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Structure of Atom:
History: For centuries scientists in the world had been trying to study different properties of matter and
arriving at some theories. But in coarse of time they were proved to be not 100% true and something was still
missing.
Dalton’s Atomic Theory: John Dalton provided the basic theory about the nature of matter. In 1808, he
presented his atomic theory which was a turning point in the study of matter. Dalton picked up the idea of
divisibility of matter, which was till then just a philosophy. He took the name ‘atoms’ as given by the Greeks
and said that the smallest particles of matter are atoms. His theory was based on the laws of chemical
combination. Dalton’s atomic theory provided an explanation for the law of conservation of mass and the law
of definite proportions.
According to Dalton’s atomic theory, all matter, whether an element, a compound or a mixture is composed of
small particles called atoms. The postulates of this theory may be stated as follows:
(i) All matter is made of very tiny particles called atoms, which participate in chemical reactions.
(ii) Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.
(iii) Atoms of a given element are identical in mass and chemical properties.
(iv) Atoms of different elements have different masses and chemical properties.
(v) Atoms combine in the ratio of small whole numbers to form compounds.
(vi) The relative number and kinds of atoms are constant in a given compound.
Dalton was the first scientist to use the symbols for elements in a very specific sense. When he used a
symbol for an element he also meant a definite quantity of that element, that is, one atom of that element.
Hydrogen Carbon Oxygen Phosphorus Sulphur Iron Copper Lead Silver Gold Mercury Platina
Fig - 1
Berzelius suggested that the symbols of elements be made from one or two letters of the name of the
element.
However Dalton’s theory was also not found perfect and further researches continued on atom. The existence
of different kinds of matter is due to different atoms constituting them. Now the questions arise:
(i) What makes the atom of one element different from the atom of another element?
(ii) Are atoms really indivisible, as per Dalton, or are there smaller constituents inside the atom?
The elucidation of the structure of atoms is based on a series of experiments.
One of the first indications that atoms are divisible comes from studying static electricity and the condition
under which electricity is conducted by different substances, which shall not happen unless there are charged
particles present in the atom.
E. Goldstein in 1886 discovered the presence of new radiations in a gas discharge and called them canal
rays. These rays were positively charged radiations which ultimately led to the discovery of a positively
charged sub-atomic particle which was later named as “Proton”.
J. J. Thomson by 1900 said that the atom may indivisible but contained at least one sub-atomic particle and
it. This particle was named as electron and had negative charge.
The charges on electron and proton were equal in magnitude but opposite to each other in nature.
In general, an electron is represented as ‘e–’ and a proton as ‘p+’. It seemed highly that an atom was
Composed of protons and electrons, mutually balancing their charges.
The mass of proton was found approximately 2000 times heavier than that of the electron. The mass of a
proton was taken as one unit and its charge as plus one. The mass of an electron is considered to be
negligible and its charge is minus one.
Thomson proposed that:
I C L S G P
3
(i) An atom may consist of a positively charged sphere and the electrons are embedded in it.
(ii) The negative and positive charges are equal in magnitude. So, the atom as a whole is electrically
neutral.
Although Thomson’s model explained that atoms are electrically neutral, the results of experiments carried
out by other scientists could not be explained by this model.
Positively Charged sphere positively charged
Negatively Charged Electrons negatively Charged
Fig - 1
Rutherford’s Experiment: Ernest Rutherford designed an experiment with fast moving alpha (α)-particles
which were made to fall on a thin gold foil.
• He selected a gold foil because he wanted as thin a layer as possible. This gold foil was about 1000 atoms
thick.
• α -particles are +2 charged helium ions. Since they have a mass of 4 u (2 protons and 2 neutrons), the fast-
moving α -particles have a considerable amount of energy.
• It was expected that α -particles would be deflected by the sub-atomic particles in the gold atoms. Since the
a-particles were much heavier than the protons, he did not expect to see large deflections.
Atoms of Gold Foil At no. 79
Atom’s Nucleus
α - Particles Rebounded
α - Particles Deflected
Fig – 3 α - Particles passed through
But, the α-particle scattering experiment gave totally unexpected results (Fig. above). The observations made
were:
 Most of the fast moving α-particles passed straight through the gold foil.
 Some of the α-particles were deflected by the foil by small angles.
 Approximately one out of every 12000 α-particles were rebounded.
