Why V3+ ion has the electronic configuration [Ar]3d2 instead of [Ar]4s2?

NetherCraft 0

Also, what is electronic configuration of Mn3+ ion?

Marco, in response to your answer “what is higher energetic levels, that is what is the type of electron orbitals will loose in cation forming” – this is the point of my question! If 3d energy level is higher than 4s level, than why V3+ ion is losing first its lower energy electrons from 4s level????????

5 Answers

  • Due to the screening of d electrons, the 4s electrons are less firmly held to the nucleus. So the first losing electrons are 4s electrons in forming V3+. But the most stable oxidation state of V is +5.

    The configuration of Mn3+ is [Ar]3d4, due to the same reason.

  • V 3 Electron Configuration

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    You are correct, the atom is Ca. I suspect that the ion is the same thing. I think the M2+ means that it simply has a 2+ charge on it. Thus [Ar] 3d2 would be Ti. Then count back 2 from there on the periodic table and that would put you at Ca.

  • The first person was correct about outermost S electrons being less firmly held to the nucleus therefore they are taken first. (FYI V= 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3)

    So yes you first lose the 4s”2″ electrons creating V2+(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3)

    YET you need to take it one step further to get to V3+ and then you must 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 [Ar]2d^2 in VEC form.

  • Lancenigo di Villorba (TV), Italy

    Dear Lady,

    your question concerns foundamental electronic configurations of vanadium(III) ions.

    As you know, vanadium is a transition metal, hence it can show a partial full’s grade of its “d electronic orbitals”. Moreover, they exist also “s orbitals” and “p orbitals” and the respective guessed electrons. Among these type of electron orbitals, what is higher energetic levels, that is what is the type of electron orbitals will loose in cation forming? Several chemical elements involve “s electrons” as “valence’s electrons”, so you estimate than “p electrons” are most loser, succeeded by “s electrons”.

    Thus, vanadium lose 3 electron forming tri-valent cation : these electrons belong “3s orbital” (two electrons) and “3d orbital” (third electron).

    About manganese(III), you can write the following :


    I hope this helps you.

    I wish you a nice night.

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