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  .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=18sp3=4×2=8 --------                    = 26Total pure orbitals from the 6 sp2carbons 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