The pair having similar hybrid geometry is :

Options

A. `BF_3,NH_3`
B. `H_2O,C_2H_2`
C. `CO_2,SO_2`
D. `CH_4H_2O`

Answer

To determine the pair having similar hybrid geometry, we need to analyze the hybridization of the given compounds. Let's break down the analysis step-by-step:

### Step 1: Identify the Compounds
We need to examine the hybridization of the following compounds:
1. BF3
2. NH3
3. H2O
4. C2H2
5. CO2
6. SO2
7. CH4

### Step 2: Determine the Hybridization of Each Compound
1. **BF3 (Boron Trifluoride)**
- Boron has 3 valence electrons and forms 3 bonds with fluorine.
- There are no lone pairs on boron.
- **Steric Number = 3 (3 bond pairs + 0 lone pairs)**
- **Hybridization = sp²**
- **Geometry = Trigonal Planar**

2. **NH3 (Ammonia)**
- Nitrogen has 5 valence electrons and forms 3 bonds with hydrogen.
- There is 1 lone pair on nitrogen.
- **Steric Number = 4 (3 bond pairs + 1 lone pair)**
- **Hybridization = sp³**
- **Geometry = Trigonal Pyramidal**

3. **H2O (Water)**
- Oxygen has 6 valence electrons and forms 2 bonds with hydrogen.
- There are 2 lone pairs on oxygen.
- **Steric Number = 4 (2 bond pairs + 2 lone pairs)**
- **Hybridization = sp³**
- **Geometry = Bent**

4. **C2H2 (Acetylene)**
- Each carbon has 4 valence electrons and forms a triple bond with another carbon and a single bond with hydrogen.
- There are no lone pairs on carbon.
- **Steric Number = 2 (2 bond pairs + 0 lone pairs)**
- **Hybridization = sp**
- **Geometry = Linear**

5. **CO2 (Carbon Dioxide)**
- Carbon has 4 valence electrons and forms double bonds with two oxygen atoms.
- There are no lone pairs on carbon.
- **Steric Number = 2 (2 bond pairs + 0 lone pairs)**
- **Hybridization = sp**
- **Geometry = Linear**

6. **SO2 (Sulfur Dioxide)**
- Sulfur has 6 valence electrons and forms 2 bonds with oxygen.
- There is 1 lone pair on sulfur.
- **Steric Number = 3 (2 bond pairs + 1 lone pair)**
- **Hybridization = sp²**
- **Geometry = Bent**

7. **CH4 (Methane)**
- Carbon has 4 valence electrons and forms 4 bonds with hydrogen.
- There are no lone pairs on carbon.
- **Steric Number = 4 (4 bond pairs + 0 lone pairs)**
- **Hybridization = sp³**
- **Geometry = Tetrahedral**

### Step 3: Compare Hybridizations
Now, we compare the hybridizations:
- **BF3**: sp²
- **NH3**: sp³
- **H2O**: sp³
- **C2H2**: sp
- **CO2**: sp
- **SO2**: sp²
- **CH4**: sp³

### Step 4: Identify Pairs with Similar Hybridization
From the analysis:
- **NH3, H2O, and CH4** all have sp³ hybridization.
- **BF3 and SO2** both have sp² hybridization.
- **C2H2 and CO2** both have sp hybridization.

### Conclusion
The pairs having similar hybrid geometry are:
- **NH3 and CH4** (both sp³)
- **BF3 and SO2** (both sp²)
- **C2H2 and CO2** (both sp)

### Final Answer
The pair having similar hybrid geometry is **NH3 and CH4** or **BF3 and SO2** or **C2H2 and CO2**.

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