Find the Oxidation Numbers Ba_2XeO_6

$B a_{2} X e O_{6}$

Step 1

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

$B a = 2 (2) = 4$

Step 2

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

$O = 6 (- 2) = - 12$

Step 3

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

$4 - 12 + 1 x = 0$

Step 4

Solve the equation for $x$ .

Simplify $4 + - 12 + 1 x$ .

Multiply $x$ by $1$ .

$4 + - 12 + x = 0$

Add $4$ and $- 12$ .

$- 8 + x = 0$

Add $8$ to both sides of the equation.

$x = 8$

Step 5

$X e$ is using an oxidation state of $8$ .

$8$

Step 6

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

$B a = 4, O = - 12, X e = 8$