Chemistry Examples

$H Z r R e$

Step 1

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

$R e = 1 (0) = 0$

Step 2

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

$H = 1 (- 7) = - 7$

Step 3

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

$0 - 7 + 1 x = 0$

Step 4

Solve the equation for $x$ .

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Step 4.1

Simplify $0 + - 7 + 1 x$ .

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Step 4.1.1

Subtract $7$ from $0$ .

$- 7 + 1 x = 0$

Step 4.1.2

Multiply $x$ by $1$ .

$- 7 + x = 0$

Step 4.2

Add $7$ to both sides of the equation.

$x = 7$

Step 5

$Z r$ is using an oxidation state of $7$ .

$7$

Step 6

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

$R e = 0, H = - 7, Z r = 7$