Shape of the compound depend on type and number of electron pair around central atom . These electron pair repel each other and stay as far as possible . The repulsion sequence is as
`L.P - L.P. gt B.P. - L.P. gt B.P - B.P.`
Choose the incorrect match

Options

A. `{:("Compound" ,, "Structure") , ((1) SnCl_(2) ,, "Linear"):}`
B. `{:("Compound" ,, "Structure") , ((2) CO_(2) ,, "Linear"):}`
C. `{:("Compound" ,, "Structure") , ((3)I_(3)^(-) ,, "Linear"):}`
D. `{:("Compound" ,, "Structure") , ((4)N_(3)^(-) ,, "Linear"):}`

Answer

To solve the question regarding the shape of compounds based on the type and number of electron pairs around the central atom, we will analyze the given compounds and their respective shapes based on the provided information.

### Step-by-Step Solution:

1. **Understanding Electron Pair Repulsion**:
- The shape of a molecule is determined by the arrangement of electron pairs (bond pairs and lone pairs) around the central atom. The repulsion between these pairs dictates their spatial arrangement.
- The order of repulsion strength is: Lone Pair - Lone Pair (L.P - L.P) > Bond Pair - Lone Pair (B.P - L.P) > Bond Pair - Bond Pair (B.P - B.P).

2. **Identifying the Compounds**:
- We will analyze the following compounds:
- SnCl₂
- CO₂
- I₃⁻
- N₃⁻

3. **Analyzing SnCl₂**:
- **Valence Electrons**: Sn has 4 valence electrons (Group 14).
- **Hybridization Calculation**:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Valence Electrons} + \text{Monovalent Atoms} + \text{Charge} \right) = \frac{1}{2} (4 + 2 + 0) = 3 \quad \text{(sp² hybridization)} \]
- **Bond Pairs and Lone Pairs**: SnCl₂ has 2 bond pairs and 1 lone pair.
- **Shape**: With sp² hybridization and 1 lone pair, the shape is bent, not linear.

4. **Analyzing CO₂**:
- **Valence Electrons**: C has 4 valence electrons.
- **Hybridization Calculation**:
\[ \text{Hybridization} = \frac{1}{2} (4 + 0 + 0) = 2 \quad \text{(sp hybridization)} \]
- **Bond Pairs**: CO₂ has 2 bond pairs and no lone pairs.
- **Shape**: The shape is linear.

5. **Analyzing I₃⁻**:
- **Valence Electrons**: I has 7 valence electrons, and with a negative charge, we add 1.
- **Hybridization Calculation**:
\[ \text{Hybridization} = \frac{1}{2} (7 + 2 + 1) = 5 \quad \text{(sp³d hybridization)} \]
- **Bond Pairs and Lone Pairs**: I₃⁻ has 2 bond pairs and 3 lone pairs.
- **Shape**: The shape is linear.

6. **Analyzing N₃⁻**:
- **Valence Electrons**: N has 5 valence electrons, and with a negative charge, we add 1.
- **Hybridization Calculation**:
\[ \text{Hybridization} = \frac{1}{2} (5 + 0 + 1) = 3 \quad \text{(sp² hybridization)} \]
- **Bond Pairs**: N₃⁻ has 3 bond pairs and no lone pairs.
- **Shape**: The shape is linear.

7. **Conclusion**:
- The incorrect match is for SnCl₂, which is stated to have a linear shape, but it actually has a bent shape due to the presence of a lone pair.

### Final Answer:
The incorrect match is SnCl₂, which is bent, not linear.

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