Q138P - Chemistry: Molecular Nature Of Matter And Change

Question

(a) Why do stable xenon fluorides have an even number of $\mathbf{\text{F}}$ atoms?

(b) Why do the ionic species  ${{\mathbf{\text{XeF}}}_{\mathbf{3}}}^{\mathbf{+}}$ and ${{\mathbf{\text{XeF}}}_{\mathbf{7}}}^{\mathbf{-}}$  have odd numbers of $\mathbf{\text{F}}$ atoms? 

(c) Predict the shape of ${{\mathbf{\text{XeF}}}_{\mathbf{3}}}^{\mathbf{+}}$.

Step-by-Step Solution

Answer

(a) To make a compound, xenon's   electrons can be promoted to the vacant $5d$ orbital. As a result, following promotion, only even electrons will be unpaired. Therefore, stable compounds must include an even number of $\text{F}$ atoms.

(b) Odd quantities of fluorine atoms are required to make an even number of paired electrons.

(c) The ${{\text{XeF}}_3}^{+}$  is  $\text{T}$-shaped, according to VSEPR theory.

A concept:

For large xenon, the attraction of electrons to the nucleus is weaker. It reacts with highly electronegative and little fluorine (and oxygen). Thus, the valence electron of $\mathbf{\text{Xe}}$ is attracted to fluorine (or oxygen). This helps in the formation of compounds.

(a) Why are there an even number of F atoms in stable xenon fluorides?

Xenon $\left(\text{Xe}\right)$ has an atomic number of   and an electron configuration of:

$\text{Xe}$ : $\left[\text{Kr}\right] 4d^{10}5s^{2}5p^6$

Xenon has an electron valence of eight. The $p$ electrons of xenon can be promoted to its unoccupied $5d$ orbital to produce a compound. As a result, only even electrons will be unpaired upon promotion. Therefore, stable compounds require an even number of $\text{F}$ atoms.

(b) Why are there odd numbers of F atoms in the ionic species role="math" localid="1663320193120" XeF 3 + and XeF 7 − ?

Xenon contains $7$ valence electrons in ${{\text{XeF}}_3}^{+}$. Xenon has $\text{9}$   valence electrons in ${{\text{XeF}}_7}^{-}$

Thus, electron promotion from p to d orbitals results in an odd number of unpaired electrons. Odd numbers of fluorine atoms are necessary to produce an even number of paired electrons.

As a result, xenon's ionic species requires an odd number of  $\text{F}$ atoms.

(c) Shape of XeF 3 + :

In ${{\text{XeF}}_3}^{+}$, there are $28$ valence electrons ($8$ electrons from xenon, $7$ electrons from each fluorine atom, and one electron subtracted to account for the positive charge from total valence electrons).

 

The following is the Lewis structure:

The core xenon atom in ${{\text{XeF}}_3}^{+}$ possesses two lone pairs of electrons linked to three fluorine atoms. As a result, there are five electron groups surrounding the core xenon atom. As a result, the class of  ${{\text{XeF}}_3}^{+}$ VSEPR is ${\text{AX}}_3{\text{E}}_2$. A stands for the center atom, $\text{X}$ for the bound atoms, and $\text{E}$ for the lone pairs.

A molecule of class ${\text{AX}}_3{\text{E}}_2$ is  $\text{T}$-shaped, according to VSEPR theory.

As a result, the ${{\text{XeF}}_3}^{+}$ is  $\text{T}$-shaped.