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The ratio between hybridized and pure orbitals in o-xylene are.
detailed solution
Correct option is C
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=13:8Talk to our academic expert!
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