Rutherford concluded from the α-particle scattering experiment that–
4
 Most of the space inside the atom is vacant because most of the α-particles passed through the gold
foil without getting deflected.
 Very few particles were deflected from their path, indicating that the positive charge of the atom
occupies very little space.
 A very small fraction of α-particles were deflected by nearly 180°, indicating that all the positive charge
and mass of the gold atoms were concentrated in a very small volume inside the atom.
From the observed results he also calculated that the diameter of the nucleus is about 10⁵ times smaller than
the diameter of the atom.
 On the basis of this experiment, Rutherford concluded the nuclear model of an atom with features like:
 There is a positively charged centre in an atom (later on it was called the nucleus of the atom). Nucleus
stores almost all the mass of an atom.
 The electrons revolve around the nucleus in circular paths called orbits.
 The nucleus is very small in size in comparison with the size of the atom.
From Rutherford’s experiment it was concluded that the atom consists of a heavy but very small positive
nucleus surrounded by negative electrons. But then what stops the electrons falling into the nucleus?
One idea could be that the atom is like a mini solar system with electrons rotating in orbits around the
nucleus, like the planets orbit the Sun. One problem with this model is that if an electron were to move in this
way it would create a changing electric and magnetic field resulting in emission of EM radiation. The
revolution of the electron in a circular orbit is not expected to be stable. Any particle in a circular orbit would
undergo acceleration and in this state charged particles would radiate energy.
If this were so, the atom should be highly unstable and matter would not exist in the current form.
But the atoms are very much stable.
Bohr’s Model of Atom:
Further Neils Bohr came out with following theory about the atomic model for overcoming the drawback in
Rutherford’s model of atom.
 Only certain special orbits known as discrete orbits of electrons are allowed inside the atom.
 While revolving in discrete orbits the electrons do not radiate energy.
He named these orbits as energy levels and represented them by the letters K, L, M, N,… or the numbers,
n=1, 2, 3, 4 etc…..
N Shell (n = 4)
M Shell (n =3)
L Shell (n = 2)
K Shell (n = 1)
Nucleus
Fig - 4
Distribution of Electrons in different Shells (Orbits):
Bohr and Bury suggested the following rules for the number of electrons to be revolving in different energy
levels or shells:
 The maximum number of electrons present in a shell is given by the formula 2n², where ‘n’ is the orbit
number or energy level index, 1,2,3,….
Hence the maximum number of electrons in different shells can be as following:
1st
orbit or K-shell will be = 2 × 1² = 2,
2nd
orbit or L-shell will be = 2 × 2² = 8,
5
3rd
orbit or M-shell will be = 2 × 3² = 18,
4th
orbit or N-shell will be = 2 × 4² = 32, and so on.
 The maximum number of electrons that can be accommodated in the outermost orbit is 8.
 Electrons are not accommodated in a given shell, unless the inner shells are filled. That is, the shells
are filled in a step-wise manner.
Below Chart shows the number of electrons in each orbit/Shell for few elements.
Elements
Symbols K
Shell
L
Shell
M
Shell
N
Shell
Total
Electrons
Valency
Hydrogen H 1 1 1
Helium He 2 2 0
Lithium Li 2 1 3 1
Beryllium Be 2 2 4 2
Boron B 2 3 5 3
Carbon C 2 4 6 4
Nitrogen N 2 5 7 3
Oxygen O 2 6 8 2
Fluorine F 2 7 9 1
Neon Ne 2 8 10 0
Sodium Na 2 8 1 11 1
Magnesium Mg 2 8 2 12 2
Aluminum Al 2 8 3 13 3
Silicon Si 2 8 4 14 4
Phosphorus P 2 8 5 15 3, 5
Sulfur S 2 8 6 16 2
Chlorine Cl 2 8 7 17 1
Argon Ar 2 8 8 18 0
Calcium Ca 2 8 8 2 20 2
Titanium Ti 2 8 10 2 22 4
Nickel Ni 2 8 16 2 28 2
Table - 1
Valency: The electrons present in the outermost shell of an atom are known as the valence electrons. From
the Bohr-Bury scheme, we also know that the outermost shell of an atom can accommodate a maximum of 8
electrons. It was observed that the atoms of elements completely filled with 8 electrons in the outermost shell
show little chemical activity. In other words, their combining capacity or valency is zero. Of these inert
elements, the helium atom has two electrons in its outermost shell and all other elements have atoms with
eight electrons in the outermost shell.
