JKBOSE 9th Class Science Solutions Chapter 10 Structure Of Atom

◆ Electron is a negatively charged particle with 1.602 x 10^-19 coulomb negative charge (- 1 unit) and mass 9.1089 × 10^-31 kg (negligible mass). It is represented by the symbol ‘e’.

◆ Proton is a positively charged particle with 1.602 x 10^-19 coulomb positive charge (+ 1 unit) and mass 1.672 x 10^-27 kg, it is represented by the symbol ‘p’. Neutron is a neutral particle with no charge and mass equal to 1.678 × 10^-27 kg, It is represented by the symbol ‘n’.

◆ Electrons were discovered by J.J. Thomson.

◆ Protons were discovered by J.J. Thomson.

◆ Neutrons were discovered by James Chadwick.

◆ Nucleus is the small, positively charged and heavy central portion in an atom which contains in it protons and neutrons.

◆ Nucleons. The neutrons and protons present in the nucleus of an atom are collectively known as nucleons.

◆ Extranuclear part contains electrons.

◆ Cathode rays consist of negatively charged particles called electrons.

◆ Anode rays consist of positively charged particles. These are also called canal rays.

◆ Atomic number of an element represents the number of protons in an atom. It is denoted by the symbol Z.

◆ Shells of an atom are designated as K, L, M, N, etc. These are also called energy levels.

◆ Elements are specified in terms of their atomic numbers. (or nucleons)

◆ Valence shell of an atom represents the outermost shell where electrons are present and the electrons are called valence electrons.

◆ Mass number of an element is the sum of the number of protons and neutrons present in the nucleus of the atom. It is denoted by A.

◆ Valence shell of an atom represents the outermost shell where electrons are present and the electrons are called valence electrons.

◆ Valency. It is the combining capacity of an atom of the element. It is equal to number of electrons in the valence shell (no. of electrons is less than 4) or 8 Number of both electrons in the valence shell (Number of electrons more than 4).

◆ Isotopes are the atoms of the same element having same atomic number but different mass numbers.

◆ Isobars are the atoms of the different elements having same mass number but different atomic numbers.

◆ Thomson’s model, Rutherford’s Nuclear Model and Bohr’s model are the different models for an atom.

◆ Rutherford’s model of an atom lead to the discovery of the nucleus.

◆ Bohr’s Model of an atom assumes quantisation of energy and quantisation of angular momentum of an electron.

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

Ans.— No, it will not carry any charge.

Ans.— Thomson’s model of atom.

1. 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.

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

Ans.— Proton.

 Ans.— Almost all the a-particles will pass undeflected and hardly any a-particle is deflected.

Ans.— These are :

(a) Electron

(b) Proton and

(c) Neutron.

Ans.— Atomic mass of Helium = 4 u

No. of protons = 2

No. of neutrons = 4 – 2 = 2

Ans.— No. of electrons in K-shell = 2

No. of electrons in L-shell = 8

Total no. of electrons = 2 + 8 = 10

Ans.— Chlorine has the electronic configuration = 2, 8, 7

∴ Valency of chlorine = 8 -7 = 1

Sulphur has the electronic configuration = 2, 8, 6

∴ Valency of sulphur = 8 – 6 = 2

Magnesium has the electronic configuration = 2, 8, 2

∴ Valency of magnesium = 2

Ans.— (i) No. of protons = 8

∴ Atomic no. of the element = 8

(ii) Zero.

Ans.—                                 P    n

Mass number of oxygen = 8 + 8 = 16

Mass number of sulphur = 16 + 16 = 32

Ans.— Pair of Isotopes : 35/17Cl and 35/17Cl

Atomic number of Cl = 17

Electronic configuration of these isotopes = K, L, M / 2,8, 7

Pair of Isobars : 40/18 Ar and 40/20 Ca

Atomic number of Ar = 18

Electronic Configuration of 40/18 Ar = K, L, M / 2, 8, 8

Atomic number of Ca = 20

Electronic Configuration of 40/20 Ca = K, L, M, N / 2, 8, 8, 2

Particle	Charge(e)	Mass (m)	Charge / Mass (e/m)
Electron	– 1.6022 × 10-19 C	9.109 × 1028 g	– 1.76 × 108
Proton	+ 1.6022 × 10-19 C	1.672 × 1024 g	+ 9.58 × 104
Neutron	Zero	1.675 × 10-24 g	Zero

