Chemistry 9701/13 — October/November 2013

Background Concept

Oxidation number (oxidation state) is a bookkeeping value assigned to an atom in a species. It represents the charge the atom would carry if all bonds were ionic, with bonding electrons transferred entirely to the more electronegative atom. Oxidation numbers track electron transfer in redox reactions: an increase in oxidation number is oxidation (loss of electrons), and a decrease is reduction (gain of electrons).

Key rules:

• An element in its elemental form has oxidation number 0.
• A monatomic ion has oxidation number equal to its charge.
• H is +1 (except in metal hydrides, where it is −1).
• O is −2 (except in peroxides, where it is −1, and in OF₂, where it is +2).
• The sum of oxidation numbers in a neutral species is 0; in a polyatomic ion it equals the ion's charge.

Understanding the Question

$\text{NH}_4\text{NO}_3$ contains two nitrogen atoms in different chemical environments: one in the ammonium cation ($\text{NH}_4^+$) and one in the nitrate anion ($\text{NO}_3^-$). These two nitrogen atoms have different oxidation numbers. The decomposition produces $\text{N}_2\text{O}$, in which the two nitrogen atoms are equivalent, each having the same oxidation number. The question asks for the change in oxidation number of each of the two original nitrogen atoms.

Approach

1. Assign the oxidation number of N in $\text{NH}_4^+$.
2. Assign the oxidation number of N in $\text{NO}_3^-$.
3. Assign the oxidation number of N in $\text{N}_2\text{O}$.
4. Compute the change for each nitrogen: change = (oxidation number in product) − (oxidation number in reactant).

Step-by-Step Reasoning

Step 1 — Ammonium ion: $\text{NH}_4^+$ has overall charge +1. Each H is +1, so the four H atoms contribute +4. Therefore $N + 4 = +1$, giving $N = -3$.

Step 2 — Nitrate ion: $\text{NO}_3^-$ has overall charge −1. Each O is −2, so the three O atoms contribute −6. Therefore $N - 6 = -1$, giving $N = +5$.

Step 3 — Product $\text{N}_2\text{O}$: This is a neutral molecule. O is −2, so $2N - 2 = 0$, giving $N = +1$ for each nitrogen.

Step 4 — Changes:

• The nitrogen that came from $\text{NH}_4^+$: −3 → +1, an increase of +4 (this nitrogen is oxidised).
• The nitrogen that came from $\text{NO}_3^-$: +5 → +1, a decrease of −4 (this nitrogen is reduced).

This matches option D: +4, −4.

Key Takeaways

• In a compound containing two identical atoms in different environments, treat each environment separately — never average them.
• Oxidation number change is always product minus reactant.
• A positive change corresponds to oxidation; a negative change corresponds to reduction.

Common Mistakes

• Averaging the two nitrogen oxidation numbers in $\text{NH}_4\text{NO}_3$ (which gives 0) and then becoming confused.
• Assigning N in $\text{NH}_4^+$ as +4 instead of −3 (forgetting the +1 charge on the ion).
• Assigning N in $\text{NO}_3^-$ as +6 instead of +5 (forgetting the −1 charge on the ion).
• Mixing up the sign of the change (using reactant − product instead of product − reactant).

Things to Be Careful About

• Always account for the overall charge of the ion when assigning oxidation numbers.
• Remember the exceptions: H is −1 in metal hydrides; O is −1 in peroxides and +2 in OF₂.
• In $\text{N}_2\text{O}$ the two nitrogen atoms are equivalent, each +1.