Ratio of hybrid and unhybrid orbitals taking part in bond formation in ethylene molecule

No of sp2 hybrid orbitals = 2 x Number of hybridized orbitals on each carbon                                 = 2 x 3                                  = 6No of pure orbitals = number of hydrogen atoms  + 2 [number of pi bonds]                              = 4 + 2 x 1                              = 6Therefore ratio of H.O and P.O = 1: 1