The exact oxidation number of iodine (I) in potassium periodate (KIO4) is +7.
Understanding Oxidation Numbers
Oxidation numbers, also known as oxidation states, are hypothetical charges assigned to atoms in a molecule or ion if all bonds were ionic. They help chemists track electron transfer in chemical reactions, particularly redox reactions. Assigning these numbers follows a specific set of rules based on an element's electronegativity and common bonding patterns.
Key rules for assigning oxidation numbers:
- Free elements: The oxidation number of an atom in a free, uncombined element is 0.
- Monatomic ions: The oxidation number of a monatomic ion is equal to its charge (e.g., Na⁺ is +1, Cl⁻ is -1).
- Alkali metals (Group 1): Always have an oxidation number of +1 in compounds (e.g., K, Na).
- Alkaline earth metals (Group 2): Always have an oxidation number of +2 in compounds (e.g., Mg, Ca).
- Oxygen: Almost always has an oxidation number of -2 in compounds. Exceptions include peroxides (e.g., H2O2, where O is -1), superoxides (e.g., KO2, where O is -1/2), and compounds with fluorine (e.g., OF2, where O is +2).
- Hydrogen: Usually has an oxidation number of +1 when bonded to nonmetals and -1 when bonded to metals.
- Fluorine: Always has an oxidation number of -1 in compounds. Other halogens (Cl, Br, I) usually have -1 unless combined with oxygen or a more electronegative halogen.
- Sum of oxidation numbers:
- For a neutral compound, the sum of all oxidation numbers must be zero.
- For a polyatomic ion, the sum of all oxidation numbers must equal the charge of the ion.
For a deeper dive into the rules of assigning oxidation numbers, you can refer to resources like ThoughtCo's guide on oxidation numbers.
Calculating the Oxidation Number of Iodine in KIO4
To determine the oxidation number of iodine in potassium periodate (KIO4), we apply the general rules:
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Identify known oxidation numbers:
- Potassium (K): As an alkali metal (Group 1), potassium always has an oxidation number of +1 in compounds.
- Oxygen (O): In most compounds, oxygen has an oxidation number of -2. Since there are four oxygen atoms, their combined contribution is 4 * (-2) = -8.
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Set up the equation:
Let the oxidation number of Iodine (I) be x.
Since KIO4 is a neutral compound, the sum of the oxidation numbers of all atoms must be equal to zero.Oxidation number of K + Oxidation number of I + (4 Oxidation number of O) = 0
(+1) + (x) + (4 -2) = 0 -
Solve for x:
+1 + x - 8 = 0
x - 7 = 0
x = +7
Therefore, the oxidation number of iodine in KIO4 is +7. This represents iodine in its highest possible oxidation state, indicative of its strong oxidizing nature in this compound.
Here's a summary table:
| Element | Oxidation Number | Quantity | Total Contribution |
|---|---|---|---|
| K | +1 | 1 | +1 |
| I | x | 1 | x |
| O | -2 | 4 | -8 |
| Total | 0 |
