electron config question

[mangos]

So i had always understood that transition metal ions lose their electrons from their "s" orbital first before their "d" orbitals. at least this was what i was taught in gen chem at school

 

however, i recently ran across a question asking the most stable electron config for Ti 2+

 

with my understanding, i had thought it would be [Ar] 3d2

 

but the answer was [Ar] 4s2

 

can anyone confirm this?

[Grey's Anatomy]

My understanding is also that ionization of transition metals results in loss of electrons from the s orbital first, so Ti 2+ should have the electron configuration [Ar] 3d2. Where does the answer say otherwise?

[MD2019_A]

3d orbital is higher in energy than the 4s orbital. When Ti loses two electrons it loses them from 3d; the highest orbital

[Grey's Anatomy]

3d orbital is higher in energy than the 4s orbital. When Ti loses two electrons it loses them from 3d; the highest orbital

 

That's what we would expect but the reverse is true. I remember it reading in the TPR books here's an online source which says:

 

Transition metal ions

 

You have already come across the fact that when the Periodic Table is being built, the 4s orbital is filled before the 3d orbitals. This is because in the empty atom, 4s orbitals have a lower energy than 3d orbitals.

 

However, once the electrons are actually in their orbitals, the energy order changes - and in all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital.

 

The reversed order of the 3d and 4s orbitals only applies to building the atom up in the first place. In all other respects, you treat the 4s electrons as being the outer electrons.

 

Note: This is another of those things that you just have to accept. The explanation again lies well beyond the level you are working at. Just remember that once you have the full electronic structure for one of these atoms, the 4s electrons are the outermost electrons.

 

http://www.chemguide.co.uk/inorganic/transition/features.html

[mangos]

My understanding is also that ionization of transition metals results in loss of electrons from the s orbital first' date=' so Ti 2+ should have the electron configuration [Ar'] 3d2. Where does the answer say otherwise?

 

i was doing prep 101 chemistry passages today and ran into this question.

 

BTW: all of prep 101 chemistry passages are collected from AAMC exams with the same questions. except each passage just has more questions than the usual 5-7 questions/passage on actual aamc

 

this is why its bothering me. the fact the question is based off an actual aamc passage..but then again..it could be an error on prep 101s part

[Grey's Anatomy]

Pretty sure it's a mistake on their part. TPR and online sources confirm that electrons in transition metals are removed from the s orbital first.

[mangos]

thanks for the confirmation!

[Gametime24]

3d orbital is higher in energy than the 4s orbital. When Ti loses two electrons it loses them from 3d; the highest orbital

 

ishaquea is right. the 3d orbital is only lower in energy after it has been filled. the 3d and 4s are very similar in energy naturally and it is not until the d shell is full that it drops below the s shell in terms of energy.

 

So the concept that you lose electrons from the highest energy level first holds here and the electrons are lost from the d shell as the aamc answer states.

 

Also, if you want to look at it from a stability and reasoning point of view, leaving the element with all full shells (by taking the d electrons) makes a much more stable element than leaving it with two free valence electrons in the d level (by taking the s electrons)

 

Hope this helped

[mangos]

ishaquea is right. the 3d orbital is only lower in energy after it has been filled. the 3d and 4s are very similar in energy naturally and it is not until the d shell is full that it drops below the s shell in terms of energy.

 

So the concept that you lose electrons from the highest energy level first holds here and the electrons are lost from the d shell as the aamc answer states.

 

Also, if you want to look at it from a stability and reasoning point of view, leaving the element with all full shells (by taking the d electrons) makes a much more stable element than leaving it with two free valence electrons in the d level (by taking the s electrons)

 

Hope this helped

 

uhh ohh...ahha..well looking at it from the stability point of view makes sense. but i had always learned that you take electrons from s first for any transition metal ion.

 

just googled this: http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/trans.php#electron

 

are there any other confirmations from other prep books that support or refute this? EK says nothing about this

[MD2019_A]

When it comes to filling up the shells, 3d fills up all 5 of its orbitals with an electron each then 4s fills its shell with one electron. The 7th electron fills up 4s orbital completing its pair and then after that 3d fills its orbitals to complete electron pairs per orbital

[Tangy]

When it comes to filling up the shells, 3d fills up all 5 of its orbitals with an electron each then 4s fills its shell with one electron. The 7th electron fills up 4s orbital completing its pair and then after that 3d fills its orbitals to complete electron pairs per orbital

 

Hello

 

I don't think that's quite true. In fact, when you use the periodic table, the row goes from 4s^2 and then into the 3d level.

 

For example, electron configuration of Sc (Atomic number 21) [Ar]4s(2)3d(1).

 

It is only during the special cases of 3d(4) and 3d(9) where you have to "promote" one of the electrons in 4s(2) to 3d(4) or 3d(9) to achieve more stability 3d(5) or 3d(10).

 

For example, electron configuration of Cr (Atomic number 24), would be:

[Ar]4s(1)3d(5), not [Ar]4s(2)3d(4).

 

Finally, when ionizing an atom, you should take it as a rule that you take the electron from the highest shell level (not subshell, but shell - "n").

 

So, if Cr is ionized, you lose the electron from the 4s level which has a higher "n" (shell level) than 3d, to become [Ar]3d(5), rather than [Ar]4s(1)3d(4).

 

In fact, here you realize that this rule is consistent, because you end up with [Ar]3d(5) which is more stable (according to the previous rule), than [Ar]4s(1)3d(4).

 

The above are rules for us to follow. There are probably explanations as to why you fill up 4s before 3d and why when ionizing, you take out from 4s first than 3d. In any case, I found the above rules to work for AAMC questions that I have done relating to electron configuration.

 

I hope this helps

 

My source would be TPR prep books and AAMC tests