Out of `CHCl_3, CH_4 and SF_4` the molecules having regular geometry are

Options

A. `CHCl_(3)` only
B. `CHCl_(3) and SF_(4)`
C. `CH_(4)`
D. `CH_(4) and SF_(4)`

Answer

To determine which of the molecules \( CHCl_3 \), \( CH_4 \), and \( SF_4 \) have regular geometry, we need to analyze each molecule based on the criteria for regular geometry, which includes having the same bonded electron pairs, a dipole moment of zero, and no lone pairs on the central atom.

### Step-by-Step Solution:

1. **Analyze \( CHCl_3 \)**:
- The central atom is carbon (C), which is bonded to three chlorine (Cl) atoms and one hydrogen (H) atom.
- The molecular geometry is tetrahedral because there are four bonded pairs of electrons.
- However, chlorine is more electronegative than hydrogen, leading to a net dipole moment in the direction of the Cl atoms. Thus, the dipole moment is not zero.
- **Conclusion**: \( CHCl_3 \) does not have regular geometry.

2. **Analyze \( SF_4 \)**:
- The central atom is sulfur (S), which is bonded to four fluorine (F) atoms.
- Sulfur has 6 valence electrons, and with 4 fluorine atoms, we have 4 bond pairs. This gives us a total of 6 (valence electrons) + 4 (bond pairs) = 10 electrons, leading to a hybridization of \( sp^3d \).
- The geometry of \( SF_4 \) is described as a seesaw shape (or trigonal bipyramidal with one equatorial position occupied by a lone pair).
- The presence of the lone pair means that the dipole moments of the fluorine bonds do not cancel out completely.
- **Conclusion**: \( SF_4 \) does not have regular geometry.

3. **Analyze \( CH_4 \)**:
- The central atom is carbon (C), which is bonded to four hydrogen (H) atoms.
- The molecular geometry is tetrahedral with four identical bond pairs (C-H).
- Since all the bonds are identical and symmetrical, the dipole moments cancel each other out, resulting in a net dipole moment of zero.
- There are no lone pairs on the central atom (C).
- **Conclusion**: \( CH_4 \) has regular geometry.

### Final Answer:
The molecule with regular geometry is \( CH_4 \).

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