NCERT Solutions for Class 9 Science Exploration Chapter 8: Journey Inside the Atom
Session 2026-27 Updated
Revise, Reflect, Refine
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.
(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 statements:
(ii) The results disproved the plum pudding model and led to the idea of a nucleus at the centre of the atom. This statement is correct. Thomson’s plum pudding model said that positive charge was spread throughout the atom. But Rutherford found that most alpha particles passed straight through, while a few were deflected sharply. This proved that the positive charge is concentrated in a tiny central region called the nucleus.
(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. This statement is correct. Since only a few alpha particles were deflected through large angles, it showed that they were coming close to a very small, dense, positively charged region. This region is the nucleus, where most of the mass of the atom is concentrated.
Incorrect statements:
(i) The experiment clearly showed the existence of neutrons in the nucleus. This statement is incorrect. Rutherford’s gold foil experiment did not show the existence of neutrons. Neutrons were discovered much later by James Chadwick. Rutherford’s experiment only showed that there is a small, dense, positively charged centre in the atom.
(iv) The way alpha particles were deflected showed that electrons move around the nucleus. This statement is incorrect. Rutherford’s experiment gave information mainly about the nucleus and the empty space in the atom. It did not directly show how electrons move around the nucleus.
Correct statements:
(ii) The results disproved the plum pudding model and led to the idea of a nucleus at the centre of the atom. This statement is correct. Thomson’s plum pudding model said that positive charge was spread throughout the atom. But Rutherford found that most alpha particles passed straight through, while a few were deflected sharply. This proved that the positive charge is concentrated in a tiny central region called the nucleus.
(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. This statement is correct. Since only a few alpha particles were deflected through large angles, it showed that they were coming close to a very small, dense, positively charged region. This region is the nucleus, where most of the mass of the atom is concentrated.
Incorrect statements:
(i) The experiment clearly showed the existence of neutrons in the nucleus. This statement is incorrect. Rutherford’s gold foil experiment did not show the existence of neutrons. Neutrons were discovered much later by James Chadwick. Rutherford’s experiment only showed that there is a small, dense, positively charged centre in the atom.
(iv) The way alpha particles were deflected showed that electrons move around the nucleus. This statement is incorrect. Rutherford’s experiment gave information mainly about the nucleus and the empty space in the atom. It did not directly show how electrons move around the nucleus.
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.
(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) Electrons lose energy while moving in fixed orbits and slowly fall into the nucleus. This statement is incorrect. According to Bohr’s model, electrons move in fixed orbits or energy levels and do not lose energy while revolving in these allowed shells.
(ii) Electrons can exist anywhere around the nucleus with no fixed energy. This statement is incorrect. Bohr proposed that electrons can exist only in certain fixed energy levels or shells. They cannot stay anywhere around the nucleus.
(iii) Electrons revolve around the nucleus in orbits of fixed energy without losing energy. This statement is correct. This is the main idea of Bohr’s model. Electrons remain stable in fixed shells and do not radiate energy while moving in those shells.
(iv) Electrons can be found between energy levels as they move around the nucleus. This statement is incorrect. According to Bohr’s model, electrons are not found between two energy levels. They can jump from one shell to another only by absorbing or releasing a fixed amount of energy.
(i) Electrons lose energy while moving in fixed orbits and slowly fall into the nucleus. This statement is incorrect. According to Bohr’s model, electrons move in fixed orbits or energy levels and do not lose energy while revolving in these allowed shells.
(ii) Electrons can exist anywhere around the nucleus with no fixed energy. This statement is incorrect. Bohr proposed that electrons can exist only in certain fixed energy levels or shells. They cannot stay anywhere around the nucleus.
(iii) Electrons revolve around the nucleus in orbits of fixed energy without losing energy. This statement is correct. This is the main idea of Bohr’s model. Electrons remain stable in fixed shells and do not radiate energy while moving in those shells.
(iv) Electrons can be found between energy levels as they move around the nucleus. This statement is incorrect. According to Bohr’s model, electrons are not found between two energy levels. They can jump from one shell to another only by absorbing or releasing a fixed amount of energy.
