[VISUAL] The complete photoelectron spectra of neutral atoms of two unknown elements, X and Y, are s — Atomic Structure Chemistry Question

STEPS:

1. Understand how to read Photoelectron Spectroscopy (PES) graphs:
* X-axis (Binding Energy): Indicates the energy required to remove an electron from a specific subshell, measured in $\text{MJ/mol}$. Note that the x-axis scale is typically reversed (decreasing from left to right). Peaks with higher binding energies correspond to inner-shell electrons that are closer to the nucleus, while peaks with lower binding energies correspond to valence-shell electrons.
* Y-axis (Relative Number of Electrons): The height of each peak is directly proportional to the number of electrons residing in that particular subshell.

2. Analyze the PES of Element X to determine its electron configuration:
* Peak 1 (at $\approx 5.4\text{ MJ/mol}$): This high-energy peak represents the innermost core subshell, $1s$. Its height corresponds to 2 electrons ($1s^2$).
* Peak 2 (at $\approx 0.4\text{ MJ/mol}$): This lower-energy peak represents the valence subshell, $2s$. Its height is exactly half that of the first peak, corresponding to 1 electron ($2s^1$).
* Total configuration for X: $1s^2 2s^1$ (3 total electrons). Element X is a neutral Lithium ($\text{Li}$) atom, which has a nuclear charge of $+3$.

3. Analyze the PES of Element Y to determine its electron configuration:
* Peak 1 (at $\approx 11.5\text{ MJ/mol}$): Represents the innermost $1s$ subshell. Its height corresponds to 2 electrons ($1s^2$).
* Peak 2 (at $\approx 0.9\text{ MJ/mol}$): Represents the valence $2s$ subshell. Its height is equal to the first peak, corresponding to 2 electrons ($2s^2$).
* Total configuration for Y: $1s^2 2s^2$ (4 total electrons). Element Y is a neutral Beryllium ($\text{Be}$) atom, which has a nuclear charge of $+4$.

4. Compare the binding energies of the two elements using Coulomb's Law:
* Coulomb's Law states that the electrostatic attractive force ($F$) between the positively charged nucleus and an electron is directly proportional to the magnitude of the charges ($q_1, q_2$) and inversely proportional to the square of the distance between them ($r$):
$$F = k \frac{q_1 q_2}{r^2}$$
* Comparing corresponding peaks (such as the $1s$ peaks: $5.4\text{ MJ/mol}$ for X vs. $11.5\text{ MJ/mol}$ for Y), we observe that every peak of Element Y is shifted to a significantly higher binding energy than the corresponding peak of Element X.
* Because both elements have their valence electrons in the same principal energy level ($n = 2$), the significantly stronger electrostatic attraction pulling Y's electrons closer to its nucleus is driven by a greater nuclear charge (more protons in the nucleus of Y than X).

5. Conclude:
* This comparative shift in binding energy provides direct evidence that Element Y has a greater nuclear charge than Element X, confirming Option C is the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: Electronegativity increases from left to right across a period. Because Element Y (Beryllium) has a greater nuclear charge than Element X (Lithium) and is further to the right in Period 2, Y is more electronegative than X, not less.
• Option B is incorrect: The first ionization energy of an element is the energy required to remove its least tightly bound valence electron. This is represented by the rightmost peak on each PES graph (the peak with the lowest binding energy). For Element X, the first ionization energy is $\approx 0.4\text{ MJ/mol}$, whereas for Element Y, it is $\approx 0.9\text{ MJ/mol}$. Therefore, Element Y has a *greater* first ionization energy than Element X.
• Option D is incorrect: Photoelectron spectroscopy measures the electronic shell structure and subshell configurations of neutral atoms. It does not provide information regarding isotopes or their relative abundances; that information is obtained via mass spectrometry.