The oxidation state of I in the ion \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}\) is (a) -1 (b) \(+1 ;(c)+7 ;(d)+8\)

Short Answer

Expert verified
The oxidation state of I in the ion H4IO6- is +7.

Step by step solution

01

Calculate Oxidation States

First, the known oxidation states of the Hydrogen (H) and Oxygen (O) atoms in the ion will be determined. Typically, the oxidation state of H in a compound is +1, while the oxidation state of O is generally -2.
02

Substitution and Calculating Iodine's Oxidation State

Substitute these values into the known equation as follows: The sum of the oxidation numbers equals the charge of the ion. Therefore, If the total charge of the ion, H4IO6-, is -1, we implement this into the equation: \(+1*4 + 1*x -2*6 = -1\) where x is the oxidation state of Iodine (I). That simplifies to \(4+ x-12 = -1\) and furthermore to \(x-8= -1\). Solving for x, x = -1 + 8 or x=7.
03

Correlate with options given

The obtained oxidation state of Iodine +7 in the ion matches option (c) in the exercise.

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