CG Board Class 9 Science Chapter 3 Solutions: Atomic Structure (परमाणु संरचना) – 2026-27

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CG SCERT (Chhattisgarh Board) Class 9 Science Solutions | 2026-27 Official Edition

Complete 100% textbook-grounded CG Board Class 9 Science Chapter 3 Solutions: Atomic Structure (परमाणु संरचना) in Bilingual Format (English & Hindi Medium).

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Part 1

In-Text Questions, Activities, & Tables

  • English Question: Is the final solution still coloured when potassium permanganate or Dettol solution is diluted repeatedly (5 to 8 times)? Why?
  • English: Yes, the final solution remains coloured/scented because matter is made up of extremely tiny particles. A single crystal of potassium permanganate contains millions of microscopic particles that keep dividing and distributing themselves throughout the water upon each dilution.
Question

InText Questions Section 35 Rutherfords Model


📝 Question:

In-Text Questions (Section 3.5: Rutherford’s Model)

💡 Solution:

Can the alpha particle scattering experiment be carried out by using silver foil or extremely thin foils of other elements rather than gold? Explain your answer with reasons.

Yes, the experiment can be performed using extremely thin foils of other heavy metals like silver or platinum. Gold was chosen by Rutherford because it is the most malleable metal, allowing it to be hammered into an ultra-thin layer (about 100 nm thick or ~1000 atoms thick) without tearing. Silver can also be beaten into very thin foils, but because its nucleus has a lower positive charge (fewer protons) than gold, the electrostatic repulsion on alpha particles will be slightly less, leading to minor quantitative differences in deflection angles.

Question 2

On what basis did Thomson challenge the theory…


📝 Question:

**: On what basis did Thomson challenge the theory of indivisibility of atom?

💡 Solution:

On what basis did Thomson challenge the theory of indivisibility of atom?

J.J. Thomson calculated the mass and charge of the negatively charged particles present in cathode rays and found that they were identical regardless of the cathode material or gas used. He identified these subatomic particles as electrons, establishing that atoms are composed of smaller constituent particles, thereby challenging Dalton’s theory of indivisibility.

Q3

InText Task Table 1 Section 36 37 BohrBury…


📝 Question:

In-Text Task & Table 1 (Section 3.6 & 3.7: Bohr-Bury Scheme)

💡 Solution:

According to the Bohr-Bury scheme, electrons fill energy shells in a stepwise manner (2n2 capacity rule and maximum 8 electrons in the outer shell)

  • Potassium (3919K): 19 electrons → K = 2, L = 8, M = 8, N = 1.
  • Calcium (4020Ca): 20 electrons → K = 2, L = 8, M = 8, N = 2.

ElementSymbolNo. of e-Orbits KOrbits LOrbits MOrbits NElectronic Config.Valence e-
HydrogenH11–––11
LithiumLi321––2, 11
CarbonC624––2, 44
OxygenO826––2, 66
SodiumNa112812, 8, 11
AluminiumAl13283–2, 8, 33
PhosphorusP15285–2, 8, 55
ChlorineCl17287–2, 8, 77
ArgonAr18288–2, 8, 88
PotassiumK1928812, 8, 8, 11
CalciumCa2028822, 8, 8, 22
Q4

InText Task Table 2 Section 38 Atomic Number…


📝 Question:

In-Text Task & Table 2 (Section 3.8: Atomic Number & Mass Number)

💡 Solution:

Lithium**: Atomic number Z = 3, Mass number A = 3 (protons) + 3 (neutrons) = 6. Symbolic form: 63Li.

  • Calcium: Atomic number Z = 20, Mass number A = 20 (protons) + 20 (neutrons) = 40. Symbolic form: 4020Ca.

ElementSymbolProtons ZMass Number ANo. of Neutrons n = A – Z
HydrogenH111 – 1 = 0
LithiumLi366 – 3 = 3
CarbonC61212 – 6 = 6
OxygenO81616 – 8 = 8
SodiumNa112323 – 11 = 12
AluminiumAl132727 – 13 = 14
PhosphorusP153131 – 15 = 16
ChlorineCl173535 – 17 = 18
ArgonAr184040 – 18 = 22
PotassiumK193939 – 19 = 20
CalciumCa204040 – 20 = 20
Question

InText Numerical Questions on Sodium Size Section 38


📝 Question:

In-Text Numerical Questions on Sodium Size (Section 3.8)

💡 Solution:

If the size of a sodium atom is 1.86 × 10-10 m and the nucleus is 105 times smaller than the atom, what will be the size of its nucleus?

  • Size of sodium atom = 1.86 × 10-10 m.
  • Ratio of atomic size to nuclear size = 105.
  • Size of sodium nucleus = (Atomic size)/(105) = (1.86 × 10-10 m)/(105) = 1.86 × 10-15 m.