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: First, calculate the atomic number and mass number of each species.
For X: Number of protons = 18, Number of neutrons = 19. Atomic number = 18, Mass number = 18 + 19 = 37.
For Y: Number of protons = 17, Number of neutrons = 18. Atomic number = 17, Mass number = 17 + 18 = 35.
For Z: Number of protons = 17, Number of neutrons = 20. Atomic number = 17, Mass number = 17 + 20 = 37.
(i) Relation between Y and Z: Y and Z are isotopes. Reason: They have the same number of protons, that is, the same atomic number (17), but different numbers of neutrons and hence different mass numbers (35 and 37).
(ii) Relation between Z and X: Z and X are isobars. Reason: They have the same mass number, 37, but different atomic numbers. X has atomic number 18, while Z has atomic number 17.
For X: Number of protons = 18, Number of neutrons = 19. Atomic number = 18, Mass number = 18 + 19 = 37.
For Y: Number of protons = 17, Number of neutrons = 18. Atomic number = 17, Mass number = 17 + 18 = 35.
For Z: Number of protons = 17, Number of neutrons = 20. Atomic number = 17, Mass number = 17 + 20 = 37.
(i) Relation between Y and Z: Y and Z are isotopes. Reason: They have the same number of protons, that is, the same atomic number (17), but different numbers of neutrons and hence different mass numbers (35 and 37).
(ii) Relation between Z and X: Z and X are isobars. Reason: They have the same mass number, 37, but different atomic numbers. X has atomic number 18, while Z has atomic number 17.
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: From the gold foil experiment, Rutherford observed that: Most alpha particles passed straight through the foil. A few alpha particles were deflected at large angles. Very few alpha particles bounced back.
Based on these observations, Rutherford drew the following conclusions about the positively charged part of the atom:
1. Position of the positive charge: The positive charge of the atom is not spread throughout the atom. Instead, it is concentrated in a very small region at the centre of the atom, called the nucleus.
2. Size of the nucleus: The nucleus is extremely small compared to the size of the atom, because only a few alpha particles were deflected.
3. Mass of the atom: Almost all the mass of the atom is concentrated in the nucleus. The large deflections and backward scattering of a few alpha particles indicated that they encountered a very dense and massive region.
4. Nature of the nucleus: The nucleus is positively charged, which causes repulsion and deflection of the positively charged alpha particles.
Based on these observations, Rutherford drew the following conclusions about the positively charged part of the atom:
1. Position of the positive charge: The positive charge of the atom is not spread throughout the atom. Instead, it is concentrated in a very small region at the centre of the atom, called the nucleus.
2. Size of the nucleus: The nucleus is extremely small compared to the size of the atom, because only a few alpha particles were deflected.
3. Mass of the atom: Almost all the mass of the atom is concentrated in the nucleus. The large deflections and backward scattering of a few alpha particles indicated that they encountered a very dense and massive region.
4. Nature of the nucleus: The nucleus is positively charged, which causes repulsion and deflection of the positively charged alpha particles.
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.
(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: Correct chronological order:
(iv) Dalton’s model
(ii) Thomson’s model
(iii) Rutherford’s model
(i) Bohr’s model
Explanation: Dalton first proposed that atoms are indivisible particles. Thomson discovered electrons and proposed the plum pudding model. Rutherford discovered the nucleus through the gold foil experiment. Bohr refined the model by introducing fixed energy levels for electrons.
(iv) Dalton’s model
(ii) Thomson’s model
(iii) Rutherford’s model
(i) Bohr’s model
Explanation: Dalton first proposed that atoms are indivisible particles. Thomson discovered electrons and proposed the plum pudding model. Rutherford discovered the nucleus through the gold foil experiment. Bohr refined the model by introducing fixed energy levels for electrons.
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 are negatively charged particles, while the nucleus contains positively charged protons. Due to electrostatic force of attraction between opposite charges: Electrons are attracted towards the nucleus. This force keeps them bound to the atom. Thus, electrons do not fly away because the electrostatic attraction between the positively charged nucleus and negatively charged electrons holds them in their orbits.
