Chemical Bonding Result Question 16
16. In compounds of type $E \mathrm{Cl}_3$, where $E=\mathrm{B}, \mathrm{P}, \mathrm{As}$ or Bi , the angles $\mathrm{Cl}-E-\mathrm{Cl}$ is in 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) $\mathrm{B}<\mathrm{P}<\mathrm{As}<\mathrm{Bi}$
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Answer:
Correct Answer: 16. ( b )
Solution:
- When $E=\mathrm{B}$ in $\mathrm{BCl}_3$, bond angle is $120^{\circ}$. When $E=\mathrm{P}$, $As$ or $Bi$ in $E \mathrm{Cl}_3$, hybridisation at $E$ will be $s p^3$. Also, if central atoms are from same group, bond angle decreases down the group provided all other things are similar. Hence, the order of bond angles is $\mathrm{BCl}_3>\mathrm{PCl}_3>\mathrm{AsCl}_3>\mathrm{BiCl}_3$
BETA
