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

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1. Feb 4, 2016

### Titan97

1. The problem statement, all variables and given/known data
Find spin only magnetic moment of $[Fe(H2O)5NO]^{2+}$

2. Relevant equations
None

3. 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?

2. Feb 5, 2016

### CrazyNinja

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. Feb 5, 2016

### Titan97

If thats the case, then there are 5 unpaired electrons. Magnetic moment won't be 3.87

4. Feb 6, 2016

### CrazyNinja

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.