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.
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: Assertion is true because discovery of electrons, protons, and neutrons helped understand atomic structure. Reason is also true because atoms are electrically neutral (protons = electrons). However, this does not explain how subatomic particles helped in understanding structure.
Explanation: Assertion is true because discovery of electrons, protons, and neutrons helped understand atomic structure. Reason is also true because atoms are electrically neutral (protons = electrons). However, this does not explain how subatomic particles helped in understanding structure.
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: Atomic number = 12, 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 (for a neutral atom)
(iv) Electronic configuration: Total electrons = 12. Distribution: K shell = 2, L shell = 8, M shell = 2. Electronic configuration = 2, 8, 2.
(i) Number of protons = Atomic number = 12
(ii) Number of neutrons = Mass number − Atomic number = 24 − 12 = 12
(iii) Number of electrons = 12 (for a neutral atom)
(iv) Electronic configuration: Total electrons = 12. Distribution: K shell = 2, L shell = 8, M shell = 2. Electronic configuration = 2, 8, 2.
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) Name of element: Helium (ii) Symbol: He (iii) Total electrons: 2 (iv) Valence electrons: 2 (v) Valency: 0 (vi) Number of protons: 2 (vii) Atomic number: 2
(b) (i) Name of element: Oxygen (ii) Symbol: O (iii) Total electrons: 8 (iv) Valence electrons: 6 (v) Valency: 2 (vi) Number of protons: 8 (vii) Atomic number: 8
(c) (i) Name of element: Calcium (ii) Symbol: Ca (iii) Total electrons: 20 (iv) Valence electrons: 2 (v) Valency: 2 (vi) Number of protons: 20 (vii) Atomic number: 20
(d) (i) Name of element: Neon (ii) Symbol: Ne (iii) Total electrons: 10 (iv) Valence electrons: 8 (v) Valency: 0 (vi) Number of protons: 10 (vii) Atomic number: 10
(a) (i) Name of element: Helium (ii) Symbol: He (iii) Total electrons: 2 (iv) Valence electrons: 2 (v) Valency: 0 (vi) Number of protons: 2 (vii) Atomic number: 2
(b) (i) Name of element: Oxygen (ii) Symbol: O (iii) Total electrons: 8 (iv) Valence electrons: 6 (v) Valency: 2 (vi) Number of protons: 8 (vii) Atomic number: 8
(c) (i) Name of element: Calcium (ii) Symbol: Ca (iii) Total electrons: 20 (iv) Valence electrons: 2 (v) Valency: 2 (vi) Number of protons: 20 (vii) Atomic number: 20
(d) (i) Name of element: Neon (ii) Symbol: Ne (iii) Total electrons: 10 (iv) Valence electrons: 8 (v) Valency: 0 (vi) Number of protons: 10 (vii) Atomic number: 10
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’s Model: According to Rutherford, electrons revolve around the nucleus similar to planets around the Sun. However, according to classical physics, a moving electron (charged particle) should continuously lose energy in the form of radiation. As a result: The electron would lose energy, its orbit would shrink, and eventually, it would fall into the nucleus. This means the atom should collapse, which does not happen in reality. Therefore, Rutherford’s model could not explain the stability of atoms.
Bohr’s Model: Bohr proposed that: 1. Electrons move in fixed circular orbits called energy levels or shells. 2. In these fixed orbits, electrons do not lose energy. 3. Energy is emitted or absorbed only when an electron jumps from one orbit to another. Thus: Electrons remain stable in their orbits, and the atom does not collapse.
Rutherford’s Model: According to Rutherford, electrons revolve around the nucleus similar to planets around the Sun. However, according to classical physics, a moving electron (charged particle) should continuously lose energy in the form of radiation. As a result: The electron would lose energy, its orbit would shrink, and eventually, it would fall into the nucleus. This means the atom should collapse, which does not happen in reality. Therefore, Rutherford’s model could not explain the stability of atoms.
