NCERT Solutions Class 9 Science Exploration Chapter 08 Journey Inside the Atom

Question 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: The correct statements are (ii) and (iii). Statement (ii) is accurate because Thomson's plum pudding model proposed that positive charge was distributed uniformly throughout the atom. Rutherford's findings revealed that most alpha particles passed straight through, while only a few were sharply deflected. This demonstrated that positive charge is not spread evenly but is instead concentrated in a very small central region known as the nucleus. Statement (iii) is also correct because when only a few alpha particles underwent large-angle deflection, it indicated they had come very close to an extremely small, dense, positively charged core - this core is the nucleus, containing virtually all the atom's mass. The incorrect statements are (i) and (iv). Statement (i) is false because Rutherford's gold foil experiment did not demonstrate the presence of neutrons; that discovery came much later through James Chadwick's work. The experiment only revealed the existence of a small, dense, positively charged centre. Statement (iv) is also false because Rutherford's experiment yielded information chiefly about the nucleus and the void within the atom; it did not provide direct evidence about electron movement around the nucleus.
In simple words: Statements (ii) and (iii) are correct - they show how the experiment proved the nucleus exists and holds most of the atom's positive charge and mass. Statements (i) and (iv) are wrong - the experiment found the nucleus but not neutrons, and it didn't directly show how electrons orbit.

Exam Tip: For Rutherford's experiment questions, always connect the particle behavior (deflection, straight passage, bouncing back) directly to the nuclear structure conclusions - this is what examiners want to see.

Question 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) This statement is incorrect. Under Bohr's model, electrons travel in defined orbits or energy levels and retain their energy while revolving in these permitted shells. (ii) This statement is incorrect. Bohr's theory states that electrons may only occupy specific fixed energy levels or shells - they cannot be positioned randomly around the nucleus. (iii) This statement is correct. This forms the core principle of Bohr's model. Electrons stay stable in fixed shells and do not emit energy as they orbit within those shells. (iv) This statement is incorrect. Bohr's model does not allow electrons to exist between two energy levels. Electrons transition from one shell to another only by taking in or giving out a specific quantity of energy.
In simple words: Statement (iii) is correct - electrons stay in fixed shells and keep their energy. The other three are wrong - electrons cannot be anywhere they want, cannot fall into the nucleus, and cannot be between shells.

Exam Tip: Remember that Bohr's model centers on fixed, stable shells where electrons don't lose energy - this concept is the model's defining feature and comes up frequently in exams.

Question 3. The composition of the nuclei of three atomic species X, Y and Z are given as follows: Explain the relation between the following: (i) Y and Z (ii) Z and X
Answer: Start by finding the atomic number and mass number of each species. For X - protons are 18, neutrons are 19, so atomic number is 18 and mass number is 18 + 19 = 37. For Y - protons are 17, neutrons are 18, so atomic number is 17 and mass number is 17 + 18 = 35. For Z - protons are 17, neutrons are 20, so atomic number is 17 and mass number is 17 + 20 = 37. (i) Y and Z are isotopes because they have the same atomic number (17), meaning the same count of protons, but their mass numbers differ (35 and 37) due to different neutron numbers. (ii) Z and X are isobars because they share the same mass number of 37, yet their atomic numbers are different - X has 18 while Z has 17.
In simple words: Y and Z are isotopes - same element, different mass. Z and X are isobars - different elements, same mass.

Exam Tip: Know the definitions: isotopes have same atomic number but different neutrons; isobars have same mass number but different protons. Always calculate both numbers before deciding which category applies.

Question 4. What conclusion did Rutherford draw about the position and characteristics of the atom's positively charged part based on the few alpha particles that bounced back or were deflected at large angles in the gold foil experiment?
Answer: Rutherford's gold foil experiment led to four key insights about the atom's positive charge. First, regarding position - the positive charge is not spread out evenly across the atom but rather packed tightly into a very small region at the atom's centre called the nucleus. Second, regarding nuclear size - the nucleus is tiny in relation to the whole atom because only a limited number of alpha particles were deflected. Third, regarding atomic mass - the nucleus holds almost all of the atom's total mass, which Rutherford concluded from observing that large deflections and bouncing back of alpha particles showed they had hit an extremely dense and heavy area. Fourth, regarding nuclear nature - the nucleus carries a positive charge, which produces a pushing force that deflects and repels the positively charged alpha particles.
In simple words: The nucleus is a tiny, dense, heavy, positively charged core at the atom's centre that repels alpha particles and holds almost all the atom's mass.

Exam Tip: List the four conclusions clearly: position (centre), size (extremely small), mass (concentrated), nature (positively charged) - examiners expect each point separately.

Question 5. 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: The proper time sequence is (iv) Dalton's model, (ii) Thomson's model, (iii) Rutherford's model, (i) Bohr's model. Dalton put forward the concept that matter is made of tiny, unbreakable units called atoms. Then Thomson found electrons and suggested the plum pudding image - a ball of positive material studded with negative electrons throughout. Rutherford later revealed the nucleus through his gold foil test, overthrowing the plum pudding idea. Finally, Bohr refined the understanding by proposing that electrons occupy defined orbits with set energy values around the nucleus, which addressed the stability problem that Rutherford's model couldn't resolve.
In simple words: Dalton said atoms exist, Thomson added electrons, Rutherford found the nucleus, and Bohr placed electrons in fixed shells - each scientist built on the previous discovery.

Exam Tip: Timeline questions require you to know not only the order but also why each model came next - focus on what each scientist discovered and how it changed thinking about atomic structure.

Question 6. 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 carry negative charge while the nucleus contains positively charged protons. An electrostatic pulling force acts between these opposite charges - the positive nucleus draws the negative electrons toward itself. This electrostatic attraction keeps the electrons bound within their orbits and prevents them from escaping the atom. The force holding them in place is strong enough to maintain their orbital motion without allowing them to break free.
In simple words: Opposite charges pull on each other. The positive nucleus pulls the negative electrons, so they stay in orbit and don't fly away.

Exam Tip: Always mention electrostatic force by name and explain that it arises from opposite charges - this shows understanding of the force, not just memorized facts.

Question 7. Assertion (A): The discovery of subatomic particles helped in understanding the atomic structure. Reason (R): The number of electrons is equal to the number of protons in an atom. Choose the correct option: (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) Both A and R are true, but R is not the correct explanation of A.
Explanation: The assertion is true because finding electrons, protons, and neutrons enabled scientists to understand how atoms are built and organized. The reason is also true - in a neutral atom, the electron count matches the proton count, keeping the atom electrically balanced. However, the reason does not explain why these discoveries helped understanding atomic structure. The equality of electrons and protons mainly shows that atoms are electrically neutral; it doesn't explain how subatomic particles revealed the structure of the atom itself.
In simple words: Both statements are right, but one doesn't explain the other. Finding particles showed us how atoms work, but equal electrons and protons just shows atoms are balanced.