The combining capacity of the atoms of elements, that is, their tendency to react and form molecules with
atoms of the same or different elements was thus explained as an attempt to attain a fully-filled outermost
shell.
An outermost-shell, which had eight electrons was said to possess an octet. Atoms would thus react, so as to
achieve an octet in the outermost shell. This is done by sharing, gaining or losing electrons. The number of
electrons gained, lost or shared so as to make the octet of electrons in the outermost shell, gives us directly
the combining capacity of the element.
For example, hydrogen/ lithium/sodium atoms contain one electron each in their outermost shell; therefore
each one of them can lose one electron. So, they are said to have valency of one. The valencies of
magnesium and aluminum are 2and 3 respectively, because magnesium has two electrons in its outermost
shell and aluminum has three electrons in its outermost shell.
If the number of electrons in the outermost shell of an atom is close to its full capacity, then valency is
determined in a different way. For example, the fluorine atom has 7 electrons in the outermost shell, and its
6
valency could be 7. But it is easier for fluorine to gain one electron instead of losing seven electrons. Hence,
its valency is determined by subtracting seven electrons from the octet and this gives us a valency of one for
fluorine. Therefore, an atom of each element has a definite combining capacity, called its valency.
ATOMIC NUMBER: We know that protons are present in the nucleus of an atom. It is the number of protons
of an atom, which determines its atomic number. It is denoted by ‘Z’. All atoms of an element have the same
atomic number Z. In fact, elements are defined by the number of protons they possess. For hydrogen, Z = 1,
because in hydrogen atom, only one proton is present in the nucleus. Similarly, for carbon, Z = 6. Therefore,
the atomic number
is defined as the total number of protons present in the nucleus of an atom.
Elements Symbols
Atomic
Number
No of
Protons
No of
Neutrons
Mass
Number
No of
Electrons
Hydrogen H 1 1 0 1 1
Helium He 2 2 2 4 2
Lithium Li 3 3 4 7 3
Beryllium Be 4 4 5 9 4
Boron B 5 5 6 11 5
Carbon C 6 6 6 12 6
Nitrogen N 7 7 7 14 7
Oxygen O 8 8 8 16 8
Fluorine F 9 9 10 19 9
Neon Ne 10 10 10 20 10
Sodium Na 11 11 12 23 11
Magnesium Mg 12 12 12 24 12
Aluminum Al 13 13 14 27 13
Silicon Si 14 14 14 28 14
Phosphorus P 15 15 16 31 15
Sulfur S 16 16 16 32 16
Chlorine Cl 17 17 18 35 17
Argon Ar 18 18 22 40 18
Table - 2
MASS NUMBER: After studying the properties of the sub atomic particles of an atom, we can conclude that
mass of an atom is practically due to protons and neutrons alone. These are present in the nucleus of an
atom. Hence
Protons and neutrons are also called nucleons. Therefore, the mass of an atom resides in its nucleus. For
example, mass of carbon is 12 u because it has 6 protons and 6 neutrons, 6 u + 6 u = 12 u. Similarly, the
mass of aluminum is 27 u (13 protons+14 neutrons). The mass number is defined as the sum of the total
number of protons and neutrons present in the nucleus of an atom. It is denoted by ‘A’. In the notation for an
atom, the atomic number, mass number and symbol of the element are to be written as:
Mass Number 14
Symbol of element for example, nitrogen is written as: N
Atomic Number 7
Isotopes:
In nature, a number of atoms of some elements have been identified, which have the same atomic number
but different mass numbers. For example, take the case of hydrogen atom; it has three atomic species,
namely protium - mass number 1 and Atomic number 1, deuterium - mass number 2 and Atomic number 1
and tritium - mass number 3 and Atomic number 1. The atomic number of each one is 1, but the mass
number is 1, 2 and 3, respectively.
7
On the basis of these examples, isotopes are defined as the atoms of the same element, having the same
atomic number but different mass numbers. Therefore, we can say that there are three isotopes of hydrogen
atom, namely protium, deuterium and tritium.