Ans.—  Limitations of J.J. Thomson model of the atom :

1. It could not explain stability of the atom.

2. It could not explain hydrogen spectrum.

Ans.— Limitations of Rutherford’s model of atom :

1. It can’t explain stability of the atom.

2. It can’t explain hydrogen spectrum.

Ans.— The main points of Bohr’s model of atom are :

1. An atom has three types of particles called fundamental particles. These are electrons, protons and neutrons.

2. The protons and neutrons are present in the nucleus, present in the centre. The electrons are present around the nucleus and at very large distances from it. There is a large vacant space between the nucleus and the electrons.

3. An atom is electrically neutral as the number of protons is equal to the number of electrons.

4. The electrons are revolving around the nucleus in fixed circular paths which are called energy levels or energy shells or orbits. The energy levels or energy shells can be counted in two ways 1, 2, 3, 4, 5, 6 or as K, L, M, N, O, P. The counting starts from the centre outwards,

5. The various energy levels are arranged in order of increasing energy. The order of energy is 1 < 2 <3 <4 <5 or K<L<M<N<O or <L<M<N<0

6. The energy of an electron in an atom is quantised.

7. There is no change in the energy of the electrons as long as they keep on revolving in the same energy level.

8. The angular momentum of an electron in an atom is quantised.

9. The change in energy can take place only when the electron jumps from one energy level to the other. If the electron gains energy from outside, it jumps from lower energy to the higher energy level. If the electron jumps from higher energy level to the lower energy level, it loses energy.

Ans.— Give Thomsons’s model, Rutherford’s model and Bohr’s model of atom.

1. 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.

2. 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 nuclear model of an atom. The main features of Rutherford’s nuclear model of an atom are :

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

2. The electrons are present at very large distances from the nucleus.

3. The total positive charge on the nucleus is equal to the total -ve charge on all the electrons so that atom on the whole is electrically neutral.

4. 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). It is different from Thomson’s model, because in Thomson’s model in an atom, the total mass, positive charge and electrons are uniformly distributed.

C. Bohr’s model of an atom. Described in Question No. 4 given above.

Ans.— The various rules for writing the electronic configuration of first elements are :

1. 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.

2. The maximum number of electrons in any shell is given as 2n² (n = No. of the shell).

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

4. 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.

Ans.— Valency. It is the combining capacity of an atom of the element and is numerically equal to the number of electrons present in the valence shell of an atom (if the no. of electrons is 1 to 4) or eight minus the number of electrons present in the valence shell of an atom (if the no. of electrons is more than 4).

e.g. Silicon (14^Si) has electronic configuration = K, L, M / 2, 8, 4

∴ its valency = 4 (No. of electrons in its valence shell)

Oxygen (16/8 O) has electronic configuration = 2,6

∴ its valency=8 – 6 = 2 (8 – No. of electrons in the valence shell)

Ans.— (i) Atomic number. It is the number of protons present in the nucleus of an atom of an element.

(ii) Mass number. It is the sum of total number of protons and neutrons present in the nucleus of an atom.

(iii) Isotopes. These are the atoms of the same element having same atomic number but different mass numbers.

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

1. 235/92 U isotope of uranium is used as a fuel in nuclear reaction.

2. 60/27 Co isotope of cobalt is used in the treatment of cancer.

3. 131/53 isotope of iodine is used in the treatment of goitre.

Ans.—Na has atomic number = 11

∴ Na atom has electrons =

Na⁺ ion has electrons = 11 – 1 = 10

The electronic configuration of Na⁺ = 2, 8 / K, L

∴ In Na⁺, K and L-shells are completely filled.

Ans.— The two isotopes of bromine are 79/35 Br and 81/35 Br

                                                                (49.7%)          (50.3%)

Average atomic mass of bromine atom = 79 × 49.7 + 81 × 50.3 / 100

= 3926.3 + 4074.3 / 100

= 80.006 u

Ans.—  Z = 3

∴ Atomic no. of element = 3

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

Its electronic configuration = K, L / 2, 1

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

The name of the element = Lithium.

