Chemistry Examples

$S O C l_{2}$

Step 1

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

$S = 1 (6) = 6$

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

$6 - 2 + 2 x = 0$

Step 4

Solve the equation for $x$ .

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Step 4.1

Add $6$ and $- 2$ .

$4 + 2 x = 0$

Step 4.2

Subtract $4$ from both sides of the equation.

$2 x = - 4$

Step 4.3

Divide each term in $2 x = - 4$ by $2$ and simplify.

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Step 4.3.1

Divide each term in $2 x = - 4$ by $2$ .

$\frac{2 x}{2} = \frac{- 4}{2}$

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.

$\frac{2 x}{2} = \frac{- 4}{2}$

Step 4.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 4}{2}$

Step 4.3.3

Simplify the right side.

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Step 4.3.3.1

Divide $- 4$ by $2$ .

$x = - 2$

Step 5

$C l$ is using an oxidation state of $- 2$ .

$- 2$

Step 6

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

$S = 6, O = - 2, C l = - 2$