Navigating through the newly revised CBSE Class 9 Science curriculum (Exploration) requires a thorough physical and chemical understanding of subatomic particles (electrons, protons, and neutrons), historical atomic models (J.J. Thomson’s Plum Pudding model, Rutherford’s Alpha-Particle Scattering Experiment, and Niels Bohr’s Quantized Energy Level model), the Bohr-Bury scheme for electron distribution across K, L, M, N shells (2n² rule and octet stability), the physical definitions of Atomic Number (Z) and Mass Number (A), valence electrons, chemical valency, fractional average atomic masses, and the industrial applications of isotopes and isobars. Chapter 8 of Class 9 Chemistry, “Journey Inside the Atom”, establishes the quantum foundation of all chemical matter. It investigates why atoms are electrically neutral despite containing charged subatomic particles; explains why Rutherford’s planetary model predicted atomic collapse under classical electrodynamics and how Bohr resolved this crisis through non-radiating stationary orbits; details the electron arrangement of the first 20 elements; and provides the mathematical formulation for calculating isotope abundances (such as chlorine-35 and chlorine-37). To help students master every aspect of this high-weightage chapter, this comprehensive solutions guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE Class 9 evaluation standards.
Every question presented in the official NCERT textbook—ranging from all introductory “Think It Over” sections and in-text “Pause and Ponder” prompts (Pages 143, 145, 149, and 150) to the complete end-of-chapter “Revise, Reflect, Refine” exercises (Questions 1 to 12 on Pages 158–160)—has been solved with exhaustive detail. Numerical problems and electronic configurations follow a step-by-step box format with explicit shell distributions and percentage calculations using clean plain-text symbols without raw LaTeX tags. Key scoring terms, official CBSE exam tags, and dynamic summary tables have been highlighted to ensure students secure maximum marks in their examinations.
Master Concept & Comparative Summary Tables
1. Fundamental Subatomic Particles Comparison Sheet
| Subatomic Particle | Discovered By (Year) | Nature of Electric Charge | Absolute Charge (Coulombs) | Absolute Mass (kg) | Relative Mass (Unified mass, u) | Location Within the Atom |
|---|---|---|---|---|---|---|
| Electron (e^-) | J.J. Thomson (1897) | Negative (-1) | -1.602 × 10⁻¹⁹ C | 9.109 × 10⁻³¹ kg | 1/1840 u (Negligible) | Revolving in discrete Energy Shells (Orbits) outside the nucleus. |
| Proton (p^+) | E. Goldstein / E. Rutherford (1886/1919) | Positive (+1) | +1.602 × 10⁻¹⁹ C | 1.672 × 10⁻²⁷ kg | 1.007 u ≈ 1 u | Packed inside the central Dense Nucleus. |
| Neutron (n^0) | James Chadwick (1932) | Neutral (0) | 0 C (Zero charge) | 1.675 × 10⁻²⁷ kg | 1.008 u ≈ 1 u | Packed inside the central Dense Nucleus (except ordinary Hydrogen). |
2. Historical Evolution of Atomic Models
| Atomic Model | Proponent & Experiment | Core Structural Postulate | Major Limitation / Reason for Failure |
|---|---|---|---|
| Dalton’s Atomic Theory | John Dalton (1808) | Atom is an indivisible, hard, solid sphere that cannot be created or destroyed. | Failed to explain the discovery of subatomic particles (electrons, protons) and isotopes. |
