Question

What is the oxidation state of sulfur in each of the following substances? a. \(\mathrm{S}_{8}\) b. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) c. \(\mathrm{NaHSO}_{4}\) d. \(\mathrm{Na}_{2} \mathrm{~S}\)

Short answer

a) In \(\mathrm{S}_{8}\), the oxidation state of sulfur is 0. b) In \(\mathrm{H}_{2} \mathrm{SO}_{4}\), the oxidation state of sulfur is +6. c) In \(\mathrm{NaHSO}_{4}\), the oxidation state of sulfur is +6. d) In \(\mathrm{Na}_{2} \mathrm{S}\), the oxidation state of sulfur is -2.

Step-by-step solution

Understanding oxidation rules

Before we begin to find the oxidation state in each of these substances, let's review a few key oxidation rules: 1. The oxidation state of a free element, like \(\mathrm{S}_{8}\), is 0. 2. The oxidation state of hydrogen is usually +1, except in metal hydrides, where it is -1. 3. The oxidation state of oxygen is usually -2, except in peroxides or superoxides. 4. Alkali metals like sodium (Na) have an oxidation state of +1; alkaline earth metals like magnesium (Mg) have an oxidation state of +2. 5. The net charge of a compound is the sum of the individual oxidation states of each element. Now, let's apply these rules to find the oxidation state of sulfur in each substance.

Sulphur in S8

Since \(\mathrm{S}_{8}\) is a free element (a molecule composed of only sulfur atoms), the oxidation state of sulfur in this substance is 0.

Sulphur in H2SO4

In \(\mathrm{H}_{2} \mathrm{SO}_{4}\), we have hydrogen, sulfur, and oxygen. First, we need to find the oxidation state of hydrogen and oxygen. Remember that hydrogen has an oxidation state of +1, and there are two hydrogen atoms; therefore, the oxidation state of hydrogen contributes a total of +2. Oxygen has an oxidation state of -2, and there are four oxygen atoms; therefore, the oxidation state of oxygen contributes a total of -8. Now we can find the oxidation state of sulfur by setting up an equation using the oxidation state rule no.5: \(+1 \cdot 2 + x + (-2) \cdot 4 = 0\) Solve for x (sulfur's oxidation state): \(2 + x -8 = 0\) \(x = 6\) The oxidation state of sulfur in \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is +6.

Sulphur in NaHSO4

In \(\mathrm{NaHSO}_{4}\), we have sodium, hydrogen, sulfur, and oxygen. Sodium has an oxidation state of +1. Hydrogen has an oxidation state of +1, and oxygen has an oxidation state of -2. Now we can find the oxidation state of sulfur by setting up an equation using the oxidation state rule no.5: \(+1 + 1 + x + (-2) \cdot 4 = 0\) Solve for x (sulfur's oxidation state): \(2 + x -8 = 0\) \(x = 6\) The oxidation state of sulfur in \(\mathrm{NaHSO}_{4}\) is +6.

Sulphur in Na2S

In \(\mathrm{Na}_{2} \mathrm{S}\), we have sodium and sulfur. Sodium has an oxidation state of +1. Since there are two sodium atoms in the compound, the oxidation state of sodium contributes a total of +2. Now we can find the oxidation state of sulfur by setting up an equation using the oxidation state rule no.5: \(+1 \cdot 2 + x = 0\) Solve for x (sulfur's oxidation state): \(2 + x = 0\) \(x = -2\) The oxidation state of sulfur in \(\mathrm{Na}_{2} \mathrm{S}\) is -2. In summary: a) \(\mathrm{S}_{8}\): Sulfur's oxidation state is 0. b) \(\mathrm{H}_{2} \mathrm{SO}_{4}\): Sulfur's oxidation state is +6. c) \(\mathrm{NaHSO}_{4}\): Sulfur's oxidation state is +6. d) \(\mathrm{Na}_{2} \mathrm{~S}\): Sulfur's oxidation state is -2.

Key concepts

Redox Chemistry

Redox chemistry is the study of chemical reactions that involve changes in oxidation states of atoms through the transfer of electrons. It involves two complementary processes known as oxidation and reduction. In an oxidation reaction, an element loses electrons, increasing its oxidation state. Conversely, during reduction, an element gains electrons, decreasing its oxidation state. The substance that donates electrons is called the reducing agent, and the one that accepts electrons is referred to as the oxidizing agent. Understanding the changes in oxidation states allows chemists to determine the direction of electron flow in a reaction, which is important for predicting the behavior of elements and compounds in various chemical processes.

For example, when sulfur in \(\mathrm{H}_{2} \mathrm{SO}_{4}\) changes to sulfur in \(\mathrm{NaHSO}_{4}\), there is no change in oxidation state. This means that in this specific case, no redox reaction is taking place since the oxidation state of sulfur remains the same. However, recognizing the constancy of sulfur's oxidation state between these compounds is crucial in determining the absence of redox action.

Chemical Bonding

Chemical bonding is the force that holds atoms together in compounds. Atoms bond by sharing or transferring valence electrons to achieve stability, often described as the complete filling of their outer electron shells. There are several types of chemical bonds, with covalent, ionic, and metallic bonds being the most common.

For instance, the sulfur atom within \(\mathrm{S}_{8}\) is bound to other sulfur atoms by covalent bonds, as they share electrons to form stable structures. \(\mathrm{NaHSO}_{4}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\), on the other hand, contain ionic bonds between the sodium ions (Na+) and sulfate anions, as well as covalent bonds within the sulfate anion itself. In the case of \(\mathrm{Na}_{2} \mathrm{S}\), the bonding between sodium ions and sulfide ions (S2-) is ionic because electrons are transferred from sodium to sulfur. Understanding these bond types helps explain how atoms in a compound affect the overall oxidation states of the elements involved.

Oxidation Rules

Oxidation rules are a set of guidelines used to assign oxidation states to atoms in a compound. These rules are essential for solving redox reactions and balancing chemical equations.

Here are some oxidation rules in a simplified form:

• The oxidation state of a pure element is always 0.
• The oxidation state of a monatomic ion equals its charge.
• Hydrogen typically has an oxidation state of +1, while oxygen generally has an oxidation state of -2.
• Alkali metals have an oxidation state of +1, and alkaline earth metals have an oxidation state of +2.
• The sum of the oxidation states in a neutral compound must be 0, and in a polyatomic ion must equal the charge of the ion.

By applying these rules to our exercise, we deduced that \(\mathrm{S}_{8}\) has an oxidation state of 0 for sulfur because it's an elemental form. For \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{NaHSO}_{4}\), sulfur has an oxidation state of +6, while in \(\mathrm{Na}_{2} \mathrm{S}\), sulfur's oxidation state is -2. These calculations are vital in identifying the nature of chemical species before and after reactions, especially in redox processes.