NCERT Solutions for Class 9 Science, Chapter 4: Structure of the atom

Class 9 is the first stepping stone for a student in the competitive world. The introduction of the CBSE Board Exam for class 10 a few years back has become an important gateway for a student. Based on the results of class 9th a student selects his future stream of Science, Commerce or Arts suiting his interest.

Takshila Learning is providing NCERT Solutions for Class 9 Chemistry as per the latest syllabus by CBSE. Class 9 is the building block for the CBSE Class 10 Board Exams, not only for your exams but also for your higher studies and career. Chemistry is the most essential subject and the knowledge in this field opens up wider career opportunities for the students.

Below you can find the NCERT solution for Class 9th Chemistry. You can get a Solution for the all-important question of Class 9 Science, Chapter 4: Structure of the atom

Intext questions (set 1)

Page no. 47

 

 

Ques 1: What are the canal rays?

Answer: Canal rays are positively charged radiations that can pass through perforated cathode plate. These rays consist of positively charged particle known as protons.

 

Ques 2: If an atom contains one electron and one proton, will it carry any charge or not?

Answer: An electron is a negatively charged particle, where is a proton is a positively charged particle. The magnitude of the 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.

 

Intext questions (set 2)

Page no. 49

 

 

Ques 1: On the basis of Thompson’s model of an atom, explain how the atom is neutral as a whole.

Answer: As per Thompson’s model of the atom, an atom consist of both the positive charges which are equal in number and magnitude. So, they balance each other as a result of which Adam as a whole is electrically neutral.

 

Ques 2: On the basis of Rutherford’s model of an atom, which subatomic particle is present in the nucleus of an atom?

Answer: On the basis of Rutherford model of an atom, protons are present in the nucleus of an atom.

 

Ques 3: Draw a sketch of Bohr’s model of an atom with three shells.

Answer:

 

 

Ques 4: What do you think would be the observation if the ∝– particle scattering experiment is carried out using a foil of a metal other than gold?

Answer:

If ∝– particle scattering experiment is carried out using a foil of any metal as thin as gold foil used by Rutherford, there would be no change in observation. But since other metals are not so malleable so, such a tin foil is difficult to obtain. If we use a thick foil, then more ∝– particle would bounce back and no idea about the location of positive marks in the atom would be available with such a certainty.

Intext questions (set 3)

Page no. 50

 

 

Ques 1: Name the three subatomic particles of an atom.

Answer: The three Sub- atomic particles of an atom are:

• Protons
• Electrons
• Neutrons

 

Ques 2: Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?

Answer: Number of neutrons =atomic mass-number of protons

Therefore, the number of neutrons in the atom = 4-2= 2

 

Intext questions (set 4)

Page no. 49

 

 

Ques 1: Write the distribution of electrons in Carbon and Sodium atoms.

Answer: The total number of electrons in a carbon atom is six. The distribution of electrons in carbon atom is given by:

First orbit or K- shell=2 electrons

Second orbit or L-shell=4 electrons

Or, we can write the distribution of electrons in a carbon atom has 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

Or, we can write distribution of electrons in a sodium atom is 2,8,1

 

Ques 2: If K and L shells of an atom are full, then what would be the total number of electrons in the atom?

Answer: The maximum capacity of K shell is two electrons and L shell can accommodate maximum eight electrons in it. Therefore, there will be 10 electrons in the atom.

 

Intext questions (set 4)

Page no. 52

 

 

Ques 1 : How will you find the valency of chlorine, sulphur and magnesium?

Answer:

If the number of electrons in the outermost shell of the atom of an element is less than or equal to 4, then the valance see of the element is equal to the number of electrons in the outer most shell. On the other hand, if the number of electrons in the outermost shell of the atom of an element is greater than four, then the valance see of that element is that in mind by subtracting the number of electrons in the outermost shell from eight.

The distribution of electrons. In chlorine, silver and magnesium atoms are 2,8,7,2,8,6 and 2,8,2 respectively.

Therefore, the number of electrons in the out Outermost shell of chlorine, sulphur, and magnesium atoms are 7,6,2 respectively

 

• Thus, the valency if chlorine = 8-7 =1

 

• Thus, the valency of sulphur = 8-6=2

 

• Thus, the valency if magnesium = 2

 

 

Intext questions (set 5)

Page no. 52

 

 

 

Ques 1:  If the number of electrons in an atom is 8 and the 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?

Answer: (i) The atomic number is equal to the number of protons. Therefore, the atomic number of the atom is 8

(ii) She is the number of both electrons and protons is equal, therefore, the charge of atom is 0

 

Ques 2: With the help of given Table, 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.