| Plum Pudding / Watermelon Model | J.J. Thomson (1898) | Atom is a sphere of uniform positive charge with negative electrons embedded inside like seeds in a watermelon. | Failed to explain the large-angle deflection of alpha particles in Rutherford’s scattering experiment. |
| Nuclear Planetary Model | Ernest Rutherford (1911) (Gold Foil Experiment) | Atom has a tiny, dense, positively charged nucleus at the center containing all mass; electrons orbit around it like planets. | According to Maxwell’s electrodynamics, accelerating electrons must radiate energy continuously, spiral inward, and collapse into the nucleus. |
| Quantized Energy Shell Model | Niels Bohr (1913) | Electrons revolve only in discrete, non-radiating stationary orbits (K, L, M, N shells) with fixed quantized energy levels. | Successfully explained atomic stability and spectral lines; forms the modern basis of Class 9 chemistry. |
3. Electronic Configuration & Valency Sheet (First 18 Elements)
| Atomic No. (Z) | Element Name | Chemical Symbol | Protons (p) | Neutrons (n) | Electrons (e) | Shell Distribution: K (2), L (8), M (8), N (18) | Valence Electrons | Valency |
|---|---|---|---|---|---|---|---|---|
| 1 | Hydrogen | H | 1 | 0 | 1 | 1 | 1 | 1 |
| 2 | Helium | He | 2 | 2 | 2 | 2 (Duplet complete) | 2 | 0 (Inert) |
| 3 | Lithium | Li | 3 | 4 | 3 | 2, 1 | 1 | 1 |
| 4 | Beryllium | Be | 4 | 5 | 4 | 2, 2 | 2 | 2 |
| 5 | Boron | B | 5 | 6 | 5 | 2, 3 | 3 | 3 |
| 6 | Carbon | C | 6 | 6 | 6 | 2, 4 | 4 | 4 |
| 7 | Nitrogen | N | 7 | 7 | 7 | 2, 5 | 5 | 8 - 5 = 3 |
| 8 | Oxygen | O | 8 | 8 | 8 | 2, 6 | 6 | 8 - 6 = 2 |
| 9 | Fluorine | F | 9 | 10 | 9 | 2, 7 | 7 | 8 - 7 = 1 |
| 10 | Neon | Ne | 10 | 10 | 10 | 2, 8 (Octet complete) | 8 | 0 (Inert) |
| 11 | Sodium | Na | 11 | 12 | 11 | 2, 8, 1 | 1 | 1 |
| 12 | Magnesium | Mg | 12 | 12 | 12 | 2, 8, 2 | 2 | 2 |
| 13 | Aluminium | Al | 13 | 14 | 13 | 2, 8, 3 | 3 | 3 |
| 14 | Silicon | Si | 14 | 14 | 14 | 2, 8, 4 | 4 | 4 |
| 15 | Phosphorus | P | 15 | 16 | 15 | 2, 8, 5 | 5 | 8 - 5 = 3, 5 |
| 16 | Sulphur | S | 16 | 16 | 16 | 2, 8, 6 | 6 | 8 - 6 = 2 |
| 17 | Chlorine | Cl | 17 | 18 | 17 | 2, 8, 7 | 7 | 8 - 7 = 1 |
| 18 | Argon | Ar | 18 | 22 | 18 | 2, 8, 8 (Octet complete) | 8 | 0 (Inert) |
NCERT In-Text Questions: “Think It Over”
Page No. 140: Think It Over (Questions 1 to 3)
Question 1 Are atoms really the smallest indivisible particles of matter as Dalton proposed? [Exam Favorite]
Answer: No, atoms are NOT indivisible particles.
- Explanation: While John Dalton’s 1808 atomic theory stated that atoms are indivisible building blocks of matter, late 19th and early 20th-century experimental discoveries proved that atoms are composed of even smaller subatomic particles:
- Electrons (discovered by J.J. Thomson in 1897 via cathode rays),
- Protons (discovered by Goldstein/Rutherford via canal rays),
- Neutrons (discovered by James Chadwick in 1932).
- Therefore, an atom can be divided into its subatomic constituent particles.
Question 2 Why did scientists continuously propose and modify atomic models across history? [Exam Favorite]
Answer: Scientists continuously modified atomic models because science is an evidence-driven, iterative process:
- Whenever new experimental observations could not be explained by an existing model, the model had to be refined or replaced.
- For example, Thomson’s plum pudding model explained electrical neutrality but failed to explain the large-angle scattering of alpha particles in Rutherford’s gold foil experiment.
