9
$\begingroup$

Is the half-full rule and full rule followed in the 6th and 7th periods? (Note: Half-full rules is Hund's rule)

Example: What is the correct electron configuration? $$\ce{W = [Xe] 6s^2 4f^{14} 5d^4}$$ doesn't use half full rule, or $$\ce{W = [Xe] 6s^1 4f^{14} 5d^5}$$ uses half full rule.

$\endgroup$
  • $\begingroup$ Just a note: here at my place, it's more common to call it "fully-filled" instead of just "full". $\endgroup$ – Gaurang Tandon Mar 31 '18 at 12:51
4
$\begingroup$

Short answer: no! The first known exception to occur in the periodic table is in period 5: niobium’s ground state electronic configuration is

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

which is not warranted by the “usual” rules for determining electronic configuration. So, the “half-full rule” is not sufficient in periods 5 and higher.


Let’s look at Wikipedia’s list of exceptions to Madelung’s rule. In period 4, the only exceptions are $\ce{Cr}$ and $\ce{Cu}$, which are accounted for by the “half-full rule”. (Don’t bother too much about the dispute for Ni.)

In period 5, the exceptions to Madelung’s rule are classified in three groups:

  • $\ce{Mo}$ and $\ce{Ag}$: they are the analogues of Cu and Cr
  • $\ce{Pd}$ is $\ce{5s^0 4d^10}$: a different type of application of the explanation that “fully filled shells are particularly stable”, because in this case two electrons from the s shell were pulled into the d shell. Not what you usually learn.
  • $\ce{Nb}$ is $\ce{5s^1 4d^4}$, and $\ce{Ru}$ is $\ce{5s^1 4d^7}$: these cannot be explained by simple rules, and one has to perform complex quantum chemistry calculations to understand these electronic configurations.

Period 6 contains further inconsistencies, because of the introduction of $\ce{f}$ orbitals into the mix.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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