Journey Inside the Atom traces how the idea of the atom changed from Kanada’s parmanus and Dalton’s indivisible atoms to a structure with electrons, protons and neutrons. It explains the Thomson, Rutherford and Bohr models and why each was replaced, then builds the tools used in all later chemistry: element symbols, atomic number, mass number, the 2n² rule, electronic configuration of the first eighteen elements, valency, isotopes, average atomic mass and isobars. School exams test both the reasoning behind the models and short numericals on particles and configurations.
Key Concepts
1. Early Ideas of the Atom and the Discovery of the Electron
Everything around us, living or non-living, is matter, and matter is made of tiny particles called atoms. Atoms are far too small to see with the naked eye.
Ancient ideas (more than 2,000 years ago):
- Acharya Kanada (India) suggested that if matter (dravya) is divided again and again, you finally reach the smallest particles that cannot be divided further. He called them parmanus. His ideas are recorded in the Sanskrit text Vaisesika Sutras. A parmanu is infinitely small and cannot be perceived by the senses. Parmanus combine to form dyads (groups of two) and triads (groups of three), and the whole material universe, including living bodies, is made of such combinations.
- Leucippus and Democritus (Greece) proposed a similar idea and called the indivisible particles atomos, which means indivisible in Greek.
Both ideas came from thinking and imagination, without experiments.
Dalton’s atomic theory (1808): John Dalton proposed that all matter is made of indivisible particles called atoms, the fundamental building blocks of matter that cannot be broken into smaller parts. His theory was based on scientific experiments of his time. It was the first scientific description of how matter is made and became the starting point for our present understanding of atomic structure.
Atoms are divisible: until the late 19th century, atoms were thought to be the smallest indivisible units of matter. Then scientists found that certain elements emit invisible energy and particles called radiation. This phenomenon is radioactivity. Particles coming out of atoms meant atoms must contain smaller particles.
J. J. Thomson and cathode rays (1897)
Thomson studied how electric current passes through gases at very low pressure. He used a glass tube with two electrodes sealed inside, connected to a high voltage supply, with a vacuum pump to remove most of the gas. He saw rays travelling from the cathode (negative electrode) to the anode (positive electrode). These are cathode rays.
- By studying cathode rays in electric and magnetic fields, he concluded they are streams of negatively charged particles with a much smaller mass than atoms.
- These particles, later called electrons, came out of atoms, so atoms have smaller subatomic parts.
- The nature of cathode rays did not depend on the material of the cathode or on the gas in the tube. So electrons are a fundamental part of every atom of every element.
- The charge of an electron is -1.602 × 10-19 C. By convention and for convenience it is taken as -1.
2. Thomson’s Model of the Atom
After finding electrons, Thomson faced a puzzle: atoms are electrically neutral, so some positive charge must balance the electrons. He proposed that the atom is a sphere of positive charge with electrons distributed throughout it.
- It was compared to a pudding with plums in it, so it is called the plum pudding model.
- A more familiar picture is a watermelon: the red pulp stands for the positively charged matter and the seeds stand for electrons spread through the atom.
- The total positive charge equals the total negative charge, so the atom as a whole is neutral.
This model was later replaced, but it was the first real attempt to explain how the positive and negative charges of an atom stay balanced.
3. Rutherford’s Gold Foil Experiment and the Nuclear Model
In 1911, Geiger and Marsden, working under Ernest Rutherford, tested Thomson’s model. They aimed a narrow beam of alpha (α) particles at an extremely thin gold foil. Alpha particles are tiny, positively charged particles emitted by certain radioactive elements. An alpha particle is the nucleus of a helium atom: 2 protons and 2 neutrons.
Expected result (from Thomson’s model): positive charge is spread out evenly, so no single spot can push strongly on an α-particle. The particles should pass straight through or be deflected only slightly.
Actual observations: most α-particles passed straight through, some were deflected through large angles, and a few bounced back. Deflection from the straight path is called scattering, so this is also called the α-ray scattering experiment. Thomson’s model could not explain the large-angle deflections, and it could not explain why most particles passed straight through.
| Observation | Conclusion drawn by Rutherford |
|---|---|
| Most α-particles passed straight through | Most of the atom is empty space |
| Some deflected through large angles | Positive charge is concentrated in a very small region, which repels the positive α-particles strongly when they come close |
| A few bounced back | This small region is dense and carries most of the mass; only a head-on approach to it turns a particle back |
Rutherford’s model (planetary model)
- The positive charge is concentrated in an extremely small central region called the nucleus.
- Most of the atom is empty space.
- The nucleus is dense, contains all the positive charge and most of the mass of the atom.
- Electrons revolve around the nucleus, somewhat like planets around the Sun. So this is called the planetary model.
Size: the nucleus is about 105 (one lakh) times smaller than the atom. The diameter of an atom is ≈ 10-10 m and the diameter of the nucleus is ≈ 10-15 m. If an atom were as large as a cricket ground (about 100 m across), the nucleus would be a black pepper grain a few millimetres across at the centre.
Limitation of Rutherford’s model
A particle moving in a circle keeps changing direction, so it is accelerating. An accelerating negatively charged electron circling the nucleus should keep losing energy. Losing energy, it would move along a spiral path inward and finally fall into the positive nucleus. Atoms would collapse. Real atoms are stable, which is why matter around us stays intact. So Rutherford’s model could not explain the stability of the atom.
Discovery of the proton
Rutherford showed that the positive charge of the nucleus comes from particles called protons, which he discovered and named. Protons are much heavier than electrons and carry a charge equal and opposite to that of an electron (+1).
For an atom to be neutral, number of protons = number of electrons. Helium has 2 protons and 2 electrons; sodium has 11 protons and 11 electrons. In each, total positive charge equals total negative charge, so the atom is neutral. All atoms are electrically neutral.
4. Bohr’s Model of the Atom
To explain why atoms are stable, Niels Bohr proposed a new model in 1913. His postulates:
- Electrons do not move randomly. They follow fixed circular paths called stationary states, orbits or shells.
- In each shell an electron has a definite amount of energy, so shells are also called energy levels.
- Shells are named K, L, M, N, … or numbered n = 1, 2, 3, 4, … starting from the nucleus.
- Electrons can revolve only in these allowed shells, never in between them.
- While moving in a fixed shell, an electron does not lose energy.
- The K shell (n = 1) is closest to the nucleus and has the least energy. Energy increases as we move away from the nucleus: an electron in L (n = 2) has more energy than one in K (n = 1).
- An electron can move to another shell by absorbing or releasing a fixed amount of energy equal to the difference between the energies of the two levels.
- Each shell can hold only a certain number of electrons.