- Subsequently, Rutherford’s nuclear model could not explain atomic stability under classical electrodynamics, which prompted Niels Bohr to introduce quantized, non-radiating electron energy levels.
Question 3 Why do electrons not collapse into the positively charged nucleus despite electrostatic attraction? [Exam Favorite]
Answer: Electrons do not collapse into the nucleus because of Niels Bohr’s Quantization Postulate:
- Electrons revolve around the nucleus only in certain special, discrete non-radiating orbits called stationary energy shells (K, L, M, N).
- While revolving within these designated orbits, electrons do not radiate or lose electromagnetic energy.
- Because their kinetic energy is conserved in these stable orbits, they maintain a stable distance from the nucleus without spiraling inward.
NCERT In-Text Questions: “Pause and Ponder”
Page No. 143: Pause and Ponder (Question 1)
Question 1 Suppose you made up your own ‘atom’, as Thomson described, using clay for the positive charge sphere and mustard seeds for electrons. If the positive charge on the clay is less than the total negative charge of the seeds, what would happen to the atom’s electrical neutrality? [Exam Favorite]
Answer:
- Effect on Neutrality: The atom would lose its electrical neutrality and become a Negatively Charged Ion (Anion).
- Scientific Reasoning: In a neutral Thomson atom, the total magnitude of the positive charge distributed across the sphere must be strictly equal to the total negative charge of all embedded electrons. If the positive charge is less than the negative charge, the net electric charge is non-zero (negative), causing the structure to behave as an anion.
Page No. 145: Pause and Ponder (Question 2)
Question 2 Why did J.J. Thomson conclude that electrons are present in all atoms regardless of the gas used in the cathode ray tube? [Exam Favorite]
Answer: J.J. Thomson observed that the charge-to-mass ratio (e/m) and physical properties of cathode rays were completely identical and independent of:
- The chemical nature of the gas enclosed inside the discharge tube (whether Hydrogen, Helium, Air, or Nitrogen), and
- The metal material used to make the cathode electrodes.
Since the emitted particles were identical under all experimental conditions, Thomson concluded that electrons are universal, fundamental subatomic constituents present in all atoms of all elements.
Page No. 149: Pause and Ponder (Question 3)
Question 3 In Rutherford’s gold foil experiment, what would happen if alpha particles (positively charged) were replaced by negatively charged particles (like fast-moving beta particles or electrons)? [Exam Favorite]
Answer: If negatively charged particles were used instead of alpha particles:
- Electrostatic Attraction Instead of Repulsion: Because the nucleus is positively charged, negatively charged particles approaching the center would experience an inward electrostatic attraction rather than a repulsive force.
- Deflection Towards the Nucleus: The particles would be pulled toward and captured by the nucleus or deflected sharply inward toward the center, rather than being repelled backward at large obtuse angles.
- No Rebounding at 180°: There would be no 180° backward rebounding because the positive nucleus attracts negative charges rather than pushing them away.
Page No. 150: Pause and Ponder (Question 4)
Question 4 Explain how Bohr’s model of the atom resolved the major objection raised against Rutherford’s atomic model. [Exam Favorite]
Answer:
- Rutherford’s Fatal Flaw: According to classical electromagnetic theory, an electron revolving in a circular orbit undergoes continuous centripetal acceleration. An accelerating charged particle must continuously radiate electromagnetic energy. Losing energy, the electron would slow down, spiral inward in a fraction of a microsecond (10⁻⁸\text{ s}), and crash into the nucleus, making all matter inherently unstable.
- Bohr’s Resolution: Niels Bohr postulated that:
- Electrons revolve only in certain discrete, non-radiating orbits (stationary shells: K, L, M, N).
- While orbiting within these designated paths, an electron does not emit any radiation.
- Energy is absorbed or emitted only when an electron jumps from one discrete orbit to another. Hence, the atom remains completely stable.
