p-Block Elements-I 1 Question 10
10. In compounds of type $E \mathrm{Cl}_{3}$, where $E=\mathrm{B}, \mathrm{P}, \mathrm{As}$ or $\mathrm{Bi}$, the angles $\mathrm{Cl}-E-\mathrm{Cl}$ for different $E$ are in the order
(1999, 2M)
(a) $\mathrm{B}>\mathrm{P}=\mathrm{As}=\mathrm{Bi}$
(b) $\mathrm{B}>\mathrm{P}>\mathrm{As}>\mathrm{Bi}$
(c) $\mathrm{B}<\mathrm{P}=\mathrm{As}=\mathrm{Bi}$
(d) B $<$ P $<$ As $<$ Bi
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Answer:
Correct Answer: 10. (b)
Solution:
- In $\mathrm{BCl}_{3}$, bond angle $=120^{\circ}$.
In $\mathrm{PCl}{3}, \mathrm{AsCl}{3}$ and $\mathrm{BiCl}_{3}$, central atom is $s p^{3}$ hybridised. Since $\mathrm{P}, \mathrm{As}$ and $\mathrm{Bi}$ are from the same group, bond angle decreases down the group. Hence, overall order of bond angle is :
$$ \mathrm{B}>\mathrm{P}>\mathrm{As}>\mathrm{Bi} $$
