CHEMISTRY · Single correct · AP · Chemistry · Atomic Structure and Properties · Atomic Structure and Properties
Problem
For the successive elements Na, Mg, and Al, photoelectron spectra reveal two key facts:
• The 3s peak shifts to higher binding energy from Na to Mg to Al.
• In Al, a new 3p peak appears at a lower binding energy than the 3s peak of Mg.
Four explanations are proposed. Which explanation is most consistent with both observations?
(a) Because each added proton is fully shielded by the added electron, the effective nuclear charge remains constant across Na, Mg, and Al; the higher 3s binding energy must therefore arise solely from growing electron–electron repulsion, while the 3p electron in Al, being farther out, experiences less repulsion and thus appears at lower binding energy than Mg’s 3s electrons.
(b) The 3p orbital penetrates closer to the nucleus than the 3s orbital, so when it begins to fill in Al its electrons feel a stronger nuclear attraction and consequently register a lower binding energy than the outer 3s electrons of Mg—even as the 3s peak continues to shift upward from Na to Al.
(c) Increasing nuclear charge outpaces additional shielding, so all valence electrons experience greater attraction; however, the 3p electron in Al is better shielded than Mg’s 3s electrons, lowering its binding energy relative to Mg’s 3s peak.
(d) Mg’s 3s subshell is exactly half-filled, a configuration that is unusually stable and therefore appears at a lower binding energy; once that special stability is lost in Al, the first 3p electron shows an even lower binding energy than the destabilised 3s electrons of Mg.