Question

The correct IUPAC name of K₂MnO₄ is

• A. Potassium tetraoxopermanganate (VI)
• B. Potassium tetraoxidomanganate (VI)
• C. Dipotassium tetraoxidomanganate (VII)
• D. Potassium tetraoxidomanganese (VI)

Answer 1

The Correct Option is B

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.

1. Firstly, let's identify the oxidation state of manganese in K₂MnO₄. In compounds, potassium (K) typically has an oxidation state of +1.
2. There are two potassium atoms, so the total contribution from potassium to the charge is +2.
3. Oxygen generally has an oxidation state of -2. With four oxygen atoms, the total contribution from oxygen is -8.
4. The compound K₂MnO₄ is neutral, so the sum of oxidation states must be zero. Let the oxidation state of manganese (Mn) be x.
5. Setting up the equation: \(2(+1) + x + 4(-2) = 0\).
6. This simplifies to: \(2 + x - 8 = 0\).
7. Solving for x gives: \(x = +6\). 
8. Manganese is in the +6 oxidation state.
9. 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.
10. 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)".