Write the electron configuration for the following ion.

Question

V3+

I got 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2. In simplified form, I got [Ar]4s^2.
I checked my answer against the answer given in the back of my chemistry textbook, and the two didn't correspond. The answer my book gives is [Ar] 3d^2.
I don't understand what I did wrong...

Answer 1

₂₃V = 1s2 2s2 2p6 3s2 3p6 3d3 4s2 = 23 electrons.
Then you remove the last 3 electrons (an this is where you probably made your error). The last three electrons are the 2 4s electrons and 1 of the 3d so the final configuration will be
1s2 2s2 2p6 3s2 3p6 3d2 = 20 electrons.

Students want to remove the 3d electrons first because the 4s fills before the 3d fills; however, the 4s electrons are the outside shell so they are pulled away first. After the two 4s electrons are gone, the next one to leave is one of the 3d electrons to leave two.

Answer 2

Ooga

Answer 3

Read it again until you understand it Anonymous.

Answer 4

tnk u

Answer 5

its [Ar]3d^2 because when we remove electrons we remove the outshell first then innershell follwos