# A particular compound of vanadium show a magnetic moment of 3.86BM the oxidation state of vanadium in the compound is( Z=23)?

##### 1 Answer

It's not entirely clear what kind of magnetic moment this is. Is it *spin-only*, or based on spin *and* orbital angular momentum? I'm guessing **spin-only**, because it is reasonable to solve.

(We would have needed to know how many unpaired electrons there were in the first place, which we should not yet know, to know how to get the total magnetic moment, if given.)

The **magnetic moment** is given by (*Miessler et al.*):

#mu = 2.0023sqrt(S(S+1) + 1/4 L(L+1))# where:

#S = sum_i |m_(s,i)|# is thetotal spinofallthe electrons in the compound. From this you could find the number of unpaired electrons only (but you will have no idea how many paired electrons!).#L = sum_i |m_(l,i)|# is thetotal orbital angular momentumof the orbitals in which each electron resides, and is found by summing the magnetic quantum number foreachorbital, foreach electronin that orbital, then taking the absolute value of the sum.#g = 2.0023# is the gyromagnetic ratio.

But the **spin-only magnetic moment** is given by

#mu_S = 2.0023sqrt(S(S+1))# .

If

#(3.86/2.0023)^2 = S(S+1) = S^2 + S#

Thus, we have a quadratic equation to solve:

#S^2 + S - 3.7163 = 0#

Solve this to obtain

#S = |(-1/2) + (-1/2) + (-1/2)| = |(+1/2) + (+1/2) + (+1/2)| = 3/2#

#= 1.5 ~~ 1.492# ,

or

This would be an issue. Since vanadium has no more than

#"V": [Ar] 3d^3 4s^2# ,

there is no other way for there to be *maybe* as

It could be either one, but is more likely the

- If any electrons were lost to the ligands, then it would have to be
#"V"^(+2)# (such as in#"VCl"_2# ). - If all the ligands are neutral and the complex is neutral as well, it could be
#"V"^(0)# . However,#"V"("CO")_6# is an example with not#3# unpaired electrons, but#1# .