Explanation

(c)
Sulphurous acid $H_{2}{SO}_{3}$
$2 + x + ( - 2 \times 3) = 0$
$x - 4 = 0$
$\therefore\ \ \ \ x = 4$
Pyrosulphuric acid ($H_{2}S_{2}O_{7})$
$2 + 2x + ( - 2 \times 7) = 0$
or $2x = 12$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore = 6$
Thiosulphuric acid ($H_{2}S_{2}O_{3})$
$2 + 2x + ( - 2 \times 3) = 0\ \ \ \ \ \ \ \ \$
$or$ $2x = 4$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 2$
Dithionous acid ($H_{2}S_{2}O_{4})$
$2 + 2x + ( - 2 \times 4) = 0$
$2x = 6$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore\ x =$ 3