Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.

Options

A. 5.9 B.M
B. 2.9 B.M
C. 6.9 B.M
D. 9.9 B.M

Explanation

To calculate the magnetic moment of a divalent ion with an atomic number of 25, we will follow these steps:

### Step 1: Identify the Element
The atomic number 25 corresponds to the element Manganese (Mn).

### Step 2: Determine the Electronic Configuration
The electronic configuration of Manganese (Mn) is:
\[ \text{Mn: } [Ar] 3d^5 4s^2 \]

### Step 3: Consider the Divalent Ion
Since we are dealing with a divalent ion (Mn²⁺), we need to remove two electrons from the electronic configuration. The electrons are removed from the outermost shell first, which is the 4s orbital:
\[ \text{Mn}^{2+}: [Ar] 3d^5 \]

### Step 4: Count the Number of Unpaired Electrons
In the \(3d^5\) configuration, all five electrons are unpaired. Therefore, the number of unpaired electrons \(n\) is:
\[ n = 5 \]

### Step 5: Use the Magnetic Moment Formula
The formula to calculate the magnetic moment (\(\mu\)) is given by:
\[ \mu = \sqrt{n(n + 2)} \, \text{BM} \]
Substituting the value of \(n\):
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \]

### Step 6: Calculate the Magnetic Moment
Calculating \(\sqrt{35}\):
\[ \mu \approx 5.92 \, \text{BM} \]
This value is approximately 5.9 BM.

### Conclusion
Thus, the magnetic moment of the divalent ion (Mn²⁺) in aqueous solution is approximately:
\[ \mu \approx 5.9 \, \text{BM} \]

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