Question

Question: The element xenon (Xe) is by no means chemically inert; it forms a number of chemical compounds with electronegative elements such as fluorine and oxygen. The reaction of xenon with varying amounts of fluorine produces${\mathbf{\text{XeF}}}_{\mathbf{\text{2}}}$and ${\mathbf{\text{XeF}}}_4$. Subsequent reaction of one or the other of these compounds with water produces (depending on conditions) ${\mathbf{\text{XeO}}}_3$, ${\mathbf{\text{XeO}}}_4$, and ${\mathbf{\text{H}}}_{\mathbf{4}}{\mathbf{\text{XeO}}}_{\mathbf{6}}$, as well as mixed compounds such as ${\mathbf{\text{XeOF}}}_{\mathbf{4}}$. Predict the structures of these six xenon compounds, using the VSEPR theory.

Short answer

The structures of these six xenon compounds are;

Step-by-step solution

The VSEPR theory.

The VSEPR theory or the valence shell electron pair repulsion theory is helpful in predicting the 3D molecular geometry of different valence shell electron bond pairs present in each atom of a molecule or ion.

Xenon chemical reactivity.

Xenon is chemically inert (noble gas); still, it forms some chemical compounds with some electronegative (EN) elements like fluorine and oxygen.

The reaction of xenon with varying amounts of fluorine produces${\text{XeF}}_2$ and ${\text{XeF}}_4$as shown below.

$\text{Xe}+{\text{F}}_{\text{2}}\underset{Ni}{\overset{400°C}{\to }}{\text{XeF}}_{\text{2}}$

Here, Xenon is${\text{sp}}^{\text{3}}\text{d}$ hybridized.

$\text{Xe}+2{\text{F}}_{\text{2}}\underset{5-6\text{atm}}{\overset{400°C}{\to }}{\text{XeF}}_4$

Here, Xenon is${\text{sp}}^{\text{3}}{\text{d}}^2$ hybridized.

The structures of the given six xenon compounds.

Here, ${\text{XeF}}_2$has linear structure because of the presence of 2 bond pairs and 3 lone pairs of electrons.

Here,${\text{XeF}}_4$is square planar because of the presence of 2 lone pairs and 4 bond pairs of electrons.

The subsequent reaction of the above two compounds with water gives ${\text{XeO}}_{\text{3}}$, ${\text{XeO}}_4$, ${\text{H}}_4{\text{XeO}}_6$and ${\text{XeOF}}_4$, as shown below.

$6{\text{XeF}}_{\text{4}}+12{\text{H}}_{\text{2}}\text{O}\to {\text{4Xe + 2XeO}}_3+24\text{HF}+3{\text{O}}_{\text{2}}$

${\text{XeF}}_6+{\text{H}}_{\text{2}}\text{O}\to {\text{XeOF}}_4+2\text{HF}$