Which of the following is (are) V-shaped?

Options

A. `S_(3)^(2-)`
B. `I_(3)^(-)`
C. `N_(3)^(-)`
D. `I_(3)^(+)`

Answer

To determine which of the given species are V-shaped, we need to analyze the molecular geometry of each compound based on their hybridization and the number of lone pairs and bond pairs around the central atom. Here’s a step-by-step solution:

### Step 1: Analyze S3 2- (Thiosulfate Ion)
1. **Identify the central atom**: Sulfur (S).
2. **Count valence electrons**: Sulfur has 6 valence electrons.
3. **Determine the number of bonds**: S3 2- has two double bonds with sulfur atoms.
4. **Calculate the effective number of electrons**:
- Valence electrons = 6
- Bonding electrons = 0 (since double bonds are not counted as monovalent)
- Charge = +2 (as it’s a 2- charge, we add 2)
- Total = 6 + 0 + 2 = 8
5. **Use the hybridization formula**:
- Hybridization = (Total electrons)/2 = 8/2 = 4 → sp3 hybridized.
6. **Determine lone pairs**:
- Total pairs = 4 (from sp3)
- Bond pairs = 2 (from double bonds)
- Lone pairs = 4 - 2 = 2.
7. **Draw the structure**:
- This results in a V-shaped structure due to the presence of 2 lone pairs.

### Step 2: Analyze I3- (Triiodide Ion)
1. **Identify the central atom**: Iodine (I).
2. **Count valence electrons**: Iodine has 7 valence electrons.
3. **Determine the number of bonds**: I3- has two bonds with other iodine atoms.
4. **Calculate the effective number of electrons**:
- Valence electrons = 7
- Bonding electrons = 2 (two bonds)
- Charge = -1 (subtract 1)
- Total = 7 + 2 - 1 = 8
5. **Use the hybridization formula**:
- Hybridization = 8/2 = 4 → sp3 hybridized.
6. **Determine lone pairs**:
- Total pairs = 4 (from sp3)
- Bond pairs = 2
- Lone pairs = 4 - 2 = 2.
7. **Draw the structure**:
- The 3 lone pairs occupy equatorial positions, leading to a linear structure, not V-shaped.

### Step 3: Analyze N3- (Azide Ion)
1. **Identify the central atom**: Nitrogen (N).
2. **Count valence electrons**: Nitrogen has 5 valence electrons.
3. **Determine the number of bonds**: N3- has a linear arrangement (N≡N-N).
4. **Calculate the effective number of electrons**:
- Valence electrons = 5
- Bonding electrons = 6 (three bonds)
- Charge = -1 (add 1)
- Total = 5 + 6 + 1 = 12
5. **Use the hybridization formula**:
- Hybridization = 12/2 = 6 → sp3d2 hybridized.
6. **Determine lone pairs**:
- Total pairs = 6
- Bond pairs = 3
- Lone pairs = 6 - 3 = 3.
7. **Draw the structure**:
- The structure is linear, not V-shaped.

### Step 4: Analyze I3+ (Triiodide Cation)
1. **Identify the central atom**: Iodine (I).
2. **Count valence electrons**: Iodine has 7 valence electrons.
3. **Determine the number of bonds**: I3+ has two bonds with other iodine atoms.
4. **Calculate the effective number of electrons**:
- Valence electrons = 7
- Bonding electrons = 2 (two bonds)
- Charge = +1 (subtract 1)
- Total = 7 + 2 - 1 = 8
5. **Use the hybridization formula**:
- Hybridization = 8/2 = 4 → sp3 hybridized.
6. **Determine lone pairs**:
- Total pairs = 4
- Bond pairs = 2
- Lone pairs = 4 - 2 = 2.
7. **Draw the structure**:
- This results in a V-shaped structure due to the presence of 2 lone pairs.

### Conclusion
The species that are V-shaped are S3 2- and I3+. Therefore, the correct options are 1 and 4.