Question

In which of the following compound, Mn has +7 oxidation state?

• A. \( MnOF \)
• B. \( MnO_2F \)
• C. \( MnO_3F_2 \)
• D. \( MnOF_2 \)
• E. \( MnO_3F \)

Hint:

$0^\circ = -2,\; F = -1$ always.

Answer 1

Answer: (E)

Step 1: Understanding Oxidation State Calculation:
The oxidation state (or oxidation number) of an atom in a compound is a hypothetical charge that the atom would have if all bonds to atoms of different elements were 100% ionic. We can calculate it by assigning known oxidation numbers to the more electronegative atoms and ensuring the sum of all oxidation numbers equals the overall charge of the molecule (which is zero for neutral compounds).
Step 2: Assigning Known Oxidation States:
In these compounds, we are dealing with manganese (Mn), oxygen (O), and fluorine (F).
Fluorine (F) is the most electronegative element, so it is always assigned an oxidation state of -1 in its compounds.
Oxygen (O) is the second most electronegative element. It is assigned an oxidation state of -2 in most of its compounds (except in peroxides, superoxides, and compounds with fluorine).
Let the oxidation state of Mn be \(x\).
Step 3: Calculating the Oxidation State of Mn in Each Option:
(A) MnOF: \(x + (-2) + (-1) = 0 \Rightarrow x = +3\)
(B) MnO\(_2\)F: \(x + 2(-2) + (-1) = 0 \Rightarrow x - 4 - 1 = 0 \Rightarrow x = +5\)
(C) MnO\(_3\)F\(_2\): This compound is unlikely as written, possibly a typo. If it exists, \(x + 3(-2) + 2(-1) = 0 \Rightarrow x - 6 - 2 = 0 \Rightarrow x = +8\), which is not possible for Mn.
(D) MnOF\(_2\): \(x + (-2) + 2(-1) = 0 \Rightarrow x - 2 - 2 = 0 \Rightarrow x = +4\)
(E) MnO\(_3\)F (Permanganyl fluoride): \(x + 3(-2) + (-1) = 0 \Rightarrow x - 6 - 1 = 0 \Rightarrow x = +7\)
Step 4: Final Answer:
Manganese has an oxidation state of +7 in the compound MnO\(_3\)F.