Goa Board Class 9 Solutions for Chemistry – Structure Of The Atom (English Medium)
Page No. 47:
Question 1:
What are canal rays?
Solution:
They are the positively charged radiations which consist of positively charged particles of atoms. Canal rays pass through perforated cathode and then travel towards another cathode in a gas discharge tube. They were given the name Canal rays by E. Goldstein in 1866 who discovered these radiations.
Concept insight: Write the proper definiton of canal rays.
Question 2:
If an atom contains one electron and one proton, will it carry any charge or not?
Solution:
An atom having one electron and one proton will not carry any charge as the positive charge on the proton will neutralise the negative charge on the electron.
Concept insight: Remember that electron carries a -1 charge and proton carries a +1 charge.
Page No. 49-I:
Question 1:
On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.
Solution:
Thomson proposed a model of an atom. According to this model of an atom, an atom consists of a sphere of positive charge. The positive charge in the atom is spread all over like the red edible part of a watermelon, while the electrons are studded in the positively charged sphere, just like the seeds in the watermelon.
The negative and positive charges are equal in magnitude. These equal and opposite charges balance each other thus the atom becomes electrically neutral as a whole.
Concept insight: Recall the charges on subatomic particles.
Question 2:
On the basis of Rutherford’s model of an atom which sub-atomic particle is present in the nucleus of an atom?
Solution:
Proton is the sub-atomic particle present in the nucleus of an atom.
Concept insight: Recall the Rutherfors’s model of an atom.
Question 3:
Draw a sketch of Bohr’s model of an atom with three shells.
Solution:
Concept insight: While answering be careful that three shells are asked in the question.
Question 4:
What do you think would be the observation if the α-particle scattering experiment is carried out using a foil of metal other than gold?
Solution:
If a foil of a light metal like lithium is used, then the observations in the alpha-particle scattering experiment would not be the same as that in the gold foil experiment.
Page No. 49-II:
Question 1:
Name the three sub-atomic particles of an atom.
Solution:
(i) Electrons
(ii) Protons and
(iii) Neutrons
Question 2:
Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?
Solution:
Atomic mass of Helium = 4 u
No. of protons = 2
As atomic mass = no. of protons + no. of neutrons
No. of neutrons = At. mass – no. of protons
= 4 – 2 = 2
Concept insight: Recall the correct formula for atomic number and mass number.
Page No. 50:
Question 1:
Write the distribution of electrons in carbon and sodium atoms.
Solution:
Element | At no | Electronic configuration | ||
K shell | M shell | L shell | ||
Carbon Sodium |
6 11 |
2 2 |
4 8 |
1 |
Concept insight: Recall that the atomic number of sodium is 11 and that of carbon is 6.
Question 2:
If K and L shell of an atom are full then what would be the total number of electrons in the atom?
Solution:
No. of electrons in K-shell = 2
No. of electrons in L-shell = 8
The total no. of electrons in the atom = 2 + 8 = 10
Concept insight: Recall that K shell has maximum of 2 electrons and L shell has a maximum of 8 electrons.
Page No. 51-I:
Question 1:
How will you find the valency of chlorine, sulphur and magnesium?
Solution:
Chlorine: At. No. of Cl = 17
Its electronic configuration = 2, 8, 7
Valency of Cl = 8° 7 = 1
Sulphur: At. no. of S = 16
Its electronic configuration = 2, 8, 6
Valency of S = 8° 6 = 2
Magnesium:
At. no. of Mg = 12
Its electronic configuration = 2, 8, 2
Valency of Mg = 2
Concept insight: As we know, when the outermost shell of an atom contains 4 or less than 4 electrons, its valency is equal to the number of valence electrons in the outermost shell and when the outermost shell contains more than 4 electrons, valency of the atom is equal to 8 – no. of valence electrons in the atom.
Page No. 52-II:
Question 1:
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?
Solution:
(i) Atomic number = Number of Protons = 8
(ii) In the given atom, total number of positive charge is equal to the total number of negative charge.
