Question

Assign oxidation states to all of the atoms in each of the following: a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) b. \(\mathrm{MnO}_{4}^{-}\) c. \(\mathrm{NO}_{3}^{-}\) d. \(\mathrm{K}_{3} \mathrm{PO}_{4}\)

Short answer

The oxidation states for each compound are: a. H₂SO₄: Hydrogen (+1), Sulfur (+6), Oxygen (-2) b. MnO₄⁻: Manganese (+7), Oxygen (-2) c. NO₃⁻: Nitrogen (+5), Oxygen (-2) d. K₃PO₄: Potassium (+1), Phosphorus (+5), Oxygen (-2)

Step-by-step solution

Find the oxidation states for H₂SO₄

We'll assign the oxidation numbers to each atom: - Hydrogen (H): +1 - Oxygen (O): -2 There is one unknown oxidation state for Sulfur (S), let's call it x. We know that the sum of the oxidation numbers should equal 0 because it's a neutral molecule. We can write an equation with x: \(x + 2(+1) + 4(-2) = 0\) Now, solve for x: \(x = 6\) The oxidation states for H₂SO₄ are: - Hydrogen: +1 - Sulfur: +6 - Oxygen: -2

Find the oxidation states for MnO₄⁻

We'll assign the oxidation numbers to each atom: - Manganese (Mn): unknown, let's call it x - Oxygen (O): -2 We know that the sum of the oxidation numbers should equal -1 because it's a negatively charged ion. We can write an equation with x: \(x + 4(-2) = -1 \) Now, solve for x: \(x = +7 \) The oxidation states for MnO₄⁻ are: - Manganese: +7 - Oxygen: -2

Find the oxidation states for NO₃⁻

We'll assign the oxidation numbers to each atom: - Nitrogen (N): unknown, let's call it x - Oxygen (O): -2 We know that the sum of the oxidation numbers should equal -1 because it's a negatively charged ion. We can write an equation with x: \(x + 3(-2) = -1\) Now, solve for x: \(x = +5\) The oxidation states for NO₃⁻ are: - Nitrogen: +5 - Oxygen: -2

Find the oxidation states for K₃PO₄

We'll assign the oxidation numbers to each atom: - Potassium (K): +1 (since it's an alkali metal) - Phosphorus (P): unknown, let's call it x - Oxygen (O): -2 We know that the sum of the oxidation numbers should equal 0 because it's a neutral molecule. We can write an equation with x: \(3(+1) + x + 4(-2) = 0\) Now, solve for x: \(x = +5\) The oxidation states for K₃PO₄ are: - Potassium: +1 - Phosphorus: +5 - Oxygen: -2

Key concepts

Oxidation Number Rules

Understanding how to assign oxidation numbers is crucial for analyzing chemical reactions, particularly redox reactions. Oxidation numbers, also known as oxidation states, help determine how electrons are transferred between atoms. While assigning these numbers, there are several key rules to follow:

• Elemental Form: Any atom in its pure elemental form has an oxidation number of 0. For instance, \( ext{N}_2\) or \( ext{O}_2\) each have oxidation numbers of 0.
• Monoatomic Ions: For ions composed of only one atom, the oxidation number is equal to the ion's charge. For example, \( ext{Na}^+\) has an oxidation number of +1.
• Group 1 Metals: These metals always have an oxidation number of +1 in compounds. Similarly, Group 2 metals have +2.
• Oxygen: In most compounds, oxygen has an oxidation number of -2. Exceptions include peroxides where it's -1.
• Hydrogen: Typically +1 with nonmetals and -1 when paired with metals.
• Halogens: Often have an oxidation number of -1, unless bonded to elements with higher electronegativities.

These rules provide a framework but must be applied in conjunction with an understanding of the molecule or ion being examined.

Neutral Molecules

Neutral molecules are those without an overall charge. This means that the sum of oxidation numbers in a neutral molecule is always zero. For example, take sulfuric acid \( ext{H}_2 ext{SO}_4\). Here's the approach:

• Assign known oxidation numbers: hydrogen is +1 and oxygen is -2.
• Identify unknown oxidation states: in this case, sulfur is unknown.
• Set up the equation based on the molecular formula: \( 2(+1) + x + 4(-2) = 0 \)
• Solve for x: \ x = +6 \

After solving, the oxidation numbers for hydrogen, sulfur, and oxygen are +1, +6, and -2, respectively. This method applies to any neutral molecule, helping to simplify complex looking formulas.

Charged Ions

Charged ions, also known as polyatomic ions, carry either a positive or negative charge. This affects the sum of their oxidation numbers. In these ions, the total of oxidation numbers equals the ion's charge. Let's consider the permanganate ion, \( ext{MnO}_4^-\):

• Firstly, assign known numbers: oxygen is -2.
• Next, find the unknown number: manganese's state is unknown.
• Set up the equation based on the charge: \( x + 4(-2) = -1 \)
• Solve for x: \ x = +7 \

The oxidation numbers for manganese and oxygen are +7 and -2, respectively. Another example is nitrate, \( ext{NO}_3^-\), where nitrogen is the unknown in the equation \(x + 3(-2) = -1\), giving nitrogen an oxidation state of +5.

Oxidation Number Calculation

Calculating oxidation numbers helps in the analysis and balancing of redox equations. By systematically applying the rules and setting equations, oxidation numbers for unknown atoms can be found:

• Identify known oxidation numbers based on typical rules.
• For unknowns, set up and solve algebraic equations using the molecular or ionic formula.
• Verify calculations by ensuring the sum of oxidation numbers equals the total charge of the molecule or ion.

For example, in potassium phosphate, \( ext{K}_3 ext{PO}_4\), known numbers are for potassium and oxygen: +1 and -2. Phosphorus is the unknown in \(3(+1) + x + 4(-2) = 0\), leading to phosphorus having an oxidation state of +5. This practice ensures all oxidation states are accurately accounted for in both neutral molecules and charged ions.