Consider the following reaction equation: $2NO_{(g)} + O_{{2}{(g)}} \to 2NO_{{2}{(g)}}$. What is the change in the oxidation number of nitrogen?

Analyze the question: We need the oxidation number of nitrogen on each side of the reaction 2NO + O₂ → 2NO₂, and then state how it changes.

Concept: In a compound, the oxidation numbers of all the atoms add up to the overall charge, which is 0 for a neutral molecule. Oxygen is usually -2.

Solution:

1. In NO (reactant): • Let the oxidation number of N be x. • x + (-2) = 0, so x = +2

2. In NO₂ (product): • Let the oxidation number of N be y. • y + 2(-2) = 0, so y - 4 = 0 and y = +4

3. Change in oxidation number of nitrogen: +2 → +4

The oxidation number increases, so nitrogen is oxidized. (Oxygen goes from 0 in O₂ to -2 in NO₂, so it is reduced. O₂ is the oxidizing agent.)

Checking the options:

• A. +2 to 0: wrong, since N in NO₂ is not 0.
• B. +2 to -2: wrong.
• C. +2 to +4: correct.
• D. +4 to +2: this is the reverse of what happens.

Answer: C.

Exam tip: Remember "OIL RIG": Oxidation Is Loss (of electrons, so the oxidation number increases) and Reduction Is Gain (the oxidation number decreases). If the oxidation number goes up, as it does here, the element is oxidized. Always check the direction, from reactants to products, because option D is a common trap.