0
$\begingroup$

I'm getting so much conflicting information!! When I practice electron configuration and google to check my answers I often see configurations that list the 3d orbital before the 4s orbital (apparantly the anomaly that the 4s is actually lower energy than 3d) and 4d written before 5s which is apparantly another anomaly. I'm assuming the order is important.

1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f

According to this diagram 5s comes before 4d and 4s comes before 3d however when I google electron configurations this rarely is shown for example Kr is shown as {Ar} 3d10 4s2 4p6. As Ar finishes at 3p6, the configuration should restart at 4s, but online it's always 3d which restarts.

Another example is Iodine 53. According to my diagram it should be {Kr} 5s2 4d10 5p5 but when I check online its always 4d10 5s2 5p5 Does this matter?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ I would suggest that you do not worry too much about it is just an exercise. Practice the simple ones. I would rather ask the teacher, how do you experimentally or theoretically determine the electron configuration even for simple elements. You will be surprised that there will be no concrete answer (I am searching for the answer for years). So why worry about Pt which is a very complex atom. $\endgroup$
    – ACR
    Commented Jan 16, 2021 at 1:38
  • 2
    $\begingroup$ The list of electron configurations is given here by NIST. math.nist.gov/DFTdata/atomdata/configuration.html $\endgroup$
    – ACR
    Commented Jan 16, 2021 at 1:39

1 Answer 1

Reset to default
2
$\begingroup$

While writing out the electronic configurations, we usually write them in the order of 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f and so on. This method of writing the electronic configuration only represents the traditional increasing order of shells ( 1 2 3 4 or K L M N ), and nested within each shell are the subshells ( s p d f).

This is only the usual method of writing the electronic configurations; it does not reflect the order of increasing energy of orbitals. The order of energy is given by the Aufbau's Principle or the building-up priciple.

In essence, according to the rule, for comparing energies of orbitals of all elements, except hydrogen, we compare the values of the sum of (n+l) ; where n is the priciple quantum number ( corresponding to K=1 L=2 N=3 etc. ) and l is the azimuthal quantum number ( corresponding to s=0 p=1 d=2 etc. ). Greater the value of this sum, more is the (relative) energy associated with that orbit. Moreover, if this sum is same for two orbitals, then we compare the value of n for those orbitals.

So, for example, you mentioned the anomaly in the energy order of 4s and 3d. Using this rule, for 4s we have n=4 & l=0, giving a sum of 4, while for 3d we have n=3 & l=2, giving a sum of 5. This gives us a greater energy for 3d, which shows us that it confirms to the rule rather than being anomalous.

Further, consider the case of 4s and 3p orbitals; the n+l sum is same for both (4 in each case), but since 4s has a greater value of n than 3p , it is more energetic.

Now, electronic configurations are important in the order 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f when we are considering cases of ionization ; so, if you want to make a +2 ion, of say, Fe, then the two electron will be removed from the outermost orbitals, i.e. 4s. However, in cases where we are concerned with the actual energies, we must subscribe to the Aufbau's Principle.

Improve this answer
$\endgroup$
5
  • $\begingroup$ I would like to understand your statement. For example, the orbitals $2p$ and $3s$ do have the same value of $n+l$. For $2p$, the sum $n+l$ is $2+1 = 3$. For $3s$, the sum $n+l$ is $3+0 = 3$. So, with your criteria, the $3s$ orbital should be more stable than the $2p$, and should be filled before the $2p$. It is not the case. What is wrong in my reasoning ? $\endgroup$
    – Maurice
    Commented Jan 16, 2021 at 11:12
  • $\begingroup$ @Maurice the 3s orbital would be more energetic; how does this imply that it would be more stable? $\endgroup$
    – user95829
    Commented Jan 16, 2021 at 11:16
  • $\begingroup$ Why is the $2p$ first filled, before the $3s$ ? Both have same $n+l$ value. In that case the principal quantum number decides. So $3s$ should be filled before $2p$. What is wrong in my development ? $\endgroup$
    – Maurice
    Commented Jan 16, 2021 at 12:22
  • $\begingroup$ @Maurice Exactly, the principal quantum number decides; as a result, 3s would be more energetic than 2p, and hence should be filled after 2p. Clearly, we would first like to fill up electrons in a manner so that first the less energetic orbital (2p, in this case) is filled, and only then the electrons should be filled in the more energetic orbital (3s, here). I don't understand as to which particular argument of my answer leads to the inverted conclusion. Could you please clarify? $\endgroup$
    – user95829
    Commented Jan 16, 2021 at 12:38
  • $\begingroup$ I misunderstood the meaning of your word "energetic". For you it means "higher in energy". For me it was "more stable" But now I understand what youweanted to state. Thank you. $\endgroup$
    – Maurice
    Commented Jan 16, 2021 at 16:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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