Question

Determine the oxidation number of each element in the following ions or compounds. (a) \(\mathrm{PF}_{6}^{-}\) (b) \(\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\) (c) \(\mathrm{UO}^{2+}\) (d) \(\mathrm{N}_{2} \mathrm{O}_{5}\) (e) \(\mathrm{POCl}_{3}\) (f) \(\mathrm{XeO}_{4}^{2-}\)

Short answer

(a) +5, (b) +5, (c) +4, (d) +5, (e) +5, (f) +6.

Step-by-step solution

Determine the Oxidation Number of Phosphorus in \(\mathrm{PF}_{6}^{-}\)

In \(\mathrm{PF}_{6}^{-}\), fluorine has an oxidation number of \(-1\). Since there are six fluorine atoms, the total contribution is \(-6\). The oxidation number of the entire ion is \(-1\), so let \(x\) be the oxidation number of phosphorus. The equation is \(x + 6(-1) = -1\). Solving gives us \(x - 6 = -1\), therefore \(x = +5\). Hence, the oxidation number of phosphorus is +5.

Determine the Oxidation Number of Arsenic in \(\mathrm{H}_{2}\mathrm{AsO}_{4}^{-}\)

In \(\mathrm{H}_{2}\mathrm{AsO}_{4}^{-}\), hydrogen has an oxidation number of \(+1\) and oxygen has \(-2\). Let \(x\) be the oxidation number of arsenic. The equation becomes \(2(1) + x + 4(-2) = -1\). Simplifying gives us \(2 + x - 8 = -1\), which leads to \(x - 6 = -1\) and therefore \(x = +5\). So, the oxidation number of arsenic is +5.

Determine the Oxidation Number of Uranium in \(\mathrm{UO}^{2+}\)

In \(\mathrm{UO}^{2+}\), oxygen has an oxidation number of \(-2\). Let \(x\) be the oxidation number of uranium. The formula gives \(x + (-2) = +2\). Solving for \(x\) gives \(x - 2 = +2\), which results in \(x = +4\). Therefore, the oxidation number of uranium is +4.

Determine the Oxidation Number of Nitrogen in \(\mathrm{N}_{2}\mathrm{O}_{5}\)

For \(\mathrm{N}_{2}\mathrm{O}_{5}\), oxygen has an oxidation number of \(-2\). There are five oxygen atoms, contributing a total of \(-10\). Let \(x\) represent the oxidation number of nitrogen. The formula becomes \(2x + 5(-2) = 0\), which simplifies to \(2x - 10 = 0\), leading to \(2x = 10\) and \(x = +5\). Therefore, the oxidation number of nitrogen is +5.

Determine the Oxidation Number of Phosphorus in \(\mathrm{POCl}_{3}\)

In \(\mathrm{POCl}_{3}\), oxygen has an oxidation number of \(-2\) and chlorine has \(-1\). There are three chlorine atoms. Let \(x\) be the oxidation number of phosphorus. The equation is \(x + (-2) + 3(-1) = 0\). Simplifying gives \(x - 2 - 3 = 0\), which leads to \(x - 5 = 0\), so \(x = +5\). The oxidation number of phosphorus is +5.

Determine the Oxidation Number of Xenon in \(\mathrm{XeO}_{4}^{2-}\)

In \(\mathrm{XeO}_{4}^{2-}\), oxygen has an oxidation number of \(-2\). There are four oxygens, yielding a total of \(-8\). Let \(x\) represent xenon's oxidation number. The equation becomes \(x + 4(-2) = -2\). This simplifies to \(x - 8 = -2\), leading to \(x = +6\). Thus, the oxidation number of xenon is +6.

Key concepts

chemical reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances, called products. These changes can involve the making and breaking of chemical bonds, leading to the reorganization of atoms. Chemical reactions are fundamental to chemical processes in everyday life, such as cooking, rusting, respiration, and photosynthesis.

In any chemical reaction, the conservation of mass holds true. This means that the total mass of the reactants equals the total mass of the products. This law applies because atoms are neither created nor destroyed during a reaction. They are simply rearranged. Balancing chemical equations is one way to ensure this conservation is reflected in our symbolic representations.

• The reactants are listed on the left side of the reaction equation.
• The products are listed on the right side.
• Only the coefficients are altered to balance the number of atoms of each element on both sides.

Understanding chemical reactions is key to mastering chemistry because they are the building blocks for understanding how substances interact and transform.

redox reactions

Redox reactions, or reduction-oxidation reactions, are a type of chemical reaction involving the transfer of electrons between two species. They play a crucial role in numerous processes, such as cellular respiration and combustion.

In a redox reaction, oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. The substance that loses electrons is said to be oxidized, and the substance that gains electrons is reduced. Redox reactions always involve both oxidation and reduction because the electrons lost by one reactant must be gained by another. This coupling is why they are collectively called redox reactions.

Key concepts to remember about redox reactions:

• They involve changes in the oxidation states of participating species.
• Oxidizing agents accept electrons and help other substances oxidize - they themselves are reduced.
• Reducing agents donate electrons and help other substances reduce - they themselves are oxidized.

The balancing of redox reactions often requires a more complex approach than balancing other types of chemical reactions, typically involving the half-reaction method that separately balances the oxidation and reduction processes.

oxidation states

Oxidation states, or oxidation numbers, are tools used by chemists to track how many electrons an atom has gained, lost, or shared during chemical bonding. This concept helps in understanding and balancing redox reactions. Every element in a compound is assigned an oxidation number that represents its actual or formal charge.

The rules for determining oxidation numbers are straightforward:

• The oxidation state of an atom in its elemental form is 0. For example, O extsubscript{2}, N extsubscript{2}, and H extsubscript{2} all have oxidation states of 0.
• For monoatomic ions, the oxidation number is the same as the charge of the ion. For instance, Na extsuperscript{+} has an oxidation number of +1.
• Hydrogen typically has an oxidation number of +1, while oxygen usually has -2.
• The sum of the oxidation states in a neutral compound must be 0, while for an ion, it must equal the ion's charge.

Being adept at determining oxidation states is crucial for identifying redox reactions and ensuring chemical equations are balanced correctly. It allows chemists to predict how compounds will interact and change during chemical reactions.