Take the 2-minute tour ×
Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. It's 100% free, no registration required.

As cited in an answer to this question, the ground state electronic configuration of Niobium is:

Nb: [Kr] 5s1 4d4

Why is that so? What factors stabilize this configuration, compared to the obvious 5s2 4d3 (Aufbau principle), or the otherwise possible 5s0 4d5 (half-filled shell).

share|improve this question
2  
it is a nice question –  New Alexandria Oct 29 '12 at 15:37
2  
5s subshell here has lower energy, then 4d subshell. However, there is non-zero energy of electron pairing on same orbital. So, we have one electron on 5s subshell and another is pushed onto 4d subshell. ||| do not overestimate these so-known 'principles', they work in most cases, true, but can fail dramatically on border cases, like this one where energy gap between two orbitals is not so big. ||| I do not post this as answer as I cannot prove my point by math or reference, it is a guess. –  permeakra Nov 3 '12 at 13:41
    
@permeakra yeah, but if you follow this simple logic, there's no reason not to push two electrons into the d shell :) Hence the reason I ask how this can explained, which will most possibly involve quantum chemistry calculations. –  F'x Nov 3 '12 at 13:57
    
@F'x It is repulsion of two electrons on same orbital that pushes one of them from 5S subshell. After one is pushed away, the other have no reason to move. –  permeakra Nov 3 '12 at 13:59
    
From what I can remember, the names come from atomic spectral line studies. S stood for Sharp, P for Principal, D for Diffused, F for Fundamental, and anything after that is alphabetical. Something about the energies and where they fall all the spectral line if I am correct. Don't know much more detail than that!! So, to lessen the workload, we write the first chunk as [noble gas]. You track back to the most recent noble gas, and write it in brackets to say: 'everything this noble gas has and then...'. Example: [Ar]4s2 (Calcium) says "everything argon has, the way argon had it, AND 2 electrons –  Susan Diaz Nov 9 '12 at 11:43
show 2 more comments

3 Answers 3

Unfortunately, the Aufbau rule cannot predict all electron configuration as it doesn't take into account electron-electron interactions. In the end the Aufbau is only a rule of thumb. Electronic levels have to be found using quantum calculations taking into account electron-electron interactions (not to mention spin orbit coupling). Therefore there is no simple or rational explanation for this.

References:

share|improve this answer
add comment

There is an explanation to this that can be generalized, which dips a little into quantum chemistry, which is known as the idea of pairing energy. I'm sure you can look up the specifics, but basically in comparing the possible configurations of Nb, we see the choice of either pairing electrons at a lower energy, or of separating them at higher energy, as seen below:

d     7 7 7 _ _         OR         7 7 7 7 _          OR        7 7 7 7 7      ^
                                                                               |
s        7L                            7                            _          Energy Gap(E)

Hopefully you're following, the top row is the d orbitals, which are higher in energy, and the bottom row is the s orbital, which is lower in energy. There is a quantifiable energy gap between the two as denoted on the side (unique for every element). As you may know, electrons like to get in the configuration that is lowest in energy. At first glance, that might suggest putting as many electrons in the s orbital (lower energy) as possible, and then filling the rest in the d-orbital. This is known as the Aufbau principal and is widely taught in chemistry classes. It's not wrong, and works most of the time, but the story doesn't end there. There is a cost to pairing the electrons in the lower orbital, two costs actually, which I will define now:

Repulsion energy: pretty simple, the idea that e- repel, and having two of them in the same orbital will cost some energy. Normally counted as 1 C for every pair of electrons.

Exchange energy: this is a little tricky, and probably the main reason this isn't taught until later in your chem education. Basically (due to quantum chem which I won't bore you with), there is a beneficial energy associated with having pairs of like energy, like spin electrons. Basically, for every pair of electrons at the same energy level (or same orbital shell in this case) and same spin (so, if you had 2 e- in the same orbital, no dice, since they have to be opposite spin), you accrue 1 K exchange energy, which is a stabilizing energy (this is very simplified, but really "stabilizing energy" is nothing more than negative energy. I hope your thermodynamics is in good shape!). The thing with exchange (or K) energy is that you get one for EVERY pair, so in the case:

7 7 7

from say a p orbital, you would get 3 K, for EACH pair, while from this example

7L 7 7 7 7

from a d6, you would get 10 K (for each unique pair, and none for the opposite spin e-)

This K is quantifiable as well (and like the repulsion energy is unique for each atom).

Thus, the combination of these two energies WHEN COMPARED TO THE BAND GAP determine the state of the electron configuration. Using the example we started with:

d     7 7 7 _ _         OR         7 7 7 7 _          OR        7 7 7 7 7      ^
                                                                               |
s        7L                            7                            _          Energy Gap(E)

PE      3K + 1C                       6K + 0C                      10K + 0C

You can see from the example that shoving 1e- up from the s to the d shell results in a loss of 1C (losing positive or "destabilizing" repulsive energy) and gaining 3K (gaining negative or "stabilizing" exchange energy). Therefore, if the sum of these two is greater than the energy gap (i.e 3K - 1C > E) then the electron will indeed be found in the d shell in Nb's ground state. Which is indeed the case for Nb.

Next lets look at perhaps exciting the second s e- up to the d shell. We gain 4 additional K but don't lose any C, and we must again overcome the energy gap for this electron to be found in the d shell.

It turns out that for Nb: 4K + 0C < E (remember that C is considered a negative value, which we're not losing any of), so Nb is ultimately found in the 5s1 4d4 configuration.

That was really long, if you have any questions, leave them in the comments below and I'll try to answer them :D

share|improve this answer
add comment

The question of anomalous electronic configurations, meaning s1 or s0 in one case (Pd) is very badly explained in textbooks. For example the anomalous configuration of Cr, 3d5 4s1 is typically explained as being due to half-filled subshell stability. This is wrong for several reasons. First of all there is nothing especially stable about half-filled subshells. Secondly it does not explain the fact that many second transition series elements show anomalous configurations even though they do not possess half-filled subshells. The possession of a half-filled subshell is neither necessary nor sufficient for there to be an anomalous configuration. There are atoms that have hfss but do not have anomalous configurations and there are atoms that have anomalous configurations but do not have hfss. Conclusion the use of the hfss explanation is ad hoc and should be avoided. It so happens that Cr and just one other atom have both hfss AND show an anomalous configuration. But it cannot be generalized.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.