In which of the following molecules no. of lonepairs and bond pairs on central atom are not equal :-

Options

A. `H_(2)O`
B. `I_(3)^(-)`
C. `O_(2)F_(2)`
D. `SCl_(2)`

Explanation

To determine which molecule has an unequal number of lone pairs and bond pairs on the central atom, we will analyze each molecule step by step.

### Step 1: Analyze the Water Molecule (H₂O)
1. **Identify the central atom**: The central atom in water is oxygen (O).
2. **Count the valence electrons**: Oxygen has 6 valence electrons.
3. **Count the surrounding atoms**: There are 2 hydrogen atoms (H), which are monovalent (1 bond each).
4. **Calculate total electrons**: Total = 6 (from O) + 2 (from 2 H) = 8 electrons.
5. **Determine hybridization**: The number of hybrid orbitals = Total electrons / 2 = 8 / 2 = 4.
6. **Calculate lone pairs**: Lone pairs = Hybrid orbitals - Bond pairs = 4 - 2 = 2 lone pairs.
7. **Count bond pairs**: There are 2 bond pairs (one for each H).
8. **Conclusion**: Water has 2 lone pairs and 2 bond pairs (equal).

### Step 2: Analyze Iodine Trifluoride (I₃⁻)
1. **Identify the central atom**: The central atom is iodine (I).
2. **Count the valence electrons**: Iodine has 7 valence electrons, and there is an additional electron due to the negative charge, making it 8 total.
3. **Count the surrounding atoms**: There are 2 iodine atoms (monovalent).
4. **Calculate total electrons**: Total = 8 (from central I) + 2 (from 2 surrounding I) = 10 electrons.
5. **Determine hybridization**: The number of hybrid orbitals = Total electrons / 2 = 10 / 2 = 5.
6. **Calculate lone pairs**: Lone pairs = Hybrid orbitals - Bond pairs = 5 - 2 = 3 lone pairs.
7. **Count bond pairs**: There are 2 bond pairs (one for each surrounding I).
8. **Conclusion**: I₃⁻ has 3 lone pairs and 2 bond pairs (not equal).

### Step 3: Analyze O₂F₂ (Dioxygen Difluoride)
1. **Identify the central atom**: The central atom is oxygen (O).
2. **Count the valence electrons**: Each oxygen has 6 valence electrons, and there are 2 fluorine atoms (F), each contributing 7 electrons.
3. **Calculate total electrons**: Total = 6 (from O) + 6 (from 2 F) = 12 electrons.
4. **Determine hybridization**: The number of hybrid orbitals = Total electrons / 2 = 12 / 2 = 6.
5. **Calculate lone pairs**: Lone pairs = Hybrid orbitals - Bond pairs = 6 - 4 = 2 lone pairs.
6. **Count bond pairs**: There are 4 bond pairs (2 for each F).
7. **Conclusion**: O₂F₂ has 2 lone pairs and 4 bond pairs (not equal).

### Step 4: Analyze SCl₂ (Sulfur Dichloride)
1. **Identify the central atom**: The central atom is sulfur (S).
2. **Count the valence electrons**: Sulfur has 6 valence electrons, and there are 2 chlorine atoms (Cl), each contributing 7 electrons.
3. **Calculate total electrons**: Total = 6 (from S) + 14 (from 2 Cl) = 20 electrons.
4. **Determine hybridization**: The number of hybrid orbitals = Total electrons / 2 = 20 / 2 = 10.
5. **Calculate lone pairs**: Lone pairs = Hybrid orbitals - Bond pairs = 10 - 2 = 8 lone pairs.
6. **Count bond pairs**: There are 2 bond pairs (one for each Cl).
7. **Conclusion**: SCl₂ has 8 lone pairs and 2 bond pairs (not equal).

### Final Conclusion
From the analysis, the only molecule where the number of lone pairs and bond pairs on the central atom are not equal is **I₃⁻**.