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

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

 

 

Intext questions (set 7)

Page no. 53

 

 

Ques 1: For the symbol H, D and T, tabulate three subatomic 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

 

 

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

Answer: (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- 2

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

 

Exercises:

Page no. 54

 

 

 

Q1. 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 particle

 

 

Q2. What are the limitations of J.J.Thomson’s model of the atom?

Answer: The limitations of J.J. Thomson’s model of the atom are:

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

It did not have any experiment support.

 

Q3. What are the limitations of Rutherford’s model of the atom?

Answer: The limitations of Rutherford model of the atom are:

It fails to explain the stability of an atom

It doesn’t explain the spectrum of hydrogen and other atoms.

 

Q4. Describe Bohr’s model of the atom.

Answer:

• The atom consists of a small positively charged nucleus and its centre.
• The whole mass of the atom is concentrated at the nucleus and the volume of the nucleus is much smaller than the volume of the atom.
• All the protons and neutrons of that Omar contain in the nucleus.
• Only certain orbit known as discrete orbits of electrons are allowed inside the atom.
• While revolving in these discreet orbit electrons do not radiate energy. These orbit or cells Are you presented by the letter K, L, M, NETC. All the numbers n= 1,2,3,4 ….. as shown in the figure below

 

 

 

Q5. Compare all the proposed models of an atom given in this chapter.

Answer:

Thomson’s model	Rutherford’s model	Bohr’s model
·       An atom consists of positively charged sphere and the electrons are embedded in it.   ·       The negative and positive charges are equal in magnitude. As a result, the atom is electrically neutral	·       An atom consists, of a positively charged centre in the atom called the nucleus. The mass of the atom is contributed mainly by the nucleus. ·       The size of the nucleus is very small as compare to the size of the atom. ·       The electrons revolve around the nucleus in well-defined orbits. ·	·       Bohr Agreed with almost all point as said by another phone except regarding the revolution of electrons in it that there are only certain orbits known as discrete orbits inside the atom which electrons revolve around the nucleus. ·       While revolving in its discrete orbits the electrons do not radiate energy.

 

 

Q6. Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.

Answer:

• 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 = 2L shell – n=2 ; 2n2 = 2(2)2 = 8M shell – n=3 ; 2n2 = 2(3)2 = 18N 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.

 

 

Q7. Define valency by taking examples of silicon and oxygen.

Answer: The valency see of an element is the combining capacity of that element. The valance see of an element is determined by the number of valance electrons present in the atom of that element.

Valency  of silicon: it has electronic configuration: 2,4,8

Thus, the valance see of silicon is 4 as these electrons can be shared with others to compete Octet

 

Valency of oxygen: it has electronic configuration: 2,6

Thus, Oxygen is 2and as it will gain 2 electrons to complete its

Octet

 

Q8. Explain with examples

(i) Atomic number,

(ii) Mass number,

(iii) Isotopes and

(iv) Isobars. 

Give any two uses of isotopes.

Answer: Atomic number: the atomic number of an element is the total number of protons present in the atom of that element. For example, nitrogen has 7 protons in its atom. thus, The atomic number of nitrogen and seven.

 

 

Mast number: the mass number of any an element is the sum of the number of protons and neutrons present in the atom of that element. For example, the atom of boron has 5protons and 6 neutrons. So, the mass number of boron is 5+6=11

 

Isotopes: these are atoms of the same element having the same atomic number, but different mass number. For example: chlorine has two isotopes with atomic number 17 but mass number 35 and 37 represented by

 

 

Isobars: these are atoms having the same as number, but the different atomic number i.e., Isobars or atoms of different elements having the same as number. For example: Ne has atomic number 10 and sodium has admin number 11 but both of them Have mass number of 22 represented by

 

 

Two uses of isotopes:

One isotope of uranium is used as a fuel in nuclear reactors.

One isotope of cobalt is used in the treatment of cancer.

 

Q9. Na+ has completely filled K and L shells. Explain.

Answer: 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.

 

 

Q10. 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 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

 

 

Q11. 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?

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%

 

 

Q12. 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.

 

 

 

Q13. 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?

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.

 

 

 

Q14. 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.

Answer: (a) Statement is False

(b)  Statement is False

(c)  Statement is True

(d)  Statement is False

 

 

Q15. 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

 

 

Q16. 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

 

 

Q17. Number of valence electrons in Cl– ion are:

(a) 16

(b) 8

(c) 17

(d) 18

Answer: (b) 8

 

Q18. 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

 

 

 

Q19. 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 – – –

 

Answer: 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