Find the Oxidation Numbers Na_2S_3O_6

$N a_{2} S_{3} O_{6}$

Step 1

Since $N a$ is in column $1$ of the periodic table, it will share $1$ electrons and use an oxidation state of $1$ .

$N a = 2 (1) = 2$

Step 2

Since $O$ is in column $6$ of the periodic table, it will share $6$ electrons and use an oxidation state of $6$ .

$O = 6 (- 2) = - 12$

Step 3

To find the oxidation state of $S$ , set up an equation of each oxidation state found earlier and set it equal to $0$ .

$2 - 12 + 3 x = 0$

Step 4

Solve the equation for $x$ .

Add $2$ and $- 12$ .

$- 10 + 3 x = 0$

Add $10$ to both sides of the equation.

$3 x = 10$

Divide each term in $3 x = 10$ by $3$ and simplify.

Divide each term in $3 x = 10$ by $3$ .

$\frac{3 x}{3} = \frac{10}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{10}{3}$

Divide $x$ by $1$ .

$x = \frac{10}{3}$

Step 5

$S$ is using an oxidation state of $\frac{10}{3}$ .

$\frac{10}{3}$

Step 6

This is the full list of oxidation states for this molecule.

$N a = 2, O = - 12, S = \frac{10}{3}$