Bohr’s Model: Bohr proposed that: 1. Electrons move in fixed circular orbits called energy levels or shells. 2. In these fixed orbits, electrons do not lose energy. 3. Energy is emitted or absorbed only when an electron jumps from one orbit to another. Thus: Electrons remain stable in their orbits, and the atom does not collapse.
11. An atom ⁷⁰X has 31 electrons. How many neutrons are there in its nucleus?
Answer: Mass number (A) = 70. Number of electrons = 31. For a neutral atom: Number of protons = Number of electrons = 31. Number of neutrons = Mass number − Atomic number = 70 − 31 = 39. Therefore, the number of neutrons = 39.
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) Number of electrons: For a neutral atom: Number of electrons = Number of protons = 79.
(i) Number of neutrons = Mass number − Number of protons = 197 − 79 = 118.
(ii) Number of electrons: For a neutral atom: Number of electrons = Number of protons = 79.
13. Complete the Table 8.5:
Answer: To complete the table, we use these facts: Atomic number = Number of protons. In a neutral atom, Number of electrons = Number of protons. Mass number = Number of protons + Number of neutrons.
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?
(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: Mass number = 35. 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) Number of protons = 17. Number of electrons = 17.
(ii) Atomic number = Number of protons = 17.
(iii) The element with atomic number 17 is Chlorine (Cl).
(iv) Electronic configuration of chlorine = 2, 8, 7.
(v) Valence electrons = 7.
(vi) New number of neutrons = 18 + 2 = 20. New mass number = Number of protons + Number of neutrons = 17 + 20 = 37. So, the new mass number will be 37.
(vii) The new atom will be an isotope of X. Since, both atoms have the same atomic number (17), but different mass numbers (35 and 37).
(i) Number of protons = 17. Number of electrons = 17.
(ii) Atomic number = Number of protons = 17.
(iii) The element with atomic number 17 is Chlorine (Cl).
(iv) Electronic configuration of chlorine = 2, 8, 7.
(v) Valence electrons = 7.
(vi) New number of neutrons = 18 + 2 = 20. New mass number = Number of protons + Number of neutrons = 17 + 20 = 37. So, the new mass number will be 37.
(vii) The new atom will be an isotope of X. Since, both atoms have the same atomic number (17), but different mass numbers (35 and 37).
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. Normally, atomic number depends only on the number of protons and mass number depends on protons and neutrons only.
(i) Atomic number will remain unchanged. Reason: Atomic number = Number of protons = 12.
(ii) Atomic mass will increase. Reason: The new particles replacing electrons are much heavier than ordinary electrons, so the total mass of the atom will become greater.
(iii) Mass number will remain unchanged. Reason: Mass number = Number of protons + Number of neutrons = 12 + 12 = 24. Electrons do not contribute to mass number.
(iv) Overall charge will remain neutral (zero), provided the number of new particles is equal to the number of protons. Reason: The new particles have the same negative charge as electrons, so their total negative charge will balance the positive charge of 12 protons.
(i) Atomic number will remain unchanged. Reason: Atomic number = Number of protons = 12.
(ii) Atomic mass will increase. Reason: The new particles replacing electrons are much heavier than ordinary electrons, so the total mass of the atom will become greater.
(iii) Mass number will remain unchanged. Reason: Mass number = Number of protons + Number of neutrons = 12 + 12 = 24. Electrons do not contribute to mass number.
(iv) Overall charge will remain neutral (zero), provided the number of new particles is equal to the number of protons. Reason: The new particles have the same negative charge as electrons, so their total negative charge will balance the positive charge of 12 protons.
Very Short Answer Type Questions
1. Who proposed the plum pudding model of the atom and what does it describe?
Answer: J. J. Thomson proposed the plum pudding model. It describes the atom as a sphere of positive charge with electrons (like plum seeds) distributed throughout it, like seeds in a watermelon or raisins in a pudding.
2. What were the key observations of the gold foil experiment?
Answer: Most alpha particles passed straight through undeflected, some were deflected at large angles, and a very few bounced straight back. This showed the atom is mostly empty space with a tiny, dense, positively charged nucleus at its centre.
