Chemistry Examples

$N a C l + K N O_{3}$

Step 1

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

$K = 1 (1) = 1$

Step 2

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

$O = 3 (- 2) = - 6$

Step 3

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

$N a = 1 (1) = 1$

Step 4

Since $C l$ is in column $7$ of the periodic table, it will share $7$ electrons and use an oxidation state of $7$ .

$C l = 1 (- 1) = - 1$

Step 5

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

$1 - 6 + 1 - 1 + 1 x = 0$

Step 6

Solve the equation for $x$ .

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Step 6.1

Simplify $1 + - 6 + 1 + - 1 + 1 x$ .

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Step 6.1.1

Combine the opposite terms in $1 + - 6 + 1 + - 1 + 1 x$ .

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Step 6.1.1.1

Subtract $1$ from $1$ .

$0 - 6 + 1 + 1 x = 0$

Step 6.1.1.2

Subtract $6$ from $0$ .

$- 6 + 1 + 1 x = 0$

Step 6.1.2

Simplify the expression.

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Step 6.1.2.1

Multiply $x$ by $1$ .

$- 6 + 1 + x = 0$

Step 6.1.2.2

Add $- 6$ and $1$ .

$- 5 + x = 0$

Step 6.2

Add $5$ to both sides of the equation.

$x = 5$

Step 7

$N$ is using an oxidation state of $5$ .

$5$

Step 8

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

$K = 1, O = - 6, N a = 1, C l = - 1, N = 5$