Many elements consist of a mixture of isotopes. Each isotope of an element is a pure substance. The
chemical properties of isotopes are similar but their physical properties are different. Chlorine occurs in nature
in two isotopic forms, with masses 35 u and 37 u in the ratio of 3:1. Obviously, the question arises: what
should we take as the mass of chlorine atom?
Let us find out. The average atomic mass of chlorine atom, on the basis of above data, will be
[(35x75/100) + (37x25/100)] = [(105/4) + (37/4)] = 142/4 = 35.5 u
The mass of an atom of any natural element is taken as the average mass of all the naturally occurring atoms
of that element. If an element has no isotopes, then the mass of its atom would be the same as the sum of
protons and neutrons in it. But if an element occurs in isotopic forms, then we have to know the percentage of
each isotopic form and then the average mass is calculated. This does not mean that any one atom of
chlorine has a fractional mass of 35.5 u. It means that if you take a certain amount of chlorine, it will contain
both isotopes of chlorine and the average mass is 35.5 u.
Applications Since the chemical properties of all the isotopes of an element are the same, normally we are not
concerned about taking a mixture. But some isotopes have special properties which find them useful in
various fields. Some of them are:
(i) An isotope of uranium is used as a fuel in nuclear reactors.
(ii) An isotope of cobalt is used in the treatment of cancer.
(iii) An isotope of iodine is used in the treatment of goitre.
ISOBARS:
Let us consider two elements — calcium, atomic number 20, and argon, atomic number 18. The number of
protons in these atoms is different, but the mass number of both these elements is 40. That is, the total
number of nucleons is the same in the atoms of this pair of elements. Atoms of different elements with
different atomic numbers, which have the same mass number, are known as isobars.
Important:
Mass of Electron = me = 9.11 x 10ˉ³¹ Kg = approximately 1/2000 mass of Proton (mp)
Mass of Proton = mp = 1.672 x 10ˉ²⁷ Kg
Mass of Neutron = mn = 1.674 x 10ˉ²⁷ Kg
Charge on Electron = -1.6 x 10 x 10ˉ¹⁹ Coulombs (-1e)
Charge on Proton = 1.6 x 10 x 10ˉ¹⁹ Coulombs (+1e)
(1 coulomb is equivalent to charge provided by approximately 6.242 x 10¹⁸ protons)
Charge of Neutron = 0
Questions and Answers:
1. What are canal rays
The positively charged radiations in a gas discharge were discovered and called canal rays by E Goldstein.
2. If an atom contains one electron and one proton, will it carry any charge or not?
No
3. On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.
Thomson proposed that:
(i) An atom may consist of a positively charged sphere and the electrons are embedded in it.
(ii) The negative and positive charges are equal in magnitude. So, the atom as a whole is electrically neutral.
4. On the basis of Rutherford’s model, which subatomic particle is present in the nucleus of an atom?
Proton as positively charged particle.
5. Draw a sketch of Bohr’s model of an atom with three shells.
See Fig 4 and draw
8
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?
If a metal foil used which has a heavier nucleus than gold then scattering will be more, but with lighter
Nucleus than gold metal foils will scatter the α-particle lesser.
7. Name the three sub-atomic particles of an atom.
Electrons, protons and neutrons.
8. Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?
2 (Two)
9. Write the distribution of electrons in carbon and sodium atoms.
i) Carbon – No. of electrons 8: K shell – 2, L Shell – 4 (1s² 2s²2p²)
ii) Sodium – No. of Electrons 11: K shell – 2, L shell – 8, M shell – 1 (1s² 2s²2p⁶ 3s¹)
10. If K and L shells of an atom are full, then what would be the total number of electrons in the atom?
K shell 2 + L shell 8 No. of electrons in the atom = 10
11. How will you find the valency of Chlorine, Sulphur and Magnesium?
Chlorine has 17 electrons: K shell = 2, L shell = 8, M shell = 7. It has 1 electron short to from octet, so
Valency of chlorine is 1 as it can accept 1 electron
Sulpher has 16 electrons: K shell = 2, L shell = 8, M shell = 6. It has 2 electrons short to from octet, so
Valency of Sulpher is 2 as it can accept 2 electrons.
Magnesium 12 electrons: K shell = 2, L shell = 8, M shell = 2. It has 2 electron in the outermost orbit so
Valency of chlorine is 1 as it can loose 1 electron.
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?
