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

Answer: (A)

3.87 and 2.84

Answer 2

[Cr(NH3)6]3+ = Cr3+
Cr3+ = 3d3 4s0

It has 3 unpaired electrons
$\mu = n \sqrt{n+2}$ BM
$\mu = \sqrt[3]{3+2}$ BM
μ = 3.87 BM

[CuF6]3- = Cu+3
Cu+3 = 3d8 4s0

It has 2 unpaired electrons
μ = $2\sqrt{2+2}$ BM
= 2.84 BM
The correct answer is option (A): 3.87 and 2.84