• # question_answer $N{{a}_{2}}{{S}_{2}}{{O}_{3}}+{{I}_{2}}\to$Product is [BHU 2003] A) $N{{a}_{2}}S$ B) $NaI$ C) $N{{a}_{2}}{{S}_{4}}{{O}_{6}}$ D) ${{S}_{2}}$

$2N{{a}_{2}}{{S}_{2}}{{O}_{3}}+{{I}_{2}}\to 2NaI+N{{a}_{2}}{{S}_{4}}{{O}_{6}}$.