Why the electronic configuration of Vanadium is 2,8,8,5 and not 2,8,11,2

The electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s2 and 3d3
1st shell has 2 electrons
2nd shell has 8 electrons
3rd shell has 11 electrons and
4th shell has 2 electrons Then why do we write 2, 8, 8, 5???!!

Electrons fill up orbitals as shown below due to some orbitals being in a lower energy level even if they are in a higher shell:

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p

There is a nice way of writing your orbitals to remember which one is filled first...see the URL below.

I agree with Adam's answer. If you see the periodic table also, it might be clear. After Argon (whose atomic number is 18), the next element is Potassium, in which the 4s is filled first. Then again in Calsium, the 4s is filled. After that, the 3d starts in Scandium, Titanium and then Vanadium. These atoms behave as if the "attachment" to the nucleus is stronger in 4s than 3d. So, 4s is filled first. And when electrons go out during chemical processes, the 3d electrons leave first, compared with the 4s ones. In other words, the electrons in the 3d behave as if they are in the farther shell. Probably that's why we write it 2,8,8,5 and not 2,8,11,2.

i don't understand who knows spdf configuration why should she bother for such a preliminary consolation like concept .
its last shell do contain 2 e
if such is the conf let it be, its just to indicate it has 3 e nos more than Ca 2,8,8,2
to clarifie the primary level concept. think by spdf only

Not exactly sure what you are asking but I can tell you that the 4s subshell is somewhat "unstable" and will loose its electrons before the 3d. Maybe that is what you are trying to say?

Hope that helps

The shells are K, L, and M. (the first three) K can max have 2 electrons, L and M can have max 8 electrons each. So it is written 2,8,8,5.

The shells are K, L, and M. (the first three) K can max have 2 electrons, L and M can have max 8 electrons each. So it is written 2,8,8,5.