NCERT Solutions Class 9th Science Chapter – 4 Structure of the Atom Question Answer

May 26, 2022

NCERT Solutions Class 9th Science Chapter 4 Structure of the Atom

Textbook	NCERT
Class	9^th
Subject	Science
Chapter	4^th
Chapter Name	Structure of the Atom
Category	Class 9^th Science
Medium	English
Source	Last Doubt

NCERT Solutions Class 9th Science Chapter – 4 Structure of the Atom Question Answer What is structure of atom Class 9, What is the structure of atoms Class 11, What are the 5 parts of an atom, What is atom full answer, What is atom Class 9 Topper, What are atoms made of, What are the 4 types of structures in chemistry, Who discovered the structure of atom, How big is atom, Who discovered atom First Class 9, What is the smallest unit of matter.

NCERT Solutions Class 9th Science Chapter 4 Structure of the Atom

Chapter – 4

Structure of Atom

Question Answer

Page: 39

Question 1. What are canal rays?

Answer – 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- The atom will be electrically neutral as one – ve charge balances one + ve charge.

Page: 41

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

Answer– 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 sub-atomic particle is present in the nucleus of an atom?

Answer– As per Rutherford’s model of an atom, the protons which are positively charged are present in the nucleus of an atom.

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

Answer – On using any metal foil, the observations of the a-particle scattering experiment would remain the same as all atoms would have same structure.

Page: 41

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

Answer – The sub-atomic particles of an atom are –
proton – positive charge
Electron – Negative Charge
Neutron – 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 – Given: Atomic mass of helium atom = 4u, 2 protons in helium nucleusAtomic mass = number of protons + number of neutrons4 = 2 + number of neutronsNumber of neutrons = 4 – 2 = 2Hence, Helium has 2 neutrons.

Page: 42

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

Answer – A carbon atom contains a total of 6 electrons. The following equation describes the electron distribution in a carbon atom: first orbit or K-shell = 2 electrons; second orbit or K-shell = 2 electrons; third orbit or K-
shell = 2 electrons; fourth orbit or K-shell
L-shell or second orbit = 4 electrons

We can also express the electron distribution in a carbon atom as 2, 4.In a sodium atom, there are 11 total electrons. The electron distribution in the sodium atom is described by: first orbit or K-shell = 2 electrons; second orbit or K-shell = 2 electrons; third orbit or K-shell = 2 electrons; fourth orbit or K-shell = 2L-shell or second orbit = 8 electrons

M-shell or third orbit = 1 electron
Alternatively, we can express the electron distribution in a sodium atom as 2, 8, 1.

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 – 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.

Page: 44

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

Answer – Chlorine : Atomic No. of Cl = 17Its electronic configuration = 2, 8, 7

Valency of Cl = 8 – 7 = 1Sulphur : Atomic no. of S = 16Its electronic configuration = 2, 8, 6
Valency of S = 8 – 6 = 2Magnesium: Atomic no. of Mg = 12Its electronic configuration = 2, 8, 2
Valency of Mg = 2

Page: 44

Question 1. If number of electrons in an atom is 8 and number of protons is also 8, then

Answer – Given: Number of electrons = 8
Number of protons = 8

(i) What is the atomic number of the atom? and
Answer – The atomic number of an atom is the same as the number of protons in that atom, hence its atomic number is 8.

(ii) What is the charge on the atom?
Answer – In an atom, the number of protons is equal to the number of electrons. Hence both the charges – positive and negative neutralize each other. Therefore, the atom does not possess any charge.

Question 2. With the help of given Table 4.1, find out the mass number of oxygen and sulphur atom.
Table: Composition of Atoms of the First Eighteen Elements with Electron Distribution in Various Shells

Element Name	Symbol	Atomic No.	Electron	Neutron	Proton	K	L	M	N	valency
Alluminium	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 –

(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, 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

Page: 45

Question 1. For the symbol H, D and T tabulate three sub-atomic particles found in each of them
Answer – 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 isobar.
Answer- Isotopes Atoms of same element having same atomic number but different mass number.

(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 the atomic number. Electronic configuration of an isobar pair is as follows,Example: Electronic configuration of ₂₀Ca⁴⁰ – K – 2; L – 8; M – 8; N – 2Electronic configuration of ₁₈Ar⁴⁰ – K – 2; L – 8; M – 8

Question 2. Write the electronic configuration of any one pair of isotopes and isobar.
Answer – Isotopes: Atoms of same element having same atomic number but different mass number.(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 the atomic number. Electronic configuration of an isobar pair is as follows,Example: Electronic configuration of ₂₀Ca⁴⁰ – K – 2; L – 8; M – 8; N – 2Electronic configuration of ₁₈Ar⁴⁰ – K – 2; L – 8; M – 8

Page: 46

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

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 par

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

Answer- 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.

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

Answer – According to Rutherford’s model of an atom the electrons are revolving in a circular orbit around the nucleus. Any such particle that revolves would undergo acceleration and radiate energy. The revolving electron would lose its energy and finally fall into the nucleus, the atom would be highly unstable. But we know that atoms are quite stable.

Question 5. Compare all the proposed Bohr’s models of an atom given in this chapter.
Answer –

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, N

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

Answer – The rules for writing of distribution of electrons in various shells for the first eighteen elements are:
(i) The maximum number of electrons present in a shell is given by the formula -2 n²
∵ n = orbit number i.e., 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

(ii) The maximum number of electrons that can be accommodated in the outermost orbit is 8.

(iii) Electrons are not accommodated in a given shell unless the inner shells are filled. (Shells are filled step-wise).

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

Answer – The valency of an atom refers to its ability to combine with other atoms. The valency of an element determines the number of bonds that an atom can make as part of a complex.

The value of silicon

Silicon has an atomic number of 14.
The element’s electronic configuration is K = 2, L= 8, M = 4.
Because silicon is a noble gas, it may either gain or lose four electrons, hence its valency is either +4 or -4 depending on the electrical configuration.
As a result, silicon has a valency of 4.

Oxygen concentration

Oxygen has an atomic number of 8.
K = 2, L = 6 is the electronic configuration.
The valency of oxygen is 6 according to the electrical arrangement, although it is far easier for oxygen to obtain 2 electrons rather than lose 6 electrons.
As a result, oxygen has a valency of – 2.

Question 8. Explain with examples:
Give any two uses of isotopes.
Answer –
(i) Atomic number
Answer – 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) Mass number,
Answer-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) Isotopes and
Answer – 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.
Answer- 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.

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

Answer – Sodium atom (Na), has atomic number =11
Number of protons =11
Number of electrons = 11
Electronic configuration of Na = K– 2, L – 8, M – 1
Sodium atom (Na) looses 1 electron to become stable and form Na⁺ ion. Hence it has completely filled K and L shells.

Question 10. 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.

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

Question 11. The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes ¹⁶₈X and ¹⁸₈X in the sample?

Answer – 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%

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

Answer – 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.

Question 13. 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 number of X and Y. What is the relation between the two species?

Answer – 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.

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.
Answer – False

(b) A neutron is formed by an electron and a proton combining together. Therefore,it is neutral.
Answer – False

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

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

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 number of neutrons
(c) different number of neutrons
(d) different atomic numbers.

Answer – (c) different number of neutrons

Question 17. Number of valence electrons in Ct ion are :

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

(d) 18
Answer – 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

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	–	–	–
10	32	–	–	–	Sulphure
–	24	–	12	–	–
–	2	–	1	–	–
–	1	0	1	0	–

Answer –

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	Sulphure
12	24	12	12	12	Magnesium
1	2	1	1	1	Deuterium
1	1	0	1	1	Protium

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