Question

Which of the following molecules has no lone pair of electrons on central atom?

• A. $SO_2$
• B. $SF_6$
• C. $NH_3$
• D. $SF_4$

Hint:

Logic Tip: A quick way to find lone pairs is simply matching the group valency to the number of monovalent atoms attached. Sulfur is in Group 16 (6 outer electrons). Attaching 6 monovalent Fluorines perfectly exhausts all 6 electrons, leaving no lone pairs and creating a perfect octahedral geometry.

Answer 1

The Correct Option is B

Concept:
To determine the number of lone pairs on the central atom, we use Valence Shell Electron Pair Repulsion (VSEPR) theory. The number of lone pairs ($lp$) is calculated by finding the total valence electrons of the central atom, subtracting the electrons shared in bonds with surrounding atoms, and dividing by 2. $$lp = \frac{\text{Valence } e^- - \text{Bonding } e^-}{2}$$
Step 1: Analyze the sulfur dioxide ($SO_2$) molecule.
Central atom: Sulfur (S) belongs to Group 16, so it has 6 valence electrons. It forms double bonds with two Oxygen atoms, utilizing 4 valence electrons ($2 \times 2 = 4$). Remaining electrons = $6 - 4 = 2$ electrons = 1 lone pair.
Step 2: Analyze the sulfur hexafluoride ($SF_6$) molecule.
Central atom: Sulfur (S) has 6 valence electrons. It forms single bonds with six Fluorine atoms, utilizing all 6 valence electrons ($6 \times 1 = 6$). Remaining electrons = $6 - 6 = 0$ electrons = 0 lone pairs.
Step 3: Analyze the ammonia ($NH_3$) and sulfur tetrafluoride ($SF_4$) molecules.
For $NH_3$: Central atom: Nitrogen (N) has 5 valence electrons. It forms single bonds with three Hydrogen atoms, using 3 electrons. Remaining electrons = $5 - 3 = 2$ electrons = 1 lone pair. For $SF_4$: Central atom: Sulfur (S) has 6 valence electrons. It forms single bonds with four Fluorine atoms, using 4 electrons. Remaining electrons = $6 - 4 = 2$ electrons = 1 lone pair. Conclusion: Only $SF_6$ has zero lone pairs on the central atom.