Question

Assign oxidation numbers to each element in (a) \(\mathrm{ClO}_{3}^{-}\) (b) \(\mathrm{H}_{2} \mathrm{SO}_{3}\) (c) \(\mathrm{K}_{2} \mathrm{O}_{2}\) (d) \(\mathrm{Na}_{3} \mathrm{~N}\)

Short answer

Question: Assign oxidation numbers to each element in the following compounds: a) ClO₃⁻, b) H₂SO₃, c) K₂O₂, d) Na₃N. Answer: a) In ClO₃⁻, Cl has an oxidation number of +5, and O has an oxidation number of -2. b) In H₂SO₃, H has an oxidation number of +1, S has an oxidation number of +4, and O has an oxidation number of -2. c) In K₂O₂, K has an oxidation number of +1, and O has an oxidation number of -1 (as it is a peroxide). d) In Na₃N, Na has an oxidation number of +1, and N has an oxidation number of -3.

Step-by-step solution

For the \(\mathrm{ClO}_{3}^{-}\) ion, we know that the oxidation number for oxygen is -2, and the overall charge of the ion is -1. #Step 2: Calculate the oxidation number of chlorine#

Using rule 2, we can set up the following equation for the sum of the oxidation numbers: \(x + 3(-2) = -1\), where x is the oxidation number of chlorine. Solving for x, we get \(x = +5\). Thus, the oxidation numbers are: Cl: +5, O: -2. #b) \(\mathrm{H}_{2} \mathrm{SO}_{3}\)# #Step 1: Identify the oxidation numbers of hydrogen and oxygen#

For \(\mathrm{H}_{2} \mathrm{SO}_{3}\), we know the oxidation number of hydrogen is +1, and that of oxygen is -2. #Step 2: Calculate the oxidation number of sulfur#

Using rule 2, we set up the following equation: \(2(+1) + x + 3(-2) = 0\), where x is the oxidation number of sulfur. Solving for x, we get \(x = +4\). Thus, the oxidation numbers are: H: +1, S: +4, O: -2. #c) \(\mathrm{K}_{2} \mathrm{O}_{2}\)# #Step 1: Identify the oxidation number of potassium#

For \(\mathrm{K}_{2} \mathrm{O}_{2}\), we know that potassium, being an alkali metal, has an oxidation number of +1. #Step 2: Calculate the oxidation number of oxygen in the peroxide ion#

Since the compound is a peroxide, the oxidation number of oxygen is -1. Thus, using rule 2, we set up the following equation: \(2(+1) + 2(-1) = 0\). This confirms that the oxidation numbers are: K: +1, O: -1. #d) \(\mathrm{Na}_{3} \mathrm{N}\)# #Step 1: Identify the oxidation number of sodium#

For \(\mathrm{Na}_{3} \mathrm{N}\), we know that sodium, being an alkali metal, has an oxidation number of +1. #Step 2: Calculate the oxidation number of nitrogen#

Using rule 2, we set up the following equation: \(3(+1) + x = 0\), where x is the oxidation number of nitrogen. Solving for x, we get \(x = -3\). Thus, the oxidation numbers are: Na: +1, N: -3.

Key concepts

Oxidation States

Understanding oxidation states, also known as oxidation numbers, is essential for delving into redox chemistry and balancing equations. Oxidation states are theoretical charges that an atom would have if electrons were completely transferred during bonding.

In our exercise, we assign oxidation numbers to several elements in compounds. The key is to use known oxidation states, such as oxygen usually being -2 (except in peroxides where it's -1), hydrogen generally +1 except when bonded to metals where it's -1, and alkali metals in compounds being +1. We then apply the rule that the sum of oxidation numbers in a compound must equal the compound's overall charge.

For complex ions like \(\mathrm{ClO}_{3}^{-}\), for example, we calculate the oxidation number of chlorine by acknowledging that the sum of the oxidation number for \(Cl\) and three oxygen atoms must equal the total charge of the ion, which is -1. Through this process, we can deduce that for \(\mathrm{ClO}_{3}^{-}\), chlorine has an oxidation number of +5.

Redox Chemistry

Redox chemistry is the branch of chemistry that studies oxidation-reduction reactions, where the transfer of electrons between chemical species takes place.

An element’s oxidation state tells you how many electrons it has lost or gained. If it has gained electrons, it has been reduced, and if it has lost electrons, it is oxidized. In our exercise examples, assigning oxidation numbers is the first step in identifying redox processes. For instance, in potassium peroxide \(\mathrm{K}_{2}\mathrm{O}_{2}\), we identify that the oxygen has an atypical oxidation state of -1, which can signify a chemical propensity for redox reactions involving oxygen species.

Every redox reaction consists of two parts: the oxidation where an element loses electrons, and the reduction where another gains electrons. In the context of learning, we can understand redox chemistry as a 'give and take' of electrons, and oxidation states serve as a tally of this electron exchange.

Balancing Chemical Equations

Balancing chemical equations is a cornerstone of chemical education, enabling students to understand the stoichiometric relationships in reactions.

A balanced equation adheres to the law of conservation of mass, which dictates that matter cannot be created or destroyed in a chemical reaction. This translates to having equal numbers of atoms for each element on both the reactants and products side of an equation. When assigning oxidation numbers, as seen in our exercise, we essentially prepare the groundwork for balancing equations in redox reactions.

By knowing the oxidation states, we can identify which elements are oxidized and which are reduced, and this allows us to balance the reaction using techniques such as the half-reaction method. Balancing equations ensures that the quantities of reactants and products are represented accurately, enabling accurate predictions of the outcomes of chemical reactions.