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

Options

A. 4.12 BM
B. 5.92 BM
C. 6.92 BM
D. 7.12 BM

Answer

To solve the problem of calculating the magnetic moment of a divalent ion with an atomic number of 25, we can follow these steps:

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

### Step 2: Determine the Divalent Ion
Since we are looking for a divalent ion, we consider the Mn ion with a +2 charge, which is Mn²⁺.

### Step 3: Write the Electronic Configuration
The electronic configuration of Mn (atomic number 25) is:
\[ \text{Mn: } [\text{Ar}] 3d^5 4s^2 \]
When Mn loses two electrons to form Mn²⁺, the electrons are removed from the 4s orbital first. Thus, the electronic configuration of Mn²⁺ is:
\[ \text{Mn}^{2+}: [\text{Ar}] 3d^5 \]

### Step 4: Count the Unpaired Electrons
In the 3d subshell of Mn²⁺, there are 5 electrons in the 3d orbitals. Since all 5 electrons can occupy different orbitals (according to Hund's rule), Mn²⁺ has:
- 5 unpaired electrons.

### Step 5: Use the Magnetic Moment Formula
The magnetic moment (μ) can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

### Step 6: Substitute the Values
Substituting \( n = 5 \) (the number of unpaired electrons):
\[ \mu = \sqrt{5(5 + 2)} \]
\[ \mu = \sqrt{5 \times 7} \]
\[ \mu = \sqrt{35} \]

### Step 7: Calculate the Value
Now, we calculate the square root:
\[ \mu \approx 5.92 \text{ Bohr magneton} \]

### Final Answer
The magnetic moment of the divalent ion (Mn²⁺) is approximately:
\[ \mu \approx 5.92 \text{ Bohr magneton} \]

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