Calculate the oxidation number of each sulfur atom in the following compounds:
$Na_2S_2O_3$
$Na_2S_4O_6$
$Na_2SO_3$
$Na_2SO_4$

(N/A) $Na_2S_2O_3$: In the structure,one $S$ atom is bonded to the central $S$ atom via a coordinate bond (acting as an acceptor),so its oxidation number is $-2$. For the central $S$ atom $(x)$:
$2(+1) + x + (-2) + 3(-2) = 0$
$2 + x - 2 - 6 = 0$
$\therefore x = +6$
Thus,the oxidation numbers are $-2$ and $+6$.
$(b)$ $Na_2S_4O_6$: In the structure,the two central $S$ atoms are bonded to each other,so their oxidation number is $0$. For the other two $S$ atoms $(x)$:
$2(+1) + 2(x) + 2(0) + 6(-2) = 0$
$2 + 2x - 12 = 0$
$2x = 10$
$\therefore x = +5$
Thus,the oxidation numbers are $0$ and $+5$.
$(c)$ $Na_2SO_3$: Using the average oxidation state method:
$2(+1) + x + 3(-2) = 0$
$2 + x - 6 = 0$
$\therefore x = +4$
$(d)$ $Na_2SO_4$: Using the average oxidation state method:
$2(+1) + x + 4(-2) = 0$
$2 + x - 8 = 0$
$\therefore x = +6$