Find the Oxidation Numbers CuOH

$C u O H$

Step 1

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

$C u = 1 (3) = 3$

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 $0$ .

$3 - 2 + 1 x = 0$

Step 4

Solve the equation for $x$ .

Simplify $3 + - 2 + 1 x$ .

Multiply $x$ by $1$ .

$3 + - 2 + x = 0$

Add $3$ and $- 2$ .

$1 + x = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

$H$ is using an oxidation state of $- 1$ .

$- 1$

Step 6

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

$C u = 3, O = - 2, H = - 1$