Final answer:
The magnetic moment of [mn(H2O)6]3+ is calculated using the formula μ = √n(n+2) given it is a high-spin complex with four unpaired electrons (μ = √24). In contrast, [Mn(CN)6]3- is a low-spin complex with one unpaired electron (μ = √3).
Step-by-step explanation:
To calculate the magnetic moments for the complexes [mn(H2O)6]3+ and [Mn(CN)6]3-, we need to determine the number of unpaired electrons in each complex. The electron pairing energy (P) is given as 28000 cm-1.
In the case of [mn(H2O)6]3+, with Δ₀ (21000 cm-1) being less than P, we can conclude that it's a high-spin complex, which means electrons will occupy the higher energy orbitals before pairing up. Considering this, Mn3+ has a d4 configuration so it will have four unpaired electrons.
For [Mn(CN)6]3-, where Δ₀ (38500 cm-1) is greater than P, it is a low-spin complex. Hence, the electrons will pair up in the lower energy orbitals, and we expect fewer unpaired electrons. In this case, there will be only one unpaired electron.
The magnetic moments (μ) can be calculated using the formula μ = √n(n+2), where n is the number of unpaired electrons. For [mn(H2O)6]3+, μ = √4(4+2) = √24, and for [Mn(CN)6]3-, μ = √1(1+2) = √3.