How Bohr explained stability: electrons still move in circles around the positive nucleus, as in Rutherford’s model. Bohr introduced stationary states as a postulate: in a stationary state the energy of the electron stays constant even though it is moving. With no energy loss, there is no inward spiral and no collapse. Bohr’s model explained many experimental observations.
Beyond Bohr: Bohr’s model also has limitations. The quantum mechanical model came next. Electrons do not follow well-defined paths like fixed Bohr orbits; they exist as electron clouds around the nucleus, and we can only predict regions where they are most likely to be. You will study this in higher classes.
| Model | Main idea | What it could not explain |
|---|---|---|
| Dalton (1808) | Atom as an indivisible particle | Radioactivity and electrons show atoms are divisible |
| Thomson (after 1897) | Positive sphere with electrons embedded | Large deflections and bounce-backs in the gold foil experiment |
| Rutherford (1911) | Tiny dense positive nucleus, electrons orbit, mostly empty space | Stability of the atom |
| Bohr (1913) | Electrons in fixed shells of definite energy, no energy loss | Has its own limitations; replaced by the quantum mechanical model |
| Quantum mechanical (modern) | Electron clouds, regions of probability | Still being studied |
5. The Neutron, Element Symbols, Atomic Number and Mass Number
Discovery of the neutron (1932)
A puzzle: hydrogen has one proton and helium has two, yet a helium atom is about four times as heavy as a hydrogen atom, not twice. Something in the nucleus must add mass without adding charge. In 1932, James Chadwick, a student of Rutherford, discovered the neutron: a particle with mass nearly equal to that of a proton and no electrical charge, symbol n. Neutrons are found in the nucleus of all atoms except hydrogen (its common form, protium, has no neutron). The mass of an atom comes mainly from its protons and neutrons; electrons are so light that their mass can be ignored.
| Subatomic particle | Symbol | Relative charge | Where found | Discovered by |
|---|---|---|---|---|
| Electron | e- | -1 | Shells around the nucleus | J. J. Thomson (1897) |
| Proton | p+ | +1 | Nucleus | Ernest Rutherford |
| Neutron | n0 | 0 | Nucleus | James Chadwick (1932) |
Neutrons and nuclear stability: protons all carry positive charge, so each repels every other. Neutrons, being neutral, reduce this repulsion by sitting between protons and increasing the distance between them. They also strengthen the nuclear force that binds all the particles of the nucleus. Heavier atoms need many more neutrons: iron has 26 protons and 30 neutrons, uranium has 92 protons and 146 neutrons.
Symbols of elements
By 1869 scientists knew about 69 elements; today we know about 118, some of them made artificially.
- Dalton (1803) introduced the first pictorial symbols for elements.
- Berzelius (1813) suggested that symbols come from the elements’ Latin names, giving alphabetic symbols.
- Today IUPAC (International Union of Pure and Applied Chemistry) approves names and symbols.
Rules for writing symbols:
- Many symbols are the first letter or first two letters of the name: hydrogen H, aluminium Al.
- The first letter is always a capital, the second (if any) is small: Al (not AL), Co for cobalt (not CO, which is carbon monoxide, a compound of two elements).
- Some use the first letter and a later letter of the name: chlorine Cl, zinc Zn.
- Some come from Latin, Greek or German names: iron Fe (Latin ferrum), mercury Hg (Greek hydrargyros), tungsten W (German wolfram).
| Element | Symbol | Name the symbol comes from |
|---|---|---|
| Copper | Cu | Cuprum |
| Gold | Au | Aurum |
| Iron | Fe | Ferrum |
| Lead | Pb | Plumbum |
| Potassium | K | Kalium |
| Silver | Ag | Argentum |
| Sodium | Na | Natrium |
Atomic number (Z)
The number of protons in the nucleus of an atom is its atomic number, symbol Z. It fixes the identity of an element and its chemical behaviour: every element has its own atomic number, and two atoms with different Z are different elements. In a neutral atom, number of electrons = number of protons = Z. Hydrogen: Z = 1 (1 proton, 1 electron). Helium: Z = 2.
Mass number (A)
The total number of protons and neutrons in the nucleus is the mass number, symbol A. Protons and neutrons together are called nucleons.
Mass number (A) = number of protons + number of neutrons, so number of neutrons = A - Z.
| Element | Protons | Neutrons | Mass number (A) |
|---|---|---|---|
| Hydrogen | 1 | 0 | 1 |
| Helium | 2 | 2 | 4 |
| Lithium | 3 | 4 | 7 |
Standard notation: write the symbol, with the mass number as a superscript on the top left and the atomic number as a subscript on the bottom left. Carbon, Z = 6, A = 12, is written 126C. Read it as: 6 protons, 6 electrons, 12 - 6 = 6 neutrons.
6. Distribution of Electrons: Bohr-Bury Rules and Electronic Configuration
Bohr and Bury gave rules for how electrons fill shells:
- The maximum number of electrons in a shell = 2n2, where n is the shell number. K (n = 1): 2 × 12 = 2. L (n = 2): 2 × 22 = 8. M (n = 3): 2 × 32 = 18.
- The outermost shell can hold at most 8 electrons (the first shell holds at most 2).
- Shells fill stepwise, from the one closest to the nucleus outward, in the order K, L, M, N. The L shell fills only after K is complete, and so on.
The distribution of electrons among the shells is the electronic configuration. Build it by adding one electron to the correct shell each time the atomic number rises by 1.
| Element | Symbol | Z | K | L | M | Configuration |
|---|---|---|---|---|---|---|
| Hydrogen | H | 1 | 1 | - | - | 1 |
| Helium | He | 2 | 2 | - | - | 2 |
| Lithium | Li | 3 | 2 | 1 | - | 2, 1 |
| Beryllium | Be | 4 | 2 | 2 | - | 2, 2 |
| Boron | B | 5 | 2 | 3 | - | 2, 3 |
| Carbon | C | 6 | 2 | 4 | - | 2, 4 |
| Nitrogen | N | 7 | 2 | 5 | - | 2, 5 |
| Oxygen | O | 8 | 2 | 6 | - | 2, 6 |
| Fluorine | F | 9 | 2 | 7 | - | 2, 7 |
| Neon | Ne | 10 | 2 | 8 | - | 2, 8 |
| Sodium | Na | 11 | 2 | 8 | 1 | 2, 8, 1 |
| Magnesium | Mg | 12 | 2 | 8 | 2 | 2, 8, 2 |
| Aluminium | Al | 13 | 2 | 8 | 3 | 2, 8, 3 |
| Silicon | Si | 14 | 2 | 8 | 4 | 2, 8, 4 |
| Phosphorus | P | 15 | 2 | 8 | 5 | 2, 8, 5 |
| Sulfur | S | 16 | 2 | 8 | 6 | 2, 8, 6 |
| Chlorine | Cl | 17 | 2 | 8 | 7 | 2, 8, 7 |
| Argon | Ar | 18 | 2 | 8 | 8 | 2, 8, 8 |
Method in three steps: (1) find the number of electrons (= Z for a neutral atom); (2) fill K up to 2, then L up to 8, then M; (3) check the outermost shell has no more than 8 and the total equals Z. Example: sulfur, Z = 16: K = 2, L = 8, leaving 16 - 10 = 6 for M. Configuration 2, 8, 6.
