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NCERT Solutions of Science for Class 9 Lesson- 4 “STRUCTURE OF THE ATOM”

TEXTUAL QUESTIONS PAGE – 47

Q.1- What are canal rays?

Ans- A beam of rays or stream of positively charged particles which travel in a direction away from anode towards cathode when any gas taken in discharge tube is subjected to the action of high voltage under low pressure are called canal rays.

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

Ans- No, it will not carry any charge.

TEXTUAL QUESTIONS PAGE – 49

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

Ans- According to Thomson’s model of an atom,

1) An atom contains a positively charged sphere in which the negatively charged electrons are embedded like the seeds in watermelon.

2) The total positive charge is equal to the total negative charge on all the electrons, so the atom on the whole is electrically neutral.

Q.2- On the basis of Rutherford’s model of an atom which sub-atomic particle is present in the nucleus of an atom?

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

Q.3- Draw a sketch of Bohr’s model of an atom with three shells.

Ans-

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Bohr’s model of an atom with three shells

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

Ans- Gold has maximum malleability. It can be beaten into thin sheets. The sheet used in experiment is about 1000 atoms thick. No other metal can be beaten so thin. So alpha scattering would not possible.

TEXTUAL QUESTIONS PAGE – 49

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

Ans- These are :

(a) Electron – negatively charged

(b) Proton – positively charged

(c) Neutron – no charge

Q.2- Helium atom has one atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?

Ans- Atomic mass of Helium = 4 u

No. of protons = 2

No. of neutrons = 4 – 2

= 2

TEXTUAL QUESTIONS PAGE – 50

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

Ans- Carbon :

Atomic number of carbon – 6

No. of protons – 6

No. of electrons – 6

Distribution of electrons –

K – 2

L – 4

Sodium :

Atomic number of sodium – 11

No. of protons – 11

No. of electrons – 11

Distribution of electrons –

K – 2

L – 8

M – 1

Q.2- If K and L shell of an atom are full, then what would be the total number of electrons in the atom?

Ans- No. of electrons in K shell = 2

No. of electrons in L shell = 8

Total number of electrons = 2 + 8 = 10

TEXTUAL QUESTIONS PAGE – 52

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

Ans- Chlorine has the electronic configuration = 2, 8, 7

Its valency = 8 – 7 = 1

Sulphur has the electronic configuration = 2, 8, 6

Its valency = 8 – 6 = 2

Magnesium has the electronic configuration = 2, 8, 2

Its valency = 8 – 2 = 6

TEXTUAL EXERCISE PAGE – 52

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

(a) What is the atomic number of the atom? (b) What is the charge on the atom?

Ans- (i) No. of protons = 8

Therefore atomic number of the element = 8

(ii) Atom is electrically neutral, means it has no charge.

Q.2- With the help of table 4.1 find out the mass number of oxygen and sulphur atom.

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Ans- Oxygen :

No. of protons = 8

No. of neutrons = 8

Atomic number = 8

Mass number of oxygen = p + n

= 8 + 8

= 16

Sulphur :

No. of protons = 16

No. of neutrons = 16

Atomic number = 16

Mass number of sulphur = p + n

= 16 + 16

= 32

TEXTUAL EXERCISE PAGE – 53

Q.1- For the symbol H, D and T tabulate three sub-atomic particles found in each of them.

Ans- Hydrogen has three isotopes. Each has same atomic number but different mass number.

IsotopeNo. of protonsNo. of electronsNo. of neutrons
Hydrogen/ Protium (H)110
Deuterium (D)111
Tritium (T)112

Q.2- Write the electronic configuration of any one pair of isotopes and isobars.

Ans- Pair of isotopes : 3517Cl and 3717Cl

Electronic configuration of these isotopes –

K L M

2, 8, 7

Pair of isobars – ⁴⁰₁₈Ar and ⁴⁰₂₀Ca

Electronic configuration of ⁴⁰₁₈Ar

= K L M

2, 8, 8

Electronic configuration of ⁴⁰₂₀Ca

= K L M N

2, 8, 8, 2

TEXTUAL EXERCISE PAGE – 54

Q.1- Compare the properties of electron, proton and neutron.