                    X    Y 

Ans.— Mass no. of X = 6 + 6 = 12

Mass no. of Y = 6 + 8 = 14

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

Ans.— (a) F (b) F (c) T (d) F.

Ans.— (a). Atomic Nucleus  (✓)

(b) Electron (x)

(c) Proton (x)

(d) Neutron (x)

Ans.—(a) the same physical properties. (✓)

(b) different chemical properties (x)

(c) different number of neutrons. (x)

(d) different atomic numbers (x)

Ans.— (a) 16 (x) (b) 8 (√) (c) 17 (x) (d) 18 (×).

Ans.— (d).

Atomic Number	Mass Number	Number of Neutrons	No of Protons	Number of Electrons	Name of the Atomic Species
9	–	10	–	–	–
16	32	–	–	–	Sulphur
–	24	–	12	–	–
–	2	–	1	–	–
–	1	0	1	0	–

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	Sulphur
12	24	12	12	12	Magnesium
1	2	1	1	1	Deuterium
1	1	0	1	0	Hydrogen

Ans.— J.J. Thomson found the presence of electrons in the cathode rays. The cathode rays were discovered with the help of discharge tube experiment as given below :

Discharge tube experiment. The discharge tube consists of a hard glass tube and has two metal plates sealed at its two ends. These plates are connected to a high voltage of battery or generator which can produce a voltage of 5000 to 10000 volts. The discharge tube is also connected to the vacuum pump which can decrease the pressure pressure to any value.

(i) When an electric discharge was passed through the discharge tube under normal atmospheric pressure, no electricity could pass because the gas inside the tube is poor conductor of electricity.

(ii) When the pressure inside the tube was reduced to 1 mm of Hg by pumping out the gas with the help of the vacuum pump, the gas inside the tube began to glow and conduct electricity.

(iii) Upon reducing the pressure further to 0.001 mm of mercury, the glow became weak but the end opposite to cathode began to glow with greenish yellow light.

It was found later on that at very low pressure (0.001 mm), some invisible rays were emitted from the cathode and upon striking the anode, they produced greenish light which is also called fluorescence. Since these rays were produced from the cathode, these are called cathode rays.

Ans.— The beam of rays or stream of particles which travel in a direction away from anode towards cathode when any gas is subjected to the action of high voltage at low pressure are called anode rays.

Goldstein used a perforated plate of metal having small holes in it as the cathode. When the pressure inside the discharge tube was reduced to 0.001 mm of Hg as in case of cathode rays, two types of rays were noticed in the discharge tube, one moving towards the anode and were called anode rays and the other moving towards the cathode and then passing through the holes. Since these rays were attracted towards the cathode, they were expected to have positive charges. They were called positive rays. As these rays moved away from the anode, they were also called anode rays.

Ans.— After the discovery of proton (one unit mass) and electron (negligible mass), it was thought that the mass of the atom is because of protons only. This is true in case of hydrogen atom. Its mass is 1 a.m.u. which is the same as the mass of the proton. But the mass of the next element helium (He) was found to be 4 a.m.u but it has only two protons. The two protons account for 2 a.m.u. mass. In order to explain the rest of the mass, it was believed that some neutral particles must be present in atom which have definite mass and no charge. These neutral particles were discovered by Chadwick in 1932 and were called neutrons. He obtained neutrons by bombarding light elements beryllium, boron with fast moving a-particles

                 9/4 Be + a → 12/6 C + 1/0 n (neutron)

Ans.— Rutherford’s a-ray scattering experiment. In 1911, Rutherford performed an experiment to establish the position of electrons and protons in an atom. This is called alpha ray scattering experiment. In this experiment, Rutherford obtained a thin stream of fast moving alpha (a) particles from radioactive source (P₀) and was allowed to strike against a thin foil of heavy metal like gold. Alpha particles are also called helium nuclei (He²⁺ with + 2 unit charge and 4 a.m.u. mass). As a result of the experiment, he made the following observations :

We know that the gold foil is made up of atoms of gold which are placed side by side. Since most of the a-particles passed through these atoms undeflected they did not come across any obstruction in their path, this shows that most of the space in the atom is light or hollow. The electrons with negligible mass were supposed to be present in this portion called extra nuclear portion. The a-particles were deflected from their path and a very few came back in the same direction. This shows that they must have met with some obstruction in their path and it must be.