The number of neutrons does not affect the configuration. Only the number of electrons matters.
7. Valency: The Combining Capacity of an Atom
The number of hydrogen or chlorine atoms with which one atom of an element combines to form a compound is its combining capacity. Hydrogen and chlorine are used as the reference because each has a combining capacity of one. In H2O, oxygen combines with 2 hydrogen atoms, so its combining capacity is 2. In NH3, nitrogen combines with 3 hydrogen atoms (combining capacity 3). In MgCl2, magnesium combines with 2 chlorine atoms (combining capacity 2).
- The outermost shell containing electrons is the valence shell; its electrons are valence electrons.
- Eight electrons in the outermost shell is an octet.
- Atoms with a complete octet (or 2 electrons in the case of helium) are largely unreactive and more stable. Atoms with incomplete valence shells are usually more reactive and lose, gain or share electrons to complete the octet.
Valency is the number of electrons gained, lost or shared to complete the octet.
- Fewer than 4 valence electrons: the atom tends to lose them. Valency = number of valence electrons. Sodium 2, 8, 1 loses 1: valency 1.
- More than 4 valence electrons: the atom tends to gain electrons. Valency = 8 - number of valence electrons. Oxygen 2, 6 gains 2: valency 2.
- Exactly 4: the atom cannot easily gain or lose four, so it shares them. Carbon 2, 4 shares 4: valency 4.
- Complete octet (or duplet for helium): no tendency to gain or lose; valency is taken as zero. Neon 2, 8 and argon 2, 8, 8 are examples.
| Element | Configuration | Valence electrons | What happens | Valency |
|---|---|---|---|---|
| Magnesium | 2, 8, 2 | 2 | Loses 2 | 2 |
| Carbon | 2, 4 | 4 | Shares 4 | 4 |
| Nitrogen | 2, 5 | 5 | Gains 3 | 3 |
| Chlorine | 2, 8, 7 | 7 | Gains 1 | 1 |
| Neon | 2, 8 | 8 | Stable octet | 0 |
Some compounds appear to break this simple valency rule. You will study them in higher classes.
8. Isotopes, Average Atomic Mass and Isobars
Isotopes
Dalton said all atoms of an element are identical and have the same mass. Later, scientists found atoms of the same element with the same number of protons (same Z) but different numbers of neutrons, and so different mass numbers. These are isotopes: atoms of the same element with the same atomic number but different mass numbers.
| Isotope | Notation | Protons | Neutrons | Electrons | Natural share |
|---|---|---|---|---|---|
| Protium | 11H | 1 | 0 | 1 | ~99.98% |
| Deuterium | 21H | 1 | 1 | 1 | ~0.015% |
| Tritium | 31H | 1 | 2 | 1 | Traces |
Carbon has three isotopes, 126C, 136C and 146C, each with 6 protons and 6 electrons but 6, 7 and 8 neutrons. 126C is the most abundant in nature.
Properties: isotopes have similar chemical properties, because they have the same number of electrons and the same electronic configuration, and chemical properties depend mainly on valence electrons. They differ in physical properties such as melting and boiling points.
Uses of isotopes:
- 23592U (uranium): fuel in nuclear reactors to generate electricity in nuclear power plants.
- 6027Co (cobalt, radioactive): radiation treatment for cancer.
- 13153I (iodine): treatment of goitre and thyroid cancer.
- 146C (carbon): finding the age of ancient fossils and artefacts in archaeology and geology.
Atomic mass unit and average atomic mass
Atoms are too tiny to weigh in grams or kilograms, so their masses are given in the unified atomic mass unit (u). Earlier this was written as amu.
Chlorine occurs as two isotopes, of mass 35 u and 37 u, in the ratio 3 : 1 (about 75% and 25%).
- Simple average (ignores abundance): (35 + 37)/2 = 36 u. This does not match nature, because the isotopes are not present in equal amounts.
- Weighted average (uses abundance): multiply each isotope’s mass by its percentage abundance/100 and add.
= 35 × 75/100 + 37 × 25/100
= 105/4 + 37/4
= 142/4 = 35.5 u
No single chlorine atom has a mass of 35.5 u. In, say, 1 million chlorine atoms there are about 7.5 lakh 3517Cl and 2.5 lakh 3717Cl atoms, and their average mass is 35.5 u.
Isobars
Calcium (Z = 20), potassium (Z = 19) and argon (Z = 18) have different numbers of protons, yet atoms of each with mass number 40 exist. They have the same total number of nucleons. Atoms of different elements with the same mass number but different atomic numbers are isobars. Because their atomic numbers differ, isobars have different numbers of electrons and different chemical properties.
| Feature | Isotopes | Isobars |
|---|---|---|
| Atomic number (Z) | Same | Different |
| Mass number (A) | Different | Same |
| Element | Same element | Different elements |
| Neutrons | Different | Different |
| Chemical properties | Similar | Different |
| Example | 3517Cl and 3717Cl | 4020Ca, 4019K, 4018Ar |
Formula and Reaction Sheet
This chapter has no chemical reactions. These are the relations and rules you will use in every numerical.
| Quantity | Formula or rule | Worked use |
|---|---|---|
| Atomic number | Z = number of protons | Chlorine: 17 protons, Z = 17 |
| Mass number | A = protons + neutrons | Lithium: 3 + 4 = 7 |
| Neutrons | Number of neutrons = A - Z | 2311Na: 23 - 11 = 12 |
| Maximum electrons in shell n | 2n2 | M shell: 2 × 9 = 18 |
| Weighted average atomic mass | Σ (mass × % abundance/100) | Chlorine: 35.5 u |
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Important Definitions
- Electron: a negatively charged subatomic particle (relative charge -1) of very small mass, present in all atoms.
- Proton: a positively charged subatomic particle (relative charge +1) in the nucleus, much heavier than an electron.
- Neutron: a subatomic particle in the nucleus with no charge and mass nearly equal to that of a proton.
- Nucleus: the tiny, dense, positively charged centre of an atom that holds most of its mass.
