Chapter 4 of 12

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Gujarat (GSEB)Class 9 Science★ 7–10 marks · CBSE Class 9 board (Unit II — Particulate Nature of Matter)

Structure of the Atom

Dalton thought atoms were indivisible. He was wrong. Class 9 Structure of the Atom covers Thomson's plum-pudding model, Rutherford's gold-foil experiment, Bohr's planetary atom, the discovery of the neutron, electron-configuration rules, isotopes & isobars, valency and 25+ exam-style problems with full step-by-step solutions.

⏱ 40 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Compare Thomson's, Rutherford's and Bohr's models of the atom and state the key experimental observation each was based on
2Identify the three sub-atomic particles (proton, neutron, electron) with mass, charge and location
3Compute atomic number, mass number, and number of neutrons for any given isotope
4Write electronic configurations of the first 20 elements using Bohr-Bury rules
5Predict valency from electronic configuration
6Distinguish isotopes (same Z, different A) from isobars (different elements, same A)
7List one application each of carbon-14, iodine-131, cobalt-60 and uranium-235

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Why this chapter matters

The electron configuration of an atom decides everything about its chemistry — its valency, what ions it forms, what compounds it can make. Master this and you can predict the chemistry of most elements without memorising case-by-case.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

🔗

Atoms and Molecules (Class 9 Ch3)

Atomic mass, mole concept, valency basics.

Structure of the Atom — Class 9 (CBSE)

Dalton said atoms were the smallest possible particles. Within a hundred years, three physicists had broken open the atom and found smaller things inside. This chapter is the story of how — and how the inside of an atom turned out to be 99.99 % empty space.

1. The story — how we learned what's inside an atom

In Dalton's time (1808), the atom was considered the smallest indivisible particle. By 1932, three landmark experiments had revealed three sub-atomic particles inside it.

1897 — J.J. Thomson discovers the electron using cathode ray tubes.
1909 — Rutherford's gold-foil experiment reveals the nucleus, a tiny dense positive core.
1932 — James Chadwick discovers the neutron, the missing-mass particle.

Each experiment overturned the previous picture of the atom. The chapter is structured around these three discoveries and the models of the atom that emerged from them.

2. Sub-atomic particles — memorise the table

Particle	Symbol	Charge	Mass (relative to H atom)	Mass (kg)	Location
Electron	e⁻	−1	1/1836 ≈ 0 (negligible)	$9.1 \times 1 0^{- 31}$	Outside nucleus, in shells
Proton	p⁺	+1	1	$1.67 \times 1 0^{- 27}$	Inside nucleus
Neutron	n	0 (neutral)	1	$1.67 \times 1 0^{- 27}$	Inside nucleus

Key takeaways:

Electrons are nearly massless compared with protons & neutrons.
Protons & neutrons together (called nucleons) account for ~ 99.9 % of an atom's mass.
Charges balance: an atom has equal protons and electrons → overall neutral.

3. Thomson's plum-pudding model (1904)

After discovering the electron, Thomson proposed:

An atom is a sphere of uniform positive charge with electrons embedded in it like plums in a pudding (or seeds in a watermelon).
The positive charge equals the total negative charge → atom is neutral.

It was the first attempt to picture the atom's interior. But it had a problem: nothing in the model predicted any specific behaviour or could be tested. That changed when Rutherford fired alpha particles at gold foil.

4. Rutherford's gold-foil experiment (1911)

The experiment

Rutherford and his students (Geiger and Marsden) shot fast-moving alpha particles (helium nuclei: 2 protons + 2 neutrons, charge +2) at a thin sheet of gold foil only a few atoms thick. Behind the foil, a fluorescent screen detected where each particle landed.

Observations

Most α-particles passed straight through with no deflection.
Some α-particles were deflected by small angles.
A very few α-particles (≈ 1 in 8000) bounced back at angles > 90°.

Rutherford famously said this last observation was "as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you."

Conclusions (the new atomic model)

Most of an atom is empty space — that's why most α-particles went straight through.
The atom has a tiny, dense, positively charged nucleus at its centre — that's what bounced α-particles back.
Electrons revolve around the nucleus like planets around the sun.

Size estimates from Rutherford's model

Atomic radius: ≈ $1 0^{- 10}$ m (1 Å).
Nuclear radius: ≈ $1 0^{- 15}$ m (1 fm).
The atom is 100,000 times bigger than its nucleus.

If a nucleus were the size of a marble, the atom would be the size of a football stadium. The rest is electrons + empty space.

