Name of Element	Symbol	Atomic number	Number of Protons	Number of Neutrons	Number of electrons	Distribution of electrons K L M N				Valency
Hydrogen Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Magnesium	H He Li Be B C N O F Ne Na Mg	1 2 3 4 5 6 7 8 9 10 11 12	1 2 3 4 5 6 7 8 9 10 11 12	– 2 4 5 6 6 7 8 10 10 12 12	1 2 3 4 5 6 7 8 9 10 11 12	1 2 2 2 2 2 2 2 2 2 2 2	– – 1 2 3 4 5 6 7 8 8 8	– – – – – – – – – – 1 2	– – – – – – – – – — – –	1 0 1 2 3 4 3 2 1 0 1 2
Aluminium Silicon Phosphorus Sulphur Chlorine Argon	Al Si P S Cl Ar	13 14 15 16 17 18	13 14 15 16 17 18	14 14 16 16 18 22	13 14 15 16 17 18	2 2 2 2 2 2	8 8 8 8 8 8	3 4 5 6 7 8	– – – – –	3 4 3,5 2 1 0

Solution:

(a) To find the mass number of Oxygen,

Number of protons = 8

Number of neutrons = 8

Atomic number = 8

Atomic mass number = Number of protons + number of neutrons = 8 + 8 = 16

Therefore, the mass number of oxygen = 16

(b) To find the mass number of Sulphur,

Number of protons = 16

Number of neutrons = 16

Atomic number = 16

Atomic mass number = Number of protons + number of neutrons = 16 + 16 = 32

Exercise-4.6 Class 9 Science Chapter 4 Structure of The Atom

1. For the symbols H, D and T, tabulate three subatomic particles found in each of them.

Solution:

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

2. Write the electronic configuration of any one pair of isotopes and isobar.

Solution:

(a) Isotopes: Isotopes are atoms which have the same number of protons, but the number of neutrons differs. This leads to the variation in mass number too.

Example: Carbon molecule exists as 6C12 and 6C14, but when their electronic configuration is noticed, both have K-2; L-4

(b) Isobars: Isobars are atoms which have the same mass number but differ in atomic number. The electronic configuration of an isobar pair is as follows:

Example: Electronic configuration of 20Ca40 – K-2; L-8; M-8; N- 2

Electronic configuration of 18Ar40 – K-2; L-8; M-8

Exercise-4.7 Class 9 Science Chapter 4 Structure of The Atom

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

Solution:

Property	Electrons	Protons	Neutrons
Charge	Negatively charged	Positively charged	No charge.
Location	Located outside the nucleus	Located within the nucleus	Located inside the nucleus of an atom
Weight	Mass is negligible	1 a.m.u	1 a.m.u
Affinity	Attracted towards positively charged	Attracted towards negatively charged	Do not get attracted to any charged particle

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

Solution: The following are the limitations of J.J. Thomson’s model of an atom:

• The model failed to explain the outcome of alpha particle scattering, which was conducted by Rutherford. The model failed to depict why the majority of these alpha particles pass through the gold foil, while some diverted through small and big angles, while some others rebound completely, returning on their path.
• It did not provide any experimental evidence and was established on imagination.

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

Solution:

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.

4. Describe Bohr’s model of the atom. \textbf

{Solution:} Bohr’s model of the atom: \begin{enumerate} \item The atom has a nucleus at the center. \item Electrons revolve around the nucleus. \item Certain special orbits, known as discrete orbits of electrons, are allowed inside the atom. \item While revolving in discrete orbits, the electrons do not radiate energy. \item These orbits or shells are called energy levels. \item These orbits or shells are represented by the letters K, L, M, N, or the numbers \( n = 1, 2, 3, 4, \ldots \). \end{enumerate}

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

Solution:

Thomson	Rutherford	Bohr
● Sphere is positively charged. ● Electrons are negatively charged and scattered all through the inside of the sphere. ● Positively charged = negatively charged ● The net charge in the atom is zero.	● The nucleus is at the centre and is positively charged, holding the entire mass. ● Electrons are negatively charged, revolving in a well-defined path ● In comparison with the nucleus, the size of the atom is very large. ● Force of attraction of the electrons towards the nucleus is balanced by centrifugal force acting away from it. As a result, electrons are not drawn close to the nucleus.	● Nucleus is present at the centre and is positively charged ● Electrons are negatively charged, revolving around but do not radiate energy. ● The distinct orbits are labelled as K, L, M, and N

