Scoped to the official blueprint · 17 questions · Module 2 (6) + Module 3 (6) + Module 4 (5)
1. An electron in H falls from n = 4 to n = 2. Is a photon absorbed or emitted, and what is |ΔE|? (R_H = 2.179×10⁻¹⁸ J)
2. For a hydrogen 4d orbital, how many radial and angular nodes are there? TRAP
3. An R(r)-vs-r plot starts at a maximum at r = 0 and crosses the r-axis twice. The orbital is: TRAP
4. Which set of quantum numbers (n, ℓ, mₗ, mₛ) is not allowed? TRAP
5. The "+" and "−" signs (or colours) on the lobes of a p-orbital boundary surface represent: TRAP
6. How many orbitals are in the n = 4 shell, and what is the most-probable radius of the 1s electron in H? (not on the data sheet — recall it)
7. The ground-state configuration of Mn²⁺ (Z of Mn = 25) is: TRAP
8. Which element has an anomalous ground-state configuration?
9. First ionization energy drops slightly from N to O mainly because: TRAP
10. How many unpaired electrons does a ground-state Fe atom have? (Fe = [Ar] 4s² 3d⁶)
11. Which correctly ranks first ionization energy (lowest → highest)? TRAP
12. Which atom is diamagnetic in its ground state?
13. How many total valence electrons are in the carbonate ion, CO₃²⁻?
14. In nitrate (NO₃⁻, drawn with one N=O and two N–O⁻), the formal charge on the central N is:
15. Which central atom has an expanded octet in its best Lewis structure?
16. For resonance to exist in a species, there must be: TRAP
17. In the cyanate ion OCN⁻, the major resonance contributor places the −1 formal charge on O rather than N because: TRAP
1. A Emitted, 4.09×10⁻¹⁹ J. ΔE = −R_H(1/2² − 1/4²) = −2.179×10⁻¹⁸(0.25 − 0.0625) = −2.179×10⁻¹⁸(0.1875) = −4.09×10⁻¹⁹ J. ΔE < 0 → photon emitted (electron drops).
2. A 1 radial, 2 angular. 4d: n=4, ℓ=2. Angular = ℓ = 2; radial = n−ℓ−1 = 4−2−1 = 1 (total = n−1 = 3). Don't swap the two rules.
3. B 3s. Max at r=0 → s (ℓ=0). Two axis crossings → 2 radial nodes → n−0−1 = 2 → n = 3 → 3s.
4. C (2, 2, 0, +½) is not allowed. ℓ must be ≤ n−1, so for n=2 the max ℓ is 1. ℓ=2 is impossible at n=2. (A, B, D all obey n>ℓ and |mₗ|≤ℓ.)
5. B Sign (phase) of Ψ. Never charge — the lobe signs show the wavefunction's phase.
6. B 16 orbitals; 52.9 pm. n=4 → n² = 16. The 1s most-probable radius of H = a₀ = 52.9 pm (memorize — it's not on the data sheet).
7. C [Ar] 3d⁵. Mn = [Ar]4s²3d⁵; remove the two 4s electrons first (highest n) → [Ar]3d⁵. (A, [Ar]4s²3d³, is neutral vanadium — the "blindly apply n+ℓ" trap.)
8. C Cu = [Ar]4s¹3d¹⁰ (anomalous; Cr is the other one). Fe, Ni, Zn are all normal.
9. B The N→O dip: O's 2p⁴ has a newly-paired orbital (extra e⁻–e⁻ repulsion → easier to remove), and N's 2s²2p³ half-filled set is extra stable. Size/Z alone would predict the opposite.
10. C 4. 3d⁶ fills ↑↓ ↑ ↑ ↑ ↑ → 4 unpaired.
11. B Na < Al < Mg < Si. IEs (kJ/mol): Na 496, Al 578, Mg 738, Si 786. The Mg→Al dip (Al removes a higher-energy 3p electron) puts Al below Mg.
12. B Zn. Zn = [Ar]4s²3d¹⁰, all paired → diamagnetic. O (2 unpaired), Mn (5), Al (1) are paramagnetic.
13. B 24. C(4) + 3·O(6) + 2 (for the 2− charge) = 4 + 18 + 2 = 24.
14. C +1. N owns 0 lone-pair e⁻ + ½(8 bonding e⁻) = 4; FC = 5 − 4 = +1.
15. C Xe in XeF₄ has 12 e⁻ (4 bonds + 2 lone pairs). N and C (period 2) never expand; B is electron-deficient, not expanded.
16. C Resonance moves π electrons (and lone pairs) — you need at least one double/triple bond. σ bonds are never broken.
17. B Electronegativity. Both contributors carry the same −1 magnitude, so FC size can't decide — the negative charge sits on the more electronegative O.
Score /17. Miss one? The reasoning names the exact rule — revisit that spot on the cram sheet. Also do the professor's own practice on LEARN (Additional Practice Material → "Review-Material for Test 2" + its solutions, and "Test 2 — test your understanding").