What is the oxidation state of nitrogen in sodium nitrite, NaNO2 a, +3 b.-1 c +2 d.none of the provided options e. +1′
The Correct Answer and Explanation is:
Correct Answer: a. +3
To determine the oxidation state of nitrogen in sodium nitrite (NaNO₂), we apply a set of oxidation number rules:
- Sodium (Na) is a Group 1 metal, so it always has an oxidation number of +1.
- Oxygen (O) in most compounds, including nitrites, has an oxidation number of –2.
- The sum of the oxidation numbers in a neutral compound must equal 0.
The chemical formula for sodium nitrite is NaNO₂. This means it contains one sodium atom (Na⁺), one nitrogen atom (N), and two oxygen atoms (O²⁻). Let us assign oxidation numbers and calculate:
- Oxidation number of sodium: +1
- Oxidation number of each oxygen: –2
- Oxidation number of nitrogen: let it be x
The total oxidation sum must be zero:(+1)+x+2(–2)=0(+1) + x + 2(–2) = 0(+1)+x+2(–2)=0+1+x–4=0+1 + x – 4 = 0+1+x–4=0x–3=0⇒x=+3x – 3 = 0 \Rightarrow x = +3x–3=0⇒x=+3
So, the oxidation state of nitrogen in sodium nitrite is +3.
This result makes chemical sense. Nitrogen forms several oxoanions with varying oxidation states. For example:
- In nitrate (NO₃⁻), nitrogen has an oxidation state of +5.
- In nitrite (NO₂⁻), nitrogen has an oxidation state of +3.
- In ammonia (NH₃), nitrogen has an oxidation state of –3.
These differences in oxidation states reflect nitrogen’s ability to form compounds in multiple valence states due to its variable bonding capabilities. Understanding oxidation numbers is important for recognizing redox reactions, predicting compound properties, and writing balanced chemical equations.
