I have seen that in many publications authors discuss the metals in the d-block of the periodic table (often called 'transition metals') separately from other metals (e.g. alkali metals and aluminum). They may also consider 'noble metals' (transition metals like Cu and Au with completely filled d-band) separately. What is the role of d-band in transition metals that make them different? In other words, what is the difference between d-band and s-,p-bands? (is it related to, for example, the differences between the shapes of wavefunctions?) Why is it important that whether the d-band is partially filled or completely filled?


$d$-states have less spatial extent than $s$- or $p$-states. Bands formed from overlapping $d$-states have less bandwidth than bands formed from overlapping $s$- or $p$-states (the more overlap, the more broadening of the originally discrete electronic spectrum of an isolated atom).

In transition metals, the more localized outer $3d$-, $4d$-, or $5d$-states screen the interaction of the half-filled outer $s$-band with other states: surfaces of transition metals to the left of the periodic table (i.e. of transition metals with few outer $d$-electrons) react more easily with e.g. oxygen than transition metals to the right with more outer $d$-electrons.

In the case of a completely filled outer $d$-shell (i.e. in the case of copper, silver, and gold), this screening is most efficient and the reason why these metals are noble. As long as the outer $d$-shell is only partially filled, Fermi-Dirac statistics dictates that the corresponding $d$-band must be pinned to the Fermi level (i.e. the filled part of this $d$-band must be below the Fermi level, the empty part above in energy). If the $d$-band is completely filled, the $d$-band is not pinned to the Fermi level anymore (no empty $d$-states), and thus in Cu, Ag, and Au, the $d$-bands lie lower in energy than in the other transition metals. $d$-electrons thus do not take part in forming the Fermi surface of Cu, Ag, and Au, rendering these metals special compared to the other transition metals.

The screening effect of the $d$-states is of quantum mechanical nature: due to the narrow $d$-bandwidth, bonding and anti-bonding states of reactants are formed at the $d$-bandedges. The more the anti-bonding states are filled, the weaker the interaction (in analogy to molecular orbital theory). In the case of Cu, Ag, and Au, the screening effect is particularly strong, because the upper $d$-bandedge lies energetically deep below the Fermi-level (due to complete $d$-band filling as explained above), and thus corresponding anti-bonding states in the vicinity of this edge are (generally) more or less completely filled leading to a strong suppression of interaction with the reactant.

Finally, the true band structure of transition metals does not consist of purely $s$-, $p$-, and $d$-states, but these states hybridize, and the bands have mixed orbital character, dominated by $s$, $p$, or $d$ depending on energy and wave vector. Nevertheless, given this dominant orbital character, a simplified picture of pure $s$- and $d$-bands as used above still makes sense.

| cite | improve this answer | |
  • $\begingroup$ Many thanks. I don't understand the physical meaning of "screen"ing in the 2nd paragraph. Do you mean electric-field screening? As far as I know, in isolated metal atoms, the stronger the screening, the lower the ionization energy, and the higher the chemical reactivity. For example, in alkali metals, Cs reacts with water much more intensely than Na does. $\endgroup$ – apadana Jun 12 '19 at 15:07
  • $\begingroup$ The electrostatic screening of the 6s states in Cs is related to the shielding of the nucleus by inner-shell electrons (leading to a weaker binding of the 6s and thus higher reactivity). In the case of the screening of interaction due to outer d shells in transition metals, the effect is a quantum mechanical one (i.e. not really of electrostatic nature). This is indeed not obvious, and I have added a paragraph about the screening effect to the answer. $\endgroup$ – v-joe Jun 12 '19 at 16:28

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.