Find the oxidation state of I in H₄IO₆⁻

The correct answer is +7.

To find the oxidation state of iodine (I) in the orthoperiodate ion, $H_4{\mathrm{IO}}_6^{-}$, we can set up an algebraic equation based on the known oxidation states of the other elements and the overall charge of the ion.

Let's recall the standard oxidation numbers for the other elements involved:
1. Hydrogen (H) generally has an oxidation state of +1 when bonded to non-metals.
2. Oxygen (O) typically has an oxidation state of -2 in most compounds (except in peroxides or when bonded to fluorine).

Let the oxidation state of iodine (I) be represented by $x$.

The sum of the oxidation states of all atoms in a polyatomic ion must equal the overall net charge of the ion. The overall charge of the $H_4{\mathrm{IO}}_6^{-}$ ion is -1.

We can write the equation as follows:
$4\left(\text{oxidation state of H}\right)+1\left(\text{oxidation state of I}\right)+6\left(\text{oxidation state of O}\right)=-1$

Substituting the known oxidation states into the equation:
$4(+1)+x+6(-2)=-1$

Simplify the terms:
$4+x-12=-1$

Combine the numerical constants:
$x-8=-1$

Solve for $x$ by adding 8 to both sides of the equation:
$x=-1+8$
$x=+7$

Therefore, the oxidation state of iodine (I) in $H_4{\mathrm{IO}}_6^{-}$ is +7.