NCERT Solutions For Class 9 Science Chapter 4 - Cbsestudyguru

NCERT Solutions For Class 9 Science Chapter 4

NCERT Solutions For Class 9 Science Chapter 4 Structure Of The Atom

1. Exercise Questions
2. Intext Questions

NCERT Solutions For Class 9 Science Chapter 4 Structure Of The Atom on this step-by-step answer guide . In some of State Boards and CBSE schools, students are taught thru NCERT books. As the chapter comes to an end, students are requested few questions in an exercising to evaluate their knowledge of the chapter. Students regularly want guidance dealing with those CBSE Solutions For Class 9 Science Chapter 4 Structure Of The Atom.

It’s most effective natural to get stuck withinside the exercises while solving them so that you could assist students rating better marks, we have provided grade by grade NCERT answers for all exercises of Class 9 Science Structure Of The Atom so you can be searching for assist from them. Students have to solve those exercises carefully as questions withinside the very last exams are requested from those so these exercises at once have an impact on students’ final score. Find all NCERT Solutions for Class nine Science Structure Of The Atom below and prepare for your tests easily.

 

NCERT Solution for Class 9 Science

Chapter 4 – Structure Of The Atom

Exercise Questions

 

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

Ans .

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

Ans . Limitations of J.J Thomson’s model of an atom are :

( a ) It could not explain the result of the scattering experiment performed by Rutherford .

( b ) It did not have any experimental evidence in its support .

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

Ans . Limitations of Rutherford’s model of the atom are :

( a ) It does not explain the stability of atom .

( b ) It does not explain the spectrum of hydrogen and other atoms .

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

Ans . Bohr proposed the following postulates for revising the Rutherford’s model .

( a ) Atom has central nucleus surrounded by electrons .

( b ) An atom consists of small heavy positively charged nucleus in the centre and the electrons revolve around it . in circular paths called orbits or shells .

( c ) Each orbit has fixed energy , so these orbits are called energy levels or energy shells .

( d ) The order of the energy of these energy shells will be :

K < L < M < N < 0 < …… or , 1 < 2 < 3 < 4 < 5 < …….

( e ) As long as an electron remains in a particular orbit , it does not lose or gain energy .

( f ) Energy is neither absorbed nor emitted when electron is moving in an orbit . But energy is absorbed when it jumps from lower orbit to higher orbit . Whereas energy is emitted when it jumps from higher orbit to lower orbit .

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

Ans . Comparison of different proposed Model :

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

Ans . The distribution of elements in different orbits is governed by a scheme called Bohr – Bury scheme .

Rules for Bohr – Bury scheme are :

( i ) The maximum number of electrons present in any shell is given by the formula Inl . Where n = no . of orbit .

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

( iii ) Electrons in an atom do not occupy a new shell unless all the inner shells are completely filled .

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

Ans . Valency is the number of electrons gained , lost or shared so as to complete the octet of electrons in valence shell .

Valency of silicon : It has electronic configuration as 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 as 2,6 Thus , It will gain 2 electrons to complete its octet . So its valency = 2

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

Ans . ( i ) Atomic number : The atomic number of an element is equal to the number of protons in the nucleus of its atom , e.g. , Oxygen has 6 protons , hence its atomic no . is 6 .

( ii ) Mass number : The mass number of an atom is equal to the number of protons and neutrons in its nucleus .

Nuclcons = number of protons + number of neutrons

E.g. Protons + Neutrons = Nucleus = Mass number i.e. Mass number of oxygen is 6 + 6 = 12

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

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

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Both calcium and argon have same mass number but different atomic number .

Two uses of isotopes are :

( i ) An isotope of iodine is used in the treatment of goitre .

( ii ) An isotope of uranium is used as a fuel in nuclear reactors

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

Ans . Sodium atom ( Na ) , has atomic number = 11

Number of protons = 11

Number of electrons = 11

Electronic configuration of Na = K L M = 2, 8, 1

Sodium atom ( Na ) looses 1 electron to become stable and form Na + ion . Hence it has completely filled K and L shells .