NCERT Chapter-End Exercises: “Revise, Reflect, Refine” (Pages 158–160)
Question 1 Choose the correct options and explain the reason for the correct and incorrect options in the following statements regarding Bohr’s atomic model: [Exam Favorite] (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:
- Correct Statement: (iii) Electrons revolve around the nucleus in orbits of fixed energy without losing energy. (Reason: This is the fundamental postulate of Bohr’s model. Electrons in stationary discrete orbits have quantized, fixed energy levels and do not radiate energy).
- Incorrect Statements Explained:
- (i) is Incorrect: Electrons in stationary orbits do not lose energy; hence, they never fall into the nucleus.
- (ii) is Incorrect: Electrons cannot exist arbitrarily anywhere; they are strictly confined to quantized orbits of specific radii and energies.
- (iv) is Incorrect: Electrons can never exist in between energy levels; they make instantaneous quantum jumps between allowed discrete shells by absorbing or emitting exact quanta of energy.
Question 2 Could an orange or a lemon be used as an analogical model for Thomson’s atomic model? Explain the similarities and limitations. [Exam Favorite]
Answer:
- Similarities (Why it works as an analogy):
- In an orange or lemon, the pulp represents the uniform sphere of positive charge.
- The embedded seeds distributed inside represent the negatively charged electrons held within the positive mass, similar to Thomson’s watermelon/plum pudding concept.
- Limitations of the Analogy:
- In an actual atom, electrons are extremely light (1/1840\text{ u}) subatomic particles with dynamic electromagnetic properties, not hard rigid seeds.
- The analogy implies positive charge is a solid tangible substance, whereas in reality, positive charge resides inside a concentrated, microscopic central nucleus.
Question 3 What conclusion did Rutherford draw about the position and characteristics of the atom’s positively charged part based on the observation that 1 in 12,000 alpha particles bounced back at 180° in the gold foil experiment? [Exam Favorite]
Answer: Rutherford drew three historic conclusions:
- Extremely Small Size: Since only a tiny fraction (1 in 12,000) was deflected by 180°, the positively charged part occupies an extremely minute fraction of the total volume of the atom (Nucleus radius \approx 10⁻¹⁵\text{ m} vs Atom radius \approx 10⁻¹⁰\text{ m}, a factor of 10⁵ smaller).
- Central Position: The entire positive charge of the atom is concentrated at the very center of the atom, in a region he named the Nucleus.
- Mass Concentration: Almost the entire mass of the atom is concentrated inside this tiny central nucleus, providing the rigid inertia needed to deflect heavy alpha particles backward.
Question 4 Arrange the following atomic models in the correct chronological order of their historical evolution: (a) Bohr’s model (b) Thomson’s Plum Pudding model (c) Rutherford’s Nuclear model (d) Dalton’s Indivisible Sphere model [Exam Favorite]
Answer: The correct chronological evolutionary sequence is: Dalton’s Model (1808) → Thomson’s Model (1898) → Rutherford’s Model (1911) → Bohr’s Model (1913). (Order: d → b → c → a).
Question 5 State the rules given by Bohr and Bury for the distribution of electrons in different energy shells of an atom. [Exam Favorite]
Answer: The Bohr-Bury scheme is governed by three fundamental rules:
- Maximum Capacity Formula (2n² Rule): The maximum number of electrons that can be accommodated in a given shell is given by the formula 2n², where n is the principal quantum shell number:
- For K-Shell (n=1): Maximum electrons = 2 × (1)² = 2 electrons.
- For L-Shell (n=2): Maximum electrons = 2 × (2)² = 8 electrons.
- For M-Shell (n=3): Maximum electrons = 2 × (3)² = 18 electrons.
- For N-Shell (n=4): Maximum electrons = 2 × (4)² = 32 electrons.
- Octet Rule for Outermost Shell: The maximum number of electrons that can be accommodated in the outermost valence shell is strictly 8 electrons, regardless of the shell’s theoretical capacity.
