Find Hybridization of Central Atom Step by Step

 

❓ Question

In $\text{SO}_2$, $\text{NO}_2^-$, and $\text{N}_3^-$, the hybridizations at the central atom are respectively:

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Hybridization
👉 Steric number
👉 VSEPR theory.

Main idea:

$$\boxed{ \text{Steric Number} = \text{Bond pairs}+\text{Lone pairs} }$$

Then:

• SN = 2 → $sp$
  
• SN = 3 → $sp^2$
  
• SN = 4 → $sp^3$

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🔷 Step 1 — Hybridization of $SO_2$ 💯

Structure of $SO_2$:

Central atom = S

Sulfur has:

✔ 2 sigma bonds
✔ 1 lone pair

Thus steric number:

$$SN=2+1=3$$

Hence:

$$\boxed{ SO_2 \rightarrow sp^2 }$$

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🔷 Step 2 — Hybridization of $NO_2^-$

Nitrite ion:

Central atom = N

Nitrogen has:

✔ 2 sigma bonds
✔ 1 lone pair

Thus:

$$SN=3$$

Hence:

$$\boxed{ NO_2^- \rightarrow sp^2 }$$

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🔷 Step 3 — Hybridization of $N_3^{-}$

Azide ion:

$$N_3^-$$

Central nitrogen forms:

✔ 2 sigma bonds
✔ 0 lone pairs

Steric number:

$$SN=2$$

Hence:

$$\boxed{ N_3^- \rightarrow sp }$$

Linear geometry confirms this.

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🔷 Step 4 — Final Matching

$$SO_2 \rightarrow sp^2$$
$$NO_2^- \rightarrow sp^2$$
$$N_3^- \rightarrow sp$$

Thus correct option:

$$\boxed{ sp^2,\ sp^2,\ sp }$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Resonance structures dekhkar hybridization confuse kar dena

❌ Multiple bonds ko multiple sigma bonds maan lena

Remember:

Only sigma bonds count in steric number.

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✅ Final Answer

$$\boxed{ sp^2,\ sp^2,\ sp }$$

(Option 4)

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