The unpaired electron of CH3 free radical occupies p-orbital. — Chemical Bonding Chemistry Question

💡 Solution & Explanation

Step 1: Apply Bent's rule: more electronegative substituents (like F) occupy orbitals with greater p-character, leaving more s-character for the S-O bond. Step 2: Greater s-character in the S-O bond leads to a shorter and stronger S-O bond. Thus, SOF2 has the shortest S-O bond. Step 3: Since Bromine is the least electronegative substituent among F, Cl, and Br, it leaves the least s-character for the S-O bond, making the S-O bond length maximum in SOBr2, corresponding to option (b).