Drawbacks of Rutherford's model

Classical physics predicted that orbiting electrons (being accelerated) should radiate energy and spiral into the nucleus within $1 0^{- 8}$ s — but atoms are stable! Something was incomplete.
Couldn't explain why hot gases emit specific colours (line spectra), not a continuous rainbow.

These problems opened the door for Bohr.

5. Bohr's model (1913)

Bohr fixed Rutherford's model with three postulates:

Electrons orbit the nucleus in fixed paths called shells or orbits (also called energy levels) — labelled K, L, M, N, … (or n = 1, 2, 3, 4, …).
Each shell has a specific energy. While an electron is in a shell, it does NOT radiate energy.
Energy is absorbed/emitted only when an electron jumps between shells — absorbed (going outward) or emitted (coming inward).

This explained why atoms are stable AND why spectral lines exist (each transition between shells corresponds to a specific energy = specific colour of light).

Maximum electrons per shell — Bohr-Bury rule

$Max electrons in shell n = 2 n^{2}$

Shell	n	Max electrons
K	1	2
L	2	8
M	3	18
N	4	32

Additional rules:

The outermost shell can have a maximum of 8 electrons (the octet rule — only applies to outermost).
An inner shell must be filled completely before electrons can fill the next outer shell.

Electronic configuration — how to write it

Example — Sodium (Z = 11):

11 electrons.
K shell: 2 (filled).
L shell: 8 (filled).
M shell: 1 (left over).
Configuration: 2, 8, 1.

Example — Chlorine (Z = 17):

K: 2, L: 8, M: 7.
Configuration: 2, 8, 7.

Example — Argon (Z = 18):

K: 2, L: 8, M: 8.
Configuration: 2, 8, 8 (filled — that's why argon is a noble gas, unreactive).

6. Atomic number and mass number

Two important integers describe any atom:

Atomic number ( $Z$ ) = number of protons in the nucleus. (Also equals number of electrons in a neutral atom.) This identifies the element. Carbon is Z = 6. Always. Change Z and you have a different element.

Mass number ( $A$ ) = number of protons + number of neutrons = total nucleons.

So: number of neutrons = $A - Z$ .

Standard notation:

$_{Z}^{A} X$

For example, $_{6}^{12} C$ means: carbon, atomic number 6, mass number 12. So 6 protons + 6 neutrons + 6 electrons.

Memorise the first 20 elements (in order)

Z	Element	Symbol	Config
1	Hydrogen	H	1
2	Helium	He	2
3	Lithium	Li	2, 1
4	Beryllium	Be	2, 2
5	Boron	B	2, 3
6	Carbon	C	2, 4
7	Nitrogen	N	2, 5
8	Oxygen	O	2, 6
9	Fluorine	F	2, 7
10	Neon	Ne	2, 8
11	Sodium	Na	2, 8, 1
12	Magnesium	Mg	2, 8, 2
13	Aluminium	Al	2, 8, 3
14	Silicon	Si	2, 8, 4
15	Phosphorus	P	2, 8, 5
16	Sulphur	S	2, 8, 6
17	Chlorine	Cl	2, 8, 7
18	Argon	Ar	2, 8, 8
19	Potassium	K	2, 8, 8, 1
20	Calcium	Ca	2, 8, 8, 2

This table is the single most useful thing to memorise in Class 9 chemistry. Most periodic-table questions, valency questions, ion-formation questions trace back here.

7. Valency from electron configuration

Valency = combining capacity of an atom = the number of electrons it needs to gain, lose, or share to reach a noble-gas configuration (full outermost shell, typically 8 electrons).

Three rules:

If outermost electrons ≤ 4 → valency = outermost electron count (atom LOSES them).
If outermost electrons > 4 → valency = 8 − outermost electron count (atom GAINS them).
If outermost shell is full (2 for He, 8 for others) → valency = 0 (inert / noble).

Examples

Sodium (2, 8, 1): outermost = 1, so valency = 1. Tends to lose 1 e⁻ → Na⁺.
Magnesium (2, 8, 2): outermost = 2, valency = 2. Loses 2 e⁻ → Mg²⁺.
Aluminium (2, 8, 3): outermost = 3, valency = 3. Loses 3 e⁻ → Al³⁺.
Carbon (2, 4): outermost = 4, valency = 4. Shares (in CH₄, CO₂).
Nitrogen (2, 5): outermost = 5, valency = 8 − 5 = 3. Gains 3 (or shares) → N³⁻ / NH₃.
Oxygen (2, 6): outermost = 6, valency = 8 − 6 = 2. Gains 2 → O²⁻.
Chlorine (2, 8, 7): outermost = 7, valency = 8 − 7 = 1. Gains 1 → Cl⁻.
Neon (2, 8): outermost = 8, valency = 0 (noble gas, inert).

