What is the oxidation number of Sulphur atom in Na_(2)S_(2)O_(3) (Na: 1 A group . 0: 6A group). A) -2 B) +4 C) +2 D) +3 E) -1

To find the oxidation number of sulfur in $Na_{2}S_{2}O_{3}$, we can use the following steps:<br /><br />1. Assign oxidation numbers to the elements based on their group numbers:<br /> - Sodium (Na) is in group 1A, so its oxidation number is +1.<br /> - Oxygen (O) is in group 6A, so its oxidation number is -2.<br /><br />2. equation based on the sum of oxidation numbers:<br /> The sum of the oxidation numbers of all atoms in a neutral compound is zero. In this case, we have:<br /> $2(Na) + 2(S) + 3(O) = 0$<br /><br />3. Substitute the known oxidation numbers into the equation:<br /> $2(+1) + 2(S) + 3(-2) = 0$<br /> $2 + 2(S) - 6 = 0$<br /> $2(S) - 4 = 0$<br /><br />4. Solve for the oxidation number of sulfur:<br /> $2(S) = 4$<br /> $S = 2$<br /><br />Therefore, the oxidation number of sulfur in $Na_{2}S_{2}O_{3}$ is +2.<br /><br />The correct answer) $+2$.