Question

The spin-only magnetic moment of \([Co({H}_2\text{O)}_6^{2+}]\) (rounded off to one decimal place) is \(\underline{\hspace{2cm}}\) BM.

Hint:

The spin-only magnetic moment formula depends on the number of unpaired electrons in the ion's electron configuration.

Answer 1

Correct Answer: 3.8

For the spin-only magnetic moment, we use the formula: \[ \mu_{\text{eff}} = \sqrt{n(n+2)} \, \text{BM} \] where \( n \) is the number of unpaired electrons.
For \( [Co(H_2O)_6]^{2+} \), cobalt has an atomic number of 27. In the \( +2 \) oxidation state, it has the electron configuration \( 3d^7 \). The number of unpaired electrons in \( 3d^7 \) configuration is \( 3 \), so \( n = 3 \).
Now, substituting \( n = 3 \) into the formula: \[ \mu_{\text{eff}} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.9 \, \text{BM} \] Thus, the spin-only magnetic moment is \( \boxed{3.9 \, \text{BM}} \).