Exam Tip: In assertion-reason questions, always check whether the reason actually explains the assertion, not just whether both facts are true - this distinction is crucial for getting full marks.

Question 8. Magnesium is essential for many biological processes, including muscle contraction. 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.
Answer: Given atomic number = 12 and mass number = 24. (i) Number of protons = atomic number = 12. (ii) Number of neutrons = mass number - atomic number = 24 - 12 = 12. (iii) Number of electrons = 12 in a neutral atom. (iv) Electronic arrangement - K shell holds 2 electrons, L shell holds 8 electrons, and M shell holds 2 electrons. The electronic configuration is 2, 8, 2.
In simple words: Magnesium has 12 protons and 12 neutrons in its nucleus. It has 12 electrons arranged as 2 in the first ring, 8 in the second ring, and 2 in the third ring.

Exam Tip: Always show the calculation for neutrons clearly (Mass number - Atomic number), and remember that electrons are distributed using the 2n² rule, filling inner shells first.

Question 9. Find the following information for the elements shown in Fig. 8.17: (i) Name of element (ii) Symbol (iii) Total electrons (iv) Valence electrons (v) Valency (vi) Number of protons (vii) Atomic number
Answer:
(a) (i) Helium (ii) He (iii) 2 total electrons (iv) 2 valence electrons (v) Valency 0 (vi) 2 protons (vii) Atomic number 2

(b) (i) Oxygen (ii) O (iii) 8 total electrons (iv) 6 valence electrons (v) Valency 2 (vi) 8 protons (vii) Atomic number 8

(c) (i) Calcium (ii) Ca (iii) 20 total electrons (iv) 2 valence electrons (v) Valency 2 (vi) 20 protons (vii) Atomic number 20

(d) (i) Neon (ii) Ne (iii) 10 total electrons (iv) 8 valence electrons (v) Valency 0 (vi) 10 protons (vii) Atomic number 10
In simple words: For each diagram, count the electron rings to identify the element, then find all the requested properties by reading the rings or using basic rules like valence electrons being those in the outermost ring.

Exam Tip: When analyzing electron diagrams, valence electrons are always the outermost ring. Elements with 8 in the outer shell (or 2 for K-shell) have valency 0 - remember this pattern.

Question 10. 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: Rutherford proposed electrons orbit the nucleus like planets around the Sun. However, classical physics states that any moving, charged object should give off energy as radiation continuously. This leads to a chain reaction - the electron loses energy, its path shrinks inward, and it spirals into the nucleus. Yet atoms don't collapse, so Rutherford's model could not account for why atoms remain stable. Bohr solved this problem by introducing three key ideas. First, electrons can only move in certain fixed circular paths or energy shells. Second, while in these fixed shells, electrons do not give off any energy. Third, energy is only emitted or taken in when an electron jumps between shells. Because of these principles, electrons stay in stable orbits and atoms do not break down - the model successfully explained atomic stability.
In simple words: Rutherford's electrons should fall into the nucleus, but they don't. Bohr fixed this by saying electrons stay in fixed shells without losing energy, so atoms stay stable.

Exam Tip: Contrast the two models clearly: Rutherford had the structure right but couldn't explain stability; Bohr kept the structure and added the energy-level concept to resolve the stability problem.

Question 11. An atom ⁷⁰X has 31 electrons. How many neutrons are there in its nucleus?
Answer: Mass number (A) = 70. Number of electrons = 31. In a neutral atom, number of protons equals number of electrons, so there are 31 protons. Number of neutrons = mass number - atomic number = 70 - 31 = 39. Therefore, the nucleus contains 39 neutrons.
In simple words: The mass number is 70. Since it has 31 electrons, it has 31 protons. Subtracting gives 70 - 31 = 39 neutrons.

Exam Tip: Always remember that for neutral atoms, electrons equal protons. Use the subtraction formula (A - Z) to find neutrons - this is the most direct path.

Question 12. An atom has 79 protons and a mass number of 197. Calculate: (i) the number of neutrons, and (ii) the number of electrons.
Answer: Number of protons = 79, mass number = 197. (i) Number of neutrons = mass number - number of protons = 197 - 79 = 118. (ii) For a neutral atom, number of electrons = number of protons = 79.
In simple words: Subtract protons from mass number to get neutrons: 197 - 79 = 118. Since it's neutral, electrons equal protons: 79.

Exam Tip: These calculations are straightforward if you apply the formulas correctly - always state that you're assuming a neutral atom to justify equal electrons and protons.

Question 13. Complete the Table 8.5:
Answer: To fill the table, apply these relationships: Atomic number equals the number of protons. In a neutral atom, the number of electrons equals the number of protons. Mass number equals the sum of protons and neutrons.

Exam Tip: Set up the relationships as column headers and work systematically - this prevents errors. Double-check that mass number equals protons plus neutrons in every row.

Question 14. Aman was discussing the structure of atom with his classmates. During the discussion, he learnt that an element X has a mass number of 35 and contains 18 neutrons. Based on this information, answer the following questions: (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?
Answer: Given mass number = 35 and number of neutrons = 18. Number of protons = mass number - number of neutrons = 35 - 18 = 17. Since the atom is neutral, number of electrons = number of protons = 17. (i) The atom has 17 protons and 17 electrons. (ii) Atomic number = number of protons = 17. (iii) The element with atomic number 17 is Chlorine (Cl). (iv) The electron arrangement is 2, 8, 7. (v) Valence electrons = 7. (vi) If two neutrons are added, new neutrons = 18 + 2 = 20. New mass number = number of protons + number of neutrons = 17 + 20 = 37. (vii) The two atoms are isotopes because both have the same atomic number (17) but different mass numbers (35 and 37).
In simple words: Element X is chlorine with 17 protons, 17 electrons, and 7 valence electrons. If we add 2 neutrons, the mass becomes 37 instead of 35, making the new atom an isotope of chlorine.

Exam Tip: Work through each sub-part systematically using the definitions and formulas given. Make sure you identify the element and then state the isotope relationship clearly at the end.

Question 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: Number of protons = 12, number of neutrons = 12. Keep in mind that atomic number is determined only by proton count, while mass number relies only on protons and neutrons. (i) Atomic number will stay the same. Reason: atomic number equals the number of protons, which is 12 and has not changed. (ii) Atomic mass will increase. Reason: the replacement particles are much heavier than normal electrons, making the overall mass of the atom greater. (iii) Mass number will stay the same. Reason: mass number equals protons plus neutrons, which equals 12 + 12 = 24, and electrons don't contribute to mass number. (iv) Overall charge will remain neutral (zero), provided that the new particles are equal in number to the protons. Reason: since the replacement particles carry the same negative charge as electrons, their combined negative charge will offset the positive charge of the 12 protons.
In simple words: Only the mass of the atom changes because the new particles weigh more. Atomic number, mass number, and charge stay the same because protons and neutrons count for those, not electrons.

