Chemistry Examples

$B a C r_{2} O_{7}$

Step 1

Since $B a$ is in column $2$ of the periodic table, it will share $2$ electrons and use an oxidation state of $2$ .

$B a = 1 (2) = 2$

Step 2

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

$O = 7 (- 2) = - 14$

Step 3

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

$2 - 14 + 2 x = 0$

Step 4

Solve the equation for $x$ .

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Step 4.1

Add $2$ and $- 14$ .

$- 12 + 2 x = 0$

Step 4.2

Add $12$ to both sides of the equation.

$2 x = 12$

Step 4.3

Divide each term in $2 x = 12$ by $2$ and simplify.

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Step 4.3.1

Divide each term in $2 x = 12$ by $2$ .

$\frac{2 x}{2} = \frac{12}{2}$

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.

$\frac{2 x}{2} = \frac{12}{2}$

Step 4.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{12}{2}$

Step 4.3.3

Simplify the right side.

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Step 4.3.3.1

Divide $12$ by $2$ .

$x = 6$

Step 5

$C r$ is using an oxidation state of $6$ .

$6$

Step 6

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

$B a = 2, O = - 14, C r = 6$