What is the change in the oxidation number of phosphorus in the following reaction? 4P(s) + 5O2(ag) → 2P2O5(s)

Analyze the question: We need the oxidation number of phosphorus before and after the reaction 4P + 5O₂ → 2P₂O₅.

Concept:

• Any element in its free, uncombined state has an oxidation number of 0.
• In most compounds, oxygen has an oxidation number of -2.
• The sum of oxidation numbers in a neutral compound is 0.

Solution:

1. Reactant side: Phosphorus, P(s), is an element in its free state, so its oxidation number is 0.

2. Product side: In P₂O₅, let the oxidation number of P be x. • 2x + 5(-2) = 0 • 2x - 10 = 0 • 2x = 10 • x = +5

3. Change: Phosphorus goes from 0 to +5. This is oxidation, because the oxidation number increases (P loses electrons), while oxygen is reduced from 0 to -2.

Answer: The oxidation number of phosphorus changes from 0 to +5, which is option B.

Tip: Always start with the rule "free element = 0." Then find the other oxidation number by making the total charge of the compound zero. In oxides, use O = -2. (Note: the "(ag)" in the question is just a typo for "(g)".)