# Why is the vanadium(3+) ion paramagnetic?

I know that the electron configuration of vanadium is $[\ce{Ar}]\mathrm{4s^2 3d^3}$.

None of the electrons in the 3d subshell are paired. Once it loses these three electrons, shouldn't the remainder of the electrons be paired? How can $\ce{V^{3+}}$ be paramagnetic if it loses all its unpaired electrons?

In addition to the general rules of how electronic configurations of atoms and ions are calculated, the elements from the $$\mathrm{d}$$-block (a.k.a. the transition metals) obey one special rule:

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

(I took this formulation from these 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 $$\mathrm{4s}$$ electrons before you remove the $$\mathrm{3d}$$-electrons. So, you have the following electronic configurations:

$$\ce{V}$$ is $$\ce{[Ar]} \mathrm{4s^2 3d^3}$$

$$\ce{V^2+}$$ is $$\ce{[Ar]} \mathrm{4s^0 3d^3}$$

$$\ce{V^3+}$$ is $$\ce{[Ar]} \mathrm{4s^0 3d^2}$$

$$\ce{V^4+}$$ is $$\ce{[Ar]} \mathrm{4s^0 3d^1}$$

$$\ce{V^5+}$$ is $$\ce{[Ar]} \mathrm{4s^0 3d^0}$$

And thus, $$\ce{V^3+}$$ is paramagnetic, because it has two unpaired $$\mathrm{3d}$$-electrons. In fact, all the ions above are paramagnetic, except $$\ce{V^5+}$$.

• is there a reason why you skipped ${V^+ }$ which I suppose is [Ar]3d3 4s1 or [Ar]4s1 3d3 ? Commented Jan 12 at 10:30
• Find me a vanadium(I) compound with strong ionic character and I will consider editing it in. The only 4th period metals that commonly show the +1 oxidation state are potassium and copper. Commented May 4 at 10:00