- Stationary state (orbit, shell, energy level): a fixed circular path around the nucleus in which an electron has a definite energy and does not lose energy.
- Atomic number (Z): the number of protons in the nucleus of an atom.
- Mass number (A): the total number of protons and neutrons in the nucleus of an atom.
- Electronic configuration: the distribution of electrons among the shells of an atom.
- Valence electrons: the electrons in the valence shell.
- Valency: the number of electrons gained, lost or shared by an atom to complete its octet.
- Isotopes: atoms of the same element with the same atomic number but different mass numbers.
- Isobars: atoms of different elements with the same mass number but different atomic numbers.
Solved Examples (NCERT-Based)
Example 1: Particles from Z and nucleons
Question: An atom with an atomic number of 26 has 56 nucleons. Find out its number of electrons, protons and neutrons.
Solution:
- Protons = Z = 26.
- Electrons in a neutral atom = protons = 26.
- Nucleons = protons + neutrons = mass number = 56. Neutrons = 56 - 26 = 30.
(This matches iron, which has 26 protons and 30 neutrons.)
Example 2: Neutrons from Z and A
Question: The nucleus of an atom contains 20 protons. If its mass number is 41, find the number of neutrons in it.
Solution: Neutrons = A - protons = 41 - 20 = 21.
Question: An atom has 18 neutrons and an atomic number of 17. What is its mass number?
Solution: A = protons + neutrons = 17 + 18 = 35.
Example 3: Magnesium atom
Question: For an atom of magnesium with a mass number of 24 and atomic number 12, determine the number of (i) protons, (ii) neutrons, (iii) electrons, and also illustrate the arrangement of electrons in a magnesium atom.
Solution:
- Protons = Z = 12.
- Neutrons = A - Z = 24 - 12 = 12.
- Electrons = 12 (neutral atom).
Arrangement: K takes 2, L takes 8, leaving 12 - 10 = 2 for M. Configuration 2, 8, 2. Magnesium has 2 valence electrons and tends to lose them, so its valency is 2.
Example 4: Neutrons and electrons of a heavy atom
Question: An atom has 79 protons and a mass number of 197. Calculate (i) the number of neutrons, and (ii) the number of electrons.
Solution: (i) Neutrons = 197 - 79 = 118. (ii) In a neutral atom electrons = protons = 79.
Question: An atom 70X has 31 electrons. How many neutrons are there in its nucleus?
Solution: A neutral atom has electrons = protons, so Z = 31. Neutrons = 70 - 31 = 39.
Example 5: Complete the table
Question: Complete Table 8.5. The given entries are: (a) atomic number 5, 6 neutrons; (b) mass number 14, 7 electrons, nitrogen; (c) mass number 24, 12 protons; (d) atomic number 15, 16 neutrons; (e) mass number 1, 0 neutrons.
Method: protons = electrons = Z; A = Z + neutrons.
| Z | A | Neutrons | Protons | Electrons | Element |
|---|---|---|---|---|---|
| 5 | 5 + 6 = 11 | 6 | 5 | 5 | Boron |
| 7 | 14 | 14 - 7 = 7 | 7 | 7 | Nitrogen |
| 12 | 24 | 24 - 12 = 12 | 12 | 12 | Magnesium |
| 15 | 15 + 16 = 31 | 16 | 15 | 15 | Phosphorus |
| 1 | 1 | 0 | 1 | 1 | Hydrogen |
Example 6: Electronic configuration and valence electrons
Question: Write the electronic configuration of the elements having atomic numbers 12, 16 and 18.
Solution:
- Z = 12 (magnesium): 2, 8, 2
- Z = 16 (sulfur): 2, 8, 6
- Z = 18 (argon): 2, 8, 8
Question: Identify the number of electrons in the outermost shell of 126C, 199F and 2814Si.
Solution: Only Z matters. Carbon (Z = 6): 2, 4, so 4. Fluorine (Z = 9): 2, 7, so 7. Silicon (Z = 14): 2, 8, 4, so 4. The mass numbers 12, 19 and 28 give neutrons, which play no part in the configuration.
Example 7: Element X (case study)
Question: An element X has a mass number of 35 and contains 18 neutrons. (i) How many electrons and protons does element X have? (ii) What is its atomic number? (iii) Identify the element X. (iv) Write its electronic configuration. (v) How many valence electrons does it have? (vi) What will be the mass number if two neutrons are added to its nucleus? (vii) What will be the relation of X with the new atom?
Solution:
- Protons = A - neutrons = 35 - 18 = 17; electrons = 17, protons = 17.
- Atomic number Z = 17.
- Z = 17 is chlorine (Cl).
- Configuration 2, 8, 7.
- Valence electrons = 7 (so it gains 1; valency 1).
- New mass number = 35 + 2 = 37.
- Same Z (17), different A (35 and 37): they are isotopes of chlorine.
Example 8: Isotopes and isobars from nuclear composition
Question: The composition of the nuclei of three atomic species X, Y and Z is: X has 18 protons and 19 neutrons; Y has 17 protons and 18 neutrons; Z has 17 protons and 20 neutrons. Explain the relation between (i) Y and Z, (ii) Z and X.
Solution: First find each mass number. X: 18 + 19 = 37. Y: 17 + 18 = 35. Z: 17 + 20 = 37.
- Y and Z: same atomic number (17), different mass numbers (35 and 37). They are isotopes of the same element (chlorine).
- Z and X: same mass number (37), different atomic numbers (17 and 18). They are atoms of different elements, so they are isobars.
Example 9: Average atomic mass of bromine
Question: If a bromine atom is available in the form of, say two isotopes, 7935Br (49.7%) and 8135Br (50.3%), calculate the average atomic mass of the bromine atom.
Solution:
Average atomic mass = 79 × 49.7/100 + 81 × 50.3/100
= (3926.3 + 4074.3)/100
= 8000.6/100 = 80.006 u ≈ 80 u
Check: the two abundances are close to equal, so the answer should be close to the midpoint of 79 and 81, slightly towards 81 because 81Br is a little more common. 80.006 u fits.
Competency-Based Questions (with answers)
Q1. Assertion-Reason (textbook Pause and Ponder 7)
Assertion (A): Rutherford concluded that most of the mass of an atom is concentrated in a small region at the centre called the nucleus.
Reason (R): According to Thomson’s model, electrons are embedded in a uniformly distributed positive charge sphere.
(i) Both A and R are true, and R is the correct explanation of A. (ii) Both A and R are true, but R is not the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true.
Answer: (ii). A is true: from the gold foil experiment Rutherford concluded that a tiny, dense nucleus holds all the positive charge and most of the mass. R is true: it correctly describes Thomson’s plum pudding model. R does not explain A, because R describes an earlier model, and Rutherford’s conclusion came from the α-particles that were deflected through large angles or bounced back, a result that disproved Thomson’s model.
