Find the Oxidation Numbers Bi(OH)_3

$B i {(O H)}_{3}$

Since $B i$ is in column $0$ of the periodic table, it will share $0$ electrons and use an oxidation state of $0$ .

$B i = 1 (0) = 0$

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

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

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

$0 - 2 + 3 x = 0$

Solve the equation for $x$ .

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Subtract $2$ from $0$ .

$- 2 + 3 x = 0$

Add $2$ to both sides of the equation.

$3 x = 2$

Divide each term by $3$ and simplify.

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Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Divide $x$ by $1$ .

$x = \frac{2}{3}$

$H$ is using an oxidation state of $\frac{2}{3}$ .

$\frac{2}{3}$

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

$B i = 0, O = - 2, H = - 2$

Find the Oxidation Numbers Bi(OH)_3

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