Atomic Structure and Properties

How this unit is tested

What you have to know

13 practice questions

1. A sample of iron(III) oxide, Fe2O3, has a mass of 15.0 g. How many moles of Fe2O3 are present? (Molar mass Fe2O3 = 159.7 g/mol)
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   Answer. 0.0939 mol

   Use $n = \dfrac{m}{M}$: $n = \dfrac{15.0\text{ g}}{159.7\text{ g/mol}} = 0.0939\text{ mol}$. Always show the division with units to demonstrate dimensional analysis.
2. Which 10.0 g sample contains the greatest number of atoms?
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   Answer. 10.0 g of Na

   Moles = mass/molar mass, so the element with the smallest molar mass produces the most moles from the same mass. Since each is monatomic, more moles means more atoms, and Na has the smallest molar mass of the four.
3. In a photoelectron spectroscopy (PES) spectrum, what do the relative heights of the peaks represent?
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   Answer. The relative number of electrons in that subshell

   Peak position encodes binding energy, but peak height (area) is proportional to how many electrons occupy that particular subshell — this is how PES spectra reveal electron configuration.
4. An element's PES spectrum shows four peaks corresponding to subshells 1s, 2s, 2p, and 3s with relative peak heights 2:2:6:2. Identify the element and write its ground-state electron configuration.
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   Answer. Magnesium (Mg): $1s^2 2s^2 2p^6 3s^2$

   The peak heights give the electron count in each subshell: 2+2+6+2 = 12 total electrons, which corresponds to atomic number 12, magnesium.
5. Write the ground-state electron configuration for chromium (Cr, Z = 24), and note how it deviates from the expected Aufbau order.
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   Answer. $[Ar]4s^1 3d^5$

   The expected configuration by simple Aufbau filling would be $[Ar]4s^2 3d^4$, but a half-filled 3d subshell combined with a half-filled 4s subshell is more stable, so one 4s electron shifts into 3d.
6. Write the ground-state electron configuration for the Fe3+ ion (Fe, Z = 26).
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   Answer. $[Ar]3d^5$

   Neutral Fe is $[Ar]4s^2 3d^6$. When forming cations, electrons are removed from the highest principal quantum number first, so both 4s electrons leave before any 3d electrons, giving Fe3+ as $[Ar]3d^5$ after removing three electrons total.
7. A compound is found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.
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   Answer. CH2O

   Assume a 100 g sample: C = 40.0 g / 12.01 g/mol = 3.33 mol; H = 6.7 g / 1.01 g/mol = 6.7 mol; O = 53.3 g / 16.00 g/mol = 3.33 mol. Dividing each by the smallest value (3.33) gives a 1:2:1 ratio, so the empirical formula is CH2O.
8. Which correctly ranks the first ionization energies from lowest to highest for Na, Mg, and Al?
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   Answer. Na < Al < Mg

   Ionization energy generally increases across a period, but removing Al's single 3p electron is easier than removing one of Mg's paired 3s2 electrons because a filled s subshell is extra stable. So Al dips below Mg, while both remain above Na.
9. Rank the following isoelectronic species in order of increasing ionic radius: Mg2+, Na+, F-, O2-.
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   Answer. Mg2+ < Na+ < F- < O2-

   All four species have the same electron configuration (10 electrons, like Ne), but they have different nuclear charges. More protons pull the same electron cloud in more tightly, so radius decreases as atomic number increases: Mg2+ (Z=12) is smallest and O2- (Z=8) is largest.
10. Based on valence electron configurations, predict the empirical formula of the ionic compound formed between calcium and nitrogen.
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    Answer. Ca3N2

    Calcium loses 2 electrons to form Ca2+ and reach a noble gas configuration; nitrogen gains 3 electrons to form N3-. To balance total charge, three Ca2+ ions (+6) combine with two N3- ions (-6), giving the formula Ca3N2.
11. Which compound would you expect to have a greater lattice energy: MgO or NaCl? Justify your answer using Coulomb's law.
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    Answer. MgO has the greater lattice energy.

    Lattice energy scales as $E \propto \dfrac{Q_1 Q_2}{r}$. Mg2+ and O2- carry charges of magnitude 2, compared to magnitude 1 for Na+ and Cl-, and their ionic radii are comparable or smaller. The larger charge product makes the electrostatic attraction, and thus the lattice energy, much greater for MgO.
12. A 10.00 g mixture contains 3.00 g of NaCl and 7.00 g of KNO3. What is the mass percent of NaCl in the mixture?
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    Answer. 30.0%

    Mass percent in a mixture is mass of the component divided by total mass of the mixture, times 100: $\dfrac{3.00\text{ g}}{10.00\text{ g}} \times 100 = 30.0\%$. This differs from percent composition of a single compound because the components here are two separate substances, not elements within one formula.
13. Consider the 2p subshell across period 2, from boron to neon. How does the PES-measured binding energy of the 2p electrons change across the period, and why?
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    Answer. Binding energy generally increases from B to Ne, with a slight dip at oxygen.

    Increasing effective nuclear charge across the period pulls the 2p electrons closer and more tightly, raising binding energy overall. The dip at oxygen occurs because pairing a fourth 2p electron introduces extra electron-electron repulsion, temporarily lowering the energy needed to remove it, mirroring the analogous ionization energy anomaly between N and O.

What people get wrong

1. Forgetting to convert percent abundances to decimals before computing a weighted average atomic mass — always divide by 100 first, or the result will be off by roughly a factor of 100.
2. Filling 3d before 4s when forming a transition metal cation — electrons are removed from the highest principal quantum number (4s) first, so Fe3+ is $[Ar]3d^5$, not $[Ar]4s^2 3d^3$.
3. Confusing PES peak height with peak position — height tells you the number of electrons in a subshell, position tells you binding energy; mixing these up leads to misidentifying elements.
4. Assuming ionization energy rises smoothly across every period — the Group 2→13 (e.g., Mg to Al) and Group 15→16 (e.g., N to O) dips are frequently tested and must be explained by subshell stability, not treated as errors.
5. Reporting an empirical formula as if it were the molecular formula — always check whether a given molar mass is a whole-number multiple of the empirical formula mass before finalizing an answer.
6. Treating mass percent of a mixture (two separate compounds) the same as percent composition of a single compound — mixture problems use total mass of all components, not moles from a single formula.