Chapter 8: Problem 10
The oxidation state of \(\mathrm{Xe}\) in \(\mathrm{Na}_{4} \mathrm{XeO}_{6}\) is (1) \(+2\) (2) \(+4\) (3) \(-8\) (4) \(+8\)
Short Answer
Expert verified
The oxidation state of Xe is +8.
Step by step solution
01
- Identify known oxidation states
In the compound \(\text{Na}_{4} \text{XeO}_{6}\), recognize that sodium (Na) typically has an oxidation state of \(+1\) and oxygen (O) typically has an oxidation state of \(-2\).
02
- Set up the oxidation state equation
The sum of the oxidation states for all elements in a neutral molecule is zero. Thus, set up the equation: \[4(\text{oxidation state of Na}) + (\text{oxidation state of Xe}) + 6(\text{oxidation state of O}) = 0 \].
03
- Plug in known oxidation states
Substitute the known oxidation states into the equation: \[4(+1) + \text{oxidation state of Xe} + 6(-2) = 0 \].
04
- Simplify the equation
Simplify the equation to solve for the oxidation state of xenon (Xe): \[4 + \text{oxidation state of Xe} - 12 = 0\] which further simplifies to \[\text{oxidation state of Xe} = 8 \].
05
- Verify the calculations
Confirm the calculation: \[4(+1) + (+8) + 6(-2) = 4 + 8 - 12 = 0 \], which holds true, confirming that the oxidation state of xenon (Xe) is indeed \(+8\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
oxidation state rules
Understanding oxidation states is essential in chemistry for tracking how electrons are transferred between atoms in compounds. Here are some fundamental rules:
• The oxidation state of a free element (not combined with others) is always 0. For example, \(\text{O}_2\) or \(H_2\).
• For monoatomic ions, the oxidation state is equal to the ion's charge. For example, \(\text{Na}^+\) has an oxidation state of \(+1\).
• Oxygen typically has an oxidation state of \(-2\) in most compounds, though there are exceptions like in peroxides where it is \(-1\).
• Hydrogen generally has an oxidation state of \(+1\) when bonded with non-metals and \(-1\) with metals.
• The sum of oxidation states in a neutral molecule must be zero, and in an ion, it must equal the ion's charge.
These rules help you set up and solve equations to find unknown oxidation states, as demonstrated in the problem where we identified the oxidation state of xenon (Xe) in \( \text{Na}_4 \text{XeO}_6 \).
• The oxidation state of a free element (not combined with others) is always 0. For example, \(\text{O}_2\) or \(H_2\).
• For monoatomic ions, the oxidation state is equal to the ion's charge. For example, \(\text{Na}^+\) has an oxidation state of \(+1\).
• Oxygen typically has an oxidation state of \(-2\) in most compounds, though there are exceptions like in peroxides where it is \(-1\).
• Hydrogen generally has an oxidation state of \(+1\) when bonded with non-metals and \(-1\) with metals.
• The sum of oxidation states in a neutral molecule must be zero, and in an ion, it must equal the ion's charge.
These rules help you set up and solve equations to find unknown oxidation states, as demonstrated in the problem where we identified the oxidation state of xenon (Xe) in \( \text{Na}_4 \text{XeO}_6 \).
xenon compounds
Xenon, though a noble gas, can form various compounds, usually with highly electronegative elements like oxygen or fluorine. These compounds often show interesting and unique oxidation states.
For instance, in \( \text{Na}_4 \text{XeO}_6 \) (sodium perxenate), Xe is in its highest known oxidation state of \( +8 \). This is quite rare and demonstrates the element's capacity to form intricate compounds despite its general chemical inertness.
Xenon compounds are vital in advanced chemistry topics and often come up in studying redox reactions and ultimately help us understand the reactivity and bonding of 'unreactive' elements.
For instance, in \( \text{Na}_4 \text{XeO}_6 \) (sodium perxenate), Xe is in its highest known oxidation state of \( +8 \). This is quite rare and demonstrates the element's capacity to form intricate compounds despite its general chemical inertness.
Xenon compounds are vital in advanced chemistry topics and often come up in studying redox reactions and ultimately help us understand the reactivity and bonding of 'unreactive' elements.
chemical equation balancing
Balancing chemical equations is a fundamental skill in chemistry. It ensures that the same number of each type of atom appears on both sides of the equation, following the law of conservation of mass. Here’s how you can approach it step-by-step:
• **Identify each element present on both sides of the equation.**
• **Count the number of atoms of each element in the reactants and products.**
• **Adjust coefficients (numbers in front of compounds) to balance each element one at a time.** Start with elements that appear in only one reactant and one product, and leave more complex compounds or elements for later.
• **Check your work:** After balancing, ensure that all elements have the same number of atoms on both sides of the equation.
Balancing equations is crucial for accurate reactions, predicting product yields, and understanding reaction mechanisms in both inorganic and organic chemistry.
• **Identify each element present on both sides of the equation.**
• **Count the number of atoms of each element in the reactants and products.**
• **Adjust coefficients (numbers in front of compounds) to balance each element one at a time.** Start with elements that appear in only one reactant and one product, and leave more complex compounds or elements for later.
• **Check your work:** After balancing, ensure that all elements have the same number of atoms on both sides of the equation.
Balancing equations is crucial for accurate reactions, predicting product yields, and understanding reaction mechanisms in both inorganic and organic chemistry.
inorganic chemistry problems
Inorganic chemistry involves studying non-organic compounds, which includes a vast array of substances such as salts, metals, minerals, and various gases. Here's what often comes up in problems:
• **Determining oxidation states:** As covered, this is vital in understanding redox reactions and the properties of compounds.
• **Balancing chemical equations:** Fundamental for predicting the outcomes of reactions.
• **Identifying compound structures and geometries:** Recognizing how atoms are arranged within a molecule and predicting the shapes of molecules.
• **Predicting properties and reactivity:** Based on periodic trends (such as electronegativity) and bonding theories.
Problems in inorganic chemistry challenge students to apply these concepts, often combining several in one complex problem, helping to cultivate a well-rounded understanding of chemical principles.
• **Determining oxidation states:** As covered, this is vital in understanding redox reactions and the properties of compounds.
• **Balancing chemical equations:** Fundamental for predicting the outcomes of reactions.
• **Identifying compound structures and geometries:** Recognizing how atoms are arranged within a molecule and predicting the shapes of molecules.
• **Predicting properties and reactivity:** Based on periodic trends (such as electronegativity) and bonding theories.
Problems in inorganic chemistry challenge students to apply these concepts, often combining several in one complex problem, helping to cultivate a well-rounded understanding of chemical principles.
