12
$\begingroup$

The d-block sub-shell is always one shell lower than the s-block and p-block sub-shells. Assuming only the electrons in the highest energy shells count toward the set of valence electrons (is that correct?), d-block is never in the highest energy shell, and so none of the electrons in the d subshells would ever count toward the valence electrons.

I suspect I'm mistaken. Am I? If so, where have gone wrong?

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ Of course it is - d-block is block of elements which have valence d electrons. $\endgroup$
    – Mithoron
    Commented May 30, 2015 at 0:48
  • $\begingroup$ @Mithoron, so then, the valence electrons are just those electrons in the outermost shell of each sub-shell? $\endgroup$
    – Hal
    Commented May 30, 2015 at 0:57
  • $\begingroup$ Just because in the 4s is a higher energy orbital than the 3d, it isn't necessarily true for other elements... Also, atomic energies are not simply sum of orbital energies. $\endgroup$
    – Greg
    Commented May 30, 2015 at 8:52
  • $\begingroup$ Valence electrons are those, which participate in forming bonds. The rules of thumb stated above are just that. The atom is not aware of your pen and paper drawings, it binds in a way which minimizes total energy of the system. And your drawings are here to help you systematically understand that. $\endgroup$
    – ssavec
    Commented Jun 1, 2015 at 8:37

1 Answer 1

Reset to default
11
+50
$\begingroup$

The answer is merely a matter of definition of "valence electrons".

Many texts define "valence electrons" in a way that explicitly includes d electrons of unfilled (but not filled) d subshells.

For example: Chemistry structure and dynamics by Spencer et al. at page 125:[1]

We can define valence electrons as electrons on an atom that are not present in the previous rare gas, ignoring filled d or f subshells.

Many books published in the last 10 years use this definition.

That d electrons may be valence electrons is also supported by the 18-Electron rule (at least to the extent that there is such a rule).

[1] James N. Spencer, George M. Bodner, Lyman H. Rickard: Chemistry: Structure and Dynamics, 5th Edition. John Wiley & Sons: 2010. ISBN: 978-0-470-58711-9

Improve this answer
$\endgroup$
3
  • $\begingroup$ The lanthanides are an interesting case, since their $4f$ subshells are "buried" and generally inaccessible for bonding. Thus, by the rule in Spencer, their $4f$ electrons are valence electrons; but if one considers valence electrons to be those that engage actively in bonding, then they would not be part of the valence. $\endgroup$
    – hBy2Py
    Commented Nov 18, 2015 at 12:30
  • $\begingroup$ If we ignore filled d subshells, how then Cu(OH)2 forms? Cu must have 1 valence electron then, and it clearly has more. There must be a simple explanation that I've overlooked. $\endgroup$
    – CowperKettle
    Commented Jan 28, 2016 at 19:19
  • 2
    $\begingroup$ @CopperKettle if you define "valence electrons" the way the book quoted in the answer does, then you would say "Cu has one valence electron, but non-valence electrons participate in bonding". And if you define "valence electrons" as electron that participate in bond, you would say Cu has more than one valence electron, let's say 11 valence electrons. It's just a matter of defining an unnecessary term. scientificamerican.com/article/… $\endgroup$
    – DavePhD
    Commented Jan 28, 2016 at 19:47

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.