Exam Tip: This question tests understanding of what determines each atomic property - remember that atomic number and mass number depend on nucleons (protons and neutrons), not on electrons.

Section-wise Notes of Exploration Chapter 8 - Journey Inside the Atom

Section 8.1 Rediscovering the Roots of Atomic Theory

Ancient ideas about the smallest unit of matter:

• Acharya Kanada (India): Called the smallest indivisible particles parmanus. Recorded in the Sanskrit text Vaisesika Sutras. Combinations of parmanus form dyads, triads and all of the material universe.
• Leucippus and Democritus (Greece): Called indivisible particles atomos (Greek for indivisible).
• John Dalton (1808): First scientific atomic theory - all matter is made of indivisible atoms; atoms are the fundamental building blocks of matter.

Key questions after Dalton's theory:

• What are atoms made up of?
• What would atoms look like if we could see them?
• What makes atoms of one element different from another?

Note: The concept of 'atom' originated as an imaginary idea, not from experimental observations.

Section 8.2 A Short Historical Journey Through Atomic Models

Discovery of Electron (1897 - J. J. Thomson):

• Studied conduction of electric current through gases at low pressure in a cathode ray tube
• Rays moved from cathode (negative) to anode (positive) - called cathode rays
• Concluded cathode rays are streams of negatively charged particles - electrons
• Nature of cathode rays was independent of cathode material or gas - electrons are present in all atoms
• Charge of electron: -1.602 × 10⁻¹⁹ C (taken as -1 by convention)
• Nobel Prize in Physics: 1906

8.2.1 Thomson's Model (Plum Pudding Model):

• Atom is a sphere of positive charge with electrons distributed throughout
• Also called watermelon model - red pulp = positive charge, seeds = electrons
• First attempt to show how positive and negative charges are balanced in an atom

8.2.2 The Gold Foil Experiment (1911 - Geiger and Marsden under Rutherford):

A narrow beam of alpha (α) particles was aimed at an extremely thin gold foil. Alpha particles are tiny, positively charged particles emitted from radioactive elements; each is a helium nucleus (2 protons + 2 neutrons). Based on Thomson's model, scientists expected all particles to pass straight through or be slightly deflected.

Actual results:

• Most alpha particles passed straight through undeflected - the atom is mostly empty space
• Some were deflected at large angles - a dense positive charge exists at the centre
• A very few bounced back - the nucleus is extremely small and dense

This experiment is also called the α-ray scattering experiment. Thomson's model failed to explain these results.

Rutherford's Model (Planetary Model):

• Positive charge is concentrated in an extremely small, dense nucleus
• Most of the atom is empty space
• Electrons revolve around the nucleus like planets around the Sun
• Diameter of atom ≈ 10⁻¹⁰ m; Diameter of nucleus ≈ 10⁻¹⁵ m
• Nucleus is 10⁵ (one lakh) times smaller than the atom

Limitation of Rutherford's Model:

• Cannot explain atomic stability
• A circularly moving electron is accelerating and should continuously lose energy
• Losing energy should cause the electron to spiral inward and fall into the nucleus
• But atoms are stable - this contradiction was not resolved by Rutherford's model

Discovery of Proton (Rutherford):

• Nucleus carries positive charge from particles called protons
• Protons are much heavier than electrons
• Charge of proton = +1 (equal and opposite to electron)
• For neutral atom: number of protons = number of electrons

8.2.3 Bohr's Model (1913 - Niels Bohr): Proposed to explain atomic stability. Key postulates:

• Electrons move in fixed circular paths called stationary states, orbits, or shells - also called energy levels
• Shells are labelled K, L, M, N... or n = 1, 2, 3, 4...
• While moving in a fixed shell, an electron does NOT lose energy
• K-shell (n = 1) is closest to nucleus and has the least energy; energy increases with distance from nucleus
• An electron can move between shells by absorbing or releasing a fixed amount of energy equal to the difference between energy levels
• Each shell can hold a limited number of electrons
• Nobel Prize: 1922

Why shells are called K, L, M, N: Named after early X-ray experiments by Charles Barkla who labelled X-ray lines starting from K, leaving room for possible earlier series. Later even Bohr's model had limitations - modern quantum mechanical model (electron clouds) replaced it. You will study this in higher grades.

Section 8.3 What Components Contribute to the Mass of an Atom?

Puzzle: Helium has 2 protons but its mass is 4 times that of hydrogen (1 proton) - not double. Why?

8.3.1 Discovery of Neutron (1932 - James Chadwick):

• Discovered a neutral particle with mass nearly equal to a proton
• Named neutron, symbol n⁰
• Present in the nucleus of all atoms except hydrogen
• Mass of an atom = mass of protons + mass of neutrons (electron mass is negligible)
• Nobel Prize in Physics: 1935

Why neutrons are needed in heavy nuclei: Protons repel each other (same positive charge). Neutrons help reduce this repulsion and strengthen nuclear force - heavier atoms need more neutrons than protons (e.g., Uranium: 92 protons, 146 neutrons).

Subatomic Particles Summary:

• Electron | e⁻ | Relative Charge -1 | Location: Orbits around nucleus
• Proton | p⁺ | Relative Charge +1 | Location: In nucleus
• Neutron | n⁰ | Relative Charge 0 | Location: In nucleus (except H)

Section 8.4 Symbols of Elements

Dalton (1803): First pictorial symbols for elements.

Berzelius (1813): Suggested alphabetic symbols derived from Latin names.

IUPAC (International Union of Pure and Applied Chemistry): Currently approves all names and symbols.

Rules for writing symbols:

• First letter is always uppercase; second letter (if present) is lowercase
• Many symbols are first one or two letters of English name (e.g., H for Hydrogen, Al for Aluminium)
• Some symbols come from the first letter and a non-adjacent letter (e.g., Cl for Chlorine, Zn for Zinc)
• Some symbols are from Latin, Greek, or German names: Fe (Ferrum = Iron), Hg (Hydrargyros = Mercury), W (Wolfram = Tungsten), Na (Natrium = Sodium), K (Kalium = Potassium), Au (Aurum = Gold), Ag (Argentum = Silver), Pb (Plumbum = Lead), Cu (Cuprum = Copper)

Currently 118 elements are known. Symbols allow international scientific communication across language barriers.

