Introduction

NCERT Solutions for Class 9 Science Chapter 12, Patterns in Life: Diversity and Classification, help students understand the remarkable variety of living organisms and the importance of organising them into meaningful groups. This chapter from the new NCERT textbook, Exploration: A Textbook of Science for Grade 9, explores the patterns of similarities and differences found among living organisms.

The solutions provide clear explanations for textbook questions and activities, helping students understand the basis of classification, identify important characteristics of different groups of organisms, and revise the concepts effectively for examinations.

Chapter Overview

Class 9 Science Chapter 12 – Patterns in Life: Diversity and Classification

The living world contains an enormous variety of organisms. This chapter explores how scientists identify patterns in this diversity and use similarities and differences to classify organisms into groups.

Students learn why classification is necessary and how organisms can be grouped according to their characteristics. The chapter helps develop an understanding of biological diversity and the relationships between different groups of organisms.

Use this section to introduce the specific classification systems, characteristics, examples, and biological groups covered in the current NCERT textbook.

NCERT Intext Solution

(For exact question see NCERT Textbook)

1. Question

(i) The positive charge on the clay is lesser than the total negative charge of the beads?
→ The atom would have an overall negative charge because the negative charge would be greater than the positive charge. It would no longer be a neutral atom.

(ii) By mistake, the clay itself carries a bit of negative charge? Would your model still represent a neutral atom?
→ No. The atom would not be neutral unless the total positive and negative charges are exactly equal. Extra negative charge on the clay would make the atom negatively charged.

2. Question

Matches Thomson’s idea:

• The soft pulp can represent the positively charged material.
• The seeds can represent electrons embedded in it.
• Seeds are spread throughout the pulp, similar to electrons in Thomson’s model.

Falls short:

• Seeds are much larger than electrons.
• The fruit does not represent electrical charges.
• Electrons are extremely tiny and have specific properties, unlike seeds.
• The distribution of seeds in a fruit is not exactly like the distribution of electrons in an atom.

3. Question

→ Thomson observed that cathode rays were produced regardless of the type of gas used or the material of the electrodes. Since the same negatively charged particles were always obtained, he concluded that these particles (electrons) are common to all atoms and are a fundamental component of matter.

4. Question

→ Negatively charged particles would be attracted towards the positively charged nucleus. As a result, many particles would be deflected more strongly, and some might even be captured by the atom. The pattern of scattering would be very different from that observed with positively charged α-particles.

5. Question

→ According to Thomson’s model, positive charge was spread uniformly throughout the atom. Therefore, α-particles should have passed through with only small deflections. The fact that a few α-particles bounced back sharply showed that the atom contains a tiny, dense, positively charged nucleus capable of strongly repelling α-particles. This observation could not be explained by Thomson’s model, so it was rejected.

6. Question

Sample Answer:
“How did you feel when you first observed some α-particles bouncing back from the gold foil, and did you immediately realize that it meant the atom had a nucleus?”

(Any thoughtful question related to his experiment is acceptable.)

7.Question

Answer: (ii) Both A and R are true, but R is not the correct explanation of A.

Explanation:

Assertion (A): Rutherford concluded that most of the mass of an atom is concentrated in a small region at the centre called the nucleus.
True

Reason (R): According to Thomson’s model, electrons are embedded in a uniformly distributed positive charge sphere.
True

However, the reason only describes Thomson’s atomic model and does not explain why Rutherford concluded that mass is concentrated in the nucleus. Rutherford’s conclusion came from the gold foil experiment, where a few α-particles were deflected sharply or bounced back.

Correct Option: (ii) Both A and R are true, but R is not the correct explanation of A.

8. Question

Answer:
If I discovered a new element, I could name it Tusharium (after Tushar).

Symbol: Tu

Justification:

• The symbol consists of one capital letter followed by one lowercase letter.
• It is derived from the element’s name.
• It follows the IUPAC convention for writing element symbols.

9. Question

Answer:
If different scientists used different symbols for the same element:

1. There would be confusion in scientific communication.
2. Scientists from different countries might misinterpret chemical formulas and equations.
3. Research findings would become difficult to understand and compare.
4. Textbooks and scientific records would lack uniformity.
5. International collaboration in science would become more difficult.

Therefore, a standard system of symbols (IUPAC) is essential for clear and universal communication.

