Question

The spin only magnetic moment of a divalent ion in aqueous solution (atomic number 29) is ________ BM.

Hint:

The formula for spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ BM. Memorize the values for n=1 to 5: n=1, $\mu$=1.73; n=2, $\mu$=2.83; n=3, $\mu$=3.87; n=4, $\mu$=4.90; n=5, $\mu$=5.92 BM. Notice the value is always slightly less than n+1.

Answer 1

Correct Answer: 2

Given the atomic number 29, which corresponds to copper (Cu). Copper in its divalent state (Cu2+) has lost two electrons. The electronic configuration of copper (atomic number 29) is [Ar] 3d10 4s1. When copper loses two electrons, it forms Cu2+, resulting in the configuration [Ar] 3d9. To find the spin-only magnetic moment (μ), use the formula: μspin=√n(n+2) BM, where n is the number of unpaired electrons. In a 3d9 configuration, there is one unpaired electron. Therefore, n=1. Substituting n=1 into the formula gives:  μspin=√1(1+2)=√3 BM. Calculating gives μspin≈1.73 BM. The expected range is given as 2,2, suggesting a value close to 2 BM. However, the calculated result of approximately 1.73 BM is accurate for a single unpaired electron in this scenario.