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Chem Chemistry

ChemistryNEB 2076 (old course)

Assign oxidation number to the underlined element in each of the following: [2] i) N 2O 2 [1] ii) NaH 2PO 4 [1]

2

Answer

Padhai · Based on the official syllabus and past papers
Updated 10 Oct 2026

i) For $\underline{N} X_{2} O X_{2}$ :

Let the oxidation number of nitrogen be $x$ .
Oxygen typically has an oxidation number of $- 2$ .
The sum of oxidation numbers in a neutral molecule is zero.
Therefore, $2 x + 2 (- 2) = 0$ .
Simplifying: $2 x - 4 = 0$ .
Solving for $x$ : $2 x = 4$ → $x = + 2$ .
Oxidation number of nitrogen in $N X_{2} O X_{2}$ is +2.

ii) For $NaH X_{2} \underline{P} O X_{4}$ :

Sodium ( $Na$ ) has an oxidation number of $+ 1$ .
Hydrogen ( $H$ ) has an oxidation number of $+ 1$ .
Oxygen ( $O$ ) has an oxidation number of $- 2$ .
Let the oxidation number of phosphorus ( $P$ ) be $x$ .
The sum of oxidation numbers in a neutral molecule is zero.
Therefore, $1 + 2 (1) + x + 4 (- 2) = 0$ .
Simplifying: $1 + 2 + x - 8 = 0$ → $x - 5 = 0$ .
Solving for $x$ : $x = + 5$ .
Oxidation number of phosphorus in $NaH X_{2} PO X_{4}$ is +5.

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