Chemistry Examples

$C O H_{2}$

Step 1

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

$H = 2 (1) = 2$

Step 2

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

$O = 1 (- 2) = - 2$

Step 3

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

$2 - 2 + 1 x = 0$

Step 4

Simplify $2 + - 2 + 1 x$ .

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Step 4.1

Add $2$ and $- 2$ .

$0 + 1 x = 0$

Step 4.2

Add $0$ and $1 x$ .

$1 x = 0$

Step 4.3

Multiply $x$ by $1$ .

$x = 0$

Step 5

$C$ is using an oxidation state of $0$ .

$0$

Step 6

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

$H = 2, O = - 2, C = 0$