Match the shape of molecules in Column 1 with the type of hybridisation in Column II.
`{:("Column I","Column II"),((a)"Tetrahedral",(i)sp^(2)),((b)"Trigonal",(ii) sp),((c)"Linear",(iii)sp^(3)),((d)"Trigonal bipyramidal",(iv)sp^(3)d):}`

Answer

To solve the problem of matching the shapes of molecules with their corresponding types of hybridization, we can follow these steps:

### Step 1: Identify the Shape and Hybridization of Each Molecule

1. **Tetrahedral (a)**:
- Example: Methane (CH₄)
- Central atom: Carbon (C)
- Valence electrons of C: 4
- Number of monovalent atoms (H): 4
- Hybridization calculation:
\[ H = \frac{1}{2} \left( \text{Valence electrons} + \text{Monovalent atoms} + \text{Charge} \right) = \frac{1}{2} (4 + 4 + 0) = \frac{8}{2} = 4 \]
- Hybridization: sp³

2. **Trigonal Planar (b)**:
- Example: Boron Trichloride (BCl₃)
- Central atom: Boron (B)
- Valence electrons of B: 3
- Number of monovalent atoms (Cl): 3
- Hybridization calculation:
\[ H = \frac{1}{2} (3 + 3 + 0) = \frac{6}{2} = 3 \]
- Hybridization: sp²

3. **Linear (c)**:
- Example: Beryllium Dichloride (BeCl₂)
- Central atom: Beryllium (Be)
- Valence electrons of Be: 2
- Number of monovalent atoms (Cl): 2
- Hybridization calculation:
\[ H = \frac{1}{2} (2 + 2 + 0) = \frac{4}{2} = 2 \]
- Hybridization: sp

4. **Trigonal Bipyramidal (d)**:
- Example: Phosphorus Pentachloride (PCl₅)
- Central atom: Phosphorus (P)
- Valence electrons of P: 5
- Number of monovalent atoms (Cl): 5
- Hybridization calculation:
\[ H = \frac{1}{2} (5 + 5 + 0) = \frac{10}{2} = 5 \]
- Hybridization: sp³d

### Step 2: Match the Shapes with Hybridizations

Now that we have determined the hybridizations for each shape, we can match them:

- (a) Tetrahedral → (iii) sp³
- (b) Trigonal Planar → (i) sp²
- (c) Linear → (ii) sp
- (d) Trigonal Bipyramidal → (iv) sp³d

### Final Matches
- a → iii
- b → i
- c → ii
- d → iv

### Summary of the Matches:
- (a) Tetrahedral → (iii) sp³
- (b) Trigonal Planar → (i) sp²
- (c) Linear → (ii) sp
- (d) Trigonal Bipyramidal → (iv) sp³d

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