Atomic No. is 8
(ii) What is the charge on the atom?
Charge on atom is 0 as atoms are neutral.
13. What is the mass number of Oxygen and Sulphur atoms?
Oxygen has 8 Protons and 8 Neutrons; therefore Mass number of Oxygen is 16.
Sulpher has 16 Protons and 16 Neutrons therefore Mass number of Sulpher is 32.
14. For the symbol H, D and T tabulate three sub-atomic particles found in each of them.
For H electrons 1, Protons 1, Neutrons 0 Total Sub atomic particles 2
For D electrons 1, Protons 1, Neutrons 1 Total Sub atomic particles 3
For T electrons 1, Protons 1, Neutrons 2 Total Sub atomic particles 4
15. Write the electronic configuration of any one pair of isotopes and isobars.
12
6C No of Protons 6 + No. of Neutrons 6 Atomic mass no. of a Carbon isotope 12.
14
6C No of Protons 6 + No. of Neutrons 8 Atomic mass no. of another Carbon isotope 14.
Argon has 18 protons and 22 neutrons and mass no. is 40. Calcium has 20 Protons + 20 neutrons and
Mass no. is 40 this is an example of Isobars.
Worked Exercises:
1. Compare the properties of electrons, protons and neutrons.
Electron – has negative charge of -1.6 x 10ˉ¹⁹ Coulombs, mass of 9.11 x 10ˉ³¹ Kg = approximately 1/2000 mass
of Proton, revolves around nucleus in definite orbits.
Protons have equal and opposite charge of +1.6x10ˉ¹⁹ coulombs, having mass of 1.672 x 10ˉ²⁷ Kg,
Remains stable inside the atom nucleus.
Neutrons are also placed in the atom nucleus along with protons, has 0 charge.
Protons and neutrons together called nucleons.
2. What are the limitations of J.J. Thomson’s model of the atom?
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.
9
3. What are the limitations of Rutherford’s model of the atom?
The limitations of Rutherford's model of the atom are It failed to explain the stability of an atom.
It doesn't explain the spectrum of hydrogen and other atoms.
4. Describe Bohr’s model of the atom.
The atom consists of a small positively charged nucleus at its center. 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 the atom are contained in the nucleus. Only certain orbits known as discrete orbits
of electrons are allowed inside the atom. While revolving in these discrete orbits electrons do not radiate energy.
These orbits or cells are represented by the letters K, L, M, N etc. or the numbers, n = 1, 2, 3, 4, . .
As shown in Fig - 4.
5. Compare all the proposed models of an atom given in this chapter.
Thomson’s model: An atom consists of a 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.
Rutherford’s model: An atom consists of a positively charged center 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
compared to the size of the atom.
Bohr’s model: Bohr agreed with almost all points as said by Rutherford except regarding the
revolution of electrons for which he added that there are only certain orbits known as discrete orbits
inside the atom in which electrons revolve around the nucleus. While revolving in its discrete orbits the
Electrons do not radiate energy.
6. Summarize the rules for writing of distribution of electrons in various shells for the first 18 elements.
The rules for writing of the distribution of electrons in various shells for the first eighteen elements are given
below. If n gives the number of orbit or energy level, then 2n² gives the maximum number of electrons possible
in a given orbit or energy level. Thus, First orbit or K-shell will have 2 electrons, Second orbit or L-shell will have
8 electrons, Third orbit or M-shell will have 18 electrons. If it is the outermost orbit, then it should have not more
than 8 electrons. There should be step-wise filling of electrons in different orbits, i.e., electrons are not
accompanied in a given orbit if the earlier orbits or shells are incompletely filled.
7. Define valency by taking examples of silicon and oxygen.
The valency of an element is the combining capacity of that element. The valency of an element is determined
by the number of valence electrons present in the atom of that element. Valency of Silicon: It has electronic
configuration: 2, 8, 4 Thus, the valency of silicon is 4 as these electrons can be shared with others to complete
octet. Valency of Oxygen: It has electronic configuration: 26 Thus, the valency of oxygen is 2 as it will gain 2
electrons to complete its octet.
8. Explain with examples (i) Atomic number, (ii) Mass number, (iii) Isotopes and iv) Isobars. Give any two uses of
isotopes.
(i) 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. So, the atomic number of nitrogen is 7.
