Find spin only magnetic moment of [Fe(H2O)5NO](2+)

In summary, the spin only magnetic moment of [Fe(H2O)5NO]^{2+} is 3.87, with a configuration of [Ar]3d^64s^1 and 5 unpaired electrons. However, there are conflicting reports on the oxidation states of Fe and NO, which may affect the accuracy of this calculation. Further research is needed to determine the correct values.
  • #1
Titan97
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Homework Statement


Find spin only magnetic moment of ##[Fe(H2O)5NO]^{2+}##

Homework Equations


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The Attempt at a Solution


When I used Google, the magnetic moment in 3.87
##Fe## is in +1 oxidation state. So configuration is $$[Ar]3d^64s^1$$
There are 5 unpaired electrons. How does magnetic moment become 3.87?
 
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  • #2
the configuration readjusts to 3d7. And there are more problems: Fe is said to be in +3 and NO is in -1. Look it up in Google.
 
  • #3
If that's the case, then there are 5 unpaired electrons. Magnetic moment won't be 3.87
 
  • #4
Yeah, I know. And it is a problem for me too, for my textbook says that it is Fe+ and NO+. But apparently that is 15 year old data. I'm not sure which one to follow, and as of now, I have no time to follow my research on the net up. I will look it up again and get back.
 
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FAQ: Find spin only magnetic moment of [Fe(H2O)5NO](2+)

What is the formula for calculating spin-only magnetic moment?

The formula for calculating spin-only magnetic moment is given by μ=√n(n+2) where n is the number of unpaired electrons.

How do I determine the number of unpaired electrons in a complex?

To determine the number of unpaired electrons in a complex, you need to first determine the electronic configuration of the central metal ion. Then, using Hund's rule, fill up the orbitals with electrons, taking into account any ligand field splitting. The number of electrons in the partially filled orbitals is equal to the number of unpaired electrons in the complex.

What is the electronic configuration of Fe(III) in [Fe(H2O)5NO](2+)?

The electronic configuration of Fe(III) is [Ar]3d5. This means that there are 5 electrons in the d-orbitals, with 3 pairs and 1 unpaired electron.

How does ligand field splitting affect the spin-only magnetic moment?

Ligand field splitting affects the spin-only magnetic moment by reducing the number of unpaired electrons. This is because the ligand field can cause some of the electrons to pair up, resulting in a lower magnetic moment.

What is the spin-only magnetic moment for [Fe(H2O)5NO](2+)?

The spin-only magnetic moment for [Fe(H2O)5NO](2+) is calculated by first determining the number of unpaired electrons. In this case, Fe(III) has one unpaired electron, so the magnetic moment is equal to √1(1+2)=√3μB.

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