3. Define atomic number and mass number.
Answer: Atomic number (Z) is the number of protons in the nucleus of an atom. Mass number (A) is the total number of nucleons (protons + neutrons). Number of neutrons = A − Z. Atomic number uniquely identifies an element.
4. What is the electronic configuration of sodium (atomic number 11)?
Answer: Sodium has 11 electrons. Using the 2n² rule: K-shell = 2, L-shell = 8, remaining 1 electron in M-shell. Electronic configuration of sodium = 2, 8, 1. It has 1 valence electron, so its valency is 1.
5. Define isotopes and give one example.
Answer: Isotopes are atoms of the same element with the same atomic number but different mass numbers due to different numbers of neutrons. Example: Hydrogen has three isotopes — protium (¹H), deuterium (²H) and tritium (³H), all with atomic number 1.
6. Define isobars and give one example.
Answer: Isobars are atoms of different elements with the same mass number but different atomic numbers. Example: Argon (Z=18), Potassium (Z=19) and Calcium (Z=20) all have mass number 40 – they are isobars of each other.
7. What is valency? How is it related to valence electrons?
Answer: Valency is the combining capacity of an atom — the number of electrons gained, lost or shared to complete its octet and become stable. Valency is determined by the number of valence electrons (electrons in the outermost shell).
8. State the main limitation of Rutherford’s atomic model.
Answer: Rutherford’s model could not explain atomic stability. A circularly moving electron is constantly accelerating, so it should lose energy, spiral inward and fall into the nucleus. But atoms are stable – this contradiction was unresolved by Rutherford’s model.
9. How did Bohr’s model overcome Rutherford’s limitation?
Answer: Bohr proposed that electrons move only in fixed allowed orbits called stationary states. While in a stationary state, an electron does NOT lose energy, even though it moves around the nucleus. This postulate explained why atoms are stable.
10. What is the maximum number of electrons that can occupy the K, L, and M shells?
Answer: Using the 2n² formula: K-shell (n=1) = 2×1² = 2 electrons; L-shell (n=2) = 2×2² = 8 electrons; M-shell (n=3) = 2×3² = 18 electrons. However, the maximum in any outermost shell is limited to 8 electrons.
11. Give four applications of isotopes used in science and medicine.
Answer: ²³⁵U — nuclear fuel in reactors for electricity generation. ⁶⁰Co — radiation therapy for cancer treatment. ¹³¹I — treatment of goitre and thyroid cancer. ¹⁴C — carbon dating to determine age of fossils and ancient artefacts.
12. Why do isotopes have identical chemical properties?
Answer: Isotopes have the same number of electrons and identical electronic configurations. Since chemical properties depend mainly on valence electrons (outermost shell electrons), and these are the same for all isotopes, their chemical behaviour is identical.
13. Who discovered the neutron and when? Why was this discovery important?
Answer: James Chadwick discovered the neutron in 1932. This discovery explained why helium (2 protons) has four times the mass of hydrogen (1 proton) – the extra mass comes from neutrons. It resolved the mystery of atomic mass that protons alone could not explain.
14. Write the standard notation for an atom with atomic number 6 and mass number 12.
Answer: Standard notation places mass number (A) above and atomic number (Z) below the element symbol. For carbon (symbol C), with Z = 6 and A = 12, the notation is ¹²₆C. Number of neutrons = 12 − 6 = 6. Electronic configuration = 2, 4.
15. State the IUPAC rules for writing chemical symbols of elements.
Answer: First letter is always uppercase; second letter (if any) is always lowercase. Most symbols use first one or two letters of English name (e.g., H, He, Al). Some symbols are from Latin, Greek or German names (e.g., Fe from Ferrum, Na from Natrium, Hg from Hydrargyros).