Q2. Assertion-Reason: Isotopes
Assertion (A): The isotopes 3517Cl and 3717Cl have similar chemical properties.
Reason (R): Isotopes of an element have the same number of electrons and the same electronic configuration.
(i) Both A and R are true, and R is the correct explanation of A. (ii) Both A and R are true, but R is not the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true.
Answer: (i). A is true: both atoms are chlorine and react in the same way. R is true: both have 17 protons and 17 electrons, arranged 2, 8, 7. R explains A, because chemical properties depend mainly on the valence electrons, and both isotopes have 7 valence electrons. The two isotopes differ only in their number of neutrons (18 and 20), which changes physical properties such as boiling point.
Q3. Case-based: Isotopes in daily life
A hospital uses one isotope to treat cancer by radiation and another to treat goitre. A museum dates an ancient wooden object. A power plant generates electricity from a nuclear fuel.
(a) Name the isotope used in each case. (b) Why do all isotopes of an element show similar chemical properties? (c) State one property in which isotopes differ.
Answer: (a) Cancer radiation treatment: cobalt-60 (6027Co). Goitre and thyroid cancer: iodine-131 (13153I). Dating: carbon-14 (146C). Nuclear fuel: uranium-235 (23592U). (b) They have the same number of electrons and the same electronic configuration, and chemical properties depend mainly on valence electrons. (c) Physical properties such as melting point and boiling point.
Q4. Source-based: Chlorine’s average mass
Chlorine occurs in nature as two isotopes of mass 35 u and 37 u, in the ratio 3 : 1.
(a) Find the simple average. (b) Find the weighted average. (c) Why is the weighted average the correct value? (d) In 1 million chlorine atoms, how many are 35Cl?
Answer: (a) (35 + 37)/2 = 36 u. (b) 35 × 3/4 + 37 × 1/4 = 105/4 + 37/4 = 142/4 = 35.5 u. (c) The isotopes are not present in equal amounts; the weighted average accounts for each one’s relative abundance, so it reflects the element as it occurs in nature. (d) 3/4 of 10 lakh = 7.5 lakh atoms.
Q5. Reasoning: Why Rutherford failed and Bohr succeeded
Both Rutherford’s and Bohr’s models have electrons orbiting the nucleus. Why did Rutherford’s model fail to explain atomic stability, while Bohr’s model succeeded?
Answer: In Rutherford’s model an electron moving in a circle is accelerating, so it should keep losing energy, spiral inward and fall into the nucleus, making atoms collapse. Rutherford’s model had nothing to prevent this. Bohr postulated stationary states: fixed shells in which the electron’s energy stays constant even though it moves. With no energy loss there is no spiral and no collapse. Electrons change energy only by jumping between shells, absorbing or releasing a fixed amount of energy.
Q6. Case-based: Heavy electrons (textbook exercise 15)
In an atom, there are 12 protons and 12 neutrons in the nucleus. Now, imagine that all the electrons are replaced with some hypothetical particles that have the same charge as electrons but are 500 times heavier. What effect will this replacement have on the atom’s: (i) Atomic number (ii) Atomic mass (iii) Mass number (iv) Overall charge
Answer: (i) No change: Z = number of protons = 12. (ii) Atomic mass increases, because each of the 12 negative particles now adds much more mass than an electron did, so electron mass can no longer be ignored. (iii) No change: A counts only protons and neutrons, 12 + 12 = 24. (iv) No change: the 12 particles carry the same total negative charge as 12 electrons, balancing the 12 protons, so the atom stays neutral.
Q7. Reasoning: Choosing the right statements (textbook exercise 1)
Choose the correct options and explain the reason for the correct and incorrect options in the context of Ernest Rutherford’s gold foil experiment: (i) The experiment clearly showed the existence of neutrons in the nucleus. (ii) The results disproved the plum pudding model and led to the idea of a nucleus at the centre of the atom. (iii) The large deflection of a few alpha particles indicated that most of the mass of the atom and positive charge are packed into a tiny centre. (iv) The way alpha particles were deflected showed that electrons move around the nucleus.
Answer: Correct: (ii) and (iii). (i) is incorrect: neutrons were discovered by Chadwick in 1932; the experiment gave no evidence of neutrons. (ii) is correct: an evenly spread positive charge cannot turn back α-particles, so the plum pudding model failed and a central nucleus was proposed. (iii) is correct: only a small, dense, positive centre can deflect heavy, fast positive particles through large angles. (iv) is incorrect: the deflections were caused by the positive nucleus; electrons are far too light to deflect α-particles. Orbiting electrons were an assumption of Rutherford’s model; the scattering pattern gave no evidence about electron motion.
Q8. Reasoning: Correct or incorrect for Bohr’s model (textbook exercise 2)
Which of the following statements are correct or incorrect according to the Bohr’s atomic model? Give a reason for each statement. (i) Electrons lose energy while moving in fixed orbits and slowly fall into the nucleus. (ii) Electrons can exist anywhere around the nucleus with no fixed energy. (iii) Electrons revolve around the nucleus in orbits of fixed energy without losing energy. (iv) Electrons can be found between energy levels as they move around the nucleus.
Answer: (i) Incorrect: in a stationary state an electron does not lose energy. (ii) Incorrect: electrons occupy only allowed shells, each with a definite energy. (iii) Correct: this is Bohr’s central postulate. (iv) Incorrect: electrons revolve only in allowed shells and not in between them.
Q9. Riddle (textbook Pause and Ponder 16)
Solve this riddle: I am an atom with a mass number of 23 and 11 protons. I am a soft metal and react vigorously with water. Who am I and how many neutrons do I have?
Answer: Z = 11 is sodium (Na). Neutrons = 23 - 11 = 12. Its configuration 2, 8, 1 shows one valence electron, which it loses easily; this is why it is so reactive.
Important Questions for Exams
1-Mark Questions
Q1. Who discovered the electron, and in which year?
Answer: J. J. Thomson, in 1897.
Q2. What is the maximum number of electrons that the M shell can hold?
Answer: 2n2 = 2 × 32 = 18.
Q3. Write the symbols of potassium, silver and tungsten.
Answer: K, Ag, W.
Q4. Define isobars and give one example.
Answer: Atoms of different elements with the same mass number but different atomic numbers, for example 4020Ca and 4018Ar.
Q5. What is the valency of an element with configuration 2, 8, 3?
Answer: 3 (it loses its 3 valence electrons; the element is aluminium).
2-Mark Questions
Q6. How did the ancient idea of the atom differ from Dalton’s atomic theory?
