A sample containing atoms of C and F was analyzed using x-ray photoelectron spectroscopy. The portio — Atomic Structure Chemistry Question

STEPS:

1. Understand the fundamental concept of X-ray Photoelectron Spectroscopy (XPS): XPS is an analytical technique used to measure the binding energies of electrons in different subshells of an atom. By bombarding a sample with high-energy X-ray photons, core-level electrons (such as those in the inner $1s$ subshell) are ejected. The instrument measures the kinetic energy of these ejected photoelectrons and calculates their binding energy—the electrostatic energy holding them to the nucleus.
2. Relate core binding energy to Coulombic attraction: The binding energy of an electron is governed by Coulomb's Law:
$$E \propto \frac{q_1 q_2}{r}$$
For core $1s$ electrons, which are closest to the nucleus, electron shielding from outer-shell electrons is negligible. Therefore, the attractive force holding these core electrons in place is directly proportional to the number of protons in the nucleus (the nuclear charge, $Z$).
3. Compare the nuclear charges of the elements:
* Carbon (C): Has an atomic number of $6$, meaning it has 6 protons in its nucleus.
* Fluorine (F): Has an atomic number of $9$, meaning it has 9 protons in its nucleus.
4. Determine which element has the higher core binding energy: Because fluorine has a greater nuclear charge ($9\text{ protons}$) than carbon ($6\text{ protons}$), its nucleus pulls on its core $1s$ electrons with a much stronger electrostatic force. As a result, significantly more energy is required to eject a $1s$ electron from a fluorine atom than from a carbon atom.
5. Identify the correct peak from the spectrum: Looking at the provided XPS graph:
* Peak X lies at a much higher binding energy of approximately $68\text{ MJ/mol}$.
* Peak Y lies at a lower binding energy of approximately $28\text{ MJ/mol}$.
Because fluorine's $1s$ electrons must have a higher binding energy than carbon's $1s$ electrons due to fluorine's greater nuclear charge, Peak X represents the $1s$ peak for the fluorine atoms. This matches Option B.

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WHY_OTHERS_WRONG:

• Option A is incorrect: While first ionization energy deals with the energy required to remove the outermost valence electron, this XPS spectrum specifically measures core $1s$ electrons. Furthermore, the premise is factually incorrect: fluorine has a *greater* first ionization energy than carbon ($1680\text{ kJ/mol}$ vs $1086\text{ kJ/mol}$), not a smaller one.
• Option C is incorrect: Although fluorine is indeed more electronegative than carbon, electronegativity is a chemical property describing how strongly an atom attracts shared bonding electrons in a covalent bond. It is not the physical cause of inner-shell core binding energy. Additionally, this option incorrectly attributes the lower binding energy peak (Peak Y) to fluorine.
• Option D is incorrect: While fluorine does have a smaller atomic radius than carbon due to its greater effective nuclear charge, atomic radius refers to the size of the valence shell. It does not dictate core $1s$ binding energy, which is determined by the distance of the $1s$ orbital to the nucleus (which is extremely small and comparable for both Period 2 elements). Additionally, this option incorrectly attributes the lower binding energy peak (Peak Y) to fluorine.