Problem 1: Determine hybridization and shape of $SF_{4}$

Step 1: Identify the central atom and count its valence electrons.

• Central atom: S (Group 16) → 6 valence electrons

Step 2: Count the number of bonds.

• $SF_{4}$ has 4 S–F bonds → 4 bond pairs

Step 3: Calculate lone pairs on S. $\text{Lone pairs} = \frac{\text{valence e}^- - \text{no. of bonds}}{2} = \frac{6 - 4}{2} = 1 \text{ lone pair}$

Step 4: Calculate steric number. $\text{SN} = 4 \text{ (bonds)} + 1 \text{ (lone pair)} = 5$

Step 5: Determine hybridization.

• SN = 5 → $sp^{3}$d hybridization

Step 6: Identify electron geometry.

• SN = 5 → Trigonal bipyramidal (TBP) electron geometry

Step 7: Place lone pair in equatorial position (preferred).

• In TBP, lone pairs go to equatorial positions (2 interactions at 90° vs 3 at 90° for axial)

Step 8: Determine molecular shape (remove lone pair).

• 2 axial F + 2 equatorial F + 1 equatorial lone pair → Seesaw (or sawhorse) shape

Step 9: Identify bond angles.

• Axial F–S–F ≈ 173° (compressed from 180° by equatorial lone pair)
• Equatorial F–S–F ≈ 102° (compressed from 120° by lone pair)

Answer: $SF_{4}$ — $sp^{3}$d hybridization, seesaw molecular shape.