The hybridizations of atomic orbitals of nitrogen in NO2+, NO3- and NH4+ respectively are
- sp, sp3 and sp2
- sp2, sp3 and sp
- sp, sp2 and sp3
- sp2, sp and sp3
Answer
(3) sp, sp2 and sp3
Solution
Number of hybrid orbitals = 12(V+M-C+A)
NO2+: 12(5+0-1+0)=2 ⇒ sp (linear, O= +N=O)
NO3-: 12(5+0+1+0)=3 ⇒ sp2 (trigonal planar)
NH4+: 12(5+4-1+0)=4 ⇒ sp3 (tetrahedral)
∴ sp, sp2 and sp3.
Answer: (3)