The shape of $SF_{4}$ molecule is:

Question

Accepted Answer

The shape of the $ SF_4 $ molecule is best described as "see-saw," which corresponds to Option 3. Here's a detailed explanation supporting this:

1. **Explanation for Option 3 (See-saw):**
   - The $ SF_4 $ molecule consists of a sulfur atom bonded to four fluorine atoms. Sulfur has six valence electrons, and when it forms bonds with four fluorine atoms, it uses four of these electrons, leaving two electrons as a lone pair.
   - The electron geometry around the sulfur atom is trigonal bipyramidal, which includes five regions of electron density (four bonding pairs and one lone pair).
   - The presence of the lone pair affects the molecular geometry, causing a distortion from the ideal trigonal bipyramidal shape. The lone pair occupies one of the equatorial positions to minimize repulsion, resulting in a "see-saw" shape for the molecule.
   - This shape is characterized by two axial and two equatorial fluorine atoms, with the lone pair occupying the third equatorial position.

2. **Explanation for why other options are incorrect:**

- **Option 1 (Tetrahedral):**
     - A tetrahedral shape occurs when there are four bonding pairs and no lone pairs around the central atom. In $ SF_4 $, there is one lone pair, which alters the geometry from tetrahedral to see-saw. Therefore, $ SF_4 $ cannot be tetrahedral.

- **Option 2 (Square planar):**
     - A square planar shape is typically observed in molecules with a central atom surrounded by four bonding pairs and two lone pairs, such as in certain transition metal complexes. $ SF_4 $ has only one lone pair, and its electron geometry is trigonal bipyramidal, not octahedral, which is required for a square planar shape.

- **Option 4 (Square pyramidal):**
     - A square pyramidal shape arises from a central atom with five bonding pairs and one lone pair, resulting in an octahedral electron geometry. In $ SF_4 $, there are only four bonding pairs and one lone pair, leading to a trigonal bipyramidal electron geometry, not octahedral. Thus, the shape cannot be square pyramidal.

In conclusion, the presence of one lone pair in $ SF_4 $ and the resulting distortion from the ideal trigonal bipyramidal geometry leads to the "see-saw" shape, making Option 3 the correct choice.