spin only magnetic moment of [Mn(H2O)6]2+
Question
spin only magnetic moment of [Mn(H2O)6]2+
Solution
The spin-only magnetic moment of a complex can be calculated using the formula:
μ = √(n(n+2)) Bohr magneton
where n is the number of unpaired electrons.
For the complex [Mn(H2O)6]2+, the Mn is in the +2 oxidation state. In this state, Mn has a 3d5 electron configuration. All five of these electrons are unpaired.
So, we substitute n = 5 into the formula:
μ = √(5(5+2)) = √35 = 5.92 Bohr magneton
So, the spin-only magnetic moment of [Mn(H2O)6]2+ is 5.92 Bohr magneton.
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