Number of Protons (8) = Number of electrons (8)
So, the charge on the atom will be zero.
Concept insight: Recall the formula of atomic number. Also, remember that the charge on proton is +1 and that on electron is -1.
Question 2:
With the help of Table 4.1, find out the mass number of oxygen and sulphur atom.
Solution:
Mass number = Number of Protons + Number of neutrons
So, Mass number of oxygen = 8 + 8 = 16
Mass number of sulphur = 16 + 16 = 32
Page No. 53:
Question 1:
For the symbol H, D and T tabulate three sub-atomic particles found in each of them.
Solution:
H, D and T are the three isotopes of hydrogen with same atomic number and different mass numbers of 1, 2 and 3 respectively.
Element | Symbols | Electrons | Protons | Neutrons |
Hydrogen Deuterium Tritium |
H D T |
1 11 |
1 1 1 |
0 1 2 |
Concept insight: Recall the mass numbers of each of the isotopes.
Question 2:
Write the electronic configuration of any one pair of isotopes and isobars.
Solution:
Isotopes | Protons | Electrons | Neutrons | ElectronicsConfiguration |
Shell | ||||
K L M N | ||||
3517Cl 3717Cl |
17 17 |
17 17 |
18 20 |
2 8 7 – 2 8 7 – |
Isobars | Protons | Electrons | Neutrons | ElectronicsConfiguration |
Shell | ||||
K L M N | ||||
4020Ca 4018Ar |
20 18 |
20 28 |
20 22 |
2 8 8 2 2 8 8 – |
Concept insight: Recall the definitions of isotopes and isobars.
Page No. 54:
Question 1:
Compare the properties of electrons, protons and neutrons.
Solution:
The properties of electrons, protons and neutrons (i.e. nature of their charge, mass and location) are given below in a tabulated form-
Particle | Nature of charge | Mass | Location |
Electron Proton Neutron |
negative (-1) or -1.6 x 10-19 C positive (+1) or +1.6 x 10-19 C No charge |
9.0 x 10-31 kg 1.672 x 10-27 kg (1 u) 1.672 x 10-27 kg (1 u) |
Extra nuclear part Nucleus Nucleus |
Concept insight: This question is very important from exam point of view.
Question 2:
What are the limitations of J. J. Thomson’s model of the atom ?
Solution:
J.J. Thomson attributed the mass of an atom due to electrons and protons which are evenly spread throughout the atom. But this did not agree with observations of Rutherford according to whom the mass is concentrated in a very small space later called nucleus.
Thomson’s model of the atom could not explain the results of alpha particle scattering experiment carried out by Rutherford.
Concept insight: Recall the features of J. J. Thomson’s model of the atom.
Question 3:
What are the limitations of Rutherford’s model of the atom ?
Solution:
The major limitation of Rutherford’s model of the atom is that it does not explain the stability of the atom. As we know now, when charged bodies move in circular motion, they emit radiations. This means that the electrons revolving round the nucleus (as suggested by Rutherford) would lose energy and come closer and closer to nucleus, and a stage will come when they would finally merge into the nucleus. This makes the atom unstable, which is clearly not the case. The electrons do not fall into the nucleus, atoms are very stable and do not collapse on their own.
Concept insight: Recall the features of Rutherford’s model of the atom.
Question 4:
Describe Bohr’s model of the atom.
Solution:
In order to overcome the objections raised against Rutherford’s model of the atom, Neils Bohr put forward his model of the atom. According to Bohr’s model of the atom,
1. An atom consists of a small positively charged nucleus at its centre.
2. The whole mass of the atom is concentrated at the nucleus.
3. The volume of nucleus is smaller than the volume of the atom (by a ratio of about 1 : 105).
4. The protons and neutrons of the atom are present in the nucleus.
5. The electrons of the atom, which are negatively charged, revolve around the nucleus in definite circular paths known as orbits or which are designated as K, L, M, N etc. or numbered as (n) = 1, 2, 3, 4 etc. (outward from the nucleus).