Answer: Kanada’s parmanus and the Greek atomos were imaginary ideas that did not come from experiments. Dalton’s theory (1808) proposed indivisible atoms on the basis of scientific experiments of his time, so it was the first scientific description of matter.
Q7. Why did Thomson conclude that electrons are present in all atoms?
Answer: The nature of cathode rays was the same whatever the material of the cathode and whatever gas filled the tube. Since the same negative particles came from every material, electrons must be a basic part of the atoms of every element.
Q8. Why does a helium atom have about four times the mass of a hydrogen atom though it has only two protons?
Answer: Helium’s nucleus also has 2 neutrons, each with mass nearly equal to a proton. So its mass is due to 4 nucleons, against 1 in hydrogen. Electron mass is negligible.
3-Mark Questions
Q9. State the Bohr-Bury rules for filling electrons in shells. Use them to write the configuration of phosphorus (Z = 15).
Answer: (1) The maximum number of electrons in a shell is 2n2. (2) The outermost shell holds at most 8 electrons. (3) Shells fill stepwise from the innermost, K, L, M, and so on. Phosphorus: K = 2, L = 8, M = 15 - 10 = 5. Configuration 2, 8, 5.
Q10. What is valency? How is the valency of sodium, oxygen and carbon decided by their configurations?
Answer: Valency is the number of electrons gained, lost or shared by an atom to complete its octet. Sodium (2, 8, 1) has fewer than 4 valence electrons and loses 1: valency 1. Oxygen (2, 6) has more than 4 and gains 2: valency 2. Carbon (2, 4) cannot easily gain or lose 4, so it shares 4: valency 4.
Q11. Explain and arrange the following statements in the correct chronological order to show how atomic models have evolved over time. (i) Bohr’s model proposed that electrons move in fixed orbits around the nucleus, each with a definite energy. (ii) Thomson’s model depicted the atom as a ‘plum pudding’ with electrons embedded in a sphere of positive charge. (iii) Rutherford’s model proposed that atoms have a dense central nucleus. (iv) Dalton’s model described atoms as indivisible particles.
Answer: Order: (iv) → (ii) → (iii) → (i). Dalton (1808) gave indivisible atoms. The electron’s discovery (1897) showed atoms are divisible, so Thomson placed electrons inside a positive sphere. The gold foil experiment (1911) disproved this, and Rutherford placed the positive charge and most of the mass in a tiny nucleus. His model could not explain stability, so Bohr (1913) proposed fixed shells in which electrons do not lose energy.
Q12. Electrons move around the nucleus in orbits. Why do they not fly away from the atom? Explain what keeps them attracted to the nucleus.
Answer: Electrons are negatively charged and the nucleus is positively charged because of its protons. Opposite charges attract, so the nucleus pulls each electron towards it and keeps it bound to the atom as it moves around.
5-Mark Questions
Q13. Describe Rutherford’s gold foil experiment. State its observations, the conclusions drawn, the model proposed and one limitation of the model.
Answer:
- Setup: in 1911 Geiger and Marsden, under Rutherford, aimed a narrow beam of positively charged α-particles (helium nuclei) at an extremely thin gold foil.
- Observations: most particles passed through undeflected; some were deflected through large angles; a few bounced back.
- Conclusions: most of the atom is empty space; the positive charge and most of the mass are concentrated in an extremely small, dense nucleus.
- Model: a tiny positive nucleus at the centre, about 105 times smaller than the atom (atom ≈ 10-10 m, nucleus ≈ 10-15 m), with electrons revolving around it like planets around the Sun (planetary model).
- Limitation: a revolving electron is accelerating and should lose energy, spiral into the nucleus and make the atom collapse. Atoms are stable, so the model could not explain stability.
Q14. (a) Define isotopes. (b) Give the three isotopes of hydrogen with their numbers of protons, neutrons and electrons. (c) State two uses of isotopes. (d) Calculate the average atomic mass of chlorine, whose isotopes of mass 35 u and 37 u occur in the ratio 3 : 1.
Answer: (a) Atoms of the same element with the same atomic number but different mass numbers. (b) Protium 11H: 1 proton, 0 neutrons, 1 electron. Deuterium 21H: 1, 1, 1. Tritium 31H: 1, 2, 1. (c) Uranium-235 as fuel in nuclear reactors; carbon-14 for finding the age of fossils and artefacts (also cobalt-60 for cancer treatment and iodine-131 for goitre). (d) 35 × 75/100 + 37 × 25/100 = 26.25 + 9.25 = 35.5 u.
Common Mistakes and Examiner Tips
- Mistake: writing that Rutherford’s experiment proved neutrons exist. Fix: neutrons were discovered by Chadwick in 1932. The gold foil experiment revealed the nucleus.
- Mistake: using mass number to write electronic configuration. Fix: configuration depends only on the number of electrons, which is Z. For 199F, use 9, never 19.
- Mistake: writing 2, 8, 9 or putting more than 8 in the outermost shell. Fix: the outermost shell holds at most 8 electrons. For the first 18 elements, fill K to 2, L to 8, then M.
- Mistake: writing symbols as AL, CO, CL. Fix: first letter capital, second letter small: Al, Co, Cl. CO is a compound, carbon monoxide.
- Mistake: writing the valency of chlorine or oxygen as 7 or 6. Fix: with more than 4 valence electrons, valency = 8 - valence electrons. Chlorine 1, oxygen 2.
- Mistake: mixing up isotopes and isobars. Fix: isoTopes share the aTomic number (same element); isobars share the mass number (different elements).
- Mistake: taking the simple average (36 u) as chlorine’s atomic mass. Fix: use the weighted average with abundances: 35.5 u.
- Mistake: saying one chlorine atom weighs 35.5 u. Fix: 35.5 u is an average over many atoms; each atom is either 35 u or 37 u.
- Mistake: saying Bohr’s electrons are not moving. Fix: they move in circles around the nucleus; what stays constant in a stationary state is their energy.
- Mistake: saying the K shell has the most energy. Fix: K (n = 1) is closest to the nucleus and has the least energy; energy increases outward.
- Tip: in every numerical, write the relation first (A = p + n, electrons = Z) and then substitute. Show units (u) for atomic mass, and state the final answer clearly.
Quick Revision Points
- Acharya Kanada: parmanus, recorded in the Vaisesika Sutras; dyads and triads.
- Dalton (1808): first scientific atomic theory; atoms indivisible.
- Thomson (1897): cathode rays in a low-pressure gas tube are negatively charged particles, electrons, present in all atoms.
- Gold foil experiment (1911, Geiger and Marsden under Rutherford): most α-particles pass, some deflect sharply, a few bounce back.
