Question

Obtain the oxidation number for the element noted in each of the following. (a) \(\mathrm{N}\) in \(\mathrm{NH}_{2}^{-}\) (b) \(\mathrm{I} \mathrm{in} \mathrm{IO}_{3}^{-}\) (c) \(\mathrm{H}\) in \(\mathrm{H}_{2}\) (d) \(\mathrm{H}\) in \(\mathrm{H}_{2}\)

Short answer

(a) -3; (b) +5; (c) 0; (d) 0

Step-by-step solution

Determine known oxidation states

Identify the oxidation numbers of the elements that often have fixed oxidation states. For example, hydrogen (H) is usually +1, oxygen (O) is usually -2.

Set up the equation for oxidation number equality

For every compound or ion, the sum of the oxidation states of all elements must equal the overall charge of the compound. Write this as an equation where the total oxidation states add up to the charge of the compound.

Calculate oxidation state for each element

Use the equation from Step 2 to solve for the unknown oxidation number of the element in question. Apply the known values and perform algebraic manipulation if necessary to isolate and solve for the unknown oxidation number.

Solve example (a): \(\mathrm{N}\) in \(\mathrm{NH}_2^-\)

The compound is \(\mathrm{NH}_2^-\). Let \(x\) be the oxidation number of nitrogen. Then, \( x + 2(+1) = -1 \) because there are two hydrogen atoms each contributing +1. Simplify to solve: \( x + 2 = -1 \) \( x = -1 - 2 = -3 \). Therefore, the oxidation number of nitrogen in \(\mathrm{NH}_2^-\) is -3.

Solve example (b): \(\mathrm{I}\) in \(\mathrm{IO}_3^-\)

The compound is \(\mathrm{IO}_3^-\). Let \(x\) be the oxidation number of iodine. Then, \( x + 3(-2) = -1 \) because there are three oxygen atoms each contributing -2. Simplify: \( x - 6 = -1 \) \( x = -1 + 6 = +5 \). Therefore, the oxidation number of iodine in \(\mathrm{IO}_3^-\) is +5.

Solve example (c): \(\mathrm{H}\) in \(\mathrm{H}_2\)

\(\mathrm{H}_2\) is a molecule with two hydrogen atoms bonded together. In each elemental state, the oxidation number is 0. Thus, the oxidation number of hydrogen in \(\mathrm{H}_2\) is 0.

Solve example (d): \(\mathrm{H}\) in \(\mathrm{H}_2\)

This repeats Example (c). The element hydrogen is in its diatomic molecular form, which is also in elemental form, so its oxidation number is 0.

Key concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are fundamental chemical processes. They involve the transfer of electrons between substances. In these reactions:

• Oxidation involves the loss of electrons.
• Reduction involves the gain of electrons.

When one substance is oxidized, another is reduced, making it a tandem process. This is key in energy production and many chemical reactions. Redox reactions change the oxidation states of the participating elements, shedding light on electron transfers which are pivotal in various biological and chemical processes. Understanding these reactions helps us to analyze how substances interact and transform.

Chemical Compounds

Chemical compounds are substances formed from two or more elements that are chemically bonded together. These can be classified into two main types:

• Ionic compounds, where elements exchange electrons and form charged ions.
• Covalent compounds, where elements share electron pairs.

Each compound has a unique set of properties and a distinct formula which represents the types and numbers of atoms it contains. For example:- In \(\mathrm{NH}_2^-\), nitrogen and hydrogen form bonds with distinct electron sharing patterns.Understanding chemical compounds involves recognizing how elements use their electrons to bond, affecting the compound's properties and behavior in reactions.

Oxidation States

Oxidation states or oxidation numbers help us understand redox reactions and track electron flow in compounds. An oxidation state is a hypothetical charge that an atom would have if all bonds to different atoms were completely ionic. Key points include:

• Elemental forms of atoms have oxidation states of 0.
• Oxygen typically has an oxidation number of -2, and hydrogen is usually +1 when bonded to non-metals.
• The sum of oxidation numbers in a neutral compound is zero, while for ions, it's the same as the ion's charge.

Let's take the compound \(\mathrm{IO}_3^-\). Iodine in this compound gains an oxidation state of +5 to balance the charge contributed by oxygen.Mastering these numbers provides insights into the roles elements play in chemical reactions and is crucial for stoichiometry calculations.