Chemistry Examples

${(N H_{4})}_{2} S$

Step 1

Since $H$ is in column $1$ of the periodic table, it will share $1$ electrons and use an oxidation state of $1$ .

$H = 42 (1) = 42$

Step 2

Since $N$ is in column $5$ of the periodic table, it will share $5$ electrons and use an oxidation state of $5$ .

$N = 1 (- 3) = - 3$

Step 3

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

$42 - 3 + 1 x = 0$

Step 4

Solve the equation for $x$ .

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Step 4.1

Simplify $42 + - 3 + 1 x$ .

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Step 4.1.1

Multiply $x$ by $1$ .

$42 + - 3 + x = 0$

Step 4.1.2

Add $42$ and $- 3$ .

$39 + x = 0$

Step 4.2

Subtract $39$ from both sides of the equation.

$x = - 39$

Step 5

$S$ is using an oxidation state of $- 39$ .

$- 39$

Step 6

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

$H = 42, N = - 3, S = - 39$