Problem 83 - Chemistry and Chemical Reactivity

Question

In which of the following species is Xe in the +4 oxidation state? (a) \(\mathrm{XeOF}_{4}\) (c) \(\mathrm{XeF}_{3}^{+}\) (b) \(\mathrm{Xe}_{2} \mathrm{F}_{3}^{+}\) (d) \(\mathrm{XeO}_{6}^{4-}\)

Step-by-Step Solution

Answer

(c) \( \mathrm{XeF}_{3}^{+} \) has xenon in the +4 oxidation state.

Step 1: Understanding Oxidation State

The oxidation state of an atom in a molecule or ion is the hypothetical charge that atom would have if all bonds to atoms of different elements were 100% ionic.

Step 2: Calculate Oxidation State in XeOF4

In the compound \( \mathrm{XeOF}_{4} \), assign oxygen an oxidation state of \(-2\) and fluorine an oxidation state of \(-1\). Let the oxidation state of xenon be \( x \). The equation becomes \( x + (-2) + 4(-1) = 0 \). Solve for \( x \) to find \( x = +6 \).

Step 3: Calculate Oxidation State in XeF3+

In the ion \( \mathrm{XeF}_{3}^{+} \), assign fluorine an oxidation state of \(-1\). Let the oxidation state of xenon be \( x \). The equation is \( x + 3(-1) = +1 \). Solving for \( x \) gives \( x = +4 \).

Step 4: Calculate Oxidation State in Xe2F3+

In the ion \( \mathrm{Xe}_{2} \mathrm{F}_{3}^{+} \), assign fluorine an oxidation state of \(-1\). Let the oxidation state of each xenon be \( x \). The equation becomes \( 2x + 3(-1) = +1 \). Solving for \( x \) gives \( x = +2 \).

Step 5: Calculate Oxidation State in XeO64-

In the ion \( \mathrm{XeO}_{6}^{4-} \), assign oxygen an oxidation state of \(-2\). Let the oxidation state of xenon be \( x \). The equation is \( x + 6(-2) = -4 \). Solving for \( x \) gives \( x = +8 \).

Step 6: Identify Correct Answer

Using the oxidation states calculated, xenon is in the \(+4\) oxidation state in \( \mathrm{XeF}_{3}^{+} \).

Key Concepts

Understanding Xenon Compounds

Xenon, a noble gas, rarely forms compounds due to its stable electronic configuration. However, in certain conditions, it can participate in chemical reactions and form compounds with oxygen and fluorine. These xenon compounds can have a variety of oxidation states, which makes them interesting in the study of chemistry. Xenon compounds like \( \mathrm{XeOF}_4 \) and \( \mathrm{XeF}_3^+ \) involve xenon bonded with other electronegative elements.

• \( \mathrm{XeOF}_4 \) is an example where xenon is bonded with both oxygen and fluorine. Here, xenon exhibits a higher oxidation state due to electronegative oxygen.
• \( \mathrm{XeF}_3^+ \) highlights xenon in a lower oxidation state as it is bonded only with fluorine.

Xenon compounds are vital in understanding not only noble gas chemistry but also oxidation state calculations. Observing xenon combining in varied oxidation states allows chemists to explore the limits of noble gas reactivity.

Oxidation State Rules Simplified

The oxidation state, also known as oxidation number, is a concept where atoms are assigned a hypothetical charge. By following specific rules, we can calculate these oxidation states, which help in understanding chemical behavior. Here are some simple oxidation state rules:

• Elements in their natural state have an oxidation state of zero. For example, \( \text{O}_2 \) or \( \text{H}_2 \).
• For a simple ion, the oxidation state is equal to its charge. For example, \( \text{Na}^+ \) has an oxidation state of +1.
• Oxygen usually has an oxidation state of -2, except in peroxides and with fluorine.
• Fluorine always has an oxidation state of -1 in compounds.
• The sum of the oxidation states in a compound or polyatomic ion must equal the overall charge.

By applying these rules, you can determine the oxidation state of xenon in various compounds. It helps to set up equations where the known oxidation states of elements assist in finding the unknown state. This method is crucial for predicting the reactivity and properties of the compounds.

Calculating Oxidation State in Molecular Ions

Molecular ions like \( \mathrm{XeF}_3^+ \) and \( \mathrm{Xe}_2 \mathrm{F}_3^+ \) present an interesting challenge for calculating oxidation states. To determine the oxidation state in these ions, follow these steps:

1. **Identify Known Oxidation States**: Use known oxidation states of elements, such as \(-1\) for fluorine, as guidance.2. **Set Up an Equation**: Use the formula that expresses the sum of all oxidation states equating to the ion's total charge.

• For \( \mathrm{XeF}_3^+ \), set up: \( x + 3(-1) = +1 \).
• For \( \mathrm{Xe}_2 \mathrm{F}_3^+ \), set: \( 2x + 3(-1) = +1 \).

3. **Solve for Unknowns**: Solve the equation to find the oxidation state of xenon \( x \).

• In \( \mathrm{XeF}_3^+ \), solving gives an oxidation state of +4 for xenon.
• In \( \mathrm{Xe}_2 \mathrm{F}_3^+ \), xenon has an oxidation state of +2.

These calculations allow us to unravel complex ionic structures and understand how different elements interact within molecular ions. Understanding these interactions is essential for further explorations in inorganic chemistry.