Class 9 Science Chapter 4 question answer Structure of the atom

Ncert Solutions for Class 9 Science Chapter 4: Structure of the atom class 9 questions and answers

Textbook	Ncert
Class	Class 9
Subject	Science
Chapter	Chapter 4
Chapter Name	Structure of the atom class 9 ncert solutions
Category	Ncert Solutions
Medium	English

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In Text Questions Page No: 39

Question 1: What are canal rays?

Answer 1: Canal rays are positively charged radiations which led to the discovery of positively charged sub-atomic particle called proton.

Question 2: If an atom contains one electron and one proton, will it carry any charge or not?

Answer 2: An electron is a negatively charged particle, whereas a proton is a positively charged particle. The magnitude of their charges is equal. Therefore, an atom containing one electron and one proton will not carry any charge. Thus, it will be a neutral atom.

In Text Questions Page No: 41

Question 1: On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.

Answer 1: According to Thomson’s model of an atom
(i) An atom consists of a positively charged sphere and the electrons are embedded in it,
(ii) The negative and positive charges are equal in magnitude. So the atom is electrically neutral.

Question 2: On the basis of Rutherford’s model of an atom, which subatomic particle is present in the nucleus of an atom?

Answer 2: As per Rutherford’s model of an atom, the positively charged protons are the ones that are present in the atom.

Question 3: Draw a sketch of Bohr’s model of an atom with three shells.

Answer 3:

Question 4: What do you think would be the observation if the α-particle scattering experiment is carried out using a foil of a metal other than gold?

Answer 4: In such a case, almost all the alpha-particles would pass without hitting the atoms and hardly any alpha particle would be deflected.

In Text Questions Page No: 41

Question 1: Name the three sub-atomic particles of an atom.

Answer 1:  An atom consists of three subatomic particles:

• Protons  – Positively charged
• Electrons – Negatively charged
• Neutrons – Neutral in nature (no charge)

Question 2: Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?

Answer 2: Atomic mass = 4u

• Atomic mass = Number of neutrons + Number of protons
• 4 = Number of neutrons + 2
• Number of neutrons = 4 − 2
• Number of neutrons = 2
• ∴ The number of neutrons in a helium atom = 2.

In Text Questions Page No: 42

Question 1: Write the distribution of electrons in carbon and sodium atoms.

Answer 1: To write the distribution of electrons in atoms, we use the electronic configuration, which shows how electrons are arranged in different energy levels (or shells) around the nucleus.

1. Carbon (Atomic number = 6)

• Number of electrons = 6
• Electron distribution = 2, 4
  (2 electrons in the first shell, 4 in the second shell)

2. Sodium (Atomic number = 11)

• Number of electrons = 11
• Electron distribution = 2, 8, 1
  (2 in the first shell, 8 in the second shell, and 1 in the third shell)

Question 2: If K and L shells of an atom are full, then what would be the total number of electrons in the atom?

Answer 2: K shell can hold 2 electrons and L shell can hold 8 electrons.When both the shells are full, there will be (8 + 2) 10 electrons in the atom.

In Text Questions Page No: 44

Question 1: How will you find the valency of chlorine, sulphur and magnesium?

Answer 1: To find the valency of an element, we need to look at the number of electrons in its outermost shell (also called the valence electrons). Valency is usually the number of electrons an atom needs to gain, lose, or share to complete its outermost shell (usually 8 electrons – the octet rule).

1. Chlorine (Atomic number = 17)

• Electron distribution = 2, 8, 7
• Outer shell electrons = 7
• To complete 8, it needs 1 more electron
  Valency = 1

2. Sulphur (Atomic number = 16)

• Electron distribution = 2, 8, 6
• Outer shell electrons = 6
• Needs 2 more electrons to complete 8
  Valency = 2

3. Magnesium (Atomic number = 12)

• Electron distribution = 2, 8, 2
• Has 2 electrons in the outer shell, which it loses easily
  Valency = 2

In Text Questions Page No: 44

Question 1: If number of electrons in an atom is 8 and number of protons is also 8, then (i) what is the atomic number of the atom? and (ii) what is the charge on the atom?

Answer 1: (i) The atomic number is equal to the number of protons. Therefore, the atomic number of the atom is 8.

(ii) Since the number of both electrons and protons is equal, therefore, the charge on the atom is 0.

Question 2: With the help of Table 4.1, find out the mass number of oxygen and sulphur atom.