10. Question

• Atomic number (Z) = 26
• Mass number (A) = 56

Protons = 26
Electrons = 26 (for a neutral atom)
Neutrons = A − Z = 56 − 26 = 30

Answer:

• Electrons = 26
• Protons = 26
• Neutrons = 30

11. Question

• Protons = 20
• Mass number = 41

Neutrons = 41 − 20 = 21

Answer: 21 neutrons

12. Question

• Atomic number = 17 (Protons = 17)
• Neutrons = 18

Mass number = Protons + Neutrons = 17 + 18 = 35

Answer: 35

13. Question

• Mass number = 23
• Electrons = 11

For a neutral atom:

Protons = Electrons = 11

Neutrons = Mass number − Protons = 23 − 11 = 12

Answer: 12 neutrons

14. Question

(i) (Carbon)
Electronic configuration = 2, 4
Outermost shell electrons = 4

(ii) (Fluorine)
Electronic configuration = 2, 7
Outermost shell electrons = 7

(iii)  (Silicon)
Electronic configuration = 2, 8, 4
Outermost shell electrons = 4

15. Question

16. Question

• Mass number = 23
• Number of protons = 11

Since atomic number = number of protons = 11, the element is Sodium (Na).

Number of neutrons = Mass number − Atomic number

= 23 − 11

= 12

Answer:

• Element: Sodium (Na)
• Number of neutrons: 12

17. Question

• Number of protons = 11 for both atoms.
• Therefore, their atomic number = 11 (same for both).

Mass number = Protons + Neutrons

• First atom:
• Second atom:

Answer:

• Atomic numbers: Same (11)
• Mass numbers: Different (23 and 24)
• They are the same element (Sodium) but are different isotopes.

18. Question

Average atomic mass

Answer:

Average atomic mass of bromine = 80.006 u (≈ 80.0 u).

NCERT Exercise Solution

(For exact question see NCERT Textbook)

1. Question

(i) Incorrect

Reason: Rutherford’s gold foil experiment did not discover neutrons. Neutrons were discovered later by James Chadwick (1932).

(ii) Correct

Reason: Large-angle deflections of a few α-particles showed that positive charge is concentrated in a small central region, disproving Thomson’s plum pudding model.

(iii) Correct

Reason: Rutherford concluded that almost all the mass and positive charge are concentrated in a very small region called the nucleus.

(iv) Incorrect

Reason: Rutherford’s experiment provided evidence only for the existence of the nucleus. It did not explain the arrangement of electrons around it.

2. Question

(i) Incorrect

Reason: According to Bohr, electrons in fixed orbits do not lose energy.

(ii) Incorrect

Reason: Electrons occupy only specific orbits with definite energies.

(iii) Correct

Reason: These stable paths are called stationary orbits.

(iv) Incorrect

Reason: Electrons cannot exist between two energy levels; they jump from one level to another by absorbing or emitting energy.

3. Question

(i) Y and Z

Number of protons = 17 (same)

Mass numbers:

• Y = 17 + 18 = 35
• Z = 17 + 20 = 37

Y and Z are isotopes because they have the same atomic number but different mass numbers.

(ii) Z and X

Mass numbers:

• Z = 17 + 20 = 37
• X = 18 + 19 = 37

Atomic numbers are different (17 and 18).

Z and X are isobars because they have the same mass number but different atomic numbers.

4. Question

Rutherford concluded that:

• The positive charge and almost all the mass of an atom are concentrated in a very small, dense central region called the nucleus.
• The nucleus is positively charged.
• Most of the atom is empty space.
• Alpha particles that came very close to the nucleus experienced strong repulsion and were deflected or bounced back.

Final Answers (in brief)

1. (i) ✗, (ii) ✓, (iii) ✓, (iv) ✗

2. (i) ✗, (ii) ✗, (iii) ✓, (iv) ✗

3.(i) Y and Z → Isotopes

• (ii) Z and X → Isobars

4. Positive charge and most of the mass are concentrated in a tiny, dense, positively charged nucleus at the centre of the atom, while most of the atom is empty space.

5. Question

Correct order of evolution of atomic models:

1. (iv) Dalton’s model described atoms as indivisible particles.
2. (ii) Thomson’s model proposed the “plum pudding” model.
3. (iii) Rutherford’s model proposed a dense central nucleus.
4. (i) Bohr’s model proposed fixed orbits with definite energies.

Chronological order:

(iv) → (ii) → (iii) → (i)

6. Question

Electrons are negatively charged and the nucleus contains positively charged protons. The electrostatic force of attraction between opposite charges keeps electrons bound to the nucleus. According to Bohr’s model, electrons revolve in fixed energy levels and do not lose energy while moving in these orbits.