- Stepwise Inward-to-Outward Filling: Electrons are not accommodated in a given shell unless the inner shells are completely filled (stepwise progressive filling).
Question 6 Write the electronic configuration, number of valence electrons, and valency of the following elements: (a) Carbon (Z = 6) (b) Nitrogen (Z = 7) (c) Neon (Z = 10) (d) Magnesium (Z = 12) (e) Chlorine (Z = 17) [Exam Favorite]
Answer:
================================================================================
ELECTRONIC CONFIGURATION & VALENCY BREAKDOWN:
--------------------------------------------------------------------------------
(a) Carbon (Atomic Number Z = 6):
• Shell Distribution : K = 2, L = 4
• Valence Electrons : 4
• Valency : 4 (Needs 4 electrons to complete octet)
(b) Nitrogen (Atomic Number Z = 7):
• Shell Distribution : K = 2, L = 5
• Valence Electrons : 5
• Valency : 8 - 5 = 3 (Needs 3 electrons to complete octet)
(c) Neon (Atomic Number Z = 10):
• Shell Distribution : K = 2, L = 8
• Valence Electrons : 8
• Valency : 8 - 8 = 0 (Completely filled octet, inert noble gas)
(d) Magnesium (Atomic Number Z = 12):
• Shell Distribution : K = 2, L = 8, M = 2
• Valence Electrons : 2
• Valency : 2 (Loses 2 valence electrons to achieve stable octet)
(e) Chlorine (Atomic Number Z = 17):
• Shell Distribution : K = 2, L = 8, M = 7
• Valence Electrons : 7
• Valency : 8 - 7 = 1 (Gains 1 electron to complete octet)
================================================================================
Question 7 Why did Rutherford’s planetary atomic model fail to explain atomic stability, and how did Niels Bohr overcome this limitation? [Exam Favorite]
Answer:
- Failure of Rutherford’s Model: Rutherford compared electrons to planets revolving around the sun. However, an electron is a charged particle. In circular motion, it experiences centripetal acceleration. According to classical electrodynamics (Maxwell’s theory), an accelerating charged particle must continuously radiate electromagnetic energy. Losing energy, its orbital radius would continuously shrink, causing the electron to spiral into the nucleus within 10⁻⁸\text{ seconds}, destroying the atom.
- Bohr’s Solution: Niels Bohr resolved this by proposing that electrons can only revolve in special discrete non-radiating orbits (stationary states). In these orbits, their orbital energy is quantized and conserved, preventing them from radiating energy or spiraling into the nucleus.
Question 8 Magnesium is essential for biological processes, including muscle contraction. For an atom of Magnesium with mass number 24 and atomic number 12, determine: (i) number of protons, (ii) number of neutrons, (iii) number of electrons, and (iv) illustrate the arrangement of electrons in its shells. [Exam Favorite]
Answer:
================================================================================
ATOMIC STRUCTURE OF MAGNESIUM (Z = 12, A = 24):
--------------------------------------------------------------------------------
GIVEN:
• Atomic Number (Z) = 12
• Mass Number (A) = 24
CALCULATIONS:
(i) Number of Protons = Z = 12
(ii) Number of Electrons = Number of Protons = 12 (in neutral atom)
(iii) Number of Neutrons = Mass Number (A) - Atomic Number (Z)
= 24 - 12 = 12 neutrons
(iv) SHELL DISTRIBUTION (BOHR-BURY SCHEME):
• K-Shell : 2 electrons
• L-Shell : 8 electrons
• M-Shell : 2 electrons
Configuration = 2, 8, 2 (Valence Electrons = 2, Valency = 2).
