The oxidation number of nitrogen in $Pb(NO_{3})_{2}$ is

Analyze the question: We need the oxidation number of nitrogen in lead(II) nitrate, Pb(NO₃)₂.

Concept: In a neutral compound, the oxidation numbers of all atoms add up to zero. For a compound containing a polyatomic ion, you can work out the charge on the ion first, then find the oxidation number of the central atom.

Solution:

1. Pb(NO₃)₂ is made of Pb²⁺ ions and nitrate ions, NO₃⁻. • The 2 nitrate ions balance the 2+ charge on lead, so each nitrate carries a charge of −1.

2. For the nitrate ion, NO₃⁻: • Let the oxidation number of N be x. • Oxygen is usually −2, so 3 oxygens give 3(−2) = −6. • x + (−6) = −1 • x = −1 + 6 = +5

3. Check using the whole compound: • Pb = +2, 2 N = 2(+5) = +10, 6 O = 6(−2) = −12 • (+2) + (+10) + (−12) = 0 ✓

Answer: The oxidation number of nitrogen is +5, which is option D.

Tip: Any nitrate ion (NO₃⁻) always has nitrogen at +5. Remember "nitrate = N is +5" and you can answer this type of question in seconds. Also, don't confuse it with nitrite (NO₂⁻), where nitrogen is +3.