Structure of the Atom — Practice

Question 1

Name the scientist who discovered the (a) electron, (b) proton, (c) neutron.

Accepted Answer

Step 1 — Recall the three discovery dates.
  Electron: J.J. Thomson, 1897 (cathode ray tube).
  Proton: E. Goldstein, 1886 (canal rays — anode rays). Later named by Rutherford.
  Neutron: James Chadwick, 1932.

✦ Answer: (a) J.J. Thomson, (b) E. Goldstein (with naming by Rutherford), (c) James Chadwick.

Question 2

Find the number of protons, neutrons and electrons in ⁴⁰₂₀Ca²⁺.

Accepted Answer

Step 1 — Read the notation.
  Z = 20 (atomic number). A = 40 (mass number). Charge = +2 → lost 2 electrons.

Step 2 — Compute each.
  Protons = Z = 20.
  Neutrons = A − Z = 40 − 20 = 20.
  Electrons = 20 − 2 = 18 (because of +2 charge).

✦ Answer: 20 protons, 20 neutrons, 18 electrons.

Question 3

Write the electronic configuration of phosphorus (Z = 15).

Accepted Answer

Step 1 — Place 15 electrons in shells via Bohr-Bury (2n²).
  K (n=1): max 2. Fill 2.
  L (n=2): max 8. Fill 8 (running total 10).
  M (n=3): the remaining 5 electrons.

✦ Answer: 2, 8, 5.

Question 4

Find the valency of nitrogen (Z = 7).

Accepted Answer

Step 1 — Configuration: 2, 5.
  Outermost electrons = 5.

Step 2 — Apply rule.
  Outermost > 4 → valency = 8 − 5 = 3.

✦ Answer: Valency of N = 3.

Verified by chemistry: N forms NH₃ (3 covalent bonds), N³⁻ in nitrides, N has trivalent compounds.

Question 5

State two postulates of Bohr's model of the atom.

Accepted Answer

Step 1 — Recall Bohr's three postulates and pick any two.

Step 2 — Write each clearly.
  (i) Electrons revolve around the nucleus in fixed circular paths called orbits or shells (labelled K, L, M, N, ...).
  (ii) While an electron is in a particular shell, it does not radiate energy — the orbit is stable.
  (Or: Energy is emitted/absorbed only when an electron jumps from one shell to another.)

✦ Answer: Any two of the above postulates with a clear sentence each.

Question 6

Define isotopes. Give one example.

Accepted Answer

Step 1 — Definition.
  Isotopes are atoms of the SAME element (same atomic number Z, same number of protons) but DIFFERENT mass numbers (different number of neutrons).

Step 2 — Example.
  Hydrogen has three isotopes: ¹H (protium, 0 n), ²H (deuterium, 1 n), ³H (tritium, 2 n). All have Z = 1.
  Or carbon: ¹²C (6 n) and ¹⁴C (8 n).

✦ Answer: Same Z, different A. Example: ¹²C and ¹⁴C.

Question 7

What were the three observations of Rutherford's gold-foil experiment, and what did each suggest about the structure of the atom?

Accepted Answer

Step 1 — List each observation and its inference.

(i) Most α-particles passed straight through the gold foil without deflection.
    → Inference: The atom is MOSTLY EMPTY SPACE.

(ii) A small fraction of α-particles were deflected by small angles.
    → Inference: There is a CONCENTRATED POSITIVE CHARGE inside the atom (deflecting some positive α-particles).

(iii) Very few α-particles (~ 1 in 8000) bounced back at angles greater than 90°.
    → Inference: The positive charge and almost all the mass are concentrated in a TINY DENSE NUCLEUS at the centre of the atom.

✦ Answer: All three observations + inferences. The combined conclusion was Rutherford's planetary model: tiny dense nucleus with electrons revolving in empty space.

Question 8

State two drawbacks of Rutherford's atomic model and explain how Bohr addressed them.

Accepted Answer

Step 1 — Drawback 1.
  An orbiting electron is constantly accelerating (in a circle). By classical electromagnetism, accelerating charges radiate energy. So Rutherford's electrons should rapidly lose energy and spiral into the nucleus in ~ 10⁻⁸ s. But atoms are stable!

Step 2 — Drawback 2.
  Rutherford's model couldn't explain why excited atoms emit specific frequencies (line spectra), not a continuous rainbow.

Step 3 — Bohr's resolution.
  Bohr proposed that electrons orbit ONLY in specific 'allowed' shells where they do NOT radiate energy. They emit/absorb energy only when JUMPING between shells. This:
    (i) Keeps atoms stable.
    (ii) Predicts discrete spectral lines corresponding to specific shell-to-shell jumps.

✦ Answer: Drawbacks: classical instability + inability to explain line spectra. Bohr fixed both with the postulate of quantised shells.

Question 9

(a) Write the electronic configurations of sulphur (Z = 16) and chlorine (Z = 17). (b) Predict the most likely ion each will form and explain.

Accepted Answer

Step 1 (a) — Configurations using Bohr-Bury.
  S: K=2, L=8, M=6 → 2, 8, 6.
  Cl: K=2, L=8, M=7 → 2, 8, 7.

Step 2 (b) — Predict ions.
  S has 6 outermost electrons → needs 2 more for octet → gains 2 e⁻ → S²⁻.
  Cl has 7 outermost electrons → needs 1 more for octet → gains 1 e⁻ → Cl⁻.

Step 3 — Both reach noble-gas Ar configuration (2, 8, 8) after gaining electrons.

✦ Answer: S (2, 8, 6) → S²⁻; Cl (2, 8, 7) → Cl⁻. Both reach octet to mimic argon's stable configuration.

Question 10

Why do ⁴⁰Ar and ⁴⁰Ca have very different chemical properties despite having the same mass number?

Accepted Answer

Step 1 — Their atomic numbers differ.
  Ar: Z = 18, config 2, 8, 8. Filled outermost shell → noble gas, INERT.
  Ca: Z = 20, config 2, 8, 8, 2. Two outermost electrons → metal, tends to lose them → Ca²⁺.

Step 2 — Chemistry depends on electron configuration (the outermost shell), not on mass number.
  Despite identical A = 40, they have completely different electron configurations and hence completely different chemistry.

✦ Answer: They are ISOBARS (same A, different Z). Chemical properties depend on the electron configuration (determined by Z), which differs entirely — Ar is inert, Ca is a reactive metal.

Question 11

Explain why neon, argon and krypton are placed in the same group of the periodic table even though they have different atomic numbers.

Accepted Answer

Step 1 — Write the configurations.
  Ne (Z = 10): 2, 8.
  Ar (Z = 18): 2, 8, 8.
  Kr (Z = 36): 2, 8, 18, 8.

Step 2 — Identify the commonality.
  All three have a FILLED OUTERMOST SHELL with 8 electrons.

Step 3 — Consequence.
  Filled outermost shell → no tendency to gain or lose electrons → CHEMICALLY INERT.
  Same chemical behaviour → same group (Group 18, noble gases).

✦ Answer: All three have 8 electrons in their outermost shell (filled, stable configuration), giving them identical chemical inertness. Same group placement reflects same chemical behaviour, not same atomic number.

Question 12

An element X has the following composition: ⁸¹Br (80 % abundance) and ⁷⁹Br (20 % abundance). Calculate the average atomic mass of bromine.

Accepted Answer

Step 1 — Use weighted average.
  Avg. atomic mass = Σ (mass × fractional abundance).

Step 2 — Convert percentages to fractions.
  ⁸¹Br: 80 % = 0.80. ⁷⁹Br: 20 % = 0.20.

Step 3 — Substitute.
  Avg = 0.80 × 81 + 0.20 × 79
      = 64.8 + 15.8
      = 80.6 u.

✦ Answer: Average atomic mass of bromine ≈ 80.6 u.

(Real-world: bromine's accepted atomic mass is 79.9 u, with a roughly 50:50 mix of ⁷⁹Br and ⁸¹Br. This problem uses different abundances for practice.)

Question 13

Why is the electronic configuration of potassium 2, 8, 8, 1 instead of 2, 8, 9?

Accepted Answer

Step 1 — Bohr-Bury rule reminder.
  Maximum electrons per shell n: 2n² (K=2, L=8, M=18, N=32).
  BUT: the outermost shell is capped at 8 electrons (octet rule), and electrons fill in order K → L → M → N.

Step 2 — For K (Z = 19):
  Place 19 electrons.
  K: 2 (filled).
  L: 8 (filled).
  M: can hold up to 18 in total, BUT if M is to be the outermost shell, only 8 are allowed.

Step 3 — Where does the 9th electron go?
  Once M reaches 8 (with 9 electrons placed in M+L+K = 18 total), the 19th electron skips to the next shell (N) rather than violate the octet rule for the outermost.

Step 4 — Hence: K = 2, 8, 8, 1.
  Once a 4th shell is started, the M shell may later expand to its full 18 (in higher-Z elements like Sc, Ti, etc.).

✦ Answer: The outermost shell of any neutral atom holds at most 8 electrons (octet rule). When M would otherwise hold 9, the 9th electron goes to N first. So K's config is 2, 8, 8, 1 — not 2, 8, 9.

Question 14

Match the isotope to its application: (a) Carbon-14, (b) Iodine-131, (c) Cobalt-60, (d) Uranium-235.

Accepted Answer

Step 1 — Recall each application.

(a) Carbon-14 → RADIOCARBON DATING.
  Ancient organic samples (wood, bone, charcoal) lose ¹⁴C at a known rate (half-life 5730 y). Measuring the remaining ¹⁴C gives age up to ~ 50,000 y.

(b) Iodine-131 → DIAGNOSIS AND TREATMENT OF HYPERTHYROIDISM.
  The thyroid absorbs iodine. Radioactive I-131 concentrates there, destroying overactive thyroid tissue with low whole-body dose.

(c) Cobalt-60 → CANCER RADIATION THERAPY.
  Emits high-energy gamma rays. Focused beams kill rapidly dividing tumour cells.

(d) Uranium-235 → NUCLEAR FUEL.
  Fissionable: a U-235 nucleus splits when hit by a slow neutron, releasing energy. Powers nuclear reactors and atom bombs.

✦ Answer: (a) Radiocarbon dating, (b) Hyperthyroidism treatment, (c) Cancer therapy, (d) Nuclear fuel for fission reactors.

Question 15

An atom X has mass number 27 and atomic number 13. (a) Identify X. (b) Write its electronic configuration. (c) State the valency. (d) Predict the ion it forms. (e) Write the formula of its oxide.

Accepted Answer

Step 1 (a) — Identify.
  Z = 13 → Aluminium (Al).

Step 2 (b) — Configuration.
  13 electrons: K=2, L=8, M=3 → 2, 8, 3.

Step 3 (c) — Valency.
  Outermost electrons = 3 (≤ 4) → valency = 3. Tendency to LOSE 3 e⁻.

Step 4 (d) — Ion.
  Al → Al³⁺ + 3e⁻. Configuration becomes 2, 8 (like neon, stable).

Step 5 (e) — Oxide formula.
  Al is +3, O is −2. Cross-multiply: Al₂O₃.
  Verify: 2 × 3 = 6 positive charge, 3 × 2 = 6 negative charge. ✓

✦ Answer: (a) Aluminium, (b) 2, 8, 3, (c) Valency 3, (d) Al³⁺, (e) Al₂O₃.

Practice — Structure of the Atom

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