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

$\ce{Nb: [Kr] 5s^1 4d^4}$

Why is that so? What factors stabilize this configuration, compared to the obvious $\ce{5s^2 4d^3}$ (Aufbau principle), or the otherwise possible $\ce{5s^0 4d^5}$ (half-filled shell)?

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    $\begingroup$ 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. $\endgroup$ – permeakra Nov 3 '12 at 13:41
  • $\begingroup$ @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. $\endgroup$ – F'x Nov 3 '12 at 13:57
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    $\begingroup$ @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. $\endgroup$ – permeakra Nov 3 '12 at 13:59
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    $\begingroup$ @F'x This question is very close to my heart. My favorite element and current chemical obsession, palladium, does this same thing to a further extreme. I've never really been able to understand the explanation I've found as they involve high level quantum stuff. I would humbly suggest that the answer to this lay more with the physicists then chemists. $\endgroup$ – Cargo Dec 8 '12 at 1:30
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    $\begingroup$ I was about to ask "Why are there exceptions to the Aufbau principle/Madelung rule?" but found this and I assume it would be considered a duplicate. Yet, I think this question deserves being re-titled since it's about many elements and not just Niobium. $\endgroup$ – Molx Jun 6 '15 at 21:50
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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 $\ce{Nb}$, we see the choice of either pairing electrons at a lower energy, or of separating them at higher energy, as seen below:

d:   ↿ ↿ ↿ _ _        ↿ ↿ ↿ ↿ _        ↿ ↿ ↿ ↿ ↿     ^
                 OR               OR               | 
s:   ⥮               ↿                _            Energy gap (E)

The top row is for the d-orbitals, which are higher in energy, and the bottom row is for 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 principle 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 chemistry education. Basically (due to quantum chemistry 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:

↿ ↿ ↿

from say a p-subshell, you would get 3 K, for each pair, while from this example:

⥮ ↿ ↿ ↿ ↿

from a $\ce{d^6}$, 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 determines the state of the electron configuration. Using the example we started with:

d:   ↿ ↿ ↿ _ _         ↿ ↿ ↿ ↿ _         ↿ ↿ ↿ ↿ ↿    ^
s:   ⥮           OR   ↿           OR   _            | 
PE:  3K + 1C          6K + 0C          10K + 0C     Energy gap (E)

You can see from the example that shoving 1 e- up from the s to the d-subshell 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 $\ce{Nb}$'s ground state. Which is indeed the case for $\ce{Nb}$.

Next, lets look at perhaps exciting the second s e- up to the d-subshell. 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-subshell.

It turns out that for $\ce{Nb}$: 4K + 0C < E (remember that C is considered a negative value, which we're not losing any of), so $\ce{Nb}$ is ultimately found in the $\ce{5s^1 4d^4}$ configuration.

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    $\begingroup$ Just in case it wasn't obvious for any reader: "from say a p-subshell, you would get 3 K" that's because there are $^3C_2=3$ electron pairs. Similarly, "from a $\ce{d6}$, you would get 10 K" that's because there are $^5C_2=10$ electron pairs. The author calculated this by simple combinatorics ;) $\endgroup$ – Gaurang Tandon Mar 31 '18 at 12:39
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The question of anomalous electronic configurations, meaning $\mathrm{s^1}$ or $\mathrm{s^0}$ in one case (Pd) is very badly explained in textbooks.

For example, the anomalous configuration of Cr ($\mathrm{3d^5~4s^1}$) 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 (henceforth abbreviated as hfss) 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 is, 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.

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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:

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Just as in a seesaw similarly in Zr equal number of 2 electrons are present each for 6s and 5d orbitals which balances each others forces; however, in Nb as the new entrant electron is bound to enter into the vacant 5d orbital it helps in pulling 1 of the 2 electrons of 6s orbital into the 5d orbital due to sheer electronic number superiority and the following electrons similarly are successively pulled into the vacant 5d orbitals of the subsequent elements of this second series of transition elements period.

This only happens as the two 6s and 5d orbitals have almost negligible energy gap and it is not possible to prevent electron entering the 5d orbital from the 6s orbital.

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