Find the Oxidation Numbers Fe(OH)_2

$F e {(O H)}_{2}$

Step 1

Since $F e$ is in column $3$ of the periodic table, it will share $3$ electrons and use an oxidation state of $3$ .

$F e = 1 (3) = 3$

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 $H$ , set up an equation of each oxidation state found earlier and set it equal to $0$ .

$3 - 2 + 2 x = 0$

Step 4

Solve the equation for $x$ .

Add $3$ and $- 2$ .

$1 + 2 x = 0$

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

Step 5

$H$ is using an oxidation state of $- \frac{1}{2}$ .

$- \frac{1}{2}$

Step 6

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

$F e = 3, O = - 2, H = - \frac{1}{2}$