Question

The calculated spin only magnetic moments of $\left[ Cr \left( NH _{3}\right)_{6}\right]^{3+}$ and $\left[ CuF _{6}\right]^{3-}$ in BM, respectively, are
(Atomic number of $Cr$ and $Cu$ are $24$ and $29$ , respectively)

• A. 3.87 and 2.84
• B. 4.90 and 1.73
• C. 3.87 and 1.73
• D. 4.90 and 2.84

Answer 1

The Correct Option is A

Step 1: Understanding the Given Complexes
We are given two complexes with their respective charges and electronic configurations:
1. \( [Cr(NH_3)_6]^{3+} \), where the central metal ion is \( \text{Cr}^{3+} \).
2. \( [CuF_6]^{3-} \), where the central metal ion is \( \text{Cu}^{3+} \).
Step 2: Electronic Configuration of \( \text{Cr}^{3+} \)
The electronic configuration of \( \text{Cr}^{3+} \) is \( 3d^3 4s^0 \), as chromium has an atomic number of 24, and when it loses 3 electrons, it achieves the configuration of \( 3d^3 4s^0 \).
This means that \( \text{Cr}^{3+} \) has 3 unpaired electrons.
Step 3: Calculate the Magnetic Moment for \( \text{Cr}^{3+} \)
The magnetic moment (\( \mu \)) for a metal ion with \( n \) unpaired electrons is given by the formula: \[ \mu = \sqrt{n(n + 2)} \text{ BM} \] where \( n \) is the number of unpaired electrons. For \( \text{Cr}^{3+} \), \( n = 3 \), so:
\[ \mu = \sqrt{3(3 + 2)} \text{ BM} = \sqrt{3 \times 5} \text{ BM} = \sqrt{15} \text{ BM} \] Taking the square root of 15, we get: \[ \mu = 3.87 \text{ BM} \] Step 4: Electronic Configuration of \( \text{Cu}^{3+} \)
The electronic configuration of \( \text{Cu}^{3+} \) is \( 3d^8 4s^0 \), as copper has an atomic number of 29, and when it loses 3 electrons, it achieves the configuration of \( 3d^8 4s^0 \).
This means that \( \text{Cu}^{3+} \) has 2 unpaired electrons.
Step 5: Calculate the Magnetic Moment for \( \text{Cu}^{3+} \)
Using the same formula for the magnetic moment as before: \[ \mu = \sqrt{n(n + 2)} \text{ BM} \] For \( \text{Cu}^{3+} \), \( n = 2 \), so: \[ \mu = \sqrt{2(2 + 2)} \text{ BM} = \sqrt{2 \times 4} \text{ BM} = \sqrt{8} \text{ BM} \] Taking the square root of 8, we get: \[ \mu = 2.84 \text{ BM} \] Step 6: Conclusion
Therefore, the magnetic moments for the two complexes are:
- For \( [Cr(NH_3)_6]^{3+} \), \( \mu = 3.87 \text{ BM} \).
- For \( [CuF_6]^{3-} \), \( \mu = 2.84 \text{ BM} \).
The correct answer is option (A): \( 3.87 \) and \( 2.84 \).