================================================================================
Question 9 An atom of an element has 79 protons and a mass number of 197. Calculate: (i) the number of neutrons, and (ii) the number of electrons. Identify the chemical element. [Exam Favorite]
Answer:
================================================================================
CALCULATION:
--------------------------------------------------------------------------------
GIVEN DATA:
• Number of Protons (Z) = 79
• Mass Number (A) = 197
STEP 1: Number of Electrons
In a neutral atom: Number of Electrons = Number of Protons = 79
STEP 2: Number of Neutrons
Formula: Number of Neutrons = Mass Number (A) - Atomic Number (Z)
Neutrons = 197 - 79 = 118 neutrons
STEP 3: Identification of Element
Element with Atomic Number Z = 79 is GOLD (Chemical Symbol: Au).
FINAL ANSWER:
• Number of Neutrons = 118
• Number of Electrons = 79
• Name of Element = Gold (Au)
================================================================================
Question 10 Define Isotopes and Isobars. Give two examples of each. Mention two important industrial or medical applications of radioisotopes. [Exam Favorite]
Answer:
- Isotopes: Atoms of the same chemical element having the same atomic number (Z) but different mass numbers (A) due to different numbers of neutrons (e.g., Protium ¹H, Deuterium ²H, Tritium ³H; Carbon-12 and Carbon-14).
- Isobars: Atoms of different chemical elements having the same mass number (A) but different atomic numbers (Z) (e.g., Calcium (Z=20, A=40) and Argon (Z=18, A=40)).
- Key Applications of Radioisotopes:
- Medical Radiotherapy: An isotope of Cobalt (Cobalt-60) is used in the treatment of cancer.
- Goitre Treatment: An isotope of Iodine (Iodine-131) is used in the treatment of thyroid goitre disease.
- Nuclear Power Fuel: An isotope of Uranium (Uranium-235) is used as fuel in nuclear fission reactors.
Question 11 Natural chlorine occurs as two isotopes: 75% of Chlorine-35 (mass 35 u) and 25% of Chlorine-37 (mass 37 u). Calculate the average atomic mass of a chlorine atom. [Exam Favorite]
Answer:
================================================================================ NUMERICAL SOLUTION (AVERAGE ATOMIC MASS OF CHLORINE): -------------------------------------------------------------------------------- GIVEN DATA: • Isotope 1: Chlorine-35, Mass = 35 u, Natural Abundance = 75% (3/4) • Isotope 2: Chlorine-37, Mass = 37 u, Natural Abundance = 25% (1/4) CALCULATION: Average Mass = [ (Mass of Isotope 1 × % Abundance) + (Mass of Isotope 2 × % Abundance) ] / 100 Average Mass = [ (35 × 75) + (37 × 25) ] / 100 Average Mass = [ 2625 + 925 ] / 100 Average Mass = 3550 / 100 = 35.5 u Alternative Fraction Method: Average Mass = (35 × 3/4) + (37 × 1/4) = (105 / 4) + (37 / 4) = 142 / 4 = 35.5 u FINAL ANSWER: The average atomic mass of chlorine is 35.5 u. ================================================================================
Question 12 Why is the chemical behavior of isotopes of an element identical, while their physical properties differ? [Exam Favorite]
Answer:
- Identical Chemical Properties: Chemical properties are determined entirely by the number of valence electrons and electronic configuration. Because all isotopes of an element possess the exact same atomic number (Z) and identical electron arrangements, they undergo identical chemical bonding and reactions.
- Different Physical Properties: Physical properties (such as density, mass, boiling point, melting point, and rate of diffusion) depend on the atomic mass and number of neutrons. Because isotopes have different numbers of neutrons and different mass numbers (A), their physical properties vary.
Frequently Asked Questions (FAQs) – Class 9 Science Chapter 8
Question 1: What are Canal Rays? Who discovered them? [Exam Favorite] Answer: Canal rays (or anode rays) are streams of positively charged subatomic ions discovered by E. Goldstein in 1886 during gas discharge tube experiments, which led to the discovery of the proton.
Question 2: What is the overall electric charge on an atom? Why? [Exam Favorite] Answer: An atom is electrically neutral (zero net charge) because the total number of positively charged protons in the nucleus is exactly equal to the total number of negatively charged electrons revolving in its shells.
