Answer :
The types of hybrid orbitals of nitrogen in [tex]N O_{2}^{+}, NO_3^-, and \ NH_4^+[/tex] respectively are expected to be [tex](ii) \ sp, \ sp^{2} \ and \ sp^{3}[/tex]
Explanation:
From the given,
[tex]N O_{2}^{+} \ is \ nitrite \ ion[/tex]
[tex]N O_{3}^{-} \ is \ nitrate \ ion[/tex]
[tex]NH_4^+ \ is \ ammonium \ ion[/tex]
Formula for hybridization is:
[tex]\frac{1}{2}(V+H-C+A)[/tex]
Where,
H = Number of "surrounding monovalent atoms"
C = "Cationic charge"
V = Number of "valance electrons" in central atom
A = "Anionic charge"
- [tex]N O_{2}^{+}=\left(\frac{1}{2}\right) \times(5+0-1+0)=2=s p[/tex]
- [tex]N O_{3}^{-}=\left(\frac{1}{2}\right) \times(5+2-0+1)=3=s p^{2}[/tex]
- [tex]N H_{4}^{+}=\left(\frac{1}{2}\right) \times(5+4-1+0)=4=s p^{3}[/tex]
Hence (ii) is correct answer.