Short Answer Type Questions
1. Describe the gold foil experiment. What were the three key observations and what did each prove?
Answer: Geiger and Marsden, working under Rutherford in 1911, directed a beam of alpha particles at a thin gold foil. Observation 1: Most particles passed straight through: proved the atom is mostly empty space. Observation 2: Some particles were deflected at large angles: proved the positive charge is concentrated, not spread throughout. Observation 3: A very few particles bounced straight back: proved there exists an extremely small, very dense, positively charged nucleus at the centre of the atom. These three observations together completely disproved Thomson’s plum pudding model.
2. Compare Thomson’s model and Rutherford’s model of the atom. How did the gold foil experiment disprove Thomson’s model?
Answer: In Thomson’s model, the positive charge is spread uniformly throughout the atom with electrons embedded in it – like raisins in a pudding. There is no central nucleus. In Rutherford’s model, the atom is mostly empty space, with all positive charge concentrated in a tiny, dense central nucleus. Electrons revolve around it like planets. The gold foil experiment disproved Thomson’s model because it showed that some alpha particles were sharply deflected or bounced back – impossible if positive charge were spread evenly. A concentrated nucleus was needed to cause such strong deflections.
3. Explain why atoms are electrically neutral. Give one example to support your answer.
Answer: Atoms are electrically neutral because the number of positively charged protons in the nucleus exactly equals the number of negatively charged electrons revolving around it. The equal and opposite charges cancel each other completely. Example: Sodium has atomic number 11, so it has 11 protons (+11 charge) and 11 electrons (−11 charge). Net charge = 11 − 11 = 0. Similarly, all atoms maintain this equality between protons and electrons and are therefore electrically neutral.
4. Distinguish between isotopes and isobars with two examples of each.
Answer: Isotopes are atoms of the same element with the same atomic number (Z) but different mass numbers (A) due to different numbers of neutrons. Chemical properties are identical; physical properties differ. Examples of isotopes: ¹H, ²H, ³H (all Z=1); ¹²C, ¹³C, ¹⁴C (all Z=6). Isobars are atoms of different elements with the same mass number (A) but different atomic numbers (Z). Examples of isobars: ⁴⁰Ar (Z=18), ⁴⁰K (Z=19), ⁴⁰Ca (Z=20), all A=40; also ¹⁴C (Z=6) and ¹⁴N (Z=7), both A=14.
5. Write the electronic configuration of chlorine (Z=17). Determine its valency and explain how you arrived at it.
Answer: Chlorine has 17 electrons. Filling shells using 2n²: K-shell: 2 electrons (full), L-shell: 8 electrons (full), M-shell: 7 electrons (remaining). Electronic configuration: 2, 8, 7. Chlorine has 7 valence electrons in its outermost (M) shell. Since this is more than 4, chlorine tends to gain electrons to complete its octet. It needs to gain 1 more electron to reach 8 (stable octet). Therefore, the valency of chlorine is 1.
6. Explain the concept of weighted average atomic mass using chlorine as an example.
Answer: Chlorine exists in nature as two isotopes – ³⁵Cl (atomic mass 35 u, abundance 75%) and ³⁷Cl (atomic mass 37 u, abundance 25%). If we took a simple arithmetic average, we would get (35+37)/2 = 36 u. But this is incorrect because the two isotopes don’t occur in equal amounts. The accurate weighted average atomic mass accounts for their relative abundance: Weighted average = (35 × 75/100) + (37 × 25/100) = 26.25 + 9.25 = 35.5 u. This means no individual chlorine atom has a mass of 35.5 u — each is either 35 or 37 u — but in a large sample, the average works out to 35.5 u.
7. What is valency? Determine the valency of sodium, oxygen, and carbon from their electronic configurations.
Answer: Valency is the number of electrons an atom gains, loses or shares to achieve a stable octet and complete the outermost shell. Sodium (Z=11): Electronic configuration 2, 8, 1. Has 1 valence electron – loses 1 to get octet. Valency = 1. Oxygen (Z=8): Electronic configuration 2, 6. Has 6 valence electrons – gains 2 to get octet. Valency = 2. Carbon (Z=6): Electronic configuration 2, 4. Has 4 valence electrons – cannot easily gain or lose, so shares 4. Valency = 4.
