## Structure Of The Atom Notes And CBSE Solutions of class 9th Science Chapter 4

• • January 6, 2021
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### Exercise Page: 54

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

Solution:

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

Solution:

The following are the limitations of the 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 majority of these alpha particles pass through gold foil while some are diverted through small and big angles, while some others rebound completely, returning back 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:

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. Matter would not exist in their known form which clearly is an assumption as atoms are highly stable.

4. Describe Bohr’s model of the atom.

Solution:

• An atom holds the nucleus at the centre.
• Negatively charged electrons revolve around the nucleus.
• The atoms in it contains distinct orbits of electrons.
• Electrons do not radiate energy when they are in their orbits.
• The distinct orbits are named as K, L, M, N orbits. Numbers used to denote them are n=1, 2, 3, 4

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

Solution:

6. Thomson’s Model of Atom

7. Rutherford’s Model of Atoms

8. Bohr’s model of the atom

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

Solution:

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

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

The distribution of electrons in silicon atom 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

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 . Mass number is 40. Atomic number is 20.

(iii) The atoms which have the same number of protons but 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 the atomic number.

Examples are,  20Ca40and  18Ar40

Uses of isotopes:

• The isotope of Iodine atom is used to treat goitre and 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 bromine atom is available in the form of, say, two isotopes  35Br79 (49.7%) and 35Br81 (50.3%), calculate the average atomic mass of Bromine atom.

Solution:

The atomic masses of two isotopic atoms are 79 (49.7%) and 81 (50.3%).

Thus, total mass = (79 * 49.7 / 100) + (81 * 50.3 / 100) = 39.263 + 40.743 = 80.006 u

13. The average atomic mass of a sample of an 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 given data,

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

1620 = 16a + 1800 – 18a

1620 = 1800 – 2a

a = 90%

Hence, the percentage of 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 as 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 proton.

(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.

Solution:

(a) Statement is False

(b)  Statement is False

(c)  Statement is True

(d)  Statement is False

17. Put a tick(✓) against correct choice and cross(x) against 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

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.

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