Final answer:
The oxidation numbers of nitrogen in N₂H₄, NO₃⁻, and HNO₂ are -2, +5, and +3, respectively.To find the oxidation numbers, we apply the rules for assigning oxidation states.
Step-by-step explanation:
The student is asking to determine the oxidation number of nitrogen in three different compounds: N₂H₄, NO₃⁻, and HNO₂.
- For N₂H₄ (hydrazine): Hydrogen is usually +1, and because the molecule is neutral, the two nitrogens must balance the hydrogens. There are four hydrogens, so the total positive charge is +4. With two nitrogens, each nitrogen must be -2 to balance this (+4 from hydrogen and -4 from nitrogen equals zero).
- For NO₃⁻ (nitrate ion): Oxygen is typically -2, and there are three oxygens for a total of -6. As the compound has a negative charge of 1, the nitrogen must be +5 to balance the oxygens and the extra negative charge (-6 from oxygen and +5 from nitrogen gives a total of -1, which is the charge on the ion).
- For HNO₂ (nitrous acid): Oxygen is -2 and there are two oxygens, so total oxygen charge is -4. Hydrogen is +1. To make the compound neutral, nitrogen must have an oxidation number of +3 (-4 from oxygens, +1 from hydrogen, +3 from nitrogen equals zero).
The oxidation numbers of nitrogen in the given compounds are -2 for N₂H₄, +5 for NO₃⁻, and +3 for HNO₂.