6. What is Thomson’s model of the atom? \textbf

{Solution:} Thomson’s model, also known as the “plum pudding model,” describes the atom as follows: \begin{itemize} \item According to this model, an atom can be considered as a sphere of uniformly distributed positive charge. \item Electrons are distributed symmetrically within this sphere. \item The electrons are held in place by the positive charge through electrostatic forces. The mutual repulsions between the electrons are balanced by the attractive forces towards the center of the sphere. \item In an atom with a single electron, such as the hydrogen atom, the electron is situated at the center of the positive sphere. \item In an atom with two electrons, like the helium atom, the two electrons are symmetrically positioned on opposite sides of the center, each at a distance equal to half the radius of the positive sphere. \item In a three-electron system, the electrons are placed at the corners of an equilateral triangle, with the side of the triangle equal to the radius of the sphere. \item Thomson’s model explains the arrangement of electrons from 1 to 100 within the positive sphere. \end{itemize} This model is often referred to as the “plum pudding model,” where the electrons are likened to plums embedded in a positively charged pudding, similar to seeds in a watermelon.

7. Explain the Rutherford model of an atom. \textbf

{Solution:} Rutherford’s model of the atom: \begin{itemize} \item \textbf{Introduction of the Nucleus:} Ernest Rutherford introduced the concept of the nucleus at the center of the atom through his famous gold foil scattering experiment. \item \textbf{Composition of the Nucleus:} The nucleus is composed of protons and neutrons, which together account for most of the atom’s mass. \item \textbf{Electron Orbits:} Rutherford proposed that electrons revolve around the nucleus in their orbits, similar to how planets revolve around the sun. \item \textbf{Electrostatic Force:} The electrostatic attractive force between the positively charged nucleus and the negatively charged electrons provides the necessary centripetal force for the electrons’ orbital motion. \item \textbf{Radiation and Energy Loss:} According to electromagnetic theory, an accelerated or decelerated electrically charged particle emits electromagnetic radiation and thus loses energy. Since the electron is continuously accelerating in its orbit due to the centripetal force, it would emit radiation, lose energy, and spiral inward. \item \textbf{Spiral Path:} As the electron loses energy, its orbit would decrease in radius, leading it to eventually spiral into the nucleus and disappear. \item \textbf{Limitations and Bohr’s Contribution:} Rutherford’s model had the drawback of predicting that electrons should continuously lose energy and spiral into the nucleus. To address this, Niels Bohr introduced the Bohr model with quantum postulates to explain stable electron orbits and prevent the collapse of the atom. \end{itemize}

Q8. Summarise the rules for the writing of the distribution of electrons in various shells for the first eighteen elements.

Solution:

• The maximum number of electrons that can be accommodated in a shell is given by the formula: 2n2, where n= 1, 2, 3…
• The maximum number of electrons in different shells are:

K shell – n=1 ; 2n2 = 2(1)2 = 2

L shell – n=2 ; 2n2 = 2(2)2 = 8

M shell – n=3 ; 2n2 = 2(3)2 = 18

N shell- n=4 ; 2n2 = 2(4)2 = 32

• The outermost orbit can be accommodated with 8 electrons at the maximum.
• The electrons are not taken in unless the inner shells are filled, which are filled step-wise; hence, the highest element has K-2; L-8; M-8 distribution of electrons.

9. Define valency by taking examples of silicon and oxygen.

Solution:

The definite combining capacity of the atoms of each element, wherein electrons are lost, gained or shared to make the octet of electrons present in the outermost shell, is defined as valency. To measure valency, we can figure out the number of electrons that are required to complete the shell in which it is contained or losing excess electrons, if present, once the filling is complete.

Example: To find the valency of silicon,

The atomic number of silicon is 14.

The number of electrons is equal to the number of protons in silicon, i.e., 14.

The distribution of electrons in silicon atoms is K – 2, L – 8, M – 4

Hence, from the distribution of silicon, it is clearly evident that to fill the M shell, 4 electrons are required. Therefore, its valency is 8-4=4

To find the valency of oxygen,

The atomic number of oxygen is 8.

