Final answer:
The magnetic moment of a divalent cation of transition metal manganese (Mn) with atomic number 25 is approximately 5.92 Bohr magnetons, calculated using the formula for spin-only magnetic moment based on the 5 unpaired electrons in its electronic configuration.
Step-by-step explanation:
The magnetic moment of a divalent cation of a transition metal with an atomic number of 25 can be calculated using the formula for spin-only magnetic moment. The atomic number 25 corresponds to the element manganese (Mn). When Mn exists as a divalent cation (Mn2+), it has a total of 5 unpaired electrons due to its electronic configuration [(Ar) 3d5].
The formula for calculating the spin-only magnetic moment (μ) is μ = √(n(n+2))×μb, where n is the number of unpaired electrons, and μb (Bohr magneton) is a physical constant.
Using the formula, we plug in the number of unpaired electrons for Mn2+:
μ = √(5(5+2))×μb = √(35)×μb = √35 μb. Since one Bohr magneton equals approximately 0.9273×10-23 A·m2, μ = 5.92μb.
Therefore, the spin only magnetic moment of the Mn2+ ion is approximately 5.92 Bohr magnetons.