Ans-

Electrons ProtonsNeutrons
1) They are negatively charged.1) They are positively charged.1) They have no charge and are neutral.
2) Their mass is negligible.2) The mass of proton is 1 a.m.u.2) The mass of neutron is 1 a.m.u.
3) They are present outside the nucleus. 3) They are present inside the nucleus.3) They are present inside the nucleus.
4) They get attracted towards positive charge.4) They get attracted towards negative charge.4) They do not get attracted as they are neutral.

Q.2- What are the limitations of J.J. Thomson model of an atom?

Ans- Limitations of J.J. Thomson model of an atom –

i) It could not explain stability of the atom.

ii) It could not explain hydrogen spectrum.

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

Ans- According to Rutherford’s model of atom, the electron revolves around the nucleus in a circular orbit. Any particle in circular orbit would undergo acceleration. During acceleration charged particles would radiate energy. Thus the revolving electron lose energy and finally fall into the nucleus. This shows that the atom should be highly unstable and hence matter would not exist in the state that we know.

Q.4- Describe Bohr’s model of an atom.

Ans– Bohr put forward the following postulates about the model of an atom –

(i) An atom has three types of particles – electrons, protons and neutrons. The protons and neutrons are present in the nucleus. The electrons are revolves around the nucleus.

(ii) Certain orbits, known as the discrete orbits of electrons are allowed inside the atom.

(iii) While revolving in discrete orbits, the electrons do not radiate energy.

(iv) These orbits or shells are called energy levels and are represented by the letters K, L, M, N….. or the numbers n = 1, 2, 3, 4…. where n is the number of orbit.

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

(vi) Electrons are not accomodated in the given shells unless the inner shells are filled.

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

Ans- (A) Thomson’s model of an atom –

(a) According to Thomson, an atom may be regarded as a uniform sphere of positive electricity (protons) in which negatively charged electrons are embedded like the seeds in a watermelon.

(b) The total positive charge is equal to the total negative charge on all the electrons so that atoms on the whole is electrically neutral.

(B) Rutherford’s model of an atom –

The main postulates of Rutherford’s model of an atom are :

(a) Atom has a dense, heavy, positively charged central part called nucleus.

(b) The electrons are present at a very large distances from the nucleus.

(c) The total positive charge on the nucleus is equal to the total negative charge on all the electrons so that the atom on the whole is electrically neutral.

(d) The electrons are revolving around the nucleus so that attractive force of the nucleus is balanced by the centrifugal force ( just like planets revolve around the sun ).

(C) Bohr’s model of an atom –

Bohr put forward the following postulates about the model of an atom –

(a) An atom has three types of particles – electrons, protons and neutrons. The protons and neutrons are present in the nucleus. The electrons are revolves around the nucleus.

(b) Certain orbits, known as the discrete orbits of electrons are allowed inside the atom.

(c) While revolving in discrete orbits, the electrons do not radiate energy.

(d) These orbits or shells are called energy levels and are represented by the letters K, L, M, N….. or the numbers n = 1, 2, 3, 4…. where n is the number of orbit.

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

(f) Electrons are not accomodated in the given shells unless the inner shells are filled.

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

Ans- The various rules for writing the electronic configuration of first eighteen elements are:

(i) The energy levels or energy shells are filled in order of increasing energies. The electrons first enter the K shell (n=1) which is closest to the nucleus. This is followed by L shell (n=2), M shell (n=3) and so on. Here n represents the number of the shell.

(ii) The maximum number of electrons in any shell is given as 2n2 (n = number of shell).

(iii) The outermost energy shell cannot have more than 8 electrons. The next inner shell called penultimate shell cannot have more than 18 electrons.

(iv) It is not necessary that an energy level or shell is fully filled before the filling in the next starts. In fact filling of electrons in a new shell starts when any shell contains 8 electrons.

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

Ans- Valency – It is the combining capacity of an atom of an element and is numerically equal to the number of electron present in the valence shell of an atom or eight minus ( if the number of electrons is 1 to 4 ) the number of electrons present in the valence shell of an atom ( if the number of electrons is more than 4).

