The concept of hybridisation has been introduced to explain the shapes of molecules. It involves the intermixing of two or more atomic orbitals belonging to same atom but in or more atomic orbitals beloging to sasme atom but in different sub-shells so as to intermix and redistibute energies to from equivalent orbitals called hybrid orbitals.Depending upon toh enumber and nature of the orbitals involved, the hybridisation may be divided into sp (linear), `sp^(2)` (trigonal), `sp^(3)` (tetrahedral), `sp^(3)d` (trigonal bipyramidal), `sp^(3)d^(3)` (octahedral) and `sp^(3)d^(3)` (pentagonal bipyramidal) types. it may be noted that the orbitals of isolated atoms never hybridise and they do so at the time of bond formation.
The hybrid state of carbon in `C_(2)H_(2)` is same as that of carbon in:

Options

A. `C_(2)H_(6)`
B. `CO_(2)`
C. Benzene
D. C (diamond).

Explanation

To determine the hybrid state of carbon in \( C_2H_2 \) (acetylene), we need to analyze the bonding and calculate the steric number for the carbon atoms involved. Here’s a step-by-step solution:

### Step 1: Identify the structure of \( C_2H_2 \)
- The molecular structure of acetylene consists of two carbon atoms connected by a triple bond, with each carbon atom also bonded to one hydrogen atom.
- The structure can be represented as:
\[ H - C \equiv C - H \]

### Step 2: Calculate the steric number for carbon in \( C_2H_2 \)
- The steric number is calculated using the formula:
\[ \text{Steric Number} = \text{Number of Sigma Bonds} + \text{Number of Lone Pairs} \]
- In \( C_2H_2 \), each carbon atom forms:
- 1 sigma bond with the hydrogen atom
- 1 sigma bond as part of the triple bond with the other carbon atom (the triple bond consists of 1 sigma and 2 pi bonds)
- Therefore, for each carbon atom:
- Number of sigma bonds = 1 (with H) + 1 (with C) = 2
- Number of lone pairs = 0
- Thus, the steric number for each carbon atom is:
\[ \text{Steric Number} = 2 + 0 = 2 \]

### Step 3: Determine the hybridization
- Based on the steric number, we can determine the hybridization:
- A steric number of 2 corresponds to \( sp \) hybridization.
- Therefore, the hybridization of carbon in \( C_2H_2 \) is \( sp \).

### Step 4: Find a molecule with the same hybridization
- We need to identify which of the given molecules has the same \( sp \) hybridization as carbon in \( C_2H_2 \).
- The options provided include various compounds. We will analyze each option to find the one with \( sp \) hybridization.

1. **\( C_2H_6 \) (Ethane)**:
- Structure: \( H_3C - C(H_3) \)
- Steric Number: 4 (4 sigma bonds, 0 lone pairs) → Hybridization: \( sp^3 \) (not a match)

2. **\( CO_2 \) (Carbon Dioxide)**:
- Structure: \( O = C = O \)
- Steric Number: 2 (2 sigma bonds, 0 lone pairs) → Hybridization: \( sp \) (a match)

3. **Benzene**:
- Structure: \( C_6H_6 \) (hexagonal ring)
- Steric Number: 3 (3 sigma bonds, 0 lone pairs) → Hybridization: \( sp^2 \) (not a match)

4. **Diamond**:
- Structure: Tetrahedral arrangement of carbon atoms
- Steric Number: 4 (4 sigma bonds, 0 lone pairs) → Hybridization: \( sp^3 \) (not a match)

### Conclusion
- The molecule that has the same hybridization as carbon in \( C_2H_2 \) is \( CO_2 \) (carbon dioxide).

### Final Answer
The hybrid state of carbon in \( C_2H_2 \) is the same as that of carbon in \( CO_2 \).

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