The number of electrons is equal to the number of protons in oxygen, i.e., 8.

The distribution of electrons in oxygen atom is K – 2, L – 6

Hence, from the distribution of oxygen, it is clearly evident that to fill the M shell, 6 more electrons are required. Therefore, its valency is 8-6=2

10. Explain with examples

(i) Atomic number,

(ii) Mass number,

(iii) Isotopes and

(iv) Isobars.

Give any two uses of isotopes.

Solution:

(i) The number of positively charged protons present in the nucleus of an atom is defined as the atomic number and is denoted by Z. Example: Hydrogen has one proton in its nucleus; hence, its atomic number is one.

(ii) The total number of protons and neutrons present in the nucleus of an atom is known as the mass number. It is denoted by A. 20Ca40 . The mass number is 40. The atomic number is 20.

(iii) The atoms which have the same number of protons but a different number of neutrons are referred to as isotopes. Hence, the mass number varies.

Example: The most simple example is the Carbon molecule which exists as 6C12 and 6C14

(iv) Isobars: Isobars are atoms which have the same mass number but differ in atomic number.

Examples are, 20Ca40and 18Ar40

Uses of isotopes

• The isotope of the Iodine atom is used to treat goitre, an iodine-deficient disease.
• In the treatment of cancer, an isotope of cobalt is used.
• Fuel for nuclear reactors is derived from the isotopes of the Uranium atom.

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

Solution:

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.

12. 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.

Solution:

The atomic mass of an element is the mass of one atom of that element. Average atomic mass takes into account the isotopic abundance.

Isotope of bromine with atomic mass 79 u = 49.7%

Therefore, Contribution of 35Br79 to atomic mass = (79 × 49.7)/100

⇒ 39.26 u

Isotope of bromine with atomic mass 81 u = 50.3%

Contribution of 35Br81 to the atomic mass of bromine = (81 × 50.3)/100

⇒ 40.64u

Hence, the average atomic mass of the bromine atom = 39.26 + 40.64 u = 79.9u

13. 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?

Solution:

Let the percentage of 8X16 be ‘a’ and that of 8X18 be ‘100-a’.

As per the given data,

16.2u = 16 a / 100 + 18 (100-a) /100

1620 = 16a + 1800 – 18a

1620 = 1800 – 2a

a = 90%

Hence, the percentage of the isotope in the sample 8X16 is 90% and that of

8X18 = 100-a = 100- 90=10%

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

Solution:

Given: Atomic number, Z = 3

The electronic configuration of the element = K-2; L-1, hence its valency = 1

The element with atomic number 3 is Lithium.

15. 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?

Solution:

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.

16. 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.

Solution:

(a) The statement is False.

(b) The statement is False.

(c) The statement is True.

(d) The statement is False.

17. Put a tick(✓) against the correct choice and cross(x) against the wrong choice in questions 15, 16 and 17.

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

(a) Atomic nucleus

(b) Electron

(c) Proton

(d) Neutron

Solution:

(a) Atomic nucleus

Isotopes of an element have

(a) The same physical properties

(b) Different chemical properties

(c) Different number of neutrons

(d) Different atomic numbers

Solution:

(c) Different number of neutrons

18. Number of valence electrons in Cl– ion are

(a) 16

(b) 8

(c) 17

(d) 18

Solution:

(b) 8

The electronic distribution of Cl is K-2, L-8, M-7. Valence electrons are 7; hence, chlorine gains one electron for the formation of Cl–. Therefore, its valency is 8.

19. 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

Solution:

(d) 2, 8, 1

Complete the following table.

Atomic Number	Mass Number	Number of Neutrons	Number of Protons	Number of Electrons	Name of the Atomic Species
9 16 – – –	– 32 24 2 1	10 – – – 0	– – 12 1 1	– – – – 0	– Sulphur – – –

Solution:

The following table depicts the missing data:

Atomic number(Z) = Number of protons

Mass number = Number of neutrons + atomic number

(or)

Mass number(A) = Number of neutrons + number of neutrons

Atomic Number	Mass Number	Number of Neutrons	Number of Protons	Number of Electrons	Name of the Atomic Species
9 16 12 1 1	19 32 24 2 1	10 16 12 1 0	9 16 12 1 1	9 16 12 1 0	Fluorine Sulphur Magnesium Deuterium Hydrogen