Question

Calculate the magnetic moment of the element with atomic number $ Z = 28 $.

• A. $ 2.828 \, \text{BM} $
• B. $ 4.90 \, \text{BM} $
• C. $ 5.92 \, \text{BM} $
• D. $ 0 \, \text{BM} $

Hint:

Magnetic moment depends only on the number of unpaired electrons. Use the formula $ \mu = \sqrt{n(n+2)} $ to calculate it in Bohr magnetons (BM).

Answer 1

The Correct Option is A

Step 1: Identify the element
The atomic number $ Z = 28 $ corresponds to the element Nickel (Ni).
Step 2: Write the electron configuration of Nickel
The ground-state electron configuration of Nickel is:
$$ \text{Ni: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^8 \, 4s^2 $$ Step 3: Determine the unpaired electrons
The $ 4s $ orbital is fully filled ($ 4s^2 $), so it contributes no unpaired electrons.
The $ 3d $ orbital has 8 electrons. Using Hund's rule, the configuration is:
$$ \uparrow \downarrow \quad \uparrow \downarrow \quad \uparrow \downarrow \quad \uparrow \quad \uparrow $$ This shows 2 unpaired electrons. Step 4: Use the formula for magnetic moment
The magnetic moment ($\mu$) is given by: \[ \mu = \sqrt{n(n+2)} \text{ BM} \] where $n$ is the number of unpaired electrons. For Nickel, $n = 2$. Substituting: \[ \mu = \sqrt{2(2+2)} = \sqrt{2 \times 4} = \sqrt{8} = 2\sqrt{2} \text{ BM} \] Step 5: Approximate the value \[ 2\sqrt{2} \approx 2 \times 1.414 = 2.828 \text{ BM} \] Thus, the magnetic moment is: \[ (1) \quad 2.828 \text{ BM} \]