# How is Vanadium's 3+ ion paramagnetic?

"Both Vanadium and its 3+ ion are paramagnetic. Use electron configuration to explain why this is so"

The electron configuration of Vanadium is [Ar] 4s2 3d3.

If drawn in a box diagram, it would be shown that the electron in the $d$ orbitals aren't paired. This explains how an uncharged Vanadium is paramagnetic, but if Vandium loses 3 electrons, the only shells left are ones with paired electrons.

How can a Vanadium 3+ ion be paramagnetic if it loses all its unpaired electrons?

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In addition to the general rules of how electronic configurations of atoms and ions are calculated, the elements from the $d$ block (aka the transition metals) obey one special rule:

In general, electrons are removed from the valence-shell s orbitals before they are removed from valence d orbitals when transition metals are ionized.

(I took this formulation from this online lecture notes, but you will find equivalent statements in your textbooks.)

So, what that does mean is that if you remove electrons from vanadium (0), you will remove the 4s electrons before you remove the 3d electrons. So, you have the following electronic configurations:

V is [Ar] 4s2 3d3
V2+ is [Ar] 4s0 3d3
V3+ is [Ar] 4s0 3d2
V4+ is [Ar] 4s0 3d1
V5+ is [Ar] 4s0 3d0

And thus, V3+ is paramagnetic, because it has two unpaired 3d electrons. In fact, all the ions above are paramagnetic, except V5+.

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Vanadium can exist in many oxidation states. In the case of molecules containing Vanadium, the Vanadium ion (in either oxidation state) can exhibit different magnetic properties.

You can have different compounds of V+3 with different magnetic properties. For example, V2O3 is antiferromagnetic, whereas vanadyl (III) iodide (VI3) appears to effectively be paramagnetic. Another vanadium compound that is paramagnetic is VF4.

To know the electronic structure of a particular vanadium compound, I would consider doing QM calculations, but maybe someone has a way of predicting it without using involved calculations.

EDIT

This answer considers Vanadium compounds, as F'x has noted. Keep in mind that there are exceptions to the rules given - notably chromium, [Ar]4$s$1 3$d$5.

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sorry, but it is wrong: each ion can exist by itself (think gas phase, or isolated ion in ion trap), and it can be either diamagnetic (no unpaired electrons in its electronic configuration) or paramagnetic (one or more unpaired electron)… –  F'x Apr 28 '12 at 21:28
and I forgot to add: you are right on one count, namely that magnetic properties of compounds are not those of the vanadium ion itself… –  F'x Apr 28 '12 at 21:36
Your example of V3+ ions having different magnetic properties is misleading. Both compounds you mentioned are paramagnetic. Ferromagnetic and antiferromagnetic both refer to paramagnetic compounds, but with different spin ordering in bulk. –  Vytenis Apr 29 '12 at 20:00