Composition of Atoms of the First Eighteen Elements with Electron Distribution in Various Shells
Name of Element	Symbol	Atomic Number	Number of protons	Number of Neutrons	Number of Electrons	Distribution of Electrons				Valency
						K	L	M	N
Hydrogen	H	1	1	–	1	1	–	–	–	1
Helium	He	2	2	2	2	2	–	–	–	0
Lithium	Li	3	3	4	3	2	1	–	–	1
Beryllium	Be	4	4	5	4	2	2	–	–	2
Boron	B	5	5	6	5	2	3	–	–	3
Carbon	C	6	6	6	6	2	4	–	–	4
Nitrogen	N	7	7	7	7	2	5	–	–	3
Oxygen	O	8	8	8	8	2	6	–	–	2
Fluorine	F	9	9	10	9	2	7	–	–	1
Neon	Ne	10	10	10	10	2	8	–	–	0
Sodium	Na	11	11	12	11	2	8	1	–	1
Magnesium	Mg	12	12	12	12	2	8	2	–	2
Aluminium	Al	13	13	14	13	2	8	3	–	3
Silicon	Si	14	14	14	14	2	8	4	–	4
Phosphorus	P	15	15	16	15	2	8	5	–	3.5
Sulphur	S	16	16	16	16	2	8	6	–	2
Chlorine	Cl	17	17	18	17	2	8	7	–	1
Argon	Ar	18	18	22	18	2	8	8		0

Answer 2: The number of protons and neutrons present in the nucleus of an atom is called mass number.

That is, mass number = number of protons + number of neutrons

Mass number = 8 + 8 = 16

Sulfur protons = 16, neutrons = 16

Mass number = 16 + 16 = 32

In Text Questions Page No: 45

Question 1: For the symbol H,D and T tabulate three sub-atomic particles found in each of them.

Answer 1: The following table depicts the subatomic particles in Hydrogen (H), Deuterium (D), and Tritium(T).

Isotope	Symbol	Mass no.	Atomic no.	No. of electrons	No. of protons	No. of neutrons
Hydrogen	H	1	1	1	1	0
Deuterium	D	2	1	1	1	1
Tritium	T	3	1	1	1	2

Question 2: Write the electronic configuration of any one pair of isotopes and isobars.

Answer 2: Isotopes Example: Hydrogen Isotopes

Isotopes are atoms of the same element with the same atomic number but different mass numbers.

Pair:

• Protium (¹H) → Atomic number = 1, Mass number = 1
• Deuterium (²H) → Atomic number = 1, Mass number = 2

✅ Electronic configuration for both:

• Atomic number = 1 → 1 electron
  👉 Electronic configuration = 1 (1 electron in the first shell)

They have the same configuration, because they have the same number of electrons.

Isobars Example: Calcium and Argon

Isobars are atoms of different elements with the same mass number, but different atomic numbers.

Pair:

• Calcium (⁴⁰Ca) → Atomic number = 20
• Argon (⁴⁰Ar) → Atomic number = 18

✅ Electronic configurations:

• Calcium (20 electrons): 2, 8, 8, 2
• Argon (18 electrons): 2, 8, 8

Same mass number, but different electronic configurations because they are different elements.

Exercises

Question 1: Compare the properties of electrons, protons and neutrons.

Answer 1:

Particle	Nature of Charge	Mass	Location
Electron	Electrons are negatively charged.	9 x 10–31 kg	Extra nuclear part distributed in different shell or orbits.
Proton	Protons are positively charged.	1.672 x 10–27 kg (1 µ) (approx. 2000 times that of the electron)	Nucleus
Neutron	Neutrons are neutral.	Equal to mass of proton	Nucleus

Question 2: What are the limitations of J.J. Thomson’s model of the atom?

Answer 2: According to J.J. Thomson’s model of an atom, the electrons are embedded all over in the positively charged spheres. But experiments done by other scientists showed that protons are present only in the centre of the atom and electrons are distributed around it.

Question 3: What are the limitations of Rutherford’s model of the atom?

Answer 3: The following are the limitations of Rutherford’s model of the atom:

• There is no expected stability in the revolution of the electron in a circular orbit.
• Charged particles radiate energy when accelerated, thus causing the revolving electrons to lose energy and would fall into the nucleus.
• Hence, atoms must be highly unstable. The matter would not exist in its known form, which clearly is an assumption as atoms are highly stable.

Question 4: Describe Bohr’s model of the atom.

Answer 4: Bohr’s Model of the Atom (by Niels Bohr, 1913)

Bohr proposed a model to explain how electrons are arranged in an atom. His model improved upon Rutherford’s model by explaining electron stability and energy levels.

