A) \[B{{H}_{3}}\]
B) \[B{{F}_{3}}\]
C) \[BC{{l}_{3}}\]
D) \[BB{{r}_{3}}\]
Correct Answer: A
Solution :
\[B{{H}_{3}}\]does not exist in free form because H-atoms in \[B{{H}_{3}}\]have no free electrons to form \[p\pi -\rho \pi \]back bonding and thus boron has incomplete octet and hence \[B{{H}_{3}}\]molecule dimerise to form \[{{B}_{2}}{{H}_{6}}.\]
You need to login to perform this action.
You will be redirected in
3 sec