8. Isotopes and isobars

Isotopes

Isotopes are atoms of the same element (same Z, same number of protons) but different mass numbers (different number of neutrons).

Examples:

Hydrogen has 3 isotopes:
- Protium: $_{1}^{1} H$ — 1 proton, 0 neutrons (most common).
- Deuterium: $_{1}^{2} H$ — 1 proton, 1 neutron (heavy water D₂O).
- Tritium: $_{1}^{3} H$ — 1 proton, 2 neutrons (radioactive, used in nuclear fusion).
Carbon has 3 main isotopes:
- $_{6}^{12} C$ — 6n (most common, ~99 %).
- $_{6}^{13} C$ — 7n (~1 %).
- $_{6}^{14} C$ — 8n (radioactive, used in carbon dating).
Chlorine has 2 isotopes:
- $_{17}^{35} C l$ — 18n (~75 %).
- $_{17}^{37} C l$ — 20n (~25 %).
- The "average atomic mass" 35.5 = 0.75 × 35 + 0.25 × 37.

Important: Isotopes have IDENTICAL chemical properties (because chemistry depends on electron config, which only depends on Z). But they have slightly different PHYSICAL properties (mass, density, BP — heavy water boils at 101.4 °C vs 100 °C for regular water).

Applications of isotopes

Carbon-14: dating fossils and archaeological samples (half-life 5730 y).
Iodine-131: treatment of hyperthyroidism (radioactive iodine concentrates in the thyroid).
Cobalt-60: cancer radiation therapy.
Uranium-235: nuclear fuel (fission reactor / atom bomb).
Deuterium: nuclear fusion research, moderator in heavy-water reactors.

Isobars

Isobars are atoms of different elements (different Z) but the same mass number (same A).

Examples:

$_{18}^{40} Ar$ and $_{20}^{40} Ca$ . Both have A = 40 but different Z (18 vs 20).
$_{6}^{14} C$ and $_{7}^{14} N$ . Both have A = 14 but different Z.

Memorising:

Isotopes: same protons (top number Z), different neutrons.
Isobars: same A (mass), different elements.

9. Closing thought

The chapter is a journey from one wrong picture to another, each less wrong than the last:

Dalton (1808): solid indivisible billiard ball.
Thomson (1904): plum pudding (positive sphere with embedded electrons).
Rutherford (1911): tiny nucleus + electrons orbiting in empty space.
Bohr (1913): electrons in fixed energy shells.
Modern (1925+): electrons as 3D probability clouds (you'll meet quantum-mechanical orbitals in Class 11).

Each model explained more experimental data than the last while still being partly wrong. That's how science actually advances — by improving the picture, not by leaping to perfect truth. This is the most important meta-lesson of the chapter.

You can now look at any element on the periodic table and, from its atomic number alone, write its electronic configuration, predict its valency, guess its likely ion, and reason about its chemistry. That's an enormous superpower for someone who couldn't tell sodium from potassium two months ago.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Atomic number

Z = number of protons = number of electrons in neutral atom

Identifies the element.

Mass number

A = number of protons + number of neutrons

Total nucleons in the nucleus.

Number of neutrons

n = A − Z

Most-tested 1-mark question.

Maximum electrons in shell n

Max = 2n²

Bohr-Bury rule. K=2, L=8, M=18, N=32.

Outermost-shell maximum

8 electrons (regardless of shell)

Octet rule. K shell tops at 2.

Valency (≤ 4 outermost)

Valency = number of outermost electrons

Element loses them (metals).

Valency (> 4 outermost)

Valency = 8 − outermost electrons

Element gains them (non-metals).

Isotope notation

ᴬ_Z X — A = mass number, Z = atomic number

e.g., ¹²C, ¹⁴C.

⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Swapping the meaning of atomic number and mass number

✓ Atomic number = protons (always Z). Mass number = protons + neutrons (always A). Memorise Z<A for most stable nuclei.

WATCH OUT

✗ Saying the K shell holds 8 electrons

✓ K (n=1) holds only 2 (per 2n² rule). 8 is the L shell maximum.

WATCH OUT

✗ Filling the M shell with 18 before completing the rule about outermost being capped at 8

✓ M can hold 18 in total, BUT if M is the OUTERMOST shell, it can have only 8. The remaining electrons go to N first.

WATCH OUT

✗ Writing the electronic configuration of K (Z=19) as 2, 8, 9

✓ After 2, 8, you'd fill M up to 8 (now outermost), then drop the 9th into N. K is 2, 8, 8, 1 — not 2, 8, 9.