- Rutherford: mostly empty space; tiny dense positive nucleus with most of the mass; electrons revolve (planetary model).
- Nucleus is about 105 times smaller than the atom: 10-15 m against 10-10 m.
- Rutherford’s limitation: could not explain the stability of the atom.
- Bohr (1913): fixed shells K, L, M, N (n = 1, 2, 3, 4) with definite energy; no energy loss in a shell.
- K has the least energy; energy increases outward. Shell jumps absorb or release a fixed amount of energy.
- Z = number of protons = number of electrons in a neutral atom.
- A = protons + neutrons (nucleons); neutrons = A - Z.
- Maximum electrons per shell = 2n2: K 2, L 8, M 18. Outermost shell at most 8.
- Valency: electrons lost, gained or shared to complete the octet. Na 1, Mg 2, Al 3, C 4, N 3, O 2, Cl 1.
- Isotopes: same Z, different A; similar chemical, different physical properties.
- Isobars: same A, different Z; 40Ca, 40K, 40Ar.
Weightage and Exam Pattern
Class 9 has no board exam. This chapter is assessed in your school’s periodic tests, half-yearly and annual exams, and the paper pattern and marks are set by your school. It is also a foundation for Class 10 Science, where valency and electronic configuration are used again and again.
How the chapter is usually tested:
- Numericals: finding protons, neutrons and electrons from Z and A; completing tables like Table 8.5; weighted average atomic mass.
- Configurations: writing electronic configurations of the first 18 elements and deducing valence electrons and valency.
- Reasoning: observations and conclusions of the gold foil experiment; why Thomson’s and Rutherford’s models were replaced; how Bohr explained stability.
- Competency formats: assertion-reason items, case studies on isotopes and their uses, and true-or-false statements with reasons.
Class 9 Science · Chapter 8 – swipe through all 10 cards to understand the whole chapter.
From Parmanu to Dalton
The atom began as an idea and became science with Dalton’s experiments.
- Kanada: parmanus, in the Vaisesika Sutras
- Greeks: atomos, meaning indivisible
- Dalton 1808: first scientific atomic theory
Thomson and Cathode Rays
Rays from the cathode in a low-pressure gas tube are negatively charged particles.
- Discovered in 1897 by J. J. Thomson
- Same rays for any cathode metal or gas
- So electrons are part of every atom
Plum Pudding Atom
A sphere of positive charge with electrons embedded throughout it.
- Watermelon: red pulp is positive, seeds are electrons
- Total positive charge equals total negative
- Failed to explain gold foil results
Rutherford’s Nuclear Model
Alpha particles fired at thin gold foil revealed a tiny, dense, positive nucleus.
- Most pass straight: atom is mostly empty
- Large deflections and bounce-backs: nucleus
- Limitation: cannot explain stability
Fixed Shells of Energy
Electrons revolve only in allowed shells K, L, M, N without losing energy.
- K is closest to the nucleus, lowest energy
- Energy rises as shells move outward
- Jumps absorb or release fixed energy
Electron, Proton, Neutron
Three subatomic particles build every atom.
- Proton: discovered and named by Rutherford
- Neutron: Chadwick, 1932, no charge
- Protons and neutrons carry the mass
Atomic and Mass Number
Two numbers tell you every particle count in an atom.
- Z = protons = electrons in a neutral atom
- 126C: 6 protons, 6 electrons, 6 neutrons
- Symbols: first letter capital, e.g. Al, Co
Bohr-Bury Rules
Electrons fill shells from the inside out with fixed limits.
- K 2, L 8, M 18 electrons at most
- Outermost shell holds at most 8
- Na 2, 8, 1; S 2, 8, 6; Ar 2, 8, 8
Combining Capacity
Atoms lose, gain or share electrons to complete an octet.
- Sodium 1, magnesium 2, aluminium 3
- Carbon shares 4 electrons: valency 4
- Oxygen 2, nitrogen 3, chlorine 1
Isotopes and Isobars
Same Z with different A gives isotopes; same A with different Z gives isobars.
- Hydrogen: protium, deuterium, tritium
- Uses: U-235 fuel, Co-60 cancer, C-14 dating
- Isobars: 40Ca, 40K, 40Ar
📝 Practice Journey Inside the Atom - 10 questions
Concept-check questions · with answers & explanations
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Why A: The textbook compares Thomson’s atom to a watermelon, so A is true. The comparison works precisely because the positively charged matter fills the whole sphere like the red pulp and the electrons are distributed throughout it like the seeds, so R is true and it is exactly why the analogy holds.
Why not B: R is not a separate fact; it is the very feature that makes the watermelon comparison valid.
Why not C: R correctly describes the model; the positive charge is spread out, not concentrated at a centre.
Why not D: The watermelon comparison is stated in the chapter along with the plum pudding name, so A is true.
Remember: Pulp is the positive charge, seeds are the electrons; that is the whole watermelon argument.
Why D: Deflection happens when an alpha particle passes close to a nucleus. A thicker foil has more layers of atoms in the beam’s path, so each alpha particle crosses more nuclei and the chance of a close encounter rises. Therefore a larger fraction of the particles would be deflected, and more would be scattered through large angles.
Why not A: Nuclear charges do not cancel each other; every extra nucleus in the path is one more chance for deflection.
Why not B: Even a thick foil is mostly empty space at the atomic scale, so many particles still pass through; bounce-back stays rare.
Why not C: The size of a nucleus is unchanged, but the number of nuclei the beam meets increases, and it is the number of encounters that matters.
Remember: More layers, more nuclei in the way, more scattering.
Why A: In Rutherford’s model an electron in circular motion is accelerating and should keep losing energy, spiral inward and collapse into the nucleus. Bohr kept the orbiting electrons but added the postulate of stationary states: while an electron moves in an allowed shell its energy stays constant and it does not radiate. That postulate removes the energy loss and therefore the collapse.
Why not B: The electron is the same particle in both models; its mass is not what changed.
Why not C: Electrons in Bohr’s model revolve around the nucleus in shells; they are not inside it.
Why not D: Bohr retained Rutherford’s positively charged nucleus at the centre.
Remember: Same orbiting electrons; Bohr simply forbade them from losing energy while in a shell.
Why D: The atomic mass is normally almost entirely the mass of the nucleus because electrons are negligibly light. If each of the 12 electrons became 500 times heavier, their combined mass would no longer be negligible, so the atomic mass of the atom would increase. Everything else is fixed by the nucleus or by charge, which are untouched.
Why not A: The atomic number is the number of protons, still 12; the electrons were replaced, not the protons.
Why not B: The mass number counts nucleons only, 12 protons plus 12 neutrons = 24; the heavy particles are outside the nucleus and are not nucleons.
