1
$\begingroup$

In transition metals, is the shell with the highest energy considered the valence shell? For example, in copper the electronic configuration is ${[Ar]\text{ } 3d^{10}\text{ } 4s^{1}}$. However, in accordance with the afbau principle the $3d$ subshell actually has higher energy than the $4s$ one. Does this mean the $3d$ subshell is considered the valence?

Second, why is gold not considered to be a reactive substance. It's electronic configuration is not stable: $[Xe]\text{ } {4f}^{14} {5d}^{10} 6s^1$, as it has an unfilled $6s$ orbital. So why is it not highly reactive like some other transition metals?

$\endgroup$
3
$\begingroup$
  1. Nope, as wiki tells us, only outermost electron shell is considered valence shell, though d-electrons for transition metals are considered as valence electrons. No logic here, only tradition.

  2. as you can see, there are three electron subshells above obviously inert Xe core. 5d-electons, that most of the time are far from the nuclei because of the form of their orbitals, feel dramatic increase in effective core charge through the third d-row. For example, for Ti effective charge, keeping its d-electrons is 4. For gold it is 11, so the electrons are well-held.

    For s-electrons the form of the orbital puts very significant part of electronic density near the nuclei. This means, that for heavy elements (staring from as far as Ga) a so-known 'inert electron pair effect' is observed: in comparison with p- and d-electrons, due the form of the s-orbital, s-electrons are much less shielded from the charge of the nuclei by underlying, leading to much more strong bonding of said electrons. As effect, Bi is highly unstable in valence state +5, perbromates are much stronger oxidizing agent than perchlorates, but same is not true for bromates and chlorates, and Ag, Au and Hg have very tightly bound s-electons, leading to relative inertness of said metals.

    For Au and Ag, however, strongly bound s-electrons participate in extremely powerful (due to relative small size of s-orbitals) s-s interactions, while for Hg only relatively weak s-p and d-p interactions are possible because of filled d- and s- subshells. Thus, metallic Ag and especially Au in metallic state have s-electrons involved into effective covalent interactions, leading to relatively inert nature. Beware, however, that they are far from unreactive: Ag have high affinity to sulfur, and both Ag and Au dissolve in cyanide solutions when oxygen is present, not saying about such agressive reagents as chlorine and bromine.

    Still, Au, Ag and Cu share one trait with Ia (I main subgroup) elements. Due to only spherical s-orbitals involvement in atom bonding in metal state, all three metals have high plasticity. Other d-elements have significant d-d bonding in metal states, and d-orbitals are not spherically symmetrical, leading to much less plastic behavior.

$\endgroup$
  • $\begingroup$ For the answer to #1, why do metals use both $d$ and $s$ subshells for bonding, in comparison to other elements? And as for #2, I don't quite understand what is meant by -electrons are much less shielded from the charge of the nuclei by underlying, leading to much more strong bonding of said electrons. Also, if the s-s interactions in Ag and Au make both effectively inert, why does silver tarnish whereas gold does not? $\endgroup$ – 1110101001 Jun 10 '14 at 4:10
  • $\begingroup$ @user2612743 #1 because d-orbitals for them arelarge enough to be comparable in size with s/p orbitals of next shells. #2 shielding is about inner shells reducing nuclei effective charge for outer shells. For Li, effective charge felt by the outer electron should be 3(nuclei) - 2 (celectrons on the inner shell). Electrons on s-orbitals are shielded less because most e-density on it is near nuclei, but for p- and d-orbitals it is far from nuclei. || Both Ag and Au has ~0 affinity to sulfur because of comparable size of orbitals, bit >0 for Ag, a bit <0 for Au. H2S is always present in air. $\endgroup$ – permeakra Jun 10 '14 at 5:02
  • $\begingroup$ How does the fact that the s orbital is less shielded and therefore more strongly bound to the nucleus result in s-s interactions? And also, if I understand correctly it is this s-s covalent-like interactions that essentially fill the s suborbital and result in the inertness? $\endgroup$ – 1110101001 Jun 10 '14 at 5:57
  • 1
    $\begingroup$ @user2612743 The fact of little shielding does not make s-s interactions possible, but it does make s-orbitals smaller and as result their bonding becomes more effective. This does not prevent Cu/Au/Ag from forming effective covalent bonds with other elements, but common elements can't form effective covalent bonds with Au because of vastly different orbitals size and strong bonding of Au with it's valence electrons makes forming of ionic bonds problematic. As result, Ag/Au oxides and halogenides are relatively unstable. $\endgroup$ – permeakra Jun 10 '14 at 10:17

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.