8. Why did scientists need to propose new atomic models repeatedly? What drove the evolution from Dalton to Bohr?
Answer: Each atomic model was revised when new experimental evidence could not be explained by the existing model. Dalton proposed atoms as indivisible – but Thomson’s cathode ray experiment showed electrons exist inside atoms, disproving indivisibility. Thomson’s plum pudding model – could not explain the gold foil experiment results, particularly sharp deflections of alpha particles. Rutherford’s nuclear model – correctly identified the nucleus, but could not explain atomic stability because an accelerating electron should spiral into the nucleus. Bohr’s model – resolved stability by proposing fixed energy shells where electrons don’t lose energy. Science always progresses this way: new experiments → new evidence → improved models.
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: Z = 15, so 15 electrons. K = 2, L = 8, M = 5. Configuration: 2, 8, 5. (b) Number of neutrons = Mass number − Atomic number = 31 − 15 = 16 neutrons. (c) Valency: 5 valence electrons in outermost shell. More than 4, so gains electrons to complete octet. Needs to gain 3 electrons. Valency = 3. (d) Element with Z = 15 is Phosphorus (P). Symbol: P, from Table 8.2.
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 were arrived at through philosophical reasoning and logical speculation about the ultimate nature of matter – without any experimental evidence. These were imaginative, intuitive ideas based on thought alone. John Dalton’s atomic theory (1808), in contrast, was based on systematic experimental observations of chemical reactions – specifically the laws of conservation of mass, constant proportions and multiple proportions. His theory could be tested and used to make predictions about chemical behaviour. This experimental foundation is what makes Dalton’s contribution scientific. Science requires evidence; philosophy requires reasoning. Both are valuable, but only experimental evidence can establish scientific validity.
Long Answer Type Questions
1. Describe the gold foil experiment. What were the observations and what conclusions did Rutherford draw? How did this disprove Thomson’s model?
Answer: In 1911, Geiger and Marsden, under Rutherford, directed a narrow beam of positively charged alpha particles at an extremely thin gold foil. Based on Thomson’s model (positive charge spread uniformly), they expected all particles to pass through with little or no deflection.
Actual observations:
Most alpha particles passed straight through – atom is mostly empty space. Some were deflected sharply – positive charge is concentrated, not spread. A very few bounced almost straight back – there exists an extremely small, dense, positively charged nucleus.
Rutherford concluded that the atom has a tiny, dense, positively charged nucleus at the centre with electrons revolving around it in mostly empty space. Thomson’s model was disproved because it predicted no sharp deflections – but many were observed.
Actual observations:
Most alpha particles passed straight through – atom is mostly empty space. Some were deflected sharply – positive charge is concentrated, not spread. A very few bounced almost straight back – there exists an extremely small, dense, positively charged nucleus.
Rutherford concluded that the atom has a tiny, dense, positively charged nucleus at the centre with electrons revolving around it in mostly empty space. Thomson’s model was disproved because it predicted no sharp deflections – but many were observed.
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 revolve around the nucleus only in fixed circular orbits called stationary states or shells (K, L, M, N). While in a fixed shell, electrons do NOT lose energy – this directly solved Rutherford’s stability problem, where an accelerating electron was expected to spiral into the nucleus. Energy increases with distance from nucleus. Electrons jump between shells only by absorbing or releasing exact amounts of energy equal to the difference between energy levels.
Electron distribution rules (Bohr-Bury):
Maximum electrons per shell = 2n². Outermost shell can hold maximum 8 electrons. Shells fill in order K, L, M, N from inside outward. These rules allow us to write electronic configurations and determine valency for any element.
Electron distribution rules (Bohr-Bury):
Maximum electrons per shell = 2n². Outermost shell can hold maximum 8 electrons. Shells fill in order K, L, M, N from inside outward. These rules allow us to write electronic configurations and determine valency for any element.
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) = number of protons in the nucleus. It uniquely identifies an element. In a neutral atom, number of electrons = number of protons = Z. Mass Number (A) = total nucleons = protons + neutrons. Since proton mass ≈ neutron mass, and electron mass is negligible, the atomic mass is approximately equal to the mass number. Number of neutrons = A − Z. Standard notation places A above and Z below the element symbol.