Section 8.5 Atomic Number

Atomic Number (Z) = Number of protons in the nucleus of an atom. Since atom is neutral: Number of protons = Number of electrons. Atomic number uniquely identifies an element. No two elements can have the same atomic number. Examples: Hydrogen Z = 1 (1 proton, 1 electron); Helium Z = 2 (2 protons, 2 electrons)

Section 8.6 Mass Number

Mass Number (A) = Number of protons + Number of neutrons = Total nucleons. Protons and neutrons together are called nucleons. Electron mass is negligible and not counted in mass number.

Standard notation: Mass number (A) is written above and atomic number (Z) is written below the element symbol. Example: Carbon is written as ¹²₆C (A = 12, Z = 6). Number of neutrons = A - Z.

Element | Protons | Neutrons | Mass Number
Hydrogen | 1 | 0 | 1
Helium | 2 | 2 | 4
Lithium | 3 | 4 | 7

Section 8.7 How Are Electrons Distributed in Different Energy Levels?

Bohr-Bury Rules for Electron Distribution:

Maximum electrons in a shell = 2n² (where n = shell number)

• K-shell (n=1): 2 × 1² = 2 electrons
• L-shell (n=2): 2 × 2² = 8 electrons
• M-shell (n=3): 2 × 3² = 18 electrons

Maximum electrons in outermost shell = 8 (except K-shell where maximum = 2). Electrons fill shells in order K → L → M → N (inner shells fill first).

Electronic Configuration = Distribution of electrons among shells.

Examples of electronic configurations:

Element | Z | K | L | M
Hydrogen | 1 | 1 | - | -
Helium | 2 | 2 | - | -
Lithium | 3 | 2 | 1 | -
Carbon | 6 | 2 | 4 | -
Neon | 10 | 2 | 8 | -
Sodium | 11 | 2 | 8 | 1
Magnesium | 12 | 2 | 8 | 2
Chlorine | 17 | 2 | 8 | 7
Argon | 18 | 2 | 8 | 8

Section 8.8 Combining Capacity of an Atom: Valency

Valence Shell: The outermost shell of an atom

Valence Electrons: Electrons present in the valence shell

Octet: When outermost shell has 8 electrons (fully stable)

Stability rule:

• Atoms with complete octet (8 electrons) or 2 electrons (helium) are stable and largely unreactive
• Atoms with incomplete valence shells are reactive and lose, gain or share electrons to complete octet

Valency = Number of electrons gained, lost or shared to complete the octet.

Rules for valency:

• Fewer than 4 valence electrons - tends to LOSE electrons (valency = number of valence electrons)
• More than 4 valence electrons - tends to GAIN electrons (valency = 8 - valence electrons)
• Exactly 4 valence electrons - tends to SHARE electrons (e.g., Carbon, valency = 4)

Examples:

• Sodium (2, 8, 1): 1 valence electron - loses 1 - valency = 1
• Oxygen (2, 6): 6 valence electrons - gains 2 - valency = 2
• Carbon (2, 4): 4 valence electrons - shares 4 - valency = 4
• Neon (2, 8): complete octet - valency = 0 (inert)

Section 8.9 A Deeper Look into Atomic Structure

8.9.1 Isotopes: Atoms of the same element with same atomic number (Z) but different mass numbers (A) - i.e., same protons, different neutrons.

Isotopes of Hydrogen:

1. Protium (¹₁H): 1 proton, 0 neutrons (~99.98%)
2. Deuterium (²₁H): 1 proton, 1 neutron (~0.015%)
3. Tritium (³₁H): 1 proton, 2 neutrons (trace amounts)

Isotopes of Carbon: ¹²₆C (most abundant), ¹³₆C, ¹⁴₆C

Properties of isotopes:

• Same chemical properties (same number of valence electrons, same electronic configuration)
• Different physical properties (different boiling and melting points)

Applications of Isotopes:

• ²³⁵₉₂U (Uranium-235): Fuel in nuclear reactors for electricity generation
• ⁶⁰₂₇Co (Cobalt-60): Radiation treatment for cancer
• ¹³¹₅₃I (Iodine-131): Treatment of goitre and thyroid cancer
• ¹⁴₆C (Carbon-14): Carbon dating to determine age of fossils and artefacts

Average Atomic Mass:

• Simple average ignores natural abundance of isotopes
• Weighted average atomic mass = sum of (mass of each isotope × its fractional abundance)

Example - Chlorine:

³⁵Cl (75%) and ³⁷Cl (25%). Weighted average = (35 × 75/100) + (37 × 25/100) = 26.25 + 9.25 = 35.5 u. This means individual chlorine atoms are either 35 u or 37 u, but the weighted average is 35.5 u.

8.9.2 Isobars: Atoms of different elements with the same mass number (A) but different atomic numbers (Z). Example: ⁴⁰₁₈Ar (Argon), ⁴⁰₁₉K (Potassium), ⁴⁰₂₀Ca (Calcium) - all have mass number 40 but different atomic numbers.

Important Formulae and Key Terms

Question 1. Who proposed the plum pudding model of the atom and what does it describe?
Answer: J. J. Thomson put forward the plum pudding model. This model shows the atom as a positively charged sphere with electrons (similar to plum seeds) spread throughout it, much like seeds inside a watermelon or raisins mixed into pudding.
In simple words: Thomson said atoms are like pudding - a positive jelly with tiny negative seeds (electrons) stuck inside it.

Exam Tip: Remember the name "plum pudding" and the fact that positive charge is spread throughout, not concentrated - this is the key difference from Rutherford's later nuclear model.

Question 2. What were the key observations of the gold foil experiment?
Answer: The majority of alpha particles went straight through without changing direction. Some particles were bent at steep angles, and a tiny fraction bounced back almost completely. These findings showed that atoms are mostly empty, with a small, dense, positively charged centre at the core.
In simple words: Most particles went through, some bounced off at angles, and a few came straight back - this told scientists atoms are mostly empty space with a hard nucleus in the middle.

Exam Tip: Always mention all three observations (straight through, deflected, bounced back) and link each to what it proves about atomic structure.

Question 3. Define atomic number and mass number.
Answer: Atomic number (Z) refers to the count of protons found in an atom's nucleus. Mass number (A) is the sum of all nucleons (protons and neutrons combined). To find neutrons, subtract the atomic number from the mass number - that is, A minus Z. Atomic number serves as the unique identifier for each element.
In simple words: Atomic number tells you how many protons an atom has. Mass number tells you the total weight-giving particles (protons plus neutrons). To find neutrons, just subtract one from the other.

Exam Tip: These are fixed definitions - write them exactly as stated. Always use the formula Neutrons = A - Z when solving numerical problems.

Question 4. What is the electronic configuration of sodium (atomic number 11)?
Answer: Sodium contains 11 electrons. By applying the 2n² rule, the K-shell can hold 2 electrons, the L-shell holds 8 electrons, and the remaining 1 electron sits in the M-shell. Therefore, the electronic configuration of sodium is 2, 8, 1. Since it has 1 electron in its outermost shell, its valency equals 1.
In simple words: Sodium has 11 electrons arranged in shells: 2 in the first, 8 in the second, and 1 in the third. The 1 electron in the last shell is the valency.

