Question

Which of the following chemical compound is planar?

• A. ${\mathrm{XeF}}_2$
• B. ${\mathrm{XeO}}_3\mathrm{F}$
• C. ${\mathrm{XeO}}_2{\mathrm{F}}_2$
• D. ${\mathrm{XeF}}_4$

Answer 1

The correct option is D

${\mathrm{XeF}}_4$

Expalantion for correct option:

D. ${\mathrm{XeF}}_4$

• ${\mathrm{XeF}}_4$ is a square planar molecule having two nonbonding electron pairs, one is located above and one is located below the plane of the molecule.
• In xenon tetrafluoride, $\mathrm{Xe}$ there are a total of six electrons in the $5\mathrm{p}$ orbital and two electrons in the $5\mathrm{s}$ orbital.
• There are $36$ valence electrons in the compound ${\mathrm{XeF}}_4$.
• The structure of the compound is square planar.
• The molecule is ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridized.
• The molecule has $4$ bond pairs and $2$ lone pairs.
• The structure of ${\mathrm{XeF}}_4$is shown below:

Explanation for incorrect options:

A. ${\mathrm{XeF}}_2$

• ${\mathrm{XeF}}_2$ is a linear molecule having two bonding electron pairs and three non-bonding electron pairs. It has ${\mathrm{sp}}^3\mathrm{d}$ hybridization.
• The two fluorine atoms occupy the axial position of the TBP structure and the three non-bonded pairs occupy the equatorial positions.
• It is a non-planar molecule.

B. ${\mathrm{XeO}}_3\mathrm{F}$

• In ${\mathrm{XeO}}_3\mathrm{F}$, three oxygen atom forms double bonds with xenon while Fluorine forms a single bond.
• A molecule has an unbonded electron and a distorted octahedron structure. It is not planar.

C. ${\mathrm{XeO}}_2{\mathrm{F}}_2$

• In ${\mathrm{XeO}}_2{\mathrm{F}}_2$, two Oxygen atom forms double bonds with xenon while the two fluorine atom forms a single bond.
• The molecule has an unbonded electron pair and a see-saw structure. It is not planar.

Therefore, the correct option is D. ${\mathrm{XeF}}_4$.