Q.In which of the following compounds, an element exhibits two different oxidation states.

The key idea is Oxidation Number Calculation — we assign oxidation numbers to each atom using standard rules (O is usually -2, H is +1, and the sum equals the charge of the species).

Step 1: For NH₂OH (hydroxylamine), N is bonded to two H atoms and one OH group. Let N’s oxidation state be $x$:

$x + 2(+1) + (-2) + (+1) = 0 \Rightarrow x + 2 - 2 + 1 = 0 \Rightarrow x = -1$. Only one oxidation state for N.

Step 2: For NH₄NO₃ (ammonium nitrate), it contains two distinct ions: NH₄⁺ and NO₃⁻.

In NH₄⁺: $x + 4(+1) = +1 \Rightarrow x = -3$.

In NO₃⁻: $y + 3(-2) = -1 \Rightarrow y = +5$.

Here, nitrogen appears in two different oxidation states: -3 and +5.

Step 3: For N₂H₄ (hydrazine), each N: $2x + 4(+1) = 0 \Rightarrow x = -2$ (both N same). …

Oxidation numbers are a bookkeeping tool that helps us track electron distribution in compounds. They are assigned based on a set of rules: the oxidation number of an atom in its elemental form is zero; for a monatomic ion, it equals the charge; oxygen is usually –2 (except in peroxides), hydrogen is usually +1 (except in metal hydrides), and the sum of oxidation numbers in a neutral compound is zero, while in a polyatomic ion it equals the ion's charge.

The trick in this question is that a single element can have different oxidation states in the same compound if the compound contains that element in two different structural environments. That happens when the compound is an ionic salt made of two different polyatomic ions, each containing the same element. Let's check each option.

1. NH₂OH (hydroxylamine)

   This is a neutral molecule. Let the oxidation number of N be $x$.

   H is +1 (three H atoms: two bonded to N, one to O).

   O is –2.

   Sum: $x + 3(+1) + (-2) = 0 \Rightarrow x + 3 - 2 = 0 \Rightarrow x = -1$.

   So nitrogen has a single oxidation state: –1. No two different states here.

2. NH₄NO₃ (ammonium nitrate)

   This is an ionic compound: NH₄⁺ and NO₃⁻.

   In NH₄⁺: let N be $x$. H is +1 each, four H atoms.

   $x + 4(+1) = +1 \Rightarrow x + 4 = +1 \Rightarrow x = -3$.

   In NO₃⁻: let N be $y$. O is –2 each, three O atoms.

   $y + 3(-2) = -1 \Rightarrow y - 6 = -1 \Rightarrow y = +5$.

   Nitrogen appears as –3 in the ammonium ion and +5 in the nitrate ion — two different oxidation states in the same compound. This is the answer.

3. N₂H₄ (hydrazine)

   Neutral molecule. Let N be $x$. H is +1 each, four H atoms.

   $2x + 4(+1) = 0 \Rightarrow 2x + 4 = 0 \Rightarrow x = -2$.

   Both nitrogens have the same oxidation state: –2.

4. N₃H (hydrogen azide, HN₃)

   HN₃ is a single neutral covalent molecule (H–N=N⁺=N⁻, a resonance hybrid), not a salt built from two separate ions. Treating the three nitrogens as equivalent gives an average oxidation number: …