If `Mn^(2+)` has 'X' number of unpaired electrons and 'Y' other positive oxidation states than `+2`. What is the sum of `X+Y` here?

Answer

To solve the problem, we need to determine the values of \( X \) and \( Y \) for the ion \( \text{Mn}^{2+} \).

### Step 1: Determine the electronic configuration of Mn
The atomic number of manganese (Mn) is 25. The electronic configuration of Mn is:
\[ \text{Mn: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^5 \]

### Step 2: Determine the electronic configuration of \( \text{Mn}^{2+} \)
When manganese loses 2 electrons to form \( \text{Mn}^{2+} \), the electrons are removed first from the 4s orbital before the 3d orbital. Therefore, the electronic configuration of \( \text{Mn}^{2+} \) is:
\[ \text{Mn}^{2+}: 3d^5 \, 4s^0 \]

### Step 3: Count the number of unpaired electrons (\( X \))
In the \( 3d^5 \) configuration, all five electrons are unpaired because each electron occupies a separate orbital before pairing occurs. Thus, the number of unpaired electrons \( X \) is:
\[ X = 5 \]

### Step 4: Determine the other positive oxidation states (\( Y \))
Manganese can exhibit several oxidation states. The common oxidation states of manganese are \( +2, +4, +5, +7 \). Since we are looking for oxidation states other than \( +2 \), we can consider \( +4, +5, \) and \( +7 \).

The oxidation states we can consider are:
- \( +4 \) (from losing 4 electrons: 2 from 4s and 2 from 3d)
- \( +5 \) (from losing 5 electrons: all from 3d)
- \( +7 \) (from losing all 7 electrons: 2 from 4s and 5 from 3d)

Thus, the number of other positive oxidation states \( Y \) is:
\[ Y = 3 \quad (\text{which are } +4, +5, +7) \]

### Step 5: Calculate \( X + Y \)
Now we can calculate the sum of \( X \) and \( Y \):
\[ X + Y = 5 + 3 = 8 \]

### Final Answer
The final answer is:
\[ \text{Sum of } X + Y = 8 \]

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