With O highest possible oxidation state of Mn is

Options

A. `+7`
B. `+4`
C. `+5`
D. `+3`

Answer

To determine the highest possible oxidation state of manganese (Mn) when combined with oxygen (O), we can follow these steps:

### Step 1: Understand the Electronic Configuration of Manganese
Manganese has an atomic number of 25. Its electronic configuration is:
\[ \text{Mn: } [\text{Ar}] 3d^5 4s^2 \]
This means manganese has 5 unpaired electrons in the 3d subshell and 2 electrons in the 4s subshell.

**Hint:** The number of unpaired electrons can help determine the maximum oxidation state.

### Step 2: Determine the Valence Electrons
Manganese has a total of 7 electrons in its valence shell (5 from 3d and 2 from 4s). The maximum oxidation state is often related to the total number of valence electrons.

**Hint:** The total number of valence electrons gives insight into how many electrons can be lost to achieve a higher oxidation state.

### Step 3: Identify the Maximum Oxidation State
The highest oxidation state of an element is typically equal to the total number of valence electrons it can lose. For manganese, this is +7.

**Hint:** The maximum oxidation state is often equal to the number of valence electrons.

### Step 4: Example with Oxygen
In compounds like potassium permanganate (KMnO4), manganese exhibits a +7 oxidation state. We can verify this by setting up the oxidation state equation:
- Let the oxidation state of Mn be \( x \).
- The oxidation state of potassium (K) is +1.
- The oxidation state of oxygen (O) is -2 for each of the 4 oxygen atoms, contributing a total of -8.

Setting up the equation:
\[ x + 1 + (-8) = 0 \]
This simplifies to:
\[ x - 7 = 0 \]
Thus, \( x = +7 \).

**Hint:** Setting up oxidation state equations can help confirm the oxidation states of elements in a compound.

### Conclusion
The highest possible oxidation state of manganese with oxygen is **+7**.

**Final Answer:** +7

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