Main Postulates of Bohr’s Model:

1. Electrons revolve around the nucleus in fixed circular paths called orbits or energy levels (shells) without radiating energy.
2. Each orbit has a fixed energy, so they are also called energy levels.
   These orbits are labeled as K, L, M, N… or n = 1, 2, 3, 4…
3. Electrons do not lose energy while moving in a particular orbit.
4. Energy is absorbed or emitted only when an electron jumps from one orbit to another:
   • Moving to a higher orbit → absorbs energy
   • Falling to a lower orbit → emits energy

Question 5: Compare all the proposed models of an atom given in this
chapter

Answer 5: Comparison of Atomic Models

Model	Scientist	Main Features	Limitations
Dalton’s Model	John Dalton (1808)	– Atom is a solid, indivisible sphere. – All atoms of an element are identical.	– Did not explain internal structure of atoms or subatomic particles.
Thomson’s Model	J.J. Thomson (1897)	– Atom is a positively charged sphere with electrons embedded in it (“plum pudding model”).	– Couldn’t explain results of Rutherford’s experiment.
Rutherford’s Model	Ernest Rutherford (1911)	– Atom has a small, dense, positively charged nucleus. – Electrons revolve around it.	– Couldn’t explain why electrons don’t fall into the nucleus.
Bohr’s Model	Niels Bohr (1913)	– Electrons revolve in fixed energy levels or orbits. – No energy loss in stable orbits.	– Could not explain complex atoms or electron behavior in magnetic fields.

Question 6: Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.

Answer 6: The rules for writing of the distribution of electrons in various shells for the first eighteen elements are given below.

• If n gives the number of orbit or energy level, then 2n² gives the maximum number of electrons possible in a given orbit or energy level. Thus,
  • First orbit or K-shell will have 2 electrons,
  • Second orbit or L-shell will have 8 electrons,
  • Third orbit or M-shell will have 18 electrons.
• If it is the outermost orbit, then it should have not more than 8 electrons.
• There should be step-wise filling of electrons in different orbits, i.e., electrons are not accompanied in a given orbit if the earlier orbits or shells are incompletely filled.

Question 7: Define valency by taking examples of silicon and oxygen.

Answer 7: Valency refers to the number of electrons obtained, lost, or shared to complete the octet of electrons in the valence shell.

Valency of silicon: It has electronic configuration → 2, 8, 4

Thus, 4 electrons are shared with other atoms to complete the octet, and so its valency = 4.

Valency of oxygen: It has electronic configuration → 2, 6

Thus, It will gain 2 electrons to complete its octet. So its valency = 2.

Question 8: Explain with examples (i) Atomic number, (ii) Mass number, (iii) Isotopes and iv) Isobars. Give any two uses of isotopes.

Answer 8: (i) Atomic Number (Z):

• Definition: The number of protons in the nucleus of an atom.
• It also equals the number of electrons in a neutral atom.
• 🔸 Example:
  • Hydrogen (H): Atomic number = 1
  • Carbon (C): Atomic number = 6

(ii) Mass Number (A):

• Definition: The sum of protons and neutrons in the nucleus.
  Mass Number = Number of Protons + Number of Neutrons
• 🔸 Example:
  • Carbon-12: 6 protons + 6 neutrons = 12
  • Oxygen-16: 8 protons + 8 neutrons = 16

(iii) Isotopes:

• Definition: Atoms of the same element with the same atomic number but different mass numbers.
• They differ in the number of neutrons.
• 🔸 Example:
  • Hydrogen Isotopes:
    • Protium (¹H), Deuterium (²H), Tritium (³H)

(iv) Isobars:

• Definition: Atoms of different elements with the same mass number but different atomic numbers.
• 🔸 Example:
  • Calcium-40 (²⁰Ca⁴⁰) and Argon-40 (¹⁸Ar⁴⁰)
    • Mass number = 40 for both, but different elements

Two Uses of Isotopes:

1. Medical Use:
   • Cobalt-60 is used in cancer treatment (radiotherapy).
2. Industrial Use:
   • Uranium-235 is used as fuel in nuclear reactors for power generation.

Question 9: Na⁺ has completely filled K and L shells. Explain.

Answer 9: The atomic number of sodium is 11. It has 11 electrons in its orbitals, wherein the number of protons is equal to the number of electrons. Hence, its electronic configuration is K-2 ; L-8 ; M-1 ; The one electron in the M shell is lost, and it obtains a positive charge since it has one more proton than electrons and obtains a positive charge, Na+ . The new electronic configuration is K-1; L-8, which is the filled state. Hence, it is very difficult to eliminate the electron from a filled state as it is very stable.

Question 10: If the bromine atom is available in the form of, say, two isotopes  ₃₅Br⁷⁹ (49.7%) and ₃₅Br⁸¹ (50.3%), calculate the average atomic mass of the Bromine atom.