7. Question

Answer: (ii) Both A and R are true, but R is not the correct explanation of A.

Explanation: The discovery of electrons, protons, and neutrons led to a better understanding of atomic structure. Although the number of electrons equals the number of protons in a neutral atom, this fact does not explain the assertion.

8. Question

(i) Number of protons

= Atomic number = 12

(ii) Number of neutrons

= Mass number – Atomic number
= 24 – 12 = 12

(iii) Number of electrons

= Number of protons = 12

Arrangement of electrons

K shell = 2 electrons
L shell = 8 electrons
M shell = 2 electrons

Electronic configuration = 2, 8, 2

9. Question

Electronic configurations:

• (a) He: 2
• (b) N: 2, 5
• (c) Al: 2, 8, 3
• (d) O: 2, 6

These are the answers according to the Bohr models shown in Fig. 8.17.

10. Question

Rutherford’s model failed because according to classical physics, revolving electrons should continuously lose energy and eventually fall into the nucleus, making the atom unstable.

Bohr’s model succeeded because it proposed that electrons revolve in fixed energy levels (orbits) without losing energy. Hence, atoms remain stable.

11. Question

Mass number = 70

Number of electrons = Number of protons = 31

Number of neutrons = Mass number – Number of protons

= 70 – 31

= 39

Answer: 39 neutrons

12. An atom has 79 protons and a mass number of 197. Calculate:

(i) Number of neutrons

= 197 – 79

= 118

(ii) Number of electrons

= Number of protons

= 79

Answer: Neutrons = 118, Electrons = 79

13. Question

14. Question

(i) Number of protons and electrons

Protons = 35 – 18 = 17

Electrons = 17

(ii) Atomic number

= Number of protons = 17

(iii) Identify the element X

Atomic number 17 corresponds to Chlorine (Cl).

(iv) Electronic configuration 2, 8, 7

(v) Number of valence electrons = 7

(vi) Mass number if two neutrons are added

New mass number = 35 + 2 = 37

(vii) Relation of X with the new atom

The new atom has the same atomic number but different mass number.

Therefore, they are isotopes of each other.

15. Question

(i) Atomic number

Atomic number depends only on the number of protons.

Atomic number = 12 (unchanged)

(ii) Atomic mass

Since the new particles are much heavier than electrons, the atomic mass will increase.

(iii) Mass number

Mass number = Number of protons + Number of neutrons = 12 + 12 = 24

Hence, mass number remains unchanged.

(iv) Overall charge

Since the replacing particles have the same negative charge and are equal in number to electrons, the atom remains electrically neutral.

Overall charge = Zero (neutral)

Common Mistakes to Avoid in Chapter 12

Students should avoid:

• Confusing classification with simply naming organisms.
• Memorising groups without learning their characteristics.
• Mixing up characteristics of different groups.
• Giving incorrect examples.
• Confusing similarities with differences used for classification.
• Drawing an incorrect classification chart.
• Writing incomplete characteristics when answering classification questions.
• Assuming that all organisms within a broad group have exactly the same features.

FAQs

FAQs on Class 9 Science Chapter 12

What is Chapter 12 of Class 9 Science?

Chapter 12 of the new Class 9 Science textbook is Patterns in Life: Diversity and Classification. It introduces students to the diversity of living organisms and the principles used to organise them into groups.

What is biological diversity?

Biological diversity refers to the variety of living organisms found in the world. Organisms differ in their structures, characteristics, habitats, and ways of life.

Why is classification important?

Classification helps scientists organise the enormous diversity of living organisms into manageable groups based on shared characteristics. It makes the study and identification of organisms easier.

What is the basis of classification?

Classification is based on similarities and differences in characteristics. The exact characteristics used depend on the classification system being discussed in the textbook.

What are the important topics in Chapter 12?

Important topics include diversity among organisms, the need for classification, characteristics used for grouping organisms, and the classification groups and examples covered in the current NCERT chapter.

How should I prepare Class 9 Science Chapter 12?

Read the chapter carefully and focus on the characteristics of different groups. Make comparison tables, practise classification questions, study diagrams or charts, and solve the textbook exercises using NCERT Solutions for checking.

Are NCERT Solutions enough to study Chapter 12?

NCERT Solutions are useful for understanding and checking textbook questions, but students should also study the complete chapter and memorise the important characteristics and examples through understanding.

Is Chapter 12 important for the Class 9 Science exam?

Yes. Students should understand the basis of classification and the characteristics of the groups discussed in the chapter. Classification-based questions and comparisons require both understanding and accurate recall.