Q103 / 200 · MHT CET - Medical 2026 (23 April Shift 2) · Chemistry
What is the bond angle Cl-B-Cl and molecular geometry of $\text{BCl}_3$ molecule?
- A$120^\circ$ and trigonal bipyramidal
- B$109^\circ 28'$ and tetrahedral
- C$107^\circ 18'$ and tetrahedral
- Correct answerD$120^\circ$ and trigonal planar
Solution
The central atom in $\text{BCl}_3$ is Boron (B).
The number of valence electrons in Boron is $3$. It forms $3$ single bonds with $3$ Chlorine atoms.
Number of bond pairs = $3$
Number of lone pairs = $0$
Steric number = Bond pairs + Lone pairs = $3 + 0 = 3$
A steric number of $3$ corresponds to $sp^2$ hybridization.
Since there are no lone pairs on the central atom, the molecular geometry is trigonal planar.
The bond angle for a regular trigonal planar geometry is $120^\circ$.
Answer: $120^\circ$ and trigonal planar
The number of valence electrons in Boron is $3$. It forms $3$ single bonds with $3$ Chlorine atoms.
Number of bond pairs = $3$
Number of lone pairs = $0$
Steric number = Bond pairs + Lone pairs = $3 + 0 = 3$
A steric number of $3$ corresponds to $sp^2$ hybridization.
Since there are no lone pairs on the central atom, the molecular geometry is trigonal planar.
The bond angle for a regular trigonal planar geometry is $120^\circ$.
Answer: $120^\circ$ and trigonal planar
Hence, the correct answer is option: 'D'.