The incorrect statement (s) regarding `overline(*)(C)X_(3)`

Options

A. If electronegativity of surrounding element X is less than 2.5, then central carbon atom used almost 33% s-character in their hybrid bonding orbital
B. If electronegativity of surrounding element X is less than 2.5, then central carbon atom used almost 25% s-character in their hybrid bonding orbital
C. If X is F, then species should be polar and pyramidal
D. If X is H, then species should be polar and planar

Explanation

To solve the question regarding the incorrect statements about the compound \( \overline{C}X_3 \), we will analyze the statements based on the hybridization, geometry, and polarity of the molecule.

### Step-by-Step Solution:

1. **Understanding the Structure**:
- The compound \( \overline{C}X_3 \) consists of a central carbon atom (C) bonded to three atoms (X).
- The presence of three X atoms indicates that the carbon is likely forming four bonds (including one bond with another atom, which we will consider as hydrogen or another group).

2. **Determining Hybridization**:
- The steric number is calculated as the number of sigma bonds plus the number of lone pairs around the central atom.
- In this case, since there are four sigma bonds (three with X and one with another atom), the steric number is 4.
- A steric number of 4 indicates that the hybridization of the carbon atom is \( sp^3 \).

3. **Calculating s-Character**:
- In \( sp^3 \) hybridization, the s-character is calculated as follows:
\[ \text{s-character} = \frac{1}{4} \times 100\% = 25\% \]
- Therefore, any statement claiming that the s-character is 33% is incorrect.

4. **Analyzing the Statements**:
- **Statement 1**: "If the electronegativity of the surrounding atom is less than 2.5, then the central carbon atom uses almost 33% of the s-character in their hybrid orbital."
- This is **incorrect** because the s-character is actually 25%.
- **Statement 2**: "If the electronegativity of the surrounding atom is less than 2.5, then the central atom uses almost 25% of the s-character in their hybrid bonding orbital."
- This is **correct** as it aligns with our calculation.
- **Statement 3**: "If X is F, then the species will be polar and pyramidal."
- This is **correct** because the presence of fluorine (high electronegativity) creates a dipole moment, resulting in a polar molecule.
- **Statement 4**: "If X is H, then the species should be polar and planar."
- This is **incorrect** because the molecule is non-polar due to the symmetrical arrangement of hydrogen atoms and is tetrahedral, not planar.

5. **Conclusion**:
- The incorrect statements regarding \( \overline{C}X_3 \) are **Statement 1** and **Statement 4**.

### Final Answer:
The incorrect statements regarding \( \overline{C}X_3 \) are Statement 1 and Statement 4.

---