(i) Heavy because a-particles are high speed particles and can remove light obstruction from their path.

(ii) Small because only small number of a-particles got deflected.

(iii) Positively charged because some a-particles were repelled by this obstruction.

This small, heavy and positively charged portion inside an atom was called nucleus. The protons with mass and positive charge were supposed to present in the nucleus. Later on, neutrons were also found to be present in the nucleus of the atom.

Ans.— The main points of Rutherford’s model atom are listed as follows :

(i) An atom has two parts. These are nucleus and extra nuclear portion. Extra nuclear portion

(ii) The nucleus contains protons while electrons are present in the extra nuclear portion.

(iii) Most of the mass of the atom is due to the nucleus because electrons have negligible mass.

(iv) The volume occupied by the nucleus is very small as compared to the volume occupied by extra nuclear portion.

(v) The electrons in the extra nuclear portion are not stationary. These are revolving around the nucleus. The electrons are called planetary electrons because they can be compared to various planets which revolve around the sun. The nucleus may be compared with the sun.

Ans.— The distribution of electrons in the energy levels is given with the help of Bohr Bury scheme. It is given as follows :

1. 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.

2. The maximum number of electrons in any shell is given as 2n² (n = number of the shell). The distribution in the first four energy shells is given in the table given below :

3. The outermost energy shell cannot have more than 8 electrons. The next inner shell called penultimate shell, cannot have more than 18 electrons.

4. 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 electrons.

Symbol of shell	Number of shell (Value of n)	Maximum number of electrons (2n²)
K	1	2 × (1)² = 2
L	2	2 × (2)² = 8
M	3	2 × (3)² = 18
N	4	2 × (4)² = 32

Ans.— On the basis of Bohr-Bury scheme, the distribution of electrons in first twenty elements (Z 1 to Z = 20) is given below :

In case of Potassium (Z = 19), the M shell has only 8 electrons and not 18. Actually, when the M shell gets 8 electrons, the filling of electrons starts in the next shell which is N-shell. When the N-shell gets 2 electrons, the electron are again filled in the M shell. Thus, electronic distribution in potassium (Z = 19) is

K    L    M    N

2    8    8     1 (M shell has 8 electrons and N shell has 1 electron).

Similarly, the electronic distribution in calcium (Z = 20) is :

K    L    M   N

2    8    8    2 (M shell has 8 electrons and N shell has 2 electrons).

Ans.— Cathode rays are produced when a gas is subjected to the action of

(i) High voltage of the order of 5000-10000 V and

(ii) Under a low pressure of the order of 10^-3 atm of Hg.

Ans.— (i) If an atom is electrically neutral, this means that it will have equal number of protons and electrons. Both protons and electrons have 1 unit charge but with opposite sign.

(ii) This means that nucleus contains in it both protons and neutrons which have mass.

Ans.— The atomic number (12) of magnesium indicates that its atom has 12 protons in the nucleus and an equal number of electrons in the extra nuclear portion. The mass number (24) points out that there are also 12 neutrons (24 – 12 = 12) present in the nucleus alongwith protons.

Ans.— The electronic configuration of phosphorus (P) shows that it has 5 valence electrons. Its valency is expected to be 5. But it is not possible for the atom to lose 5 electrons present in the valence shell because the energy required to remove these electrons will be very high. Therefore, phosphorus atom has valency equal to 3 (8- No. of valence electrons).

Ans.— This is because gold is highly malleable and can be beaten into very fine sheets or foils.

Ans.— The atom of different elements having same mass numbers but different atomic numbers are called isobars.

e.g. (i) 40/18 Ar and 40/19 K (ii) 86/36 Kr and 86/38Sr.

Ans.— He atom has only one shell (K-shell) which can have a maximum of 2 electrons only. It cannot have more than 2 electrons. As a result, the combining capacity of He is zero. Therefore, its valency is also zero.

