Question

Manganese forms oxide and fluoride with highest oxidation state. The difference in the highest oxidation state of Mn in the oxide and fluoride is:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is D

Manganese (Mn) is a transition metal and can exhibit a wide range of oxidation states from +2 to +7. These oxidation states are observed in various manganese compounds, depending on the nature of the ligands and the compound structure. Let’s examine the two compounds mentioned in the question: manganese oxide (MnO₂) and manganese fluoride (MnF₄). 1. Manganese Oxide (MnO₂): Manganese oxide, specifically \( \text{MnO}_2 \), is a well-known compound where manganese exists in the +4 oxidation state. The oxidation state of oxygen in oxides is typically -2. So, to balance the charges in MnO₂: \[ \text{Oxidation state of Mn} + 2(-2) = 0 \] \[ \text{Oxidation state of Mn} - 4 = 0 \quad \Rightarrow \quad \text{Oxidation state of Mn} = +4 \] Thus, in \( \text{MnO}_2 \), the oxidation state of manganese is +4. 2. Manganese Fluoride (MnF₄): Manganese fluoride, specifically \( \text{MnF}_4 \), is another compound where manganese exhibits its highest oxidation state. The oxidation state of fluorine in fluorides is -1. So, to balance the charges in MnF₄: \[ \text{Oxidation state of Mn} + 4(-1) = 0 \] \[ \text{Oxidation state of Mn} - 4 = 0 \quad \Rightarrow \quad \text{Oxidation state of Mn} = +4 \] Thus, in \( \text{MnF}_4 \), the oxidation state of manganese is also +4. 3. Manganese in Other Compounds: Manganese can also exist in other oxidation states, such as +2, +3, +6, and +7. For example:
- In \( \text{Mn}_2\text{O}_7 \), manganese has an oxidation state of +7.
- In \( \text{MnCl}_2 \), manganese has an oxidation state of +2.
- In \( \text{Mn}_2\text{O}_3 \), manganese has an oxidation state of +3.
But in the context of the question, we are concerned with the highest oxidation states of manganese in the oxide and fluoride compounds specifically. 4. Difference in Oxidation States: The question asks for the difference in the highest oxidation states of manganese in the oxide and fluoride. From the calculations above: - In \( \text{MnO}_2 \), the highest oxidation state of Mn is +4.
- In \( \text{MnF}_4 \), the highest oxidation state of Mn is also +4.
Therefore, the difference in the highest oxidation states of Mn in these two compounds is: \[ \text{Difference} = 4 - 4 = 0 \] Thus, the correct answer based on the highest oxidation states of manganese in these two compounds would be \( 4 - 4 = 0 \), and the correct answer is option (D). \[ \boxed{4} \]