Class 9th Science Chapter-4 Exercises

Structure of Atom

Exercise – Solutions


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

Property Electron Proton Neutron
Symbol e⁻ p⁺ n
Charge –1 +1 0
Mass

Very small (≈ 1/2000 of proton)

1 u 1 u
Position in atom Outside nucleus (shells) Inside nucleus

Inside nucleus


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

Answer:

  • It could not explain the results of Rutherford’s α-particle scattering experiment.

  • It failed to explain how positive charge is distributed in the atom.

  • It did not explain the stability of the atom.


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

Answer:

  • According to classical physics, revolving electrons should lose energy and fall into the nucleus.

  • Hence, the atom should be unstable, but atoms are stable in reality.

  • It failed to explain the arrangement of electrons in shells.


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

Answer:

Bohr proposed the following postulates:

  1. Electrons revolve around the nucleus in fixed circular orbits called energy levels or shells.

  2. Each orbit has a definite energy.

  3. Electrons do not radiate energy while revolving in these shells.

  4. Energy is emitted or absorbed only when an electron jumps from one shell to another.

This model successfully explained the stability of the atom.


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

Model Scientist Main idea Limitation
Thomson’s model J.J. Thomson Atom is a positive sphere with electrons embedded

Could not explain scattering experiment

Rutherford’s model E. Rutherford

Atom has a central nucleus with electrons revolving

Could not explain stability
Bohr’s model Niels Bohr

Electrons revolve in fixed energy shells

Valid only for simple atoms

Question 6. Summarise the rules for distribution of electrons in shells (Bohr–Bury rules).

Answer:

  1. Maximum electrons in a shell = 2n², where n is shell number.

  2. Maximum electrons in the outermost shell = 8.

  3. Electrons fill shells step by step, inner shells first.


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

Answer:

Valency is the combining capacity of an atom.

  • Silicon (Atomic no. 14):
    Electronic configuration = 2, 8, 4
    Valency = 4 (needs 4 electrons to complete octet)

  • Oxygen (Atomic no. 8):
    Electronic configuration = 2, 6
    Valency = 2 (needs 2 electrons to complete octet)

Question 8

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

Answer:

(i) Atomic number (Z):
Atomic number is the number of protons present in the nucleus of an atom.

Example:
Carbon has 6 protons → Atomic number = 6


(ii) Mass number (A):
Mass number is the sum of protons and neutrons present in the nucleus.

Example:
Carbon has 6 protons and 6 neutrons → Mass number = 12


(iii) Isotopes:
Isotopes are atoms of the same element having the same atomic number but different mass numbers.

Example:
¹₁H, ²₁H, ³₁H (isotopes of hydrogen)


(iv) Isobars:
Isobars are atoms of different elements having the same mass number but different atomic numbers.

Example:
⁴⁰₂₀Ca and ⁴⁰₁₈Ar

Two uses of isotopes:

  1. Cobalt-60 is used in the treatment of cancer.

  2. Uranium-235 is used as fuel in nuclear reactors.


Question 9

Na⁺ has completely filled K and L shells. Explain.

Answer:

Atomic number of sodium = 11
Electronic configuration of Na = 2, 8, 1

Sodium loses one electron to form Na⁺ ion.

Electronic configuration of Na⁺ = 2, 8

Hence, K and L shells are completely filled.


Question 10

If bromine atom is available as two isotopes ⁷⁹₃₅Br (49.7%) and ⁸¹₃₅Br (50.3%), calculate the average atomic mass of bromine.

Answer:

Average atomic mass=(79×49.7)+(81×50.3)100
=3926.3+4074.3100=8000.6100=80.0 u


Question 11

The average atomic mass of an element X is 16.2 u. What are the percentages of isotopes ¹⁶₈X and ¹⁸₈X in the sample?

Answer:

Let percentage of ¹⁶₈X = x
Then percentage of ¹⁸₈X = 100 − x

16x+18(100x)100=16.2
16x+180018x=1620
2x=180x=90

Percentages:

  • ¹⁶₈X = 90%

  • ¹⁸₈X = 10%


Question 12

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

Answer:

Atomic number Z = 3 → Element is Lithium (Li)
Electronic configuration = 2, 1

Valency = 1


Question 13

Composition of the nuclei of two atomic species X and Y is given below:

Species Protons Neutrons
X 6 6
Y 6 8

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

Answer:

  • Mass number of X = 6 + 6 = 12

  • Mass number of Y = 6 + 8 = 14

Relation:
X and Y are isotopes of the same element.


Question 14

For the following statements, write T for True and F for False:

(a) J.J. Thomson proposed that the nucleus contains only nucleons.
F

(b) A neutron is formed by an electron and a proton combining together.
F

(c) The mass of an electron is about 1/2000 times that of proton.
T

(d) An isotope of iodine is used for making tincture iodine.
F


Question 15

Rutherford’s alpha-particle scattering experiment was responsible for the discovery of:

(a) Atomic nucleus
(b) Electron
(c) Proton
(d) Neutron

Answer:

(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:

(c) different number of neutrons


Question 17

Number of valence electrons in Cl⁻ ion are:

(a) 16
(b) 8
(c) 17
(d) 18

Answer:

(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:

(d) 2,8,1

Explanation (for understanding):
Sodium has atomic number 11, so it has 11 electrons.
These are distributed as:

  • K shell = 2

  • L shell = 8

  • M shell = 1

Hence, the correct electronic configuration is 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 (Completed Table):

Atomic 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 (Hydrogen isotope)

1 1 0 1 0

Proton (H⁺ ion)


Key rules used (for exam clarity):

  • Atomic number = number of protons

  • Mass number = protons + neutrons

  • For neutral atoms: electrons = protons

  • If electrons = 0: it is a positive ion

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