Ans.— 1. The isotopes of an element have same atomic number (Z) i.e. the same number of electrons and same number of protons. Hence, they have identical chemical properties.

2. Isotopes of an element have same number of valence electrons and hence same valency.

3. Isotopes of an element have different mass numbers (or different number of neturons) and hence they have different physical properties such as mass, density, melting point, boiling point etc.

Ans.— Isotopes. The atoms of the same element having the same atomic number but different mass numbers are called isotopes.

1. Isotopes of hydrogen, oxygen, carbon are used to study the mechanisms of organic reactions.

2. Some radioactive isotopes are used for the treatment of cancer and other diseases.

The isotopes of hydrogen are :

(a) Protium, 1/1 H (b) Deuterium, 2/1 H (c) Tritium, 3/1H

The atomic structures of isotopes of hydrogen are

1/1 H or Protium	2/1 H or Deuterium	3/1 H or Tritium
Atomic number = 1	Atomic number = 1	Atomic number = 1
Mass number = 1	Mass number = 2	Mass number = 3
No. of protons = 1	No. of protons = 1	No. of protons = 1
No. of electrons = 1	No. of electrons = 1	No. of electrons = 1
No. of neutrons = 0	No. of neutrons = 1	No. of neutrons = 2

Ans.— (i) nature, discharge

(ii) hydrogen

(iii) proton

(iv) 10^-5 times.

Ans.— The elements have fractional atomic masses due to the existence of isotopes (having different mass numbers). e.g. chlorine has two isotopes 35/17 Cl and 37/17 Cl and are present in the ratio 3 : 1.

∴  Average atomic mass of chlorine = 35 × 3 + 37 × 1 / 4 = 105 + 37 / 4 = 35.5 a.m.u.

Ans.— Atomic number of S = 16

Mass number of S = 32

∴ No. of protons = 16

No. of neutrons = 32- 16 = 16

No. of electrons = 16

Proton	Electron
It has one unit of positive charge.	It has one unit of negative charge.
It has mass 1.6 × 10-24 g.	It has mass 9.1 × 10-28 g.
It is present in the nucleus of an atom.	It is present in the extranuclear part of an atom.

Ans.— The number of neutrons = Mass no. – Atomic no. = 27 – 13 = 14.

Ans.—  Na atom has 11 electrons while Na⁺ ion has 10 electrons.

Ans.— The electronic distribution of element X (Z = 15) is 2, 8, 5 and element Y (Z = 17) is 2, 8, 7.

Ans.— If M-shell has 4 electrons, this means that K and L shells are complete and its configuration is 2, 8, 4. Total number of electrons are 14.

Ans.— He²⁺ ion (also called a-particle) has two electrons less than He atom. He atom has 2p, 2n and 2e and He²⁺ ion has 2p and 2n and no electrons.

Ans.— Its two shells are completely filled (2, 8).

Ans.— X will have unit positive charge and the ion is represented as X⁺.

Ans.— It shows that the element is an inert gas in nature and its atom has either 2 or 8 valence electrons.

Ans.— 0-18 isotopes has two neutrons more than 0-16 isotope.

Ans.— It is the fixed circular path traced by an electron revolving around the nucleus as postulated by Bohr.

Ans.— Hydrogen has two types of fundamental particles i.e. electron and proton.

(A) proton and electron

(B) neutron and electron

(C) proton and neutron

(D) proton and positrons.

(A) an atom of hydrogen

(B) a molecule of hydrogen

(C) the nucleus of helium

(D) the nucleus of hydrogen atom.

Ans.— (D) the nucleus of hydrogen atom.

(A) Thomson

(B) Goldstein

(C) Chadwick

(D) Rutherford

(A) 2,7,8

(B) 2,8,7

(C) 8, 7, 2

(D) 2,8,6

Ans.—(B) 2,8,7

(A) Thomson

(B) Goldstein

(C) Chadwick

(D) Rutherford

(A) 16

(B) 17

(C) 18

(D) 10

Ans.— (C) 18

(A) different chemical properties

(B) different atomic number

(C) different number of neutrons

(D) same physical properties.

Ans.— (C) different number of neutrons