  • English: Keeping in mind the ratio of the size of the nucleus to the atomic size, how will you show a sodium atom? Can you pictorially depict a sodium atom accurately?
  • English:

No, it is impossible to draw a sodium atom accurately to scale on paper. If we draw the nucleus as a small dot of diameter 1 mm (10-3 m), the outer atomic boundary would have to be drawn at a diameter of 105 × 1 mm = 100,000 mm = 100 metres (roughly the size of a football field). Therefore, atomic diagrams are always symbolic schematics, not true-to-scale representations.

  • Hindi:
Question

InText Questions Section 39 Isotopes Isobars


📝 Question:

In-Text Questions (Section 3.9: Isotopes & Isobars)

💡 Solution:

If the atomic number of an atom is 15 and mass number is 31 then what is the number of sub-atomic particles present in the atom?

  • Atomic number (Z) = 15 ⇒ Protons (p^+) = 15.
  • For a neutral atom, Electrons (e^-) = Protons = 15.
  • Mass number (A) = 31 ⇒ Neutrons (n) = A – Z = 31 – 15 = 16.
  • Total sub-atomic particles = Protons + Electrons + Neutrons = 15 + 15 + 16 = 46.

Question 2

If the K and L shells of an…


📝 Question:

**: If the K and L shells of an atom are full and the M shell has 2 electrons then what is the atomic number of the atom?

💡 Solution:

If the K and L shells of an atom are full and the M shell has 2 electrons then what is the atomic number of the atom?

  • Full K shell = 2 electrons.
  • Full L shell = 8 electrons.
  • M shell = 2 electrons.
  • Total electrons = 2 + 8 + 2 = 12.
  • Since the atom is neutral, Atomic Number (Z) = Total electrons = 12 (Magnesium, Mg).

Question 3

Write the electronic configuration of the following sup23supsub11subNa…


📝 Question:

**: Write the electronic configuration of the following: 2311Na, 126C, 3517Cl.

💡 Solution:

Write the electronic configuration of the following: 2311Na, 126C, 3517Cl.

  1. Sodium (2311Na): Z = 11 ⇒ Electronic configuration: 2, 8, 1 (K = 2, L = 8, M = 1).
  2. Carbon (126C): Z = 6 ⇒ Electronic configuration: 2, 4 (K = 2, L = 4).
  3. Chlorine (3517Cl): Z = 17 ⇒ Electronic configuration: 2, 8, 7 (K = 2, L = 8, M = 7).

Part 2

End-of-Chapter Exercises

Exercise Question 1

Choose the correct option

(b) Thomson /

  • Correct Option / : (b) Thomson / .

(iii) Which of the following statements is true for 3919K atom: / 3919K :

Exercise Question 2

Fill in the blanks

–

Answer / : similar (same) / .

–

*(iii) 146C and 147N are …………………………………. (isotopes/isobars) of each other. / 146C 147N – ………………………………….

–

SECTION / EXERCISE

Exercise Question 3

  • English:
  • Dalton’s Atom: Dalton proposed that the atom is a solid, hard, sphere that is indivisible and has no internal subatomic structure.
  • Thomson’s Atom: Thomson proved that the atom is divisible and consists of subatomic particles. He proposed the plum pudding (watermelon) model, where positive charge is spread over a sphere with negatively charged electrons embedded inside it.
  • Hindi:
SECTION / EXERCISE

Exercise Question 4

168O and 167N are isobars. Use this example to explain isobars. / 168O 167N

  • English:

Isobars are atoms of different chemical elements that have different atomic numbers (Z) but the same mass number (A). In 168O and 167N:

  1. Their atomic numbers are different (Z = 8 for Oxygen, Z = 7 for Nitrogen), meaning they have different numbers of protons and different chemical properties.
  2. Their mass numbers are identical (A = 16), meaning the total sum of protons and neutrons in their nuclei is the same.
  • Hindi:
SECTION / EXERCISE

Exercise Question 5

Step-by-step Numerical Solution /

  • English:
  • Isotope 1: Mass = 79 u, Abundance = 50.69\%
  • Isotope 2: Mass = 81 u, Abundance = 49.31\%

Average Atomic Weight = ((Mass1 × Abundance1) + (Mass2 × Abundance2))/(100)

Average Atomic Weight = ((79 × 50.69) + (81 × 49.31))/(100)

79 × 50.69 = 4004.51

81 × 49.31 = 3994.11

Sum = 4004.51 + 3994.11 = 7998.62

Average Atomic Weight = (7998.62)/(100) = 79.9862 u ≈ 79.99 u (or 80 u)

  • Hindi:

79 × 50.69 = 4004.51

81 × 49.31 = 3994.11

= (7998.62)/(100) = 79.9862 u ≈ 79.99 u

SECTION / EXERCISE

Exercise Question 6

Find out the number of valence electrons in 168O and 147N. / 168O 147N

Step-by-step Solution /

  • English:
  1. Oxygen (168O): Atomic number Z = 8. Electronic configuration = 2, 6. Electrons in the outermost shell = 6. Valence electrons = 6.
  2. Nitrogen (147N): Atomic number Z = 7. Electronic configuration = 2, 5. Electrons in the outermost shell = 5. Valence electrons = 5.
  • Hindi:
SECTION / EXERCISE

Exercise Question 7

  • English:

Niels Bohr proposed his atomic model to overcome the drawbacks of Rutherford’s model:

  1. Discrete Orbits/Shells: Electrons revolve around the nucleus only in certain permitted, non-radiating circular orbits known as energy shells, designated by letters K, L, M, N… or integers n = 1, 2, 3, 4….
  2. Constant Energy: While revolving in these discrete orbits, electrons do not radiate energy.
  3. Bohr-Bury Rules:
  • Maximum electron capacity of a shell is given by formula 2n2.
  • Outermost shell can hold a maximum of 8 electrons.
  • Shells are filled step-wise from inner to outer orbits.
  • Hindi:
SECTION / EXERCISE

Exercise Question 8

  • English:

These atoms (168O, 178O, 188O) are isotopes of oxygen.

Reason: All three atoms belong to the same element, oxygen, and have the same atomic number (Z = 8, meaning 8 protons), but possess different mass numbers (A = 16, 17, 18) due to varying numbers of neutrons in their nuclei (8, 9, and 10 neutrons respectively).

  • Hindi:

(168O, 178O, 188O) (isotopes)

SECTION / EXERCISE

Exercise Question 9

  • English:

Main Postulates of Dalton’s Atomic Theory:

  1. All matter is made up of tiny indivisible particles called atoms.
  2. Atoms can neither be created nor destroyed in a chemical reaction.
  3. Atoms of a given element are identical in mass and chemical properties.
  4. Atoms of different elements have different masses and chemical properties.
  5. Atoms combine in simple whole number ratios to form compounds.

Limitations:

  1. Atom is divisible: Discovery of electrons, protons, and neutrons proved that atoms are divisible.
  2. Existence of Isotopes: Atoms of the same element can have different mass numbers (e.g., C-12 and C-14).
  3. Existence of Isobars: Atoms of different elements can have the same mass number (e.g., C-14 and N-14).
  • Hindi:

(Limitations)

SECTION / EXERCISE

Exercise Question 10

  • English:

Rutherford’s Experiment: Rutherford bombarded high-energy positively charged alpha particles (He2+) onto an extremely thin gold foil (100 nm thick).

Observations & Conclusions:

  1. Most alpha particles passed straight undeflected: Proved that most of the space inside the atom is empty.
  2. A few alpha particles deflected by small angles: Showed that the positive charge of the atom occupies a very small volume.
  3. Very few particles (1 in 20,000) bounced back by 180 degrees: Proved that almost all the mass and positive charge of the atom is densely concentrated in an extremely small central core called the nucleus.
  • Hindi:
SECTION / EXERCISE

Exercise Question 11

Write the rules proposed for electron distribution under the Bohr-Bury scheme. Use the rules to write the electronic configuration of given atoms. Also give the number of neutrons present in each atom: 199F, 2412Mg, 2814Si, 3115P, 3517Cl. / – : 199F, 2412Mg, 2814Si, 3115P, 3517Cl

Step-by-step Solution /

  • English:

Bohr-Bury Rules:

  1. The maximum number of electrons in a shell is 2n2, where n is shell number (K=2, L=8, M=18, N=32).
  2. Outermost shell can accommodate a maximum of 8 electrons.
  3. Inner shells are filled completely before filling outer shells.

Electronic Configurations and Neutrons (n = A – Z):

  1. Fluorine (199F): Z = 9, A = 19. Electronic Configuration: 2, 7. Neutrons = 19 – 9 = 10.
  2. Magnesium (2412Mg): Z = 12, A = 24. Electronic Configuration: 2, 8, 2. Neutrons = 24 – 12 = 12.
  3. Silicon (2814Si): Z = 14, A = 28. Electronic Configuration: 2, 8, 4. Neutrons = 28 – 14 = 14.
  4. Phosphorus (3115P): Z = 15, A = 31. Electronic Configuration: 2, 8, 5. Neutrons = 31 – 15 = 16.
  5. Chlorine (3517Cl): Z = 17, A = 35. Electronic Configuration: 2, 8, 7. Neutrons = 35 – 17 = 18.
  • Hindi:

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