A) \[BC{{l}_{3}}\]
B) \[BB{{r}_{3}}\]
C) \[B{{F}_{3}}\]
D) same for all
Correct Answer: D
Solution :
Key Idea: \[B{{F}_{3}},\,BC{{l}_{3}},\,BB{{r}_{3}}\] all have boron in \[s{{p}^{2}}\] hybrid state. \[\because \] Hybridisation of boron is Same in \[B{{F}_{3}}\,,\,BC{{l}_{3}}\] and \[BB{{r}_{3}}\] and there is no lone, pair of electron. \[\therefore \] All of them have trigonal planar shape and bond angle of \[{{120}^{o}}\].
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