Exam Tip: Show your work using the 2n² rule step-by-step, then identify the outermost shell to find valency - examiners award marks for clear working.

Question 5. Define isotopes and give one example.
Answer: Isotopes are forms of the same element where atoms share identical atomic numbers but possess different mass numbers due to varying numbers of neutrons. Hydrogen illustrates this perfectly: it exists as three isotopes - protium (¹H), deuterium (²H), and tritium (³H) - each with atomic number 1 but increasing numbers of neutrons.
In simple words: Isotopes are copies of the same element with different weights because they have different numbers of neutrons, even though they have the same number of protons.

Exam Tip: Give the hydrogen isotopes as your example since they are the most commonly asked. Remember: same atomic number, different mass number.

Question 6. Define isobars and give one example.
Answer: Isobars are atoms belonging to different elements that share the same mass number but have different atomic numbers. A clear example is the trio of Argon (Z = 18), Potassium (Z = 19), and Calcium (Z = 20) - all three have a mass number of 40 and are isobars of one another.
In simple words: Isobars are atoms from different elements that weigh the same (same mass number) but have different numbers of protons.

Exam Tip: Contrast isobars with isotopes - isotopes are same element, different mass; isobars are different elements, same mass. This comparison helps fix the distinction.

Question 7. What is valency? How is it related to valence electrons?
Answer: Valency represents an atom's ability to combine with other atoms - it is the number of electrons that an atom either gains, loses, or shares so that it achieves an octet and reaches stability. The number of valence electrons (electrons sitting in the outermost shell) directly determines an atom's valency.
In simple words: Valency is how many electrons an atom wants to gain, lose, or share. It depends on the electrons in the outermost shell.

Exam Tip: Always connect valency to the outermost shell - this is the key link that examiners test. Give examples (sodium loses 1, oxygen gains 2) to show understanding.

Question 8. State the main limitation of Rutherford's atomic model.
Answer: Rutherford's model failed to account for why atoms remain stable. According to his theory, an electron moving in a circular orbit is always accelerating and should therefore shed energy, gradually spiral inward, and eventually crash into the nucleus. Yet atoms do not collapse - this contradiction could not be resolved within Rutherford's framework.
In simple words: Rutherford said electrons move around the nucleus like planets, but moving charged particles should lose energy and fall in. Atoms don't fall apart, so his model was incomplete.

Exam Tip: This is a frequent exam question. State the problem clearly: accelerating electron should lose energy, yet atoms are stable - Rutherford had no answer.

Question 9. How did Bohr's model overcome Rutherford's limitation?
Answer: Bohr introduced the concept that electrons move only in specific, fixed orbits termed stationary states. Within these stationary states, an electron retains its energy even as it orbits the nucleus - it does not radiate or lose power. This principle directly explained why atoms do not collapse, solving the stability problem Rutherford could not address.
In simple words: Bohr said electrons move in fixed orbits where they don't lose energy. This is why atoms stay together and don't fall apart.

Exam Tip: Emphasize the term "stationary states" and the fact that electrons do NOT lose energy in these orbits - this is the core of Bohr's solution.

Question 10. What is the maximum number of electrons that can occupy the K, L, and M shells?
Answer: Using the 2n² formula: K-shell (where n = 1) can hold 2 × 1² = 2 electrons; L-shell (where n = 2) can hold 2 × 2² = 8 electrons; and M-shell (where n = 3) can hold 2 × 3² = 18 electrons. However, in practice, the maximum capacity of any outermost shell is restricted to 8 electrons.
In simple words: K-shell holds 2 electrons, L-shell holds 8, and M-shell can theoretically hold 18, but the outermost shell never has more than 8.

Exam Tip: Apply the 2n² rule accurately and note the 8-electron cap for outer shells - many students forget this practical limit.

Question 11. Give four applications of isotopes used in science and medicine.
Answer: Uranium-235 (²³⁵U) serves as nuclear fuel in power reactors for electricity production. Cobalt-60 (⁶⁰Co) is utilized in radiation therapy to treat cancer. Iodine-131 (¹³¹I) is used to treat goitre and thyroid cancers. Carbon-14 (¹⁴C) enables carbon dating, allowing scientists to establish the age of fossils and historical artifacts.
In simple words: U-235 makes nuclear power. Co-60 treats cancer with radiation. I-131 cures thyroid disease. C-14 helps date old objects.

Exam Tip: Write out all four with their symbols and uses - this is a factual list that demands accuracy. Do not skip any application.

Question 12. Why do isotopes have identical chemical properties?
Answer: Isotopes possess the same count of electrons and maintain identical electron arrangements around the nucleus. Chemical behaviour is governed primarily by the electrons in the outermost layer, and since these are identical across all isotopes of an element, they all exhibit the same chemical responses and interactions.
In simple words: All isotopes of an element have the same number of electrons and electron arrangement, so they behave the same way chemically.

Exam Tip: Stress that chemistry depends on electrons (especially valence electrons), not on neutrons - this is why isotopes behave identically in chemical reactions.

Question 13. Who discovered the neutron and when? Why was this discovery important?
Answer: James Chadwick made the neutron discovery in 1932. This breakthrough was vital because it resolved a fundamental puzzle: helium, with only 2 protons, has approximately four times the mass of hydrogen, which has 1 proton. The missing mass is explained by the presence of neutrons. Without the neutron's discovery, scientists could not explain why atomic masses were so much larger than proton numbers alone would suggest.
In simple words: Chadwick found the neutron in 1932. This solved the mystery of why atoms weigh more than just their protons can explain.

Exam Tip: Always include the year 1932 and explain the helium-hydrogen mass discrepancy - this shows why the discovery mattered.

Question 14. Write the standard notation for an atom with atomic number 6 and mass number 12.
Answer: Standard notation positions the mass number (A) as a superscript above and the atomic number (Z) as a subscript below the element's symbol. For carbon (symbol C), with Z = 6 and A = 12, this notation reads ¹²₆C. The neutron count is found by subtracting: 12 - 6 = 6 neutrons. Its electron arrangement is 2, 4.
In simple words: The mass number (12) goes on top, the atomic number (6) goes on the bottom, and the element symbol (C) is in the middle. This carbon has 6 neutrons and 4 electrons in its outer shell.

Exam Tip: Always show the notation clearly with both superscript and subscript placed correctly - presentation matters. Then calculate neutrons and configuration step-by-step.