6. As each orbit is associated with a fixed amount of energy, these orbits are also known as energy levels.
7. While revolving in diserete orbits, the electrons do not radiate energy. But when an electron jumps from one energy level to another, the energy of the atom changes.
Question 5:
Compare all the proposed models of an atom given in this chapter.
Solution:
Feature | Thomson’s model of an atom | Rutherford’s model of an atom | Bohr’s model of an atom |
1. Positive Charge (Protons) 2. Negative charge (electrons)3. Diagrammatic representation4. Limitation: |
As per Thomson’s model of an atom, an atom consists of a positively charged sphere.
The electrons are embedded in the positively charged sphere of an atom, like the seeds in a watermelon.
This model could not explain the results of alpha particle scattering experiment carried out by rutherford. |
The positive charge is concentrated at the core of the atom, which is called nucleus. The nucleus is surrounded by electrons, and the electrons and the nucleus are held together by electrostatic force of attraction. This model could not explain the stability of the atom. |
The positive charge is present in the core of the atom, called nucleus. The electrons move in discrete orbits, and each orbit is associ-ated with a definite amount of energy. Advantage: This model explains the stability of atoms. |
Question 6:
Summarize the rules for writing of distribution of electrons in various shells for the first eighteen elements.
Solution:
The Bohr and Bury scheme for the distribution of electrons in an atom is based on the following rules :
1. The maximum number of electrons which a shell can have is represented by 2n2, where n is the quantum number of that particular energy shell. Thus, the maximum number of electrons in the first four shells are :
1st (K) shell 2 x 12 = 2
2nd (L) shell 2 x 22 = 8
3rd (M) shell 2 x 32 = 18
4th (N) shell 2 x 42 = 32
2. The outermost shell, which is also called valence shell, can have a maximum of 8 electrons.
3. The shell next to (or inner to) the outermost shell, which is called the penultimate shell, can accommodate a maximum of 18
electrons, (if permitted by rule 1).
4. Electrons are not accommodated in a given shell unless the inner shells are filled, i.e., the shells are filled in a step-wise manner.
Question 7:
Define valency by taking examples of silicon and oxygen.
Solution:
Valency is defined as the combining capacity of an atom of an element. If an atom has 4 or less than 4 electrons in its valence shell, then valency is equal to the no. of valence electrons. But if it has more than 4 valence electrons, then valency is equal to 8 – no. of valence electrons.
Silicon has atomic number 14 and its electronic configuration is:
K | L | M |
2 | 8 | 4 |
So, valency of silicon = 8 – 4 = 4
Oxygen has atomic number 8 and its electronic configuration is:
K | L |
2 | 6 |
So, valency of oxygen = 8 – 6 = 2
Concept insight: Remember that the atomic number of oxygen is 8 and Si is 14.
Page No. 55:
Question 8:
Explain with examples
(i) Atomic number
(ii) Mass number
(iii) Isotopes and
(iv) Isobars.
Give any two uses of isotopes.
Solution:
(i) Atomic number: Atomic number of an atom is the total number of protons present within the nucleus of an atom is known as atomic number.
Example: As sodium atom has 11 protons in its nucleus, its atomic number is 11.
(ii) Mass number: Mass number of an atom is the sum total of the masses of all the nucleons present in the nucleus of an atom, i.e.,
Mass Number = No. of Protons + No. of Neutrons
Example: As a sodium atom has 11 protons and 12 neutrons in its nucleus, its mass number = 11 + 12 = 23.
(iii) Isotopes: Isotopes are the atoms of the same element having same atomic number but different mass number.
Example: Hydrogen has three isotopes 11H, 12H, 13H. The atomic number of all the three is 1, but their mass numbers are 1, 2 and 3 respectively.
(iv) Isobars: Isobars are the atoms of different elements having the same mass number but different atomic numbers.
