MCQ
Which one of the following is the correct set with respect to molecule, hybridization and shape?
- A$BeCl_2 , sp^2$, linear
- B$BeCl_2 , sp^2$ , triangular planar
- C$BCl_3 , sp^2$, triangular planar
- D$BCl_3 , sp^3,$ tetrahedral
✓
Answer
In $\mathrm{BCl}_{3}$, boron is sp $^{2}$ hybridised as there are 3 bond pairs and 0 lone pairs and thus, $\mathrm{BCl}_{3}$ is triangular planar is shape.
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