Question 15. State the IUPAC rules for writing chemical symbols of elements.
Answer: The opening letter of a symbol is invariably uppercase, and if a second letter exists, it is invariably lowercase. Most symbols derive from the first one or two letters of the English name of the element (examples: H for hydrogen, He for helium, Al for aluminium). Some symbols come from names in Latin, Greek, or German languages (examples: Fe comes from the Latin "Ferrum" for iron, Na from the Latin "Natrium" for sodium, Hg from the Greek "Hydrargyros" for mercury).
In simple words: First letter is always big, second letter (if any) is always small. Most come from English names, but some come from Latin or Greek.

Exam Tip: Know the Latin-origin symbols (Fe, Na, Hg, Au, Ag, Sn, Pb, Cu) since these appear in exams - simply memorizing them is faster than trying to derive them.

Short Answer Type Questions

Question 1. Describe the gold foil experiment. What were the three key observations and what did each prove?
Answer: In 1911, Geiger and Marsden, working in Rutherford's laboratory, aimed a stream of alpha particles at a very thin gold foil. The first observation was that the vast bulk of particles travelled in a straight path - proving the atom is largely empty. The second observation was that certain particles bent sharply to the side - proving that positive charge is concentrated in one location rather than spread out. The third observation was that a tiny fraction of particles reversed direction completely - proving that an extremely compact, extremely dense, positively charged centre exists at the core of the atom. Taken as a whole, these three observations completely disproved Thomson's plum pudding concept, which predicted no sharp deflections.
In simple words: Most particles went straight (atoms are mostly empty). Some bent sharply (charge is concentrated). A few bounced back (there's a hard nucleus). Thomson's pudding model was wrong.

Exam Tip: Present each of the three observations separately with its conclusion - do not combine them. Explicitly state that Thomson's model predicted something different and was therefore disproven.

Question 2. Compare Thomson's model and Rutherford's model of the atom. How did the gold foil experiment disprove Thomson's model?
Answer: Thomson's model depicted the positive charge as distributed uniformly throughout the entire atom, with electrons embedded within - similar to raisins scattered through dough. This model contained no central nucleus. In contrast, Rutherford's picture showed the atom as predominantly empty, with all positive charge packed into a small, dense core at the centre, and electrons orbiting around it like planets orbiting the sun. The gold foil results contradicted Thomson because they revealed that some alpha particles underwent extreme bending or bounced backward - this could not happen if the positive charge were evenly dispersed. A concentrated nucleus was necessary to exert the powerful forces that would bend or repel the alpha particles so forcefully.
In simple words: Thomson said positive charge fills the whole atom. Rutherford said it's all crammed in the middle. The gold foil showed some particles bent sharply, which only happens if there's a hard nucleus in the centre.

Exam Tip: Draw a simple diagram comparing the two models - Thomson's smooth pudding vs. Rutherford's nuclear model. Then explain why gold foil results fit Rutherford but not Thomson.

Question 3. Explain why atoms are electrically neutral. Give one example to support your answer.
Answer: Atoms achieve electrical neutrality because the quantity of positively charged protons within the nucleus is exactly matched by the quantity of negatively charged electrons surrounding it. These opposite charges offset one another completely, producing a net charge of zero. Consider sodium: it has an atomic number of 11, meaning it contains 11 protons (each with a +1 charge) and 11 electrons (each with a -1 charge). The calculation 11 minus 11 equals 0, confirming zero net charge. This equality holds true for all atoms - protons balance electrons - and this is why every atom remains electrically neutral.
In simple words: Atoms have the same number of positive protons and negative electrons, so the charges cancel out and the atom is neutral.

Exam Tip: Show the balance explicitly (protons = electrons = atomic number) and give a numerical example with a specific element.

Question 4. Distinguish between isotopes and isobars with two examples of each.
Answer: Isotopes are atoms of a single element possessing an identical atomic number (Z) but varying mass numbers (A) as a result of differing neutron counts. Their chemical responses remain identical; however, their physical traits diverge. Two isotope examples: protium (¹H), deuterium (²H), and tritium (³H) are all hydrogen with Z = 1; carbon-12 (¹²C), carbon-13 (¹³C), and carbon-14 (¹⁴C) are all carbon with Z = 6. Isobars are atoms of differing elements that share an identical mass number (A) but possess different atomic numbers (Z). Two isobar examples: argon-40 (⁴⁰Ar with Z = 18), potassium-40 (⁴⁰K with Z = 19), and calcium-40 (⁴⁰Ca with Z = 20) all have A = 40; carbon-14 (¹⁴C with Z = 6) and nitrogen-14 (¹⁴N with Z = 7) both have A = 14.
In simple words: Isotopes are the same element with different weights (different neutrons). Isobars are different elements with the same weight (different protons).

Exam Tip: Always list isotope examples with the same Z and isobar examples with the same A - write out the symbols with superscripts and subscripts for clarity.

Question 5. Write the electronic configuration of chlorine (Z=17). Determine its valency and explain how you arrived at it.
Answer: Chlorine possesses 17 electrons. When we fill the shells using the 2n² rule: the K-shell receives 2 electrons and becomes full; the L-shell receives 8 electrons and becomes full; the M-shell gets the remaining 7 electrons. The electronic configuration reads 2, 8, 7. Chlorine has 7 valence electrons in its outermost (M) shell. Since this exceeds 4, chlorine tends to accept electrons rather than shed them, so as to complete its octet. It requires just 1 additional electron to reach 8 (a stable octet). Therefore, chlorine's valency is 1.
In simple words: Chlorine has 17 electrons: 2, 8, 7. Its outer shell has 7 electrons. It needs to gain 1 more to have 8, so its valency is 1.

Exam Tip: Show each shell filling in order (K, L, M), identify the outermost electrons, then determine whether the atom gains or loses electrons to complete its octet - this logic is essential.

Question 6. Explain the concept of weighted average atomic mass using chlorine as an example.
Answer: In nature, chlorine appears in two forms - chlorine-35 (³⁵Cl) with an atomic mass of 35 u occurring at 75% frequency, and chlorine-37 (³⁷Cl) with an atomic mass of 37 u occurring at 25% frequency. If we calculate a simple mean, (35 + 37) / 2 equals 36 u, but this is incorrect because the two isotopes are not equally abundant. The correct weighted average atomic mass considers their actual proportions: (35 × 75/100) + (37 × 25/100) = 26.25 + 9.25 = 35.5 u. This value means that in the real world, no individual chlorine atom weighs exactly 35.5 u - each one is either 35 or 37 - but when you weigh a large sample of chlorine, the average works out to 35.5 u.
In simple words: Chlorine comes as two types - the lighter and the heavier. We can't just average them equally. We must weight them by how common each type is. That gives us 35.5 u as the average.

Exam Tip: Work through the calculation step-by-step and emphasize that the weighted average does not mean individual atoms have that mass - it is the average of the whole sample.