Example – (a) Silicon

Atomic number of silicon is 14

Electronic configuration of silicon is – K-2, L-8, M-4

To fill the M shell 4 electrons are required.

So valency of silicon is 8 – 4 = 4

(b) Oxygen

Atomic number of oxygen is 8

Electronic configuration of oxygen is – K-2, L-6

To fill the L shell 2 electrons are required.

So valency of oxygen is 8 – 6 = 2

Q.8- Explain with examples (i) Atomic number (ii) Mass number (iii) Isotopes (iv) Isobars.

Give any two uses of isotopes.

Ans- (i) Atomic number – It is number of protons present in the nucleus of an atom. It is denoted by Z. For eg. For hydrogen, Z = 1, because in hydrogen atom only one proton is present in the nucleus.

(ii) Mass number – It is the sum of total number of protons and neutrons present in the nucleus of an atom. For eg. Mass of Carbon is 12 u because it has 6 protons and 6 neutrons, 6 u + 6 u = 12 u.

(iii) Isotopes – These are the atoms of the same element having same atomic number but different mass numbers. Since their atomic numbers is same, their valency and electronic configuration is same, so the chemical reactivity is same. For eg. Hydrogen atom has three atomic species, namely protium, deuterium and tritium. The atomic number of each one is one, but the mass number is 1, 2 and 3 respectively.

Uses of Isotopes :

(i) Isotope of Uranium is used as a fuel in nuclear reaction.

(ii) Isotope of cobalt is used in the treatment of cancer.

(iii) Isotope of iodine is used in the treatment of goitre.

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

Ans- Atomic number of Na = 11

No. of electrons in Na atoms = 11

No. of electrons in Na+ ion = 11 – 1 = 10

The electronic configuration of Na+ = 2, 8

In Na+ K and L shells are completely filled.

Q.10- If bromine atom is available in the form of say two isotopes 35Br79 (49.7%) and 35Br81 (50.3%), then calculate the average mass of bromine atom.

Ans- The two isotopes of bromine are 35Br79 (49.7%) and 35Br81 (50.3%)

Average atomic mass of bromine atom

= 79 × 49.7 + 81 × 50.3 / 100

= 3926.3 + 4074.3 / 100

= 80.006 u

Q.11- 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?

Ans- Let %age of 8X16 = p

%age of 8X18 = 100 – p

therefore p × 16 + ( 100 – p ) 18 / 100 = 16.2

16p – 18p + 1800 = 1620

– 2p = – 180

p = 90

%age of 8X16 = 90%

%age of 8X18 = 100 – 90 = 10%

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

Ans- Z = 3

Atomic number of element = 3

No. of electrons in one atom of the element= 3

Atomic number of element (Z) = 3

The name of element = Lithium

Its electronic configuration = K-2, L-1

Hence its valency = 1 ( No. of electrons in the valence shell)

Q.13- Composition of the nuclei of two atomic species X and Y are given as under :

XY
Protons66
Neutrons68

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

Ans- Mass no. of X = 6 + 6 = 12 amu

Mass no. of Y = 6 + 8 = 14 amu

The two species are isotopes because they have same atomic number but different mass number.

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 electron is about 1/2000 times that of proton.

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

Ans- (a) F

(b) F

(c) T

(d) F

Put tick (✓) against correct choice and cross (×) 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)

(b) ×

(c) ×

(d) ×

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- (a) ×

(b) ×

(c)

(d) ×

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

(a) 16

(b) 8

(c) 17

(d) 18

Ans- (a) ×

(b)

(c) ×

(d) ×

Q.18- Which 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)

Q.19- Complete the following table :

Atomic No.Mass No.No. of neutrons No. of protons No. of electrons Name of atomic species
910
1632Sulphur
2412
21
1010

Ans-

Atomic No.Mass No.No. of neutronsNo. of protonsNo. of electronsName of atomic species
9191099Fluorine
1632161616Sulphur
1224121212Magnesium
12111Hydrogen Deuterium
11010Hydrogen