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) \[B{{F}_{3}}=N{{F}_{3}}<P{{F}_{3}}<Cl{{F}_{3}}\]
D) \[B{{F}_{3}}<N{{F}_{3}}<P{{F}_{3}}<Cl{{F}_{3}}\]
Correct Answer: B
Solution :
[b] \[B{{F}_{3}}\]: planar\[(s{{p}^{2}}),\] \[120{}^\circ \] bond angle (maximum),Higher the electro negativity 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 no. of lone pair of electrons, lower the bond angle.You need to login to perform this action.
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