Practicing Success
If the spin-only magnetic moment of \(Cu^{2+}\) is 1.73 BM, then the number of unpaired electrons present are: |
2 3 0 1 |
1 |
The correct answer is option 4. 1. To determine the number of unpaired electrons in \(Cu^{2+}\) given its spin-only magnetic moment, we can use the formula for calculating the spin-only magnetic moment: \(\mu_s = \sqrt{n(n+2)}\) where \( n \) is the number of unpaired electrons. Given: \(\mu_s = 1.73 \text{ BM}\) Let's solve for \( n \): \(1.73 = \sqrt{n(n+2)}\) Square both sides to eliminate the square root: \(⇒ 3 = n^2 + 2n\) [\(1.73^2 \approx 3\)] \(⇒ n^2 + 2n - 3 = 0\) \(⇒ n^2 + (3 - 1)n - 3 = 0\) \(⇒ n^2 + 3n - n - 3 = 0\) \(⇒ n(n + 3) -1(n + 3)= 0\) \(⇒ (n - 1)(n + 3) = 0\) Thus, either \(n - 1 = 0\), then \(n = 1\) or, \(n + 3 = 0\), then \(n = -3\), which is not possible. Since \( n \) must be a non-negative integer (representing the number of unpaired electrons), we take \( n = 1 \). Therefore, the number of unpaired electrons present in \(Cu^{2+}\) ion, given its spin-only magnetic moment of 1.73 BM, is 1. |