In the structure of `H_(2)CSF_(4)`, which of the following statement is/are correct?

Options

A. Two `C-H` bonds are in the same plane of axial `S-F` bonds
B. Two `C-H` bonds are in the same plane of equitorial `S-F` bonds
C. Total six atoms are in the plane
D. Equitorial `S-F` plane is perpendicular to the nodal plane of `pi-` bonds

Explanation

To determine which statements about the structure of \( H_2CSF_4 \) are correct, we will analyze the molecular geometry and bonding of the compound step by step.

### Step 1: Determine the Molecular Geometry
1. **Identify the central atom**: In \( H_2CSF_4 \), carbon (C) is the central atom.
2. **Count the surrounding atoms**: The molecule has two hydrogen (H) atoms, one sulfur (S) atom, and four fluorine (F) atoms.
3. **Determine the hybridization of carbon**: Carbon in \( H_2CSF_4 \) forms a double bond with sulfur and single bonds with two hydrogen atoms. The hybridization of carbon is \( sp^3 \) due to the presence of four regions of electron density (two C-H bonds, one C=S double bond, and two C-F bonds).

### Step 2: Draw the Lewis Structure
1. **Draw the structure**: Carbon is bonded to sulfur with a double bond and to two hydrogen atoms. The sulfur atom is bonded to four fluorine atoms.
2. **Identify the geometry**: The molecular geometry around the carbon atom is tetrahedral due to the \( sp^3 \) hybridization.

### Step 3: Analyze the Statements
1. **Statement 1**: "2 CH bonds are in the same plane of axial SF bond."
- **Analysis**: The two C-H bonds are oriented in a way that they are in the same plane as the axial S-F bonds. This statement is **correct**.

2. **Statement 2**: "2 CH bonds are in the same plane of equatorial SF bond."
- **Analysis**: The C-H bonds are not in the same plane as the equatorial S-F bonds. This statement is **incorrect**.

3. **Statement 3**: "Total 6 atoms are in the plane."
- **Analysis**: The total number of atoms (2 H, 1 C, 1 S, and 2 F) that are in the same plane is indeed 6. This statement is **correct**.

4. **Statement 4**: "Equatorial SF plane is perpendicular to the nodal plane of pi-bond."
- **Analysis**: The equatorial S-F bonds are indeed perpendicular to the nodal plane of the π-bond formed between C and S. This statement is **correct**.

### Conclusion
Based on the analysis:
- **Correct Statements**: 1, 3, and 4.
- **Incorrect Statement**: 2.

### Final Answer
The correct statements about the structure of \( H_2CSF_4 \) are:
- Statement 1: Correct
- Statement 2: Incorrect
- Statement 3: Correct
- Statement 4: Correct