Why not C: The new particles carry the same charge as electrons, so 12 negative charges still balance 12 protons and the atom stays neutral.
Remember: Z and A are nuclear bookkeeping; only the actual mass feels the heavier electrons.
Why A: Sodium, 2,8,1, has one valence electron and loses it, so its valency is 1. Chlorine, 2,8,7, has seven valence electrons and gains one to complete its octet, so its valency is also 1. Although one loses and the other gains, the number of electrons involved is the same, so Na and Cl have the same valency.
Why not B: Magnesium, 2,8,2, loses two, valency 2, while chlorine’s valency is 1.
Why not C: Oxygen, 2,6, gains two, valency 2, while aluminium, 2,8,3, loses three, valency 3.
Why not D: Oxygen’s valency is 2, while nitrogen, 2,5, gains three, valency 3.
Remember: Valency is the count of electrons moved, not the count of valence electrons: 2,8,1 and 2,8,7 both give 1.
Why B: Work out atomic and mass numbers: X has Z = 18, A = 18 + 19 = 37; Y has Z = 17, A = 17 + 18 = 35; Z has Z = 17, A = 17 + 20 = 37. Y and Z share the atomic number 17 with different mass numbers, so they are isotopes of the same element. Z and X have the same mass number 37 but different atomic numbers, 17 and 18, so they are isobars.
Why not A: This swaps the two relationships; same protons makes isotopes, same nucleon total makes isobars.
Why not C: X has 18 protons while Y and Z have 17, so X belongs to a different element and cannot be their isotope.
Why not D: X has mass number 37 and Y has 35; with different atomic numbers as well, they are neither isobars nor isotopes.
Remember: Compare protons first (same means isotopes), then compare protons plus neutrons (same means isobars).
Why D: In Thomson’s picture the atom is neutral only because the positive charge of the sphere exactly equals the total negative charge of the embedded electrons. If the positive charge is smaller than the total negative charge of the beads, the negatives are left over, so the whole model carries a net negative charge and no longer represents a neutral atom.
Why not A: Charges do not cancel just by being in the same ball; they cancel only when their amounts are equal.
Why not B: Charge and mass are separate quantities; reducing the positive charge changes the net charge, not the mass.
Why not C: A net positive charge would need more positive charge than negative, the opposite of the mistake described.
Remember: Neutral means equal amounts of positive and negative charge, not merely both present.
Why A: Step 1 (ratio of diameters): atom to nucleus is 10⁻¹⁰ ÷ 10⁻¹⁵ = 10⁵, so the nucleus is one lakh times smaller than the atom. Step 2 (scaling the model): 100 m divided by 10⁵ is 10⁻³ m, which is 1 mm, the size of a grain of black pepper placed at the centre of the ground.
Why not B: 1 cm would need a ratio of only 10⁴, one power of ten short of the real 10⁵.
Why not C: 1 m corresponds to a ratio of just 100, which ignores the huge difference between the two exponents.
Why not D: 10 m would make the nucleus a large part of the atom, contradicting the finding that the atom is mostly empty space.
Remember: Atom 10⁻¹⁰ m, nucleus 10⁻¹⁵ m, a gap of 10⁵: cricket ground to pepper grain.
Why B: The L-shell has more energy than the K-shell. To move outward from K to L the electron must gain exactly the energy difference between the two levels, so it absorbs a fixed amount of energy equal to that difference. Moving inward from L to K would release the same fixed amount.
Why not A: Energy is released when the electron moves to a lower level, that is from L down to K, not when it climbs to L.
Why not C: Being stationary states means the energy inside each shell is constant, but the two shells have different energies, so changing shells changes the energy.
Why not D: Bohr’s electrons can exist only in allowed shells, never in between; with less than the full difference the electron simply does not move.
Remember: Jump out means absorb the exact gap; jump in means release the exact gap; no half jumps.
Why B: Step 1 (neutral atom, protons equal electrons): the atom has 11 electrons, so it has 11 protons and Z = 11. Step 2 (mass number equals protons plus neutrons): neutrons = A – Z = 23 – 11 = 12. So the atom has 12 neutrons; it is sodium.
Why not A: 11 is the number of protons (and electrons), not neutrons.
Why not C: 23 is the mass number, the total of protons and neutrons, not the neutron count alone.
Why not D: 34 comes from adding 23 and 11 instead of subtracting.
Remember: Neutrons = mass number minus atomic number, and electrons tell you the atomic number in a neutral atom.
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Frequently Asked Questions
Most alpha particles passed straight through the thin gold foil, some were deflected through large angles, and a few bounced back. Rutherford concluded that most of the atom is empty space and that all the positive charge and most of the mass sit in a tiny, dense centre called the nucleus, about 10⁵ times smaller than the atom.
An electron moving in a circle keeps changing direction, so it is accelerating. An accelerating charged electron should lose energy, spiral inward and fall into the positive nucleus, so atoms would collapse. Real atoms are stable, so the model had to be modified. Bohr solved this with stationary states in which electrons do not lose energy.
The maximum number of electrons in a shell is 2n², so K holds 2, L holds 8 and M holds 18. The outermost shell can hold at most 8 electrons. Shells fill stepwise from the innermost, in the order K, L, M, N, and a shell starts filling only after the one before it is complete.
Subtract the atomic number from the mass number: neutrons = A − Z. The atomic number gives the number of protons, and the mass number is protons plus neutrons. For example, an atom with A = 197 and 79 protons has 197 − 79 = 118 neutrons and, being neutral, 79 electrons.
Valency is the number of electrons an atom gains, loses or shares to complete its octet. With fewer than 4 valence electrons the atom loses them, so sodium (2, 8, 1) has valency 1. With more than 4 it gains 8 minus that number, so oxygen (2, 6) has valency 2. Carbon (2, 4) shares 4 electrons, so its valency is 4.
Isotopes are atoms of the same element with the same atomic number but different mass numbers, such as chlorine-35 and chlorine-37. Isobars are atoms of different elements with the same mass number but different atomic numbers, such as calcium-40 and argon-40. Isotopes have similar chemical properties; isobars do not, because their electron numbers differ.
Chlorine is a mix of two isotopes, 35 u and 37 u, in the ratio 3 : 1 (75% and 25%). The weighted average is 35 × 75/100 + 37 × 25/100 = 35.5 u. No single atom weighs 35.5 u; it is the average over many atoms. The simple average of 36 u is wrong because it ignores abundance.
Isotopes of an element have the same number of protons and therefore the same number of electrons and the same electronic configuration. Chemical properties depend mainly on the valence electrons, so they are similar. Isotopes differ in the number of neutrons, which changes physical properties such as melting and boiling points.