For the given atom (Z=17, A=35):
Number of protons = 17. Number of electrons = 17 (neutral atom). Number of neutrons = 35 − 17 = 18.
Electronic configuration: K=2, L=8, M=7 (i.e., 2, 8, 7). Valence electrons = 7, Valency = 1. This is Chlorine (Cl).
For the given atom (Z=17, A=35):
Number of protons = 17. Number of electrons = 17 (neutral atom). Number of neutrons = 35 − 17 = 18.
Electronic configuration: K=2, L=8, M=7 (i.e., 2, 8, 7). Valence electrons = 7, Valency = 1. This is Chlorine (Cl).
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 atoms of the same element with identical atomic numbers (same protons) but different mass numbers, due to different numbers of neutrons.
Same chemical properties: Isotopes have the same number of electrons and identical electronic configurations. Since chemical behaviour depends on valence electrons, all isotopes of an element react identically with other substances.
Different physical properties: Physical properties like boiling point, melting point and density depend on atomic mass. Since isotopes differ in neutron number and hence mass, their physical properties differ.
Applications of isotopes:
1. ²³⁵U (Uranium-235): Nuclear fuel in reactors to generate electricity.
2. ⁶⁰Co (Cobalt-60): Radiation therapy for treating cancer tumours.
3. ¹³¹I (Iodine-131): Treatment of goitre and thyroid cancer.
4. ¹⁴C (Carbon-14): Archaeological carbon dating to determine age of fossils and artefacts.
Same chemical properties: Isotopes have the same number of electrons and identical electronic configurations. Since chemical behaviour depends on valence electrons, all isotopes of an element react identically with other substances.
Different physical properties: Physical properties like boiling point, melting point and density depend on atomic mass. Since isotopes differ in neutron number and hence mass, their physical properties differ.
Applications of isotopes:
1. ²³⁵U (Uranium-235): Nuclear fuel in reactors to generate electricity.
2. ⁶⁰Co (Cobalt-60): Radiation therapy for treating cancer tumours.
3. ¹³¹I (Iodine-131): Treatment of goitre and thyroid cancer.
4. ¹⁴C (Carbon-14): Archaeological carbon dating to determine age of fossils and artefacts.
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 is the combining capacity of an atom — the number of electrons an atom gains, loses, or shares to achieve a stable electronic configuration (complete octet or duplet for hydrogen/helium).
Sodium (Z=11): Electronic configuration 2, 8, 1. Has 1 valence electron. Fewer than 4, so loses 1 electron to achieve the octet of L-shell. Valency = 1.
Oxygen (Z=8): Electronic configuration 2, 6. Has 6 valence electrons. More than 4, gains 2 electrons to complete octet. Valency = 2.
Carbon (Z=6): Electronic configuration 2, 4. Has exactly 4 valence electrons — neither easy to gain 4 nor lose 4. So it shares 4 electrons with other atoms. Valency = 4.
Noble gases (Neon Z=10: 2,8; Argon Z=18: 2,8,8): Both have complete octets in their outermost shells. No electrons need to be gained, lost, or shared to achieve stability. Therefore they are largely unreactive and have valency = 0.
Sodium (Z=11): Electronic configuration 2, 8, 1. Has 1 valence electron. Fewer than 4, so loses 1 electron to achieve the octet of L-shell. Valency = 1.
Oxygen (Z=8): Electronic configuration 2, 6. Has 6 valence electrons. More than 4, gains 2 electrons to complete octet. Valency = 2.
Carbon (Z=6): Electronic configuration 2, 4. Has exactly 4 valence electrons — neither easy to gain 4 nor lose 4. So it shares 4 electrons with other atoms. Valency = 4.
Noble gases (Neon Z=10: 2,8; Argon Z=18: 2,8,8): Both have complete octets in their outermost shells. No electrons need to be gained, lost, or shared to achieve stability. Therefore they are largely unreactive and have valency = 0.