A) \[B{{F}_{3}}<N{{F}_{3}}<P{{F}_{3}}<Cl{{F}_{3}}\]
B) \[Cl{{F}_{3}}<P{{F}_{3}}<N{{F}_{3}}<B{{F}_{3}}\]
C) \[B{{F}_{3}}\approx N{{F}_{3}}<P{{F}_{3}}<Cl{{F}_{3}}\]
D) \[B{{F}_{3}}<N{{F}_{3}}<P{{F}_{3}}<Cl{{F}_{3}}\]
Correct Answer: B
Solution :
\[B{{F}_{3}}\] has \[s{{p}^{2}}\] hybridization and \[{{120}^{o}}\] bond angle (maximum). Its structure is planar. Higher the electronegativity of central atom (in \[s{{p}^{3}}\] hybridized) larger the bond angle but presence of lone pair of electrons decreases the bond angle. Greater the number of lone pair of electrons lower the bond angle.
You need to login to perform this action.
You will be redirected in
3 sec
