Which type of hybridisation is shown by carbon atoms from left to right in the given compound :
` CH_(2) = CH - C -= N`?

Options

A. `sp^(2),sp^(2),sp`
B. `sp^(2),sp,sp`
C. `sp,sp^(2),sp`
D. `sp^(3),sp^(2),sp`

Explanation

To determine the hybridization of the carbon atoms in the compound \( CH_2 = CH - C \equiv N \), we will analyze each carbon atom from left to right.

### Step 1: Identify the Structure
The compound can be represented as follows:
- The first carbon (C1) is part of the \( CH_2 \) group.
- The second carbon (C2) is part of the double bond with C1.
- The third carbon (C3) is connected to C2 and is part of the triple bond with nitrogen (N).

### Step 2: Analyze the First Carbon (C1)
- C1 has two hydrogen atoms and forms a double bond with C2.
- The bonds formed by C1:
- 2 sigma bonds (with 2 H atoms)
- 1 sigma bond (with C2)
- 1 pi bond (with C2)
- **Total bond pairs**: 3 (2 from H and 1 from C2)
- **Lone pairs**: 0
- **Steric number** = Number of bond pairs + Number of lone pairs = 3 + 0 = 3
- **Hybridization**: Since the steric number is 3, the hybridization of C1 is \( sp^2 \).

### Step 3: Analyze the Second Carbon (C2)
- C2 is connected to C1 and C3.
- The bonds formed by C2:
- 1 sigma bond (with C1)
- 1 sigma bond (with C3)
- 1 pi bond (with C1)
- **Total bond pairs**: 3 (1 with C1, 1 with C3, and the pi bond is not counted as a sigma bond)
- **Lone pairs**: 0
- **Steric number** = 3 + 0 = 3
- **Hybridization**: Since the steric number is 3, the hybridization of C2 is also \( sp^2 \).

### Step 4: Analyze the Third Carbon (C3)
- C3 is connected to C2 and nitrogen (N).
- The bonds formed by C3:
- 1 sigma bond (with C2)
- 2 pi bonds (with N)
- **Total bond pairs**: 2 (1 with C2 and 2 with N)
- **Lone pairs**: 0
- **Steric number** = 2 + 0 = 2
- **Hybridization**: Since the steric number is 2, the hybridization of C3 is \( sp \).

### Final Summary
- C1 (first carbon): \( sp^2 \)
- C2 (second carbon): \( sp^2 \)
- C3 (third carbon): \( sp \)

### Answer
The hybridization of the carbon atoms from left to right in the compound \( CH_2 = CH - C \equiv N \) is \( sp^2, sp^2, sp \).

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