(ii) (ii) Mass number: The mass number of 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 5 protons and 6 neutrons. So,
the mass number of boron is 5 + 6 = 11.
(iii) Isotopes: These are atoms of the same element having the same atomic number, but different mass
numbers. For example, chlorine has two isotopes with atomic number 17 but mass numbers 35 and 37
represented by
35 37
Cl and Cl
17 17
(iv) Isobars: These are atoms having the same mass number, but different atomic numbers i.e., isobars are
atoms of different elements having the same mass number. For example, Ne has atomic number 10 and
sodium has atomic number 11 but both of them have mass numbers as 22 represented by
22 22
Ne and Na
10 11
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.
10
9. Na+ has completely filled K and L shells. Explain.
The atomic number of sodium is 11. So, neutral sodium atom has 11 electrons and its electronic configuration is
2, 8, and 1. But Na+ has 10 electrons. Out of 10, K-shell contains 2 and L-shell 8 electrons respectively. Thus,
Na+ has completely filled K and L shells.
10. If bromine atom is available in the form of, say, two isotopes
79 81
Br (.7%) and Br (50.3%), calculate the average atomic mass of bromine atom.
35 35
It is given that two isotopes of bromine are 79 / 35Br (49.7%) and 81 / 35Br (50.3%). Then, the average
atomic mass of bromine atom is calculated as below:
(79x 49.7/100) + (81x50.3/100) = (3926.3/100) + (4074.3/100) = 8000.6/100 = 80.006 u.
11. The average atomic mass of a sample of an element X is 16.2 u.
16 18
What are the percentages of isotopes 8X and 8X in the sample?
Given that the average atomic mass of the sample of element X is 16.2 u.
Let the percentage of isotope 18 / 8 X is y%. So the percentage of isotope 16 / 8 X will be (100 - y)%.
Hence 18 x y/100 + 16 x (100 – y)/100 =16.2 or 18y/100 + (1600 – 16y)/100 = 16.2
Or (18y + 1600 -16y)/100 = 16.2 or 2y = 1620 – 1600 or y = 20/2 =10
Therefore, the percentage of isotope 18 / 8 X is 10%.
And, the percentage of isotope 16 / 8 X is (100 - 10) % = 90%.
12. If Z = 3, what would be the valency of the element? Also, name the element.
If Z = 3, that means the atomic number of the element is 3. Its electronic configuration is 2 and 1. Therefore the
valency of the element is 1 (since the outermost shell has only one electron). The element with Z = 3 is lithium.
13. Composition of the nuclei of two atomic species X and Y are given as X Y
Protons = 6 6, Neutrons = 6 8, Give the mass numbers of X and Y. Establish the relation between them
Mass number of X = Number of protons + Number of neutrons = 6 + 6 = 12
Mass number of Y = Number of protons + Number of neutrons = 6 + 8 = 14. These atomic species
X and Y have the same atomic number, but different mass numbers. So they are isotopes.
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. - F
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral. – F
(c) The mass of an electron is about 1/2000 times that of proton. – T
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine. – F
15. Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) Atomic Nucleus, (b) Electron, (c) Proton (d) Neutron - √ Atomic nucleus.
(b)
16. Isotopes of an element have (a) the same physical properties, (b) different chemical properties, (c)
Different number of neutrons, (d) different atomic numbers. - √ different numbers of neutrons.
17. Complete the following table.
Sl.
No.
