The ratio between hybridized and pure orbitals in o-xylene are.

Let us draw the structure of o-xylene as follows and observe the number of hybridized carbons first.1,2,3,4,5,6 Carbons, Are sp2 hybridization,  Due to which we can say that around each carbon they are three sigma bonds,'C'3 σ  bonds. Carbon number 7 and 8 carbons are sp3 hybridization, Therefore around them all are sigma bonds ‘C’ 4σ.Calculation:Each sp2  carbon have 3 hybrid orbitalsEach  sp3  carbon have 4 hybrid orbitalsEach  sp3 carbon have zero pure orbitalsAll hydrogen’s have pure orbitalsTotal hybrid orbitals:  sp2=3×6=18                                   sp3=4×2=8                                 ------------------------                                                                                                 = 26Total pure orbitals from the 6 sp2 carbons are (one p-orbital is pure from each carbon) = 6                       All hydrogens are pure, 10 Hydrogens are present =10                                                     Total pure orbitals are=16Now ratio:  2616=138=13:8