Answer 10: To calculate the average atomic mass of bromine based on its isotopes, we use the formula:

Average atomic mass = \(\frac{(m_1 \times \%_1) + (m_2 \times \%_2)}{100}\)

Where:

• m₁ and m₂ = mass numbers of the isotopes
• %₁ and %₂ = their respective percentages

Given:

• Isotope 1: \( ^{79}_{35}Br \) → mass = 79, abundance = 49.7%
• Isotope 2: \( ^{81}_{35}Br \) → mass = 81, abundance = 50.3%

Calculation:

Average atomic mass = \(\frac{(79 \times 49.7) + (81 \times 50.3)}{100}\)
= \(\frac{3936.3 + 4074.3}{100}\)
= \(\frac{8010.6}{100}\)
= \(\boxed{80.1 \, \text{u}}\)

The average atomic mass of bromine is 80.1 u (atomic mass units).

Question 11: The average atomic mass of a sample of element X is 16.2 u. What are the percentages of isotopes  ₈X¹⁶ and ₈X¹⁸ in the sample?

Answer 11: To find the percentages of isotopes of element X (likely oxygen) with mass numbers 16 and 18, we use the concept of weighted average.

Given:

• Average atomic mass of element X = 16.2 u
• Isotopes:
• \( ^{16}_8X \): mass = 16 u
• \( ^{18}_8X \): mass = 18 u

Let the percentage of \( ^{16}X \) be x%,
then the percentage of \( ^{18}X \) will be (100 − x)%

Use the formula:

• Average mass = \(\frac{(16 \times x) + (18 \times (100 – x))}{100}\)
• \(16.2 = \frac{16x + 1800 – 18x}{100}\)
• \(16.2 = \frac{-2x + 1800}{100}\)
• \(16.2 \times 100 = -2x + 1800\)
• \(1620 = -2x + 1800\)
• \(-2x = 1620 – 1800 = -180\)
• \(x = \frac{180}{2} = 90\)

• Percentage of \( ^{18}X \) = 10%
• Percentage of \( ^{16}X \) = 90%

Question 12: If Z = 3, what would be the valency of the element? Also, name the element.

Answer 12: By Z = 3, we mean that the atomic number of the element is 3. Its electronic configuration is 2, 1. Hence, the valency of the element is 1 (since the outermost shell has only one electron).
Therefore, the element with Z = 3 is lithium.

Question 13: The composition of the nuclei of two atomic species X and Y are given as under

	X	Y
Protons	6	6
Neutrons	6	8

Give the mass numbers of X and Y. What is the relation between the two species?

Answer 13: Mass number of X: Protons + neutrons = 6+6 = 12

Mass number of Y: Protons + neutrons = 6+8 = 14

They are the same element, and their atomic numbers are the same.

They are isotopes, as they differ in the number of neutrons and hence their mass numbers.

Question 14: For the following statements, write T for true and F for false.
(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
(c) The mass of an electron is about 1/2000 times that of a proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.

Answer 14:

• (a) The statement is False.
• (b)  The statement is False.
• (c)  The statement is True.
• (d)  The statement is False.

Put tick (¸) against correct choice and cross (×) against wrong choice in questions 15, 16 and 17

Question 15: Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) Atomic nucleus
(b) Electron
(c) Proton
(d) Neutron

Answer 15: (a) Atomic nucleus

Question 16: Isotopes of an element have
(a) the same physical properties
(b) different chemical properties
(c) different number of neutrons
(d) different atomic numbers

Answer 16: (c) different number of neutrons

Question 17: Number of valence electrons in Cl ^–ion are:
(a) 16
(b) 8
(c) 17
(d) 18

Answer 17: (b) 8

Question 18: Which one of the following is a correct electronic configuration of sodium?
(a) 2, 8
(b) 8, 2, 1
(c) 2, 1, 8
(d) 2, 8, 1

Answer 18: (d) 2, 8, 1

Question 19: Complete the following table.

Atomic number	Mass number	Number of Neutrons	Number of protons	Number of electrons	Name of the Atomic species
9	–	10	–	–	–
16	32	–	–	–	Sulphur
–	24	–	12	–	–
–	2	–	1	–	–
–	1	0	1	0	–

Answer 19:

Mass number	Mass number	Number of Neutrons	Number of protons	Number of electrons	Name of the Atomic species
9	19	10	9	9	Fluorine
16	32	16	16	16	Sulphur
12	24	12	12	12	Magnesium
1	2	1	1	1	Deuterium
1	1	0	1	0	Hydrogen ion