If the spin-only magnetic moment of \(Cu^{2+}\) is 1.73 BM, then the number of unpaired electrons present are:
Answer & explanation
Correct answer: option 4
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:
\((1.73)^2 = n(n+2)\)
\(⇒ 3 = n^2 + 2n\) [\(1.73^2 \approx 3\)]
\(⇒ n^2 + 2n - 3 = 0\)
\(⇒ n^2 + (3 - 1)n - 3 = 0\)
\(⇒ n^2 + 3n - n - 3 = 0\)
\(⇒ n(n + 3) -1(n + 3)= 0\)
\(⇒ (n - 1)(n + 3) = 0\)
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.