free radicals and optical isomers

my last optical isomer doubt at least for a few days from now on!!![3][3]

we tell that in free radical substitution reaction takes place, the alkyl radical formed is flat in nature...and that the attack might take place equally likely from either front side or back side...

exactly this was assumed by Brown Khrasch and Chao to prove the mechanism of free radical substitution is

a) R-H + Cl ^. → R ^. + H-Cl
b) R ^. + Cl₂ → R-Cl + Cl ^.

they assumed that free radicals are sp² hybridised and thus flat..so if the above reaction goes on the optically active (S)-(+)-CH₃CH₂CH(CH₃)CH₂Cl will give a racemic mixture of CH₃CH₂CCl(CH₃)CH₂Cl as one of the products and thus the product will be optically inactive...

experimentally the product was found to be optically inactive and thus the free radical concept was established...

now my question is that HOW THEY BECAME SO MUCH SURE THAT THE FREE RADICAL IS SP² HYBRIDISED? I SAY IT IS SP³ HYBRIDISED AND HAS A BIT PYRAMIDAL STRUCTURE AND THAT IT IS NOT FLAT...NOW HOW WUD THE CONSISTENCE OF THE ASSUMPTIONS OF THE ABOVE PROOF BE VALIDATED IF THE FREE RADICAL IS NOT FLAT???

P.S.:: THE FREE RADICALS ARE INDEED PYRAMIDAL IS SHAPE AS SUGGESTED BY X-RAY DIFFRACTION!!! SO EXPLAIN NOW!!