Electron orbital diagram of vanadium I am learning about electron configurations and I came across the "exceptions" of $\ce{Cr}$ and $\ce{Cu}$ where electrons actually half fill the $\mathrm d$ orbitals instead of the expected $\mathrm s$ orbitals. I thought maybe vanadium would do the same, but my textbook gives the answer of an electron configuration exactly following the Aufbau principle. 

I've attached a picture of my drawing for the textbook answer and what I had thought might happen — could someone explain why vanadium is not an "exception" like chromium or copper?
Electrons always fill in the lowest energy configuration possible.

Cr and Cu, as well as Cu and Ag, are exceptions in the "typical" filling order.  In the case of Cr and Cu, they are stabilized by having 2 half filled orbitals, which maximizes exchange energy and minimizes electron repulsion.  In their case, the energy to promote an s electron to the d orbitals is compensated for by the effects of exchange energy and no repulsion.  This effect is called "the stability of half filled subshells", or something to that effect, in most textbooks. 

The energy cost to promote 2 electrons from the s subshell, as would be the case for Vanadium, is too much to be compensated for by the exchange energy gained and the repulsion lost, so the lower energy configuration is to have two electrons in the s subshell and 3 in the d.

Here's a nice website that shows the electron configuration for all elements: with http://www.ptable.com/#Orbital