To determine the oxidation state of sulfur in sodium sulfite Na2SO3 , we will use the oxidation state rules:1. The oxidation state of an atom in a free element is zero.2. The oxidation state of a monatomic ion is equal to its charge.3. The sum of the oxidation states of all atoms in a neutral molecule is zero, and in an ion, it is equal to the ion's charge.In sodium sulfite Na2SO3 , we have two sodium Na atoms, one sulfur S atom, and three oxygen O atoms. Sodium is an alkali metal, and its oxidation state in compounds is usually +1. Oxygen is a non-metal, and its oxidation state in most compounds is -2.Let's assign oxidation states to the atoms in Na2SO3:- Sodium Na : +1 each - Oxygen O : -2 each - Sulfur S : x unknown Now, we will apply rule 3. Since sodium sulfite is a neutral molecule, the sum of the oxidation states of all atoms should be zero.2 +1 + x + 3 -2 = 0Solve for x:2 + x - 6 = 0x - 4 = 0x = 4So, the oxidation state of sulfur S in sodium sulfite Na2SO3 is +4.