Problem 41 - Chemistry The Central Science

Question

Write the chemical formula for each of the following compounds, and indicate the oxidation state of the group 6 \(\mathrm{A}\) element in each: (a) selenous acid, (b) potassium hydrogen sulfite, ( c) hydrogen telluride, (d) carbon disulfide, (e) calcium sulfate, (f) cadmium sulfide, (g) zinc telluride.

Step-by-Step Solution

Answer

The chemical formulas and oxidation states of the group 6A elements in each compound are as follows: (a) Selenous Acid: H2SeO3, Se oxidation state = +4. (b) Potassium Hydrogen Sulfite: KHSO3, S oxidation state = +4. (c) Hydrogen Telluride: H2Te, Te oxidation state = -2. (d) Carbon Disulfide: CS2, S oxidation state = -2. (e) Calcium Sulfate: CaSO4, S oxidation state = +6. (f) Cadmium Sulfide: CdS, S oxidation state = -2. (g) Zinc Telluride: ZnTe, Te oxidation state = -2.

Step 1: Compound (a) - Selenous Acid

Selenous acid is H2SeO3. The group 6A element in selenous acid is selenium (Se). To find the oxidation state of selenium in selenous acid, we will first assign oxidation states to all elements in the formula: - Hydrogen (H) = +1 - Oxygen (O) = -2 Then, we set up an equation to find the oxidation state of selenium (Se): 2(+1) + x + 3(-2) = 0 Solving for x, the oxidation state of selenium in selenous acid is +4.

Step 2: Compound (b) - Potassium Hydrogen Sulfite

Potassium Hydrogen Sulfite is KHSO3. The group 6A element in potassium hydrogen sulfite is sulfur (S). To find the oxidation state of sulfur in potassium hydrogen sulfite, we will first assign oxidation states to all elements in the formula: - Potassium (K) = +1 - Hydrogen (H) = +1 - Oxygen (O) = -2 Then, we set up an equation to find the oxidation state of sulfur (S): (+1) + (+1) + x + 3(-2) = 0 Solving for x, the oxidation state of sulfur in potassium hydrogen sulfite is +4.

Step 3: Compound (c) - Hydrogen Telluride

Hydrogen Telluride is H2Te. The group 6A element in hydrogen telluride is tellurium (Te). To find the oxidation state of tellurium in hydrogen telluride, we will first assign oxidation states to all elements in the formula: - Hydrogen (H) = +1 Then, we set up an equation to find the oxidation state of tellurium (Te): 2(+1) + x = 0 Solving for x, the oxidation state of tellurium in hydrogen telluride is -2.

Step 4: Compound (d) - Carbon Disulfide

Carbon Disulfide is CS2. The group 6A element in carbon disulfide is sulfur (S). To find the oxidation state of sulfur in carbon disulfide, we will first assign oxidation states to all elements in the formula: - Carbon (C) = +4 Then, we set up an equation to find the oxidation state of sulfur (S): (+4) + 2x = 0 Solving for x, the oxidation state of sulfur in carbon disulfide is -2.

Step 5: Compound (e) - Calcium Sulfate

Calcium Sulfate is CaSO4. The group 6A element in calcium sulfate is sulfur (S). To find the oxidation state of sulfur in calcium sulfate, we will first assign oxidation states to all elements in the formula: - Calcium (Ca) = +2 - Oxygen (O) = -2 Then, we set up an equation to find the oxidation state of sulfur (S): (+2) + x + 4(-2) = 0 Solving for x, the oxidation state of sulfur in calcium sulfate is +6.

Step 6: Compound (f) - Cadmium Sulfide

Cadmium Sulfide is CdS. The group 6A element in cadmium sulfide is sulfur (S). To find the oxidation state of sulfur in cadmium sulfide, we will first assign oxidation states to all elements in the formula: - Cadmium (Cd) = +2 Then, we set up an equation to find the oxidation state of sulfur (S): (+2) + x = 0 Solving for x, the oxidation state of sulfur in cadmium sulfide is -2.

Step 7: Compound (g) - Zinc Telluride

Zinc Telluride is ZnTe. The group 6A element in zinc telluride is tellurium (Te). To find the oxidation state of tellurium in zinc telluride, we will first assign oxidation states to all elements in the formula: - Zinc (Zn) = +2 Then, we set up an equation to find the oxidation state of tellurium (Te): (+2) + x = 0 Solving for x, the oxidation state of tellurium in zinc telluride is -2.

Key Concepts

Oxidation States

An oxidation state refers to the degree of oxidation (loss of electrons) of an atom in a chemical compound. It is a useful concept for understanding the electron transfer in reactions and the composition of compounds. To determine the oxidation state, you can follow these simple rules:

• The oxidation state of an element in its elemental form is always zero.
• The oxidation state of a monatomic ion is equal to its charge.
• In compounds, hydrogen is usually +1 and oxygen is usually -2.

By applying these rules, you can find the oxidation state of particular elements within a compound. Take selenous acid (\( \text{H}_2\text{SeO}_3 \)) as an example: by setting up an equation based on the usual oxidation states of hydrogen and oxygen, we can solve for selenium's oxidation state, which is +4. The ability to calculate oxidation states helps with balancing reactions and predicting compound formation.

Group 6 Elements

The group 6 elements include chromium (Cr), molybdenum (Mo), tungsten (W), and the chalcogens: oxygen (O), sulfur (S), selenium (Se), and tellurium (Te). Chalcogens are essential for life and have high chemical reactivity.
For instance:

• Oxygen is vital for respiration and makes up a significant portion of Earth's atmosphere.
• Sulfur is found in amino acids and vitamins for biological processes.
• Selenium and tellurium, while not as common, play roles in various biochemical activities and industrial applications.

In the given exercise, compounds like selenous acid, calcium sulfate, and cadmium sulfide all include group 6 chalcogens. Understanding the characteristics of these elements helps elucidate their roles in compounds and reactions.

Compound Naming

Naming chemical compounds requires understanding the types of elements present and how they combine. Correct compound naming follows specific conventions to convey information about the composition:

• Ionic compounds combine metals with non-metals. The metal (cation) name remains unchanged, while the non-metal (anion) typically ends in "-ide." For example, cadmium sulfide is named because it combines cadmium (Cd) and sulfide (S^2-).
• Covalent compounds are named by using prefixes to denote the number of atoms. Water (\( \text{H}_2\text{O} \)) doesn't follow this rule as it's a common name, but carbon disulfide (\( \text{CS}_2 \)) is named this way as it combines carbon with two sulfur atoms.
• Acids are named based on their anions: acids without oxygen in the anion are named with "hydro-" plus the anion's root and "-ic" (e.g., hydrochloric acid), while acids with oxygen follow the pattern based on their anion suffix, like "-ic" for "-ate" anions and "-ous" for "-ite" anions, as in selenous acid (\( \text{H}_2\text{SeO}_3 \)).

Using these naming principles accurately conveys the chemical structure and composition, aiding in communication and further study of chemical reactions.