Chemistry Examples

$N a_{2} S_{2} O_{4}$

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 $6$ .

$O = 4 (- 2) = - 8$

Step 3

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

$2 - 8 + 2 x = 0$

Step 4

Solve the equation for $x$ .

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Step 4.1

Add $2$ and $- 8$ .

$- 6 + 2 x = 0$

Step 4.2

Add $6$ to both sides of the equation.

$2 x = 6$

Step 4.3

Divide each term in $2 x = 6$ by $2$ and simplify.

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Step 4.3.1

Divide each term in $2 x = 6$ by $2$ .

$\frac{2 x}{2} = \frac{6}{2}$

Step 4.3.2

Simplify the left side.

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Step 4.3.2.1

Cancel the common factor of $2$ .

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Step 4.3.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{6}{2}$

Step 4.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{6}{2}$

Step 4.3.3

Simplify the right side.

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Step 4.3.3.1

Divide $6$ by $2$ .

$x = 3$

Step 5

$S$ is using an oxidation state of $3$ .

$3$

Step 6

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

$N a = 2, O = - 8, S = 3$