Answer: Two equivalent resonance structures: S central with one S=O double bond and one S–O single bond; the singly bonded O carries a −1 charge and S carries a +1 charge. (Resonance average → both S–O bonds have bond order 1.5.) SO2 is bent (≈119°), S has one lone pair (electron geometry trigonal planar, molecular shape bent). Total valence electrons = 18.
Explanation: Count valence electrons: S (6) + 2×O (6 each) = 18. Place S in center, connect to both O with single bonds (uses 4 e). Complete octets on oxygens, then place remaining electrons on S; to reduce formal charges move one lone pair from S to form a double bond with one O. That yields two resonance forms (double and single O swapped). Formal charges in each resonance form: double-bond O = 0, single-bond O = −1, S = +1. Alternatively you can draw a structure with two S=O double bonds and a lone pair on S (expanded octet) giving all formal charges 0; quantum/experimental data show the actual bonds are intermediate, consistent with the resonance (average) picture and bond order ≈1.5. Hybridization on S ≈ sp2 (one lone pair + two bonding regions).