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Q. 10 . If bromine atom is available in the form of , say , two isotopes calculate the average atomic mass of bromine atom .

Ans . The average atomic mass of bromine atom

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The average atomic mass of a sample of an element X is 16.2 u .

Q. 11. What are the percentages of isotopes A picture containing text, clipartDescription automatically generated in the Sample ?

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

Ans . Z = 3 , ( i.e , atomic number → z )

Electronic configuration = 2 , 1

Valency = 1

Name of the element is lithium

Composition of the nuclei of two atomic species X and Y are given as under

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Q. 13. Give the mass numbers of X and Y. What is the relation between the two species ?

Ans . Mass number of X = Protons + Neutrons = 6 + 6 = 12

Mass number of Y = Protons + Neutrons = 6 + 8 = 14

As the atomic number is same i.e. , = 6 .

[ atomic number = number of protons ] .

Both X and Y are isotopes of same element .

Q. 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 proton .

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

Ans . ( a ) False ( b ) False ( c ) True ( d ) False

Directions : Put tick ( 3 ) against correct choice and cross ( 7 ) against wrong choice in questions 15,16 and 17

Q. 15. Rutherford’s alpha – particle scattering experiment was responsible for the discovery of

( a ) Atomic Nucleus ( b ) Electron

( c ) Proton ( d ) Neutron

Ans . ( a ) Atomic nucleus

Q.16 . Isotopes of an element have

( a ) the same physical properties

( b ) different chemical properties

( c ) different number of neutrons

( d ) different atomic numbers .

Ans . ( c ) different number of neutrons

Q.17 . Number of valence electrons in Cl ion are :

( a ) 16 ( b ) 8

( c ) 17 ( d ) 18

Ans . ( b ) 8

Q. 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 .

Ans . ( d ) 2 , 8 , 1

Q. 19. Complete the following table .

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Ans .

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Intext Exercise

 

Page 47

Q. 1. What are canal rays ? [ NCERT Q. 1 , Page 47 ]

Ans . In 1886 , Goldstein discovered canal rays . They are positively charged radiations that consists of positively charged particles known as protons .

Q. 2. If an atom contains one electrons and one proton , will it carry any charge or not ? [ NCERT Q. 2 , Page 47 ]

Ans . A proton is a positively charged particle whereas an electron is a negatively 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 .

Page 49

Q. 3. On the basis of Thomson’s model of an atom , explain how the atom is neutral as a whole . [ NCERT Q. 1 , Page 49 – I ]

Ans . According to Thomson’s model of the atom , an atom contains both positively and negatively charged particles . The negatively charged particles are studded in the positively charged sphere , like the seeds in a watermelon . These negative and positive charges are equal in magnitude . Thus , by counter balancing each other’s effect , they make an atom neutral .

Q. 4. On the basis of Rutherford’s model of an atom , which subatomic particle is present in the nucleus of an atom ? [ NCERT Q. 2 , Page 49 – I ]

Ans . On the basis of Rutherford’s model of an atom , protons ( positively – charged particles ) are present in the nucleus of an atom .

Q. 5. Draw a sketch of Bohr’s model of an atom with three shells . [ NCERT Q. 3 , Page 49 – I ]

Ans .

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Q. 6. 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 ? [ NCERT Q. 4 , Page 49 – I ]

Ans . If the α – particle scattering experiment is carried out using a foil of a metal rather than gold , there would be no change in the observation . In the a – scattering experiment , a gold foil was taken because gold is malleable and can be fold easily or a thin foil of gold can be easily made . Also , it is difficult to make such foils from other metals .

Q. 7 . Name the three sub – atomic particles of an atom . [ NCERT Q. 1 , Page 49 – II ]

Ans . The three sub – atomic particles of an atom are :

( i ) Protons ( ii ) Electrons ( iii ) Neutrons .

