Question

Value of magnetic moment of a divalent ion in aqueous solution having atomic number 25 will be 5.92 B.M. (True/False)

Hint:

You can verify these using the empirical relation: $Mode = 3(Median) - 2(Mean)$. $3(3.75) - 2(4.2) = 11.25 - 8.4 = 2.85$, which is close to the given mode of 3.286.

Answer 1

Concept:
Magnetic moment of transition metal ions depends on the number of unpaired electrons and is calculated using the spin-only formula: \[ \mu = \sqrt{n(n+2)} \text{ B.M.} \] where \( n \) = number of unpaired electrons.
Step 1: Identify the Element
Atomic number 25 → Manganese (Mn) \[ \text{Mn}: [Ar]\,3d^5 4s^2 \]
Step 2: Divalent Ion Formation
\[ \text{Mn}^{2+}: [Ar]\,3d^5 \] All five d-electrons remain unpaired. \[ n = 5 \]
Step 3: Calculate Magnetic Moment
\[ \mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92 \text{ B.M.} \]
Conclusion:
The given statement is correct. \[ \boxed{\text{True}} \]