question_answer

Hybridisation of central atom in \[N{{F}_{3}}\] is

A)  \[s{{p}^{3}}\]                                   

B)  \[sp\]

C)  \[s{{p}^{2}}\]                                   

D)  \[ds{{p}^{2}}\]

Correct Answer: A

Solution :

                 Number of hybrid orbitals = number of \[\sigma \] bonds + number of lone pairs \[\therefore \] Hybridisation of N is \[s{{p}^{3}}\]and geometry of the molecule is pyramidal.