The comcept of redistribution of energy in different orbitals of an atom associated with different energies to give new orbitals of equal (or somethimes it may be non-equal) energy oriented in space in definite directions is called hybridization and formed new orbitals are called hybrid orbitals. The bonds formed by such orbitals are called hybrid bonds. The process of mixing of orbitals itself requires some energy. Thus, some additional energy, is needed for the hybridisation (mixing) of atomic orbitals.
Q. Select from each set the molecule or ion having the smallest bond angle :
(i) `H_(2)Se, H_(2)Te and PH_(3)`
(ii) `NO_(2)^(-) and NH_(2)^(-)`
(iii) `POF_(3) and POCl_(3)` (`X-P-X` angle)
(iv) `OSF_(2)Cl_(2) and SF_(2)(CH_(3))_(2)` (`F-S-F` angle)

Options

A. `H_(2)Se, NH_(2)^(-) POF_(3) and OSF_(2)Cl_(2)`
B. `H_(2)Te, NO_(2)^(-), POF_(3) and SF_(2)(CH_(3))_(2)`
C. `PH_(3) ,NH_(2)^(-), POCl_(3) and SF_(2)(CH_(3))_(2)`
D. `H_(2)Te, NH_(2)^(-), POF_(3) and SF_(2)(CH_(3))_(2)`

Explanation

To solve the question of selecting the molecule or ion with the smallest bond angle from each set, we will analyze the hybridization and molecular geometry of each compound. Here’s a step-by-step solution:

### Step 1: Analyze the first set - `H2Se`, `H2Te`, and `PH3`
1. **Determine Hybridization**:
- All three molecules have a central atom with a steric number of 4 (2 bonds + 2 lone pairs for H2Se and H2Te, and 3 bonds + 1 lone pair for PH3).
- Therefore, they all exhibit **sp³ hybridization**.

2. **Draw Structures**:
- **PH3** has one lone pair, leading to a bond angle slightly less than 109.5°.
- **H2Se** and **H2Te** both have two lone pairs on the central atom, which will cause greater repulsion and a smaller bond angle.

3. **Compare Bond Angles**:
- As we move down the group from Se to Te, the size of the central atom increases, leading to a decrease in bond angle.
- Therefore, **H2Te** will have the smallest bond angle due to the larger size of Te compared to Se.

**Answer for Set 1**: **H2Te**

### Step 2: Analyze the second set - `NO2^(-)` and `NH2^(-)`
1. **Determine Hybridization**:
- For **NO2^(-)**: The nitrogen has 5 valence electrons + 1 (negative charge) = 6. The steric number is 3 (1 lone pair + 2 bonds), resulting in **sp² hybridization**.
- For **NH2^(-)**: The nitrogen has 5 valence electrons + 2 (2 bonds + 1 negative charge) = 7. The steric number is 4 (2 bonds + 2 lone pairs), resulting in **sp³ hybridization**.

2. **Compare Bond Angles**:
- **NO2^(-)** has a bond angle of approximately 120° due to sp² hybridization.
- **NH2^(-)** has a bond angle of approximately 109.5° due to sp³ hybridization.

**Answer for Set 2**: **NH2^(-)**

### Step 3: Analyze the third set - `POF3` and `POCl3`
1. **Determine Hybridization**:
- Both molecules have a phosphorus atom with a steric number of 4 (1 double bond + 3 single bonds), resulting in **sp³ hybridization**.

2. **Compare Bond Angles**:
- The presence of highly electronegative fluorine atoms in **POF3** will cause the bond angles to be smaller due to the stronger bond character compared to the chlorine atoms in **POCl3**.
- Thus, **POF3** will have a smaller bond angle than **POCl3**.

**Answer for Set 3**: **POF3**

### Step 4: Analyze the fourth set - `OSF2Cl2` and `SF2(CH3)2`
1. **Determine Hybridization**:
- Both molecules have a central atom with a steric number of 4 (considering lone pairs and bonds), resulting in **sp³ hybridization**.

2. **Compare Bond Angles**:
- In **OSF2Cl2**, the presence of two lone pairs will cause a significant reduction in the bond angle.
- In **SF2(CH3)2**, the presence of bulky CH3 groups will also affect the bond angle, but the lone pairs in **OSF2Cl2** will have a more pronounced effect.

**Answer for Set 4**: **SF2(CH3)2**

### Final Answers:
1. **H2Te**
2. **NH2^(-)**
3. **POF3**
4. **SF2(CH3)2**