In compound `O_(2)SC(NH_2)_(2)`, the geometry around S and N are respectively.

Options

A. trigonal planar, trigonal pyramidal
B. tetrahedral,pyramidal
C. trigonal planar, tetrahedral
D. linear, pyramidal

Answer

To determine the geometry around sulfur (S) and nitrogen (N) in the compound \( O_2SC(NH_2)_2 \), we will follow these steps:

### Step 1: Identify the Structure of the Compound
The compound \( O_2SC(NH_2)_2 \) consists of a sulfur atom (S) bonded to two oxygen atoms (O) via double bonds and a carbon atom (C) which is further bonded to two amine groups (NH2).

### Step 2: Determine the Hybridization of Sulfur (S)
1. **Count the Valence Electrons for Sulfur**: Sulfur has 6 valence electrons.
2. **Identify the Bonds**: Sulfur forms two double bonds with oxygen and one single bond with carbon. This gives a total of 3 bonds.
3. **Apply the Hybridization Formula**:
\[ H = \frac{1}{2} \left( V + M - C + A \right) \]
Where:
- \( V \) = number of valence electrons of the central atom (S = 6)
- \( M \) = number of monovalent atoms (0 in this case)
- \( C \) = cationic charge (0)
- \( A \) = anionic charge (0)

Plugging in the values:
\[ H = \frac{1}{2} \left( 6 + 0 + 0 + 0 \right) = \frac{6}{2} = 3 \]

4. **Determine Hybridization**: A hybridization value of 3 corresponds to \( sp^2 \) hybridization.

### Step 3: Determine the Geometry around Sulfur
With \( sp^2 \) hybridization, the geometry around sulfur is **trigonal planar**.

### Step 4: Determine the Hybridization of Nitrogen (N)
1. **Identify the Bonds**: The nitrogen atom is bonded to two hydrogen atoms and one carbon atom, and it has one lone pair of electrons.
2. **Count the Sigma Bonds and Lone Pairs**:
- Number of sigma bonds = 3 (2 from H and 1 from C)
- Number of lone pairs = 1

3. **Apply the Hybridization Formula**:
\[ H = \text{Number of sigma bonds} + \text{Number of lone pairs} = 3 + 1 = 4 \]

4. **Determine Hybridization**: A hybridization value of 4 corresponds to \( sp^3 \) hybridization.

### Step 5: Determine the Geometry around Nitrogen
With \( sp^3 \) hybridization, the geometry around nitrogen is **tetrahedral**.

### Final Answer
Thus, the geometries around sulfur and nitrogen in the compound \( O_2SC(NH_2)_2 \) are respectively:
- Sulfur (S): **Trigonal Planar**
- Nitrogen (N): **Tetrahedral**