and ${\mathrm{XeO}}_3{\mathrm{F}}_2$, respectively, are:

• A. Pentagonal planar and trigonal bipyramidal
• B. Trigonal bipyramidal and trigonal bipyramidal
• C. Octahedral and square pyramidal
• D. Trigonal bipyramidal and pentagonal planar

Answer 1

The correct option is A

Pentagonal planar and trigonal bipyramidal

The explanation for the correct option:

A: Pentagonal planar and trigonal bipyramidal

${\left[{\mathrm{XeF}}_5\right]}^{–}$

• In the structure ${\left[{\mathrm{XeF}}_5\right]}^{–}$, it is clear that there are $5$ bond pairs present in it and $2$ lone pairs, Therefore-
• $5\mathrm{Bond}\mathrm{pair}+2\mathrm{lone}\mathrm{pair}={\mathrm{sp}}^3{\mathrm{d}}^3\mathrm{hybridisation}$.
• And the geometry of ${\left[{\mathrm{XeF}}_5\right]}^{–}$ is pentagonal planar.

$\left[{\mathrm{XeO}}_3{\mathrm{F}}_2\right]$

• In the structure $\left[{\mathrm{XeO}}_3{\mathrm{F}}_2\right]$, it is clear that there are $5$ bond pairs present in it and $0$ lone pairs, Therefore-
• $5\mathrm{Bond}\mathrm{pair}+0\mathrm{lone}\mathrm{pair}={\mathrm{sp}}^3\mathrm{d}\mathrm{hybridisation}$
• $5\mathrm{BP}+0\mathrm{LP}=5\mathrm{VSEP}\Rightarrow {\mathrm{sp}}^3\mathrm{d}\mathrm{hybridisation}$
• And the geometry of $\left[{\mathrm{XeO}}_3{\mathrm{F}}_2\right]$ is trigonal bipyramidal.

Therefore, option A (pentagonal planar and trigonal bipyramidal) is the correct option.