Question

Magnetic moment of a bivalent ion in aqueous solution will be, if its atomic number is 25:

• A. 1.73 BM
• B. 2.83 BM
• C. 4.96 BM
• D. 5.92 BM

Hint:

The magnetic moment is calculated using \( \mu = \sqrt{n(n + 2)} \), where \( n \) is the number of unpaired electrons. For \( \text{Mn}^{2+} \), \( n = 5 \), leading to a magnetic moment of 2.83 BM.

Answer 1

The Correct Option is B

The magnetic moment \( \mu \) of an ion is calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \, \text{BM} \] Where \( n \) is the number of unpaired electrons in the ion. For a bivalent ion with atomic number 25, we need to determine the electron configuration and the number of unpaired electrons. The electron configuration for an atomic number 25 element (Manganese) is:
\[ \text{Mn}: [Ar] 3d^5 4s^2 \] In the bivalent state \( \text{Mn}^{2+} \), the electron configuration becomes:
\[ \text{Mn}^{2+}: [Ar] 3d^5 \] In this configuration, there are 5 unpaired electrons in the \( 3d \) orbitals. Therefore, \( n = 5 \). Now, we can substitute \( n = 5 \) into the formula for magnetic moment:
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} = 2.83 \, \text{BM} \] Hence, the magnetic moment of the bivalent ion in aqueous solution is 2.83 BM. Thus, the correct answer is option (B) 2.83 BM.