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.
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