Atomic
Number
Mass
Number
Number of
Protons
Number of
Neutron
Number of
Electron
Name of Atomic Species
1 9 18 9 9 9 Fluorine
2 16 32 16 16 16 Sulpher
3 12 24 12 12 12 Magnesium
4 1 2 1 1 1 Deuterium
5 1 1 0 1 0 Hydrogen ion
Compiled for with relevant questions and For
Classes/grades 9 - 10 (MYP 4 – 5)
Base on Chapter Structure of Atom

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408283346-Structure-of-Atom.pdf

  • 1. 1
  • 2. 2 Structure of Atom: History: For centuries scientists in the world had been trying to study different properties of matter and arriving at some theories. But in coarse of time they were proved to be not 100% true and something was still missing. Dalton’s Atomic Theory: John Dalton provided the basic theory about the nature of matter. In 1808, he presented his atomic theory which was a turning point in the study of matter. Dalton picked up the idea of divisibility of matter, which was till then just a philosophy. He took the name ‘atoms’ as given by the Greeks and said that the smallest particles of matter are atoms. His theory was based on the laws of chemical combination. Dalton’s atomic theory provided an explanation for the law of conservation of mass and the law of definite proportions. According to Dalton’s atomic theory, all matter, whether an element, a compound or a mixture is composed of small particles called atoms. The postulates of this theory may be stated as follows: (i) All matter is made of very tiny particles called atoms, which participate in chemical reactions. (ii) Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction. (iii) Atoms of a given element are identical in mass and chemical properties. (iv) Atoms of different elements have different masses and chemical properties. (v) Atoms combine in the ratio of small whole numbers to form compounds. (vi) The relative number and kinds of atoms are constant in a given compound. Dalton was the first scientist to use the symbols for elements in a very specific sense. When he used a symbol for an element he also meant a definite quantity of that element, that is, one atom of that element. Hydrogen Carbon Oxygen Phosphorus Sulphur Iron Copper Lead Silver Gold Mercury Platina Fig - 1 Berzelius suggested that the symbols of elements be made from one or two letters of the name of the element. However Dalton’s theory was also not found perfect and further researches continued on atom. The existence of different kinds of matter is due to different atoms constituting them. Now the questions arise: (i) What makes the atom of one element different from the atom of another element? (ii) Are atoms really indivisible, as per Dalton, or are there smaller constituents inside the atom? The elucidation of the structure of atoms is based on a series of experiments. One of the first indications that atoms are divisible comes from studying static electricity and the condition under which electricity is conducted by different substances, which shall not happen unless there are charged particles present in the atom. E. Goldstein in 1886 discovered the presence of new radiations in a gas discharge and called them canal rays. These rays were positively charged radiations which ultimately led to the discovery of a positively charged sub-atomic particle which was later named as “Proton”. J. J. Thomson by 1900 said that the atom may indivisible but contained at least one sub-atomic particle and it. This particle was named as electron and had negative charge. The charges on electron and proton were equal in magnitude but opposite to each other in nature. In general, an electron is represented as ‘e–’ and a proton as ‘p+’. It seemed highly that an atom was Composed of protons and electrons, mutually balancing their charges. The mass of proton was found approximately 2000 times heavier than that of the electron. The mass of a proton was taken as one unit and its charge as plus one. The mass of an electron is considered to be negligible and its charge is minus one. Thomson proposed that: I C L S G P
  • 3. 3 (i) An atom may consist of a positively charged sphere and the electrons are embedded in it. (ii) The negative and positive charges are equal in magnitude. So, the atom as a whole is electrically neutral. Although Thomson’s model explained that atoms are electrically neutral, the results of experiments carried out by other scientists could not be explained by this model. Positively Charged sphere positively charged Negatively Charged Electrons negatively Charged Fig - 1 Rutherford’s Experiment: Ernest Rutherford designed an experiment with fast moving alpha (α)-particles which were made to fall on a thin gold foil. • He selected a gold foil because he wanted as thin a layer as possible. This gold foil was about 1000 atoms thick. • α -particles are +2 charged helium ions. Since they have a mass of 4 u (2 protons and 2 neutrons), the fast- moving α -particles have a considerable amount of energy. • It was expected that α -particles would be deflected by the sub-atomic particles in the gold atoms. Since the a-particles were much heavier than the protons, he did not expect to see large deflections. Atoms of Gold Foil At no. 79 Atom’s Nucleus α - Particles Rebounded α - Particles Deflected Fig – 3 α - Particles passed through But, the α-particle scattering experiment gave totally unexpected results (Fig. above). The observations made were:  Most of the fast moving α-particles passed straight through the gold foil.  Some of the α-particles were deflected by the foil by small angles.  Approximately one out of every 12000 α-particles were rebounded. Rutherford concluded from the α-particle scattering experiment that–
  • 4. 4  Most of the space inside the atom is vacant because most of the α-particles passed through the gold foil without getting deflected.  Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space.  A very small fraction of α-particles were deflected by nearly 180°, indicating that all the positive charge and mass of the gold atoms were concentrated in a very small volume inside the atom. From the observed results he also calculated that the diameter of the nucleus is about 10⁵ times smaller than the diameter of the atom.  On the basis of this