Question 3: Define Atomic Number (Z) and Mass Number (A). [Exam Favorite] Answer:
- Atomic Number (Z): The total number of protons present in the nucleus of an atom.
- Mass Number (A): The total sum of protons and neutrons (nucleons) present in the nucleus (A = Z + n).
Question 4: What are Nucleons? [Exam Favorite] Answer: Protons and neutrons collectively reside inside the central nucleus of an atom and are therefore jointly referred to as nucleons.
Question 5: What is Valency? How is it determined for metals and non-metals? [Exam Favorite] Answer: Valency is the combining capacity of an atom:
- For Metals (Valence electrons 1, 2, 3):
Valency = Number of Valence Electrons. - For Non-Metals (Valence electrons 5, 6, 7):
Valency = 8 - Number of Valence Electrons.
Question 6: Why do noble gases like Helium, Neon, and Argon have zero valency? [Exam Favorite] Answer: Noble gases have completely filled outermost valence shells (Helium has a complete duplet of 2 electrons; Neon and Argon have stable octets of 8 electrons). They have no tendency to gain, lose, or share electrons, making their valency zero.
Question 7: Which subatomic particle is not present in an ordinary Hydrogen atom? [Exam Favorite] Answer: The Neutron is completely absent in an ordinary Hydrogen (Protium, ¹H) atom, which contains 1 proton and 1 electron only.
Question 8: State the 2n² rule of electron distribution. [Exam Favorite] Answer: The maximum number of electrons that can be accommodated in an energy shell is given by 2n², where n is the shell number (K = 2, L = 8, M = 18, N = 32).
Question 9: What is the Octet Rule? [Exam Favorite] Answer: The octet rule states that atoms combine by gaining, losing, or sharing valence electrons in order to achieve a stable electronic configuration of 8 electrons in their outermost shell (duplet of 2 for helium).
Question 10: Differentiate between Isotopes and Isobars. [Exam Favorite] Answer:
- Isotopes: Same atomic number (Z), different mass numbers (A), same element (e.g., ¹²C and ¹⁴C).
- Isobars: Different atomic numbers (Z), same mass number (A), different elements (e.g., ⁴⁰Ar and ⁴⁰Ca).
Question 11: An atom has 8 electrons in total. What is its atomic number and valency? [Exam Favorite] Answer:
- Atomic Number Z = 8 (Oxygen).
- Electronic configuration = 2, 6.
- Valency = 8 – 6 = 2.
Question 12: Why is the mass of an electron neglected when calculating the atomic mass of an atom? [Exam Favorite] Answer: An electron has an extremely tiny mass (9.1 × 10⁻³¹\text{ kg} \approx 1/1840\text{ u}), which is roughly 2000 times lighter than a proton or neutron. Hence, its contribution to the total mass of the atom is negligible.
Question 13: What is the maximum number of electrons that can be held in the M-shell? [Exam Favorite] Answer: For M-shell (n = 3), the maximum theoretical capacity is 2n² = 2 × (3)² = 18 electrons (although when it acts as the outermost shell, it can hold a maximum of 8 electrons due to the octet rule).
Question 14: What was the main gold foil material used by Rutherford and why? [Exam Favorite] Answer: Rutherford used Gold foil because gold is the most malleable metal known and could be beaten into an extremely thin sheet (only about 1000 atoms thick, \approx 100\text{ nm}), ensuring alpha particles interacted with minimal layers of atoms.
Question 15: Name the radioisotope used to determine the geological age of ancient fossils. [Exam Favorite] Answer: Carbon-14 (¹⁴C), used in the radiometric dating technique known as Carbon Dating.
Mastering the NCERT Solutions for Class 9 Science Chapter 8 (Exploration), “Journey Inside the Atom”, equips students with the atomic models, subatomic particle properties, electron configurations, and isotope mathematics required for top performance in CBSE chemistry evaluations. Review the numerical box solutions, the master summary tables, and the 15 high-yield FAQs above to secure full marks in your examinations.