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.You need to login to perform this action.
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