Question 7. What is valency? Determine the valency of sodium, oxygen, and carbon from their electronic configurations.
Answer: Valency represents the number of electrons an atom must acquire, release, or share in order to fill its outer shell and reach stability. For sodium (Z = 11), the electronic configuration is 2, 8, 1. Having 1 valence electron in its outer layer (which is less than 4), sodium releases this 1 electron to achieve a complete 8-electron configuration. Its valency is 1. For oxygen (Z = 8), the configuration is 2, 6. With 6 valence electrons (more than 4), oxygen accepts 2 electrons to bring its outer shell to 8. Its valency is 2. For carbon (Z = 6), the configuration is 2, 4. With exactly 4 valence electrons, it is neither convenient to gain 4 nor shed 4, so carbon shares 4 electrons with neighbouring atoms. Its valency is 4.
In simple words: Sodium has 1 electron in its outer shell - it loses it (valency 1). Oxygen has 6 - it gains 2 (valency 2). Carbon has 4 - it shares 4 (valency 4).

Exam Tip: Always show the configuration, count the valence electrons, then decide whether the atom loses (less than 4), gains (more than 4), or shares (equals 4) electrons.

Question 8. Why did scientists need to propose new atomic models repeatedly? What drove the evolution from Dalton to Bohr?
Answer: Every atomic model underwent revision when new experimental results contradicted or could not be explained by the current model. Dalton claimed atoms were indivisible - however, Thomson's cathode ray study revealed that electrons exist inside atoms, contradicting the indivisibility claim. Thomson's plum pudding picture could not account for the results of the gold foil test, especially the pronounced deflections of alpha particles. Rutherford's nuclear design accurately identified the nucleus but failed to explain how atoms could be stable, since an electron undergoing acceleration should lose its energy and fall inward. Bohr resolved the stability problem by suggesting that electrons inhabit fixed energy shells in which they keep their energy without radiating. This is the nature of scientific advancement: fresh investigations uncover new facts, those facts reveal shortcomings in the prevailing theory, and an improved model addresses those gaps. This cycle has repeated throughout the history of atomic theory.
In simple words: Each new experiment showed that the old model was wrong. Thomson disproved Dalton. Gold foil disproved Thomson. Rutherford had a stability problem. Bohr solved it. Science improves by fixing what experiments prove wrong.

Exam Tip: This is a theory question - show the progression of models and explain what each new experiment revealed and what problem it solved.

Question 9. An atom has atomic number 15 and mass number 31. Write its (a) electronic configuration, (b) number of neutrons, (c) valency and (d) name of the element.
Answer: (a) Electronic configuration: Since Z = 15, the atom has 15 electrons. Using the 2n² rule: K = 2, L = 8, M = 5. Configuration: 2, 8, 5. (b) Number of neutrons: Apply the formula Neutrons = Mass number minus Atomic number = 31 minus 15 = 16 neutrons. (c) Valency: The atom possesses 5 valence electrons in its outer shell. Since 5 exceeds 4, the atom tends to absorb electrons to achieve an octet. It requires 3 additional electrons to reach 8. Valency = 3. (d) Name: An element with Z = 15 is phosphorus (P), as shown in the periodic table (Table 8.2).
In simple words: This atom has 15 electrons arranged as 2, 8, 5. It has 16 neutrons. Its outer shell has 5 electrons, so it gains 3 to make 8 - valency is 3. It's phosphorus.

Exam Tip: Work through each part methodically - configuration first using 2n², then neutrons using A minus Z, then valency from the outer shell, then identify the element.

Question 10. Explain why the concept of atom originated as an imaginary idea but Dalton's theory was considered scientific. What made Dalton's contribution more significant than Kanada's?
Answer: Acharya Kanada's parmanu concept and Democritus's atomos notion both emerged from philosophical speculation and logical thinking about what matter might be made of at its most basic level - they had no backing from tests or measurements. These were intuitive, creative ideas based purely on reasoning. In contrast, John Dalton's atomic theory (formulated in 1808) rested upon methodical laboratory investigations of chemical reactions - most significantly, the principles of conservation of mass, constant ratios, and multiple proportions. His framework could be put to the test in the lab and could forecast how chemicals would interact. It was this experimental grounding that makes Dalton's work genuinely scientific. Science distinguishes itself by demanding proof through measurement and testing; philosophy works through argument and contemplation. Both have merit, yet only facts from experiments can establish something as scientifically valid.
In simple words: Kanada and Democritus just guessed at atoms using thinking alone. Dalton measured real chemicals and found rules that proved atoms exist. Real science needs proof from tests, not just ideas.

Exam Tip: Clearly separate philosophical thinking (no proof needed) from scientific theory (needs experimental evidence). This distinction is crucial for understanding the evolution of atomic theory.

Long Answer Type Questions

Question 1. Describe the gold foil experiment. What were the observations and what conclusions did Rutherford draw? How did this disprove Thomson's model?
Answer: During 1911, Geiger and Marsden, operating under the direction of Rutherford, fired a concentrated beam of positively charged alpha particles onto a layer of gold foil that was extremely thin. Based on Thomson's plum pudding model (which proposed that positive charge is spread evenly across the atom), they anticipated that every alpha particle would pass straight through with only slight or no bending.

Actual observations:
- The overwhelming majority of alpha particles continued in a straight path - demonstrating that atoms are chiefly empty space
- A number of particles were sharply deflected - showing that positive charge is concentrated in one spot, not scattered throughout
- A very small number of particles reversed almost completely - proving there exists an extremely tiny, incredibly dense, positively charged core

Rutherford's conclusion was that atoms contain a small, dense, positively charged core at the centre, with electrons orbiting in the mostly empty space around it. Thomson's theory was disproven by the results because it predicted that no sharp deflections would occur - yet the experiment clearly showed many particles being strongly bent away from their original paths, something only a concentrated nucleus could accomplish.
In simple words: Scientists shot tiny particles at gold foil. Most went straight through (atoms are mostly empty). Some bent sharply (there's a hard centre). Thomson said the centre was fuzzy - wrong. Rutherford said there's a hard nucleus - right.

Exam Tip: Lay out the setup, the three observations, the conclusions, and the disproof of Thomson clearly and separately - this is a standard long-answer, and organization matters for marks.

Question 2. State Bohr's model of the atom. How did it overcome Rutherford's limitation? What are the rules for electron distribution in shells?
Answer: Niels Bohr proposed in 1913 that electrons orbit the nucleus in specific, unchanging circular paths termed stationary states or shells (named K, L, M, N). While occupying a fixed shell, electrons retain their energy - they do not radiate it away - and this principle directly addressed Rutherford's unsolved problem, in which an accelerating electron was supposed to shed energy and spiral down into the nucleus.

Energy becomes larger as distance from the nucleus increases. Electrons move between shells only when they absorb or emit precise amounts of energy matching the gap between the two energy levels involved.

