Question

Write a chemical equation for the hydrolysis of \(\mathrm{XeF}_{4}\) that yields \(\mathrm{XeO}_{3}, \mathrm{Xe}, \mathrm{O}_{2},\) and \(\mathrm{HF}\) as products.

Short answer

The balanced chemical equation for the hydrolysis of \(XeF_4\) that yields \(XeO_3\), \(Xe\), \(O_2\), and \(HF\) is \n\n \(2XeF_4 + 3H_2O \rightarrow XeO_3 + Xe + O_2 + 6HF\)

Step-by-step solution

Write Down the Unbalanced Equation

The unbalanced chemical equation for the hydrolysis of \(XeF_4\) should be written as follows:\n\n\(XeF_4 + H_2O \rightarrow XeO_3 + Xe + O_2 + HF\)

Balance the Oxygen Atoms

Begin balancing the equation starting from the Oxygen atoms on both sides. As there are three Oxygen atoms in \(XeO_3\) and two in \(O_2\), add three water molecules to balance the reactant side. The equation thus becomes:\n\n\(XeF_4 + 3H_2O \rightarrow XeO_3 + Xe + O_2 + HF\)

Balance the Hydrogen and Fluorine Atoms

Six Hydrogen atoms in the reactant side must be balanced by adding six molecules of \(HF\) at the product side. This also balances out the Fluorine atoms. The equation then becomes:\n\n\(XeF_4 + 3H_2O \rightarrow XeO_3 + Xe + O_2 + 6HF\)

Balance the Xenon Atoms

Finally balance the Xenon atoms. As there are two Xenon atoms on the product side (where one is from \(XeO_3\) and the other is from \(Xe\)), add one more molecule of \(XeF_4\) to make two Xenon atoms on the reactant side. This leads to the final balanced equation:\n\n\(2XeF_4 + 3H_2O \rightarrow XeO_3 + Xe + O_2 + 6HF\)

Key concepts

Xenon Tetrafluoride

Xenon tetrafluoride, represented chemically as \( \text{XeF}_4 \), is a noble gas compound. It consists of a xenon atom bonded to four fluorine atoms. Xenon sulfates like \( \text{XeF}_4 \) are unique because most elements in the noble gas group are naturally non-reactive. However, xenon can form stable compounds under specific conditions. This is due to xenon's ability to expand its octet, a feature not common to many elements.
\( \text{XeF}_4 \) is synthesized under high-temperature conditions. Because of its stability and reactivity, it serves as a vital reagent in high-energy chemistry. This compound is used mainly for research purposes and does not have extensive practical applications, making its understanding more relevant to students and researchers rather than industrial uses.
When reacting with water, as seen in hydrolysis reactions, \( \text{XeF}_4 \) undergoes a transformation involving its component elements – xenon shifts from bonding with fluorine to forming other compounds such as \( \text{XeO}_3 \), highlighting its ability to engage in complex chemical reactions.

Hydrolysis Reaction

Hydrolysis is a reaction between a molecule and water, often resulting in the breaking of chemical bonds. In the case of \( \text{XeF}_4 \), it undergoes hydrolysis to form a mix of products, which include \( \text{XeO}_3 \), \( \text{Xe} \), \( \text{O}_2 \), and \( \text{HF} \). This specific reaction showcases how \( \text{XeF}_4 \) interacts with water and rearranges to produce various compounds.
During hydrolysis, water molecules split into hydrogen and hydroxyl ions. These ions then participate in breaking and making of new bonds with \( \text{XeF}_4 \). This process can lead to the creation of oxyhalides like \( \text{XeO}_3 \), where water contributes oxygen atoms, and hydrofluoric acid (HF), where water transfers hydrogen atoms to fluorine.
Reactions like these are important to understand because they demonstrate the principles of bond breaking and forming, which are central concepts in chemistry. Hydrolysis reactions are also crucial in many biological processes, indicating their broad applicability beyond simple chemical equations.

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry, ensuring that the law of conservation of mass is respected—all atoms present in the reactants must be accounted for in the products. In the hydrolysis of \( \text{XeF}_4 \), this involves making sure that each type of atom present balances out on both sides of the equation.
To balance the hydrolysis equation of \( \text{XeF}_4 \), we start by identifying the number of each type of atom present in the unbalanced equation. Then follow the steps to equalize them:

• First, balance oxygen atoms, which often involve adding water molecules to either side.
• Next, adjust hydrogen atoms by considering water's contribution and balancing it through molecules like HF.
• Lastly, check the xenon and fluorine atoms, ensuring all are balanced without leaving any sides short of an atom count.

Balancing equations is like solving a puzzle. It's about finding the right coefficients to ensure each element is present equally in reactants and products, thus maintaining the integrity of the law of conservation of mass. This skill not only fosters analytical thinking but is crucial for all chemical calculations.