Oxidation Number Method for Nitric Acid Reaction

❓ Question

Balance the equation:

$$Mg + HNO_3 \rightarrow Mg(NO_3)_2 + N_2O + H_2O$$

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🖼 Question Image

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✍️ Short Explanation

This is a redox balancing problem.

👉 Magnesium gets oxidised
👉 Nitrogen in nitrate gets reduced
👉 Use oxidation number method carefully.

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🔷 Step 1 — Oxidation States 💯

For magnesium:

$$Mg^0 \rightarrow Mg^{2+}$$

Loss of electrons:

$$2e^-$$

So Mg is oxidised.

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Nitrogen in:

$$HNO_3$$

has oxidation state:

$$+5$$

In:

$$N_2O$$

average oxidation state of nitrogen:

$$+1$$

Reduction per nitrogen:

$$+5 \rightarrow +1$$

Gain:

$$4e^-$$

Since:

$$N_2O$$

contains 2 nitrogen atoms,

total electrons gained:

$$8e^-$$

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🔷 Step 2 — Electron Balance

Each Mg loses:

$$2e^-$$

To supply:

$$8e^-$$

Required Mg atoms:

$$4$$

So:

$$4Mg$$

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🔷 Step 3 — Skeleton Equation

$$4Mg + HNO_3 \rightarrow 4Mg(NO_3)_2 + N_2O + H_2O$$

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🔷 Step 4 — Balance Nitrogen

Right side nitrogen:

From:

$$4Mg(NO_3)_2$$

Nitrogen atoms:

$$4\times2=8$$

Plus:

$$N_2O$$

contains:

$$2$$

Total:

$$10$$

So left side:

$$10HNO_3$$

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🔷 Step 5 — Balance Hydrogen and Oxygen

Hydrogen on left:

$$10$$

So water:

$$5H_2O$$

Now oxygen check:

Left:

$$10\times3=30$$

Right:

$$4\times6+1+5$$
$$=24+1+5$$
$$=30$$

Balanced.

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✅ Final Balanced Equation

$$\boxed{ 4Mg+10HNO_3 \rightarrow 4Mg(NO_3)_2+N_2O+5H_2O }$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ $N_2O$ mein nitrogen oxidation state galat nikal dena

❌ Electron balance skip kar dena

❌ Oxygen directly trial method se balance karna

Remember:

$$\text{First balance electron transfer}$$

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📚 Related Topics

• Balance Redox Reaction Using Oxidation Number Method
• Zn and Nitrate Reaction Ion Electron Method
• Ion Electron Method for Alkaline Redox Reaction