Example: Mass numbers of calcium and argon atoms are 40, but different atomic numbers 20 and 18 respectively.
Two uses of isotopes are:
(i) In isotope of uranium is used as a fuel in nuclear reactors.
(ii) An isotope of cobalt is used in the treatment of cancer.
Concept insight: Definitions and examples are important from exam point of view.
Question 9:
Na+ has completely filled K and L shells. Explain.
Solution:
Atomic number of Na = 11
No. of electrons in Na atom = 11
So, No. of electrons in Na+ ion = 1 ° 1 = 10
Hence, electronic configuration of Na+ = 2, 8
In Na+, K and L shells are completely filled since K shell can have a maximum of 2 electrons and L shell can have a maximum of 8 electrons.
Concept insight: Remember that when a cation is formed, an electron is removed from the outermost shell of the atom
Question 10:
If bromine atom is in the form of say two isotopes 7935Br (49.7%) and 8135Br (50.3%), then calculate the average mass of bromine atom.
Solution:
Question 11:
The average atomic mass of a sample of an element X is 16.2u. What are the percentages of isotopes 168X and 188X in the sample?
Solution:
Concept insight: This numerical is important from exam point of view.
Question 12:
If Z = 3, what would be the valency of the element? Also, name the element.
Solution:
Atomic number, Z = 3
Distribution of electrons : K = 2, L = 1
So, Valency = 1
The element is Lithium.
Concept insight: This numerical is important from exam point of view.
Question 13:
Composition of the nuclei of two atomic spherical 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 ?
Solution:
As we know, mass number of an atom = No. of protons + No. of Neutrons
So, Mass number of X = 6 + 6 = 12
Mass number of Y = 6 + 8 = 14
As both X and Y have the same atomic number (6) but different numbers (i.e., 12 and 14 repectively), so they are isotopes.
Concept insight: This numerical is important from exam point of view.
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.
(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 of iodine which is used as a medicine.
Solution:
(a) F
(b) F
(c) T
(d) T
Question 15:
Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) Atomic nucleus
(b) Electron
(c) Proton
(d) Neutron
Solution:
(a) correct
(b) x
(c) x
(d) x
Question 16:
Isotopes of an element have:
(a) the same physical properties
(b) different chemical properties
(c) diferent number of neutrons
(d) diferent atomic numbers
Solution:
(a) x
(b) x
(c) correct
(d) x
Question 17:
Number of valence electrons in Cl– ion are
(a) 16
(b) 8
(c) 17
(d) 18
Solution:
(a) x
(b) correct
(c) x
(d) x
Page No. 56:
Question 18:
Which one of the followiong is a correct electronic configuration of sodium
(a) 2, 8
(b) 8, 2, 1
(c) 2, 1, 8
(d) 2, 8, 1
Solution:
(d) 2, 8, 1
Concept insight: Recall that the atomic number of sodium is 11.
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 16— |
– 32 24 2 1 |
10 – – – 0 |
– – 12 1 1 |
– 0 |
– sulphur – |
Solution:
First row:
Since atomic no. is 9 so, the element is Fluorine.
Atomic no. = No. of protons = no. of electrons = 9
Mass number = no. of protons + no. of neutrons = 9 + 10 = 19
Second row:
Since atomic no. is 16 so, no. of protons = no. of electrons = 16
No. of neutrons = Mass no. – no. of protons = 32 – 16 = 16
Third row:
No. of protons = Atomic no. = 12
So, the element is Magnesium.
No. of electrons = no. of protons = 12
No. of neutrons = Mass no. – no. of protons = 24 – 12 = 12
Fourth row:
No. of protons = Atomic no. = 1
So, the element is Deuterium.
No. of electrons = no. of protons = 1
No. of neutrons = Mass no. – no. of protons = 2 – 1 = 1
Fifth row:
No. of protons = Atomic no. = 1
The element is Protium since the mass number is 1.
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 Protium |
Concept insight: For asnwering this question, recall the definitions of atomic number and mass numbe