Electron distribution rules (Bohr-Bury):
- Maximum electrons per shell = 2n²
- The outermost shell can accommodate at most 8 electrons
- Shells populate in the sequence K, L, M, N from inner to outer

These guidelines permit the determination of electronic configurations and valency for all elements.
In simple words: Bohr said electrons move in fixed shells where they don't lose energy. This fixed them (no spiraling in). Outer shell never has more than 8. We fill from inside out.

Exam Tip: Begin with Bohr's key idea (fixed shells, no energy loss), then state how it solved Rutherford's stability issue, then list the distribution rules clearly with examples.

Question 3. Define atomic number and mass number. How are protons, neutrons and electrons calculated from these? Solve: An atom has Z = 17 and A = 35.
Answer: Atomic Number (Z) is the total of protons present in an atom's nucleus. This number uniquely determines which element the atom is. For a neutral atom, the electron count equals the proton count, so electrons = Z. Mass Number (A) represents the entire collection of nucleons (protons plus neutrons combined). Since protons and neutrons have virtually identical masses and electrons are nearly weightless, the total atomic mass is roughly equivalent to A. To calculate neutrons, subtract Z from A: Neutrons = A minus Z. The conventional notation puts A as a superscript and Z as a subscript alongside the element's symbol.

For the atom with Z = 17, A = 35:
- Protons = 17
- Electrons = 17 (in a neutral atom)
- Neutrons = 35 minus 17 = 18
- Electronic configuration: K = 2, L = 8, M = 7 (i.e., 2, 8, 7)
- Valence electrons = 7, so Valency = 1
- Identification: This is Chlorine (Cl)
In simple words: Z tells you protons and electrons. A tells you total weight. Subtract Z from A to get neutrons. For this atom, it's chlorine with 17 protons, 17 electrons, and 18 neutrons.

Exam Tip: Always set up the problem with clear definitions, then apply the formulas step-by-step. Show calculations for neutrons, configuration, and element identification.

Question 4. What are isotopes? Explain why they have the same chemical properties but different physical properties. Give four applications of isotopes in daily life.
Answer: Isotopes are different varieties of the same element that have matching atomic numbers (identical numbers of protons) but distinct mass numbers, resulting from variations in neutron count.

Same chemical properties: Isotopes possess identical numbers of electrons and matching electron arrangements in their shells. Since an atom's behaviour in chemical reactions depends fundamentally on its valence electrons (electrons in the outer shell), and these are the same for all isotopes, all varieties of an element perform identically in reactions with other substances.

Different physical properties: Physical characteristics such as the temperature at which a substance changes state (boiling and melting points) and density are dependent on the mass of the atoms. Since isotopes carry different numbers of neutrons, they have different total masses, and consequently their physical properties vary.

Applications of isotopes:
1. Uranium-235 (²³⁵U) - functions as nuclear fuel in power plants to manufacture electricity
2. Cobalt-60 (⁶⁰Co) - employed in radiation therapy to eliminate cancer cells
3. Iodine-131 (¹³¹I) - utilized for treating goitre and thyroid malignancies
4. Carbon-14 (¹⁴C) - applied in archaeological carbon dating to establish the age of fossils and ancient objects
In simple words: Isotopes are heavier and lighter versions of the same element. They react the same way because they have the same electrons. They have different melting/boiling points because they weigh differently.

Exam Tip: Separate the explanation into three parts: definition, why they are chemically identical, why they are physically different. List all four applications with their uses.

Question 5. What is valency? Explain using electronic configuration how the valency of sodium, oxygen and carbon is determined. Why are noble gases like neon and argon unreactive?
Answer: Valency refers to the combining capacity of an atom - the count of electrons it must obtain, shed, or share in order to reach a stable electron configuration (a full octet, or a pair of electrons for hydrogen and helium). Sodium (Z = 11) possesses an electronic configuration of 2, 8, 1. It has a single electron in its outer shell. Since this is fewer than 4, sodium surrenders 1 electron to complete the 8-electron shell below it. Thus, Valency of Sodium = 1. Oxygen (Z = 8) has an electronic configuration of 2, 6. It possesses 6 electrons in its outer layer. Since this exceeds 4, oxygen collects 2 electrons to complete its octet. Thus, Valency of Oxygen = 2. Carbon (Z = 6) displays an electronic configuration of 2, 4. It has precisely 4 electrons in its outer shell - not easy to obtain 4 nor simple to release 4. Instead, it transfers 4 electrons with adjacent atoms. Thus, Valency of Carbon = 4.

Noble gases (Neon: Z = 10, config 2, 8; Argon: Z = 18, config 2, 8, 8) already possess filled octets in their outer shells. They require no extra electrons, nor do they need to give any up or exchange any to maintain stability. Consequently, they experience minimal chemical activity and are assigned a valency of 0 (or are considered to have zero combining power).
In simple words: Valency is how many electrons an atom gives, takes, or shares. Sodium gives 1 (valency 1). Oxygen takes 2 (valency 2). Carbon shares 4 (valency 4). Noble gases already have full shells, so they don't react (valency 0).

Exam Tip: Work through each element's configuration, identify the valence electrons, then show the logic for why it gains, loses, or shares. Use the octet rule as the reasoning backbone.

Quick Revision for Exam Day

• Know all four atomic models - Dalton, Thomson, Rutherford, Bohr - with their characteristics and shortcomings
• Gold foil experiment: 3 observations + 3 findings + why Thomson's model was incorrect
• Subatomic particles: electron (−1 charge), proton (+1 charge), neutron (0 charge) - locations within the atom
• Z = atomic number = proton count = electron count (in neutral atoms)
• A = mass number = proton count + neutron count; Neutrons = A minus Z
• Electronic configuration: 2n² formula; outermost shell max = 8; fill K then L then M in sequence
• Know electronic configurations of the first 18 elements
• Valency = electrons gained (if more than 4 valence electrons) or released (if fewer than 4 valence electrons) or exchanged (if exactly 4)
• Isotopes: same atomic number, different mass number; same chemical reactions; different physical traits
• Isobars: same mass number, different atomic number; different elements
• Isotope applications: U-235 (nuclear power), Co-60 (cancer therapy), I-131 (thyroid treatment), C-14 (archaeological dating)
• Weighted average atomic mass = total of (mass multiplied by fractional abundance)
• IUPAC symbol rules: opening letter uppercase, second letter lowercase; Latin origin symbols must be memorized
• Bohr's shells: K, L, M, N = n = 1, 2, 3, 4; energy rises with distance from nucleus
• Atom diameter is approximately 10⁻¹⁰ m; Nucleus diameter is approximately 10⁻¹⁵ m; nucleus is 10⁵ times smaller

Evolution of Atomic Models - Quick Timeline