What is the stereochemistry of free radical mechanism

The reaction of n-butane with Br2 under light forms 2-bromobutane in the form of

(A) R isomer

(B) S isomer

(C) 1:1 mixture of R and S isomer

(D) 2:3 mixture of R and S isomer

You should get a 1:1 mixture.
Br-Br bond cleavage forms 2 Br radicals. One radical reacts with n-butane, and a n-butane radical is formed. This n-butane radical abstracts a Br atom from another Br-Br molecule, forming 2-bromobutane and a new Br radical (this step propagates again and again).

The carbon with the single electron in the n-butane radical is sp2 hybridized (planar symmetry). The single electron is in an unhybridized p orbital and the new bond to Br can form on either side of carbon (top or bottom of the plane), thus giving a 1:1 racemic mixture of the R and S isomer.