The mass spectrum of element X is presented in the diagram above. [VISUAL] Based on the spectrum, wh — Atomic Structure Chemistry Question

STEPS:

1. Understand what a mass spectrum of an element represents: In a mass spectrum of a pure element, the peaks along the horizontal axis (representing mass-to-charge ratio, $m/z$, which simplifies to mass in amu for singly charged ions) correspond directly to the individual isotopes of that element. The height of each peak along the vertical axis represents the relative abundance of each isotope in nature.
2. Read the isotopic masses and their approximate relative abundances from the graph:
* Isotope 1: mass $\approx 90\text{ amu}$, relative abundance $\approx 51\%$ (slightly above the 50% grid line)
* Isotope 2: mass $\approx 91\text{ amu}$, relative abundance $\approx 11\%$
* Isotope 3: mass $\approx 92\text{ amu}$, relative abundance $\approx 17\%$
* Isotope 4: mass $\approx 94\text{ amu}$, relative abundance $\approx 17\%$
* Isotope 5: mass $\approx 96\text{ amu}$, relative abundance $\approx 4\%$
3. Recall how the average atomic mass of an element is determined: The average atomic mass of an element reported on the periodic table is a weighted average of the masses of all its naturally occurring isotopes:
$$\text{Average Atomic Mass} = \sum (\text{fractional abundance} \times \text{isotopic mass})$$
4. Estimate the weighted average based on the peak distribution:
* Since the isotope at $90\text{ amu}$ is the most abundant ($\approx 51\%$), the average atomic mass must be heavily weighted toward $90$.
* However, because nearly half of the atoms in the sample ($\approx 49\%$) are heavier than $90\text{ amu}$ (with masses of $91$, $92$, $94$, and $96$), these heavier isotopes will pull the weighted average above 90.
* Since the mass 90 isotope represents more than half of the total abundance, and the rest of the isotopes are relatively close to it in mass, this weighted average is mathematically prevented from reaching or exceeding $92$:
$$\text{Average Atomic Mass} \approx (0.51 \times 90) + (0.11 \times 91) + (0.17 \times 92) + (0.17 \times 94) + (0.04 \times 96)$$
$$\text{Average Atomic Mass} \approx 45.9 + 10.0 + 15.6 + 16.0 + 3.8 \approx \mathbf{91.3\text{ amu}}$$
*(Note: This element is Zirconium, $\text{Zr}$, which has an actual atomic mass of $91.22\text{ amu}$.)*
5. Select the matching option: Because the average atomic mass is approximately $91.3\text{ amu}$, it falls between $90$ and $92$, verifying Option D as the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect: In a mass spectrum of a pure element, the peaks represent isotopes (atoms of the same element with different numbers of neutrons and therefore different masses), not oxidation states. Oxidation states describe the chemical charges of atoms in molecules or polyatomic ions.
• Option B is incorrect: A mass spectrum does not provide information about electron configurations or sublevels. A photoelectron spectrum (PES) is the specific analytical tool where peaks correspond to the binding energies of different electron sublevels (such as $1s$, $2s$, $2p$, etc.).
• Option C is incorrect: While $90$ is the mass of the most abundant *individual isotope* in the sample, the average atomic mass of the element must factor in all of the naturally occurring isotopes. The presence of heavier isotopes ($91$, $92$, $94$, and $96$) pulls the overall weighted average above $90$.