Q. 8 . Helium atom has an atomic mass of 4 u and two protons in its nucleus . How many neutrons does it have ? [ NCERT Q. 2 , Page 49 – II ]

Ans . Helium has two neutrons . The mass of an atom is equal to the sum of the masses of protons and neutrons present in its nucleus . Mean while , helium atom has two protons , mass contributed by the two protons is ( 2 x 1 ) u = 2u . Then , the remaining mass ( 4 – 2 ) u is contributed by 2u/1u = 2 neutrons .

Page 50

Q. 9. Write the distribution of electrons in carbon and sodium atoms ? [ NCERT Q. 1 , Page 50 ]

Ans . The total number of electrons in a carbon atom is 6. The distribution of electrons in carbon atom is given by :

First orbit or K – shell = 2 electrons

Second orbit or L – shell = 4 electrons

Thus , the distribution of electrons in a carbon atom will be 2 , 4 .

The total number of electrons in a sodium atom is 11. The distribution of electrons in sodium atom is given by :

First orbit or K – shell = 2 electrons

Second orbit or L – shell = 8 electrons

Third orbit or M – shell = 1 electron

Thus , the distribution of electrons in a sodium atom as 2 , 8 , 1 .

Q. 10. If K and L shells of an atom are full , then what would be the total number of electrons in the atom ? [ NCERT Q. 2 , Page 50 ]

Ans . The maximum number of electrons that can occupy K and L shells of an atom are 2 and 8 , respectively . Thus , if K and L shells of an atom are full , then the total number of electrons in the atom would be ( 2 + 8 ) = 10 .

Page 52

Q. 11. How will you find the valency of chlorine , sulphur and magnesium ? [ NCERT Q. 1 , Page 52 – I ]

Ans . If the number of electrons in the outermost shell of the atom of an element is less than or equal to 4 , then the valency of the element is equal to the number of electrons present in its outermost shell and if the number of electrons in the outermost shell of the atom of an element is greater than 4 , then the valency of that element is determined by subtracting the number of electrons in the outermost shell from 8 .

The distribution of electrons in

Chlorine is 2 , 8 , 7 .

Sulphur is 2 , 8 , 6 .

Magnesium atom is 2 , 8 , 2 .

Thus , the number of electrons in the outermost shells of chlorine , sulphur and magnesium atoms are 7 , 6 , and 2 , respectively .

Therefore ,

The valency of chlorine = 8 – 7 = 1

The valency of sulphur = 8 – 6 = 2

The valency of magnesium = 2 .

Q. 12. 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 ? [ NCERT Q. 1 , Page 52 – II ]

Ans . ( i ) The atomic number is equal to the number of protons . Thus , the atomic number of the atom is 8 .

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

Q.13 . With the help of the following table , find out the mass number of oxygen and sulphur atom . [ NCERT Q. 2 , Page 52 – II ]

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Ans . Mass number of oxygen = Number of protons + Number of neutrons = 8 + 8 = 16

Mass number of sulphur = Number of protons + Number of neutrons = 16 + 16 = 32

Page 53

Q. 14. For the symbol H , D and T tabulate three sub atomic particles found in each of them . [ NCERT Q. 1 , Page 53 ]

Ans .

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Q. 15. Write the electronic configuration of any pair of isotopes and isobars . [ NCERT Q. 2 , Page 53 ]

Ans . Isotopes have same electronic configuration as the atoms have the same atomic number but different mass numbers .

Isotopes of chlorine are 35Cl₁7 and 37C1₁7

Since Isotopes of elements have same atomic number . So , there is no change in electronic configuration = 2,8,7

Isotopes of carbon are ¹2C6 , 1³C6 and ¹4C6

Electronic configuration = 2 , 4

40Ca20 and 40Ar18 are a pair of isobars having atomic number 20 and 18

The electronic configuration of 40Ca20 = 2 , 8 , 8 , 2 .

The electronic configuration of 40Ar18 = 2 , 8 , 8 .

Since all the isotopes of an element have identical electronic configuration containing the same number of valance electrons therefore all the isotopes of an element show identical chemical properties . Isobars have different chemical properties because they have different atomic number and different electronic configurations .

 

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