The correct IUPAC name of K2MnO4 is
- Potassium tetraoxopermanganate (VI)
- Potassium tetraoxidomanganate (VI)
- Dipotassium tetraoxidomanganate (VII)
- Potassium tetraoxidomanganese (VI)
The Correct Option is B
Solution and Explanation
To determine the correct IUPAC name for the compound K2MnO4, we need to consider the oxidation state of the manganese and naming conventions according to IUPAC rules.
- Firstly, let's identify the oxidation state of manganese in K2MnO4. In compounds, potassium (K) typically has an oxidation state of +1.
- There are two potassium atoms, so the total contribution from potassium to the charge is +2.
- Oxygen generally has an oxidation state of -2. With four oxygen atoms, the total contribution from oxygen is -8.
- The compound K2MnO4 is neutral, so the sum of oxidation states must be zero. Let the oxidation state of manganese (Mn) be x.
- Setting up the equation: \(2(+1) + x + 4(-2) = 0\).
- This simplifies to: \(2 + x - 8 = 0\).
- Solving for x gives: \(x = +6\).
- Manganese is in the +6 oxidation state.
- According to IUPAC nomenclature, the anion is named by using the root of the element name plus the suffix '-ate', and the oxidation state is indicated in Roman numerals in parentheses.
- Thus, the anion MnO42− is named as 'tetraoxidomanganate(VI)' because 'tetra' indicates four oxygens, and manganese is in a +6 oxidation state.
Putting it all together, the correct IUPAC name for K2MnO4 is:
Potassium tetraoxidomanganate (VI).
This rules out the other options as they either incorrectly specify the oxidation state or use incorrect prefixes. The correct option is clearly "Potassium tetraoxidomanganate (VI)".
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