Find the Oxidation Numbers Oxidation number Mn(IO)_2

Oxidation number $M n {(I O)}_{2}$

Since $M n$ is in column $3$ of the periodic table, it will share $3$ electrons and use an oxidation state of $3$ .

$M n = 1 (3) = 3$

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

$O = 2 (- 2) = - 4$

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

$3 - 4 + 1 x = 0$

Solve the equation for $x$ .

Simplify $3 + - 4 + 1 x$ .

Multiply $x$ by $1$ .

$3 + - 4 + x = 0$

Add $3$ and $- 4$ .

$- 1 + x = 0$

Add $1$ to both sides of the equation.

$x = 1$

$I$ is using an oxidation state of $1$ .

$1$

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

$M n = 3, O = - 4, I = 1$