Q137P - Chemistry: Molecular Nature Of Matter And Change

Question

Explain why $\mathbf{\text{Xe}}$, and to a limited extent $\mathbf{Kr}$, form compounds, whereas $\mathbf{\text{He}}$, $\mathbf{Ne}$, and $\mathbf{Ar}$ do not.

Step-by-Step Solution

Answer

Because the initial ionization energy of oxygen is extremely close to that of xenon; as a result, xenon may create a variety of compounds depending on their oxidation state. The $d$ orbitals of   are empty. As a result, $\text{Kr}$ only forms a limited number of compounds.

However, the initial ionization energy of $\text{He}$, $\mathrm{Ne}$, and $\mathrm{Ar}$ are all higher than that of oxygen. As a result, all   and   electrons cannot be promoted to the unoccupied   orbitals

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.

Xenon ( Xe ) forms several compound:

The powerful oxidizing agent ${\text{PtF}}_6$ may oxidize xenon to create ${\text{XePtF}}_6$, an orange-yellow solid, since the ionization energy of oxygen $\left(1175 \raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\right)$ is extremely near to the initial ionization energy of xenon $\left(1170 \raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\right)$. Other compounds such as ${\text{XeF}}_2{\text{, XeF}}_4{\text{, XeO}}_3\text{,}$ and  ${\text{XeO}}_4$ are also created. As a result, xenon generates compounds with oxidation states ranging from $+2$ to $+8$.

The size of the xenon is the same as that of oxygen. With fluorine and oxygen, xenon produces a variety of compounds.

Helium's initial ionization energies, Neon's $\left(2380 \raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\right)$, and Argon's $\left(1520 \raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\right)$ are all higher than oxygen's first ionization energy.

It is possible to promote all $s$ and $p$ electrons to the vacant $d$ orbitals. However, in the case of $\text{He}$ and  $\mathrm{Ne}$, the $d$ orbital is absent.

Krypton ( Kr ) forms limited compound:

The initial ionization energy of $\text{Kr}$ is $1351 \raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$ which is possesses empty $d$ orbitals.

As a result, $\text{Kr}$ only forms a limited number of compounds.