Assigning an Element from Photoelectron Spectroscopy
A neutral unknown element is analyzed by photoelectron spectroscopy (PES). The spectrum shows five peaks with relative areas 2:2:6:2:1, corresponding to electrons removed from the 1s, 2s, 2p, 3s, and 3p subshells. The approximate binding energies are 1560 eV, 120 eV, 75 eV, 12 eV, and 6 eV, respectively. The element is in Period 3. A student uses the spectrum to assign the electron configuration and then compares the element's first ionization energy with that of the element immediately to its left in the same period. The student is puzzled that the first ionization energy is lower despite the higher nuclear charge. The student also notices that the third and fourth ionization energies differ by a factor of about four and wants to know whether that jump is consistent with the PES assignment.
1. Identify the element, write its ground-state electron configuration, and explain how the peak areas support your assignment.
Read the worked answer
The peak areas sum to 13 electrons, so Z = 13 and the element is aluminum. Its ground-state configuration is 1s2 2s2 2p6 3s2 3p1. The five areas match the subshell occupancies exactly: 2, 2, 6, 2, 1. The 3p peak has the lowest binding energy, consistent with the single valence electron being easiest to remove. The 1s peak has by far the highest binding energy, consistent with the deepest core electrons being most strongly attracted to the nucleus.
2. The first ionization energy of this element is lower than that of the element immediately to its left in the same period, even though nuclear charge is higher. Explain this apparent anomaly using orbital penetration and shielding, and state which electron is removed first.
Read the worked answer
The element to the left is Mg (Z = 12, ending in 3s2). Al has one more proton, but its first electron is removed from 3p, not 3s. The 3p orbital has higher n and is shielded by the filled 3s and core electrons; it is also less penetrating than 3s. The 3p electron therefore feels a lower effective nuclear charge and is easier to remove. This is why Al's first ionization energy (about 577 kJ/mol) is lower than Mg's (about 738 kJ/mol). The PES assignment is consistent because the 3p peak is the lowest-binding-energy peak.
3. A large jump in successive ionization energies occurs after the third electron is removed. Explain why, identifying the shell from which the fourth electron is removed.
Read the worked answer
After removing the 3p electron and the two 3s electrons, Al3+ has the neon core: 1s2 2s2 2p6. The fourth electron must come from the 2p subshell, which is much closer to the nucleus and much less shielded than the n = 3 valence electrons. Breaking into the n = 2 core requires far more energy, so IE4 is much larger than IE3. A factor of about four is consistent with the PES binding energies: the 2p peak is near 75 eV, while the 3s peak is near 12 eV. The jump is therefore expected, not anomalous.