WATCH OUT

✗ Saying isotopes have different chemical properties

✓ Isotopes have IDENTICAL chemical properties (same electrons, same valency). They differ only in physical properties (mass, density, BP).

WATCH OUT

✗ Confusing isotopes and isobars

✓ Iso-TOP-es: same TOP-side count (protons / Z). Isobars: same A. Memorise the mnemonic.

WATCH OUT

✗ Treating Rutherford's α-bounceback as proof that atoms are mostly nuclei

✓ It's the opposite — most α went straight through, proving atoms are MOSTLY empty space. Only a few bounced back, proving the nucleus is TINY and DENSE.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Section 4.1 (in-text)

Sub-atomic particles: discovery, charge, mass

Questions

Section 4.2 (in-text)

Atomic models: Thomson, Rutherford, Bohr

Questions

Section 4.3 (in-text)

Distribution of electrons in shells (Bohr-Bury)

Questions

Section 4.4 (in-text)

Valency from electron configuration

Questions

Section 4.5 (in-text)

Atomic number, mass number, neutron count

Questions

Section 4.6 (in-text)

Isotopes and isobars; applications

Questions

End-of-chapter

Mixed: configurations, valency, isotope properties, model comparisons

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Sub-atomic

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

▶ Show solution

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.

Q2EASY· Notation

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

▶ Show solution

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.

Q3EASY· Config

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

▶ Show solution

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.

Q4EASY· Valency

Find the valency of nitrogen (Z = 7).

▶ Show solution

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.

Q5EASY· Concepts

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

▶ Show solution

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.

Q6EASY· Isotopes

Define isotopes. Give one example.

▶ Show solution

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.

Q7MEDIUM· Rutherford

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

▶ Show solution

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.

Q8MEDIUM· Drawbacks

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

▶ Show solution

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.

Q9MEDIUM· Config + ion

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

▶ Show solution

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.

Q10MEDIUM· Isobars

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

▶ Show solution

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.

Q11MEDIUM· Periodicity

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

▶ Show solution

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.

Q12HARD· Iso problem

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

▶ Show solution

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.)

Q13HARD· K shell

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

▶ Show solution

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.

Q14HARD· Applications

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

▶ Show solution

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.

Q15HARD· Multistep

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.

▶ Show solution

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.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Electron: m ≈ 1/1836 of proton, charge −1. Proton: m = 1, charge +1. Neutron: m = 1, charge 0.
•Thomson: plum-pudding (positive sphere with embedded electrons). First model.
•Rutherford: α-scattering → tiny dense nucleus + electrons in empty space (atom is ~10⁵ × larger than nucleus).
•Bohr: electrons in fixed shells; no energy radiated while in a shell; jumps between shells emit/absorb specific energies.
•Bohr-Bury rule: max electrons per shell = 2n². K=2, L=8, M=18, N=32. Outermost capped at 8.
•Atomic number Z = protons. Mass number A = protons + neutrons. Neutrons = A − Z.
•Valency: ≤ 4 outermost → valency = outermost. > 4 → valency = 8 − outermost. Filled → valency = 0.
•Isotopes: same Z, different A (e.g., ¹²C and ¹⁴C). Same chemistry, slightly different physics.
•Isobars: same A, different Z (e.g., ⁴⁰Ar and ⁴⁰Ca). Completely different chemistry.
•Key applications: C-14 dating, I-131 thyroid, Co-60 cancer therapy, U-235 nuclear fuel.

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The real ones — pulled from the Q&A community and tutor sessions.

Classical physics can't explain this — but Bohr's postulate says electrons in 'allowed' shells DON'T radiate energy. The deeper reason (quantum mechanics, Class 11+) is that electron 'orbits' are actually standing-wave patterns that are stable by quantum rules.

The octet rule applies to the outermost shell only. While M is outermost (in elements like Na, Mg, Al through Ar), it caps at 8. Once an N-shell electron exists (potassium and beyond), M can expand further (this happens in transition metals — Class 11).

Mass number A is a count of nucleons (integer). Atomic mass is a measured weighted average (often non-integer), e.g., Cl = 35.5 because of isotope mixing.

Because most α went straight through — the atom had to be mostly empty. The rare big deflections required a tiny, very dense, very positive object — too small to be the whole atom, so it had to be a tiny nucleus inside the atom.

Mostly yes for Class 9 purposes. With modern techniques, compounds of Xe, Kr and Rn with very electronegative elements (F, O) have been made (e.g., XeF₂, XeF₄). But these are extreme conditions; under ordinary conditions noble gases don't react.

Chemistry happens via electrons (especially the outermost). All isotopes of an element have the same Z → same number of electrons → same electronic configuration → same chemistry.

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Related chapters

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