oxidation no of nitrogen in nano3

Key Ideas You Need

1. Oxidation Number (O.N.) Rules

   • Alkali metals ($Li$, $Na$, $K$…) in compounds: +1.
   • Oxygen in most compounds: −2.
   • The algebraic sum of oxidation numbers of all atoms in a neutral compound is zero.

2. How To Use the Rules
   a. Identify every known oxidation number in the formula.
   b. Assign an unknown (say $x$) to the element whose O.N. you need.
   c. Write an equation adding up all oxidation numbers, set the total to the net charge (zero for neutral molecules).
   d. Solve the simple algebra for $x$.

3. Why Each Step Is Logical

   • We treat oxidation numbers like book-keeping charges; the molecule overall must be financially balanced (net zero).
   • By substituting known values, the only mystery left is the element you’re interested in — solve for it just like solving for $x$ in basic algebra.

Imagine Lego Bricks

Think of a molecule like $NaNO_3$ as a small Lego house made from different coloured bricks.
Each colour has a fixed score:

• The yellow brick (sodium, $Na$) always gives +1 point.
• Each red brick (oxygen, $O$) always gives −2 points.

All the points added together must be zero because the whole Lego house is neutral (no extra charge).
So if three red bricks already give $3 \times (−2) = −6$ points, and one yellow brick gives +1 point, we need to find out how many points the blue brick (nitrogen) must give so everything balances to zero.

That balancing act is exactly how we find the oxidation number of nitrogen!

Step-by-Step Calculation for $NaNO_3$

1. Assign Known Oxidation Numbers

   • Sodium ($Na$): +1 (alkali metal rule).
   • Oxygen ($O$): −2 each.

2. Let Nitrogen’s O.N. Be $x$.

3. Write Charge-Balance Equation

4. Solve for $x$

5. Final Answer
   Oxidation number of nitrogen in $NaNO_3$ is +5.