Dear colleagues,

I would like to ask your thoughts about a question that I am trying to explain to a group of students whom I am helping to prepare for the Chemistry Olympiad.

Right answer: B

My explanation (I am trying to link it to periodic trends):

Reduction is the inverse of ionization energy.

In ionization energy, we have M(g) → M⁺(g) + e; the reduction runs backward: M⁺(g) + e⁻ → M(g).

The half-reaction in this question is M²⁺(aq) + 2e⁻ → M(s). Strip away the “aq” and “s” and look at the electrons: the cation M²⁺ gains two electrons to go from M²⁺ to neutral M (reduction process). It is the reverse of the element M losing two electrons to go from neutral M to M²⁺ (ionization energy).

Therefore, an element with a high ionization energy (IE) is hard to pull electrons off; this same element has a high reduction potential (E°), meaning electrons want to come back on. So, the element with the highest ionization energy has the higher reduction potential.

Now, let's analyze how ionization energy varies among Ni, Pd, and Pt.

In general, ionization energy increases from bottom to top; however, because we are dealing with transition metals, we must account for the anomalous behavior of d and f orbitals.

The reason why ionization energy increases going up a group is because each added shell brings more shielding while Zeff stays roughly the same. By that logic, we’d expect Ni to have the highest IE of the three (fewest shells, least shielding) and Pt to have the lowest (most shells, most shielding), with Pd in between.

That’s not what happens. Ni is the lowest; Pd and Pt are both higher than Ni. The reason is that between each pair, a subshell fills in that shields unusually poorly (d and f orbitals).

Between Ni and Pd, Ni’s 3d subshell is filling right before Pd. d electrons already shield worse than s or p electrons, so Pd’s valence electrons feel more of the nuclear charge.

Between Pd and Pt, the effect repeats, but worse. Before you reach Pt, all 14 lanthanides have to fill their 4f subshell, and f electrons shield even more poorly than d electrons. This lanthanide contraction raises Pt’s Zeff further still, which is why Pt’s IE stays elevated instead of dropping the way you’d expect from adding an entire extra shell.

Since IE and E° are linked, Ni’s low IE becomes Ni’s low E° (least willing to be reduced), and Pd and Pt’s anomalously high IE values become their anomalously high E° values.

9 hours ago, Scienc said:

Dear colleagues,

I would like to ask your thoughts about a question that I am trying to explain to a group of students whom I am helping to prepare for the Chemistry Olympiad.

Right answer: B

My explanation (I am trying to link it to periodic trends):

Reduction is the inverse of ionization energy.

In ionization energy, we have M(g) → M⁺(g) + e; the reduction runs backward: M⁺(g) + e⁻ → M(g).

The half-reaction in this question is M²⁺(aq) + 2e⁻ → M(s). Strip away the “aq” and “s” and look at the electrons: the cation M²⁺ gains two electrons to go from M²⁺ to neutral M (reduction process). It is the reverse of the element M losing two electrons to go from neutral M to M²⁺ (ionization energy).

Therefore, an element with a high ionization energy (IE) is hard to pull electrons off; this same element has a high reduction potential (E°), meaning electrons want to come back on. So, the element with the highest ionization energy has the higher reduction potential.

Now, let's analyze how ionization energy varies among Ni, Pd, and Pt.

In general, ionization energy increases from bottom to top; however, because we are dealing with transition metals, we must account for the anomalous behavior of d and f orbitals.

The reason why ionization energy increases going up a group is because each added shell brings more shielding while Zeff stays roughly the same. By that logic, we’d expect Ni to have the highest IE of the three (fewest shells, least shielding) and Pt to have the lowest (most shells, most shielding), with Pd in between.

That’s not what happens. Ni is the lowest; Pd and Pt are both higher than Ni. The reason is that between each pair, a subshell fills in that shields unusually poorly (d and f orbitals).

Between Ni and Pd, Ni’s 3d subshell is filling right before Pd. d electrons already shield worse than s or p electrons, so Pd’s valence electrons feel more of the nuclear charge.

Between Pd and Pt, the effect repeats, but worse. Before you reach Pt, all 14 lanthanides have to fill their 4f subshell, and f electrons shield even more poorly than d electrons. This lanthanide contraction raises Pt’s Zeff further still, which is why Pt’s IE stays elevated instead of dropping the way you’d expect from adding an entire extra shell.

Since IE and E° are linked, Ni’s low IE becomes Ni’s low E° (least willing to be reduced), and Pd and Pt’s anomalously high IE values become their anomalously high E° values.

I stress I am very rusty on my electrochemistry but this strikes me as a potentially risky oversimplification in the context of teaching students. While your explanation of the apparently anomalous order of relative ionisation energies seems reasonable, electrode potential seems to me to be a resultant of that and other factors.

Ionisation energy is the energy required for the simple process M(g) -> M+(g) + e-(g), whereas the reduction of the aqueous ion to the solid metal also involves two other very significant energy contributions, viz. the lattice energy of the solid metal i.e. for M(g) -> M(s), equivalent to the latent heat of vaporisation, and the solvation energy of the cation, M+ (g) -> M+(aq).

If you look for instance at the order of electrode potentials for reduction of alkali metal cations, the order is Li ~ Cs < Rb ~ K < Fr < Na. So evidently even for elements with the simplest electronic configurations, electrode potential does not align with ionisation energy.

I would have thought you need to show the students an energy level diagram with 3 steps: Aqueous ion to gaseous ion, gaseous ion to neutral gaseous atom and then neutral gaseous atom to solid metal. The electrode potential would be the resultant of all 3 steps in combination.

Unfortunately this would mean there is no simple way to predict the order of electrode potentials in the Ni-Pd-Pt series, unless you make the assumption that latent heats of vaporisation and solvation energies of all three cations will be very similar, which seems doubtful to me.

But, as I say, I am very rusty on this stuff. Maybe someone else here can comment.

18 hours ago, Scienc said:

Between Ni and Pd, Ni’s 3d subshell is filling right before Pd. d electrons already shield worse than s or p electrons, so Pd’s valence electrons feel more of the nuclear charge.

Between Pd and Pt, the effect repeats, but worse. Before you reach Pt, all 14 lanthanides have to fill their 4f subshell, and f electrons shield even more poorly than d electrons. This lanthanide contraction raises Pt’s Zeff further still, which is why Pt’s IE stays elevated instead of dropping the way you’d expect from adding an entire extra shell.

Your proposed explanation seems consistent with the Aufbau principle. Just for clarity, are you teaching students who should generally be expected to understand this area and its many wrinkles (Hund's Rule etc.) which apply particularly to Group X?

8 hours ago, exchemist said:

would have thought you need to show the students an energy level diagram with 3 steps: Aqueous ion to gaseous ion, gaseous ion to neutral gaseous atom and then neutral gaseous atom to solid metal. The electrode potential would be the resultant of all 3 steps in combination.

Unfortunately this would mean there is no simple way to predict the order of electrode potentials in the Ni-Pd-Pt series, unless you make the assumption that latent heats of vaporisation and solvation energies of all three cations will be very similar, which seems doubtful to me.

Good points. Also, are simple bivalent solutions of Pd, Pt real things? (Seen something about Pd(II) acetate somewhere or other)

58 minutes ago, sethoflagos said:

Your proposed explanation seems consistent with the Aufbau principle. Just for clarity, are you teaching students who should generally be expected to understand this area and its many wrinkles (Hund's Rule etc.) which apply particularly to Group X?

Good points. Also, are simple bivalent solutions of Pd, Pt real things? (Seen something about Pd(II) acetate somewhere or other)

There are quoted electrode potentials for reduction of the respective cations to the metal, certainly: https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)

I should have also clarified that the energy level diagram should really consider Gibbs free energy not just enthalpy, since entropy changes will come into play regarding solvation shells and the ordered array of a solid crystal.

But this poster is a bit weird. I don’t believe he or she is teaching, to judge by the posting history. But he or she has popped up at intervals for several years, so doesn’t look like a student either. Oh well, let’s see if there is a response.

Edited October 9Oct 9 by exchemist