Marco, in response to your answer "what is greater energetic levels, the is what is the form of electron orbitals will loose in cation forming" - this is the allude of my question! If 3d power level is greater than 4s level, than why V3+ ion is losing first its lower energy electrons from 4s level????????
Due to the screening of d electrons, the 4s electron are much less firmly held to the nucleus. So the an initial losing electrons space 4s electrons in developing V3+. However the many stable oxidation state of V is +5.
The an initial person was correct about outermost S electrons being less firmly organized to the nucleus thus they room taken first. (FYI V= 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3)
So correct you very first lose the 4s"2" electrons producing V2+(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3)
YET you have to take it one step further to get to V3+ and then you need to take away a valence electrons in the outermost shell (3d) making it 1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 or
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Lancenigo di Villorba (TV), Italy
your question comes to foundamental digital configurations the vanadium(III) ions.
As you know, vanadium is a change metal, thus it can display a partial full"s grade of the "d electronic orbitals". Moreover, castle exist additionally "s orbitals" and "p orbitals" and also the corresponding guessed electrons. Among these type of electron orbitals, what is greater energetic levels, that is what is the form of electron orbitals will loose in cation forming? number of chemical elements involve "s electrons" as "valence"s electrons", so you estimate 보다 "p electrons" are many loser, thrived by "s electrons".
Thus, vanadium lose 3 electron creating tri-valent cation : this electrons belong "3s orbital" (two electrons) and also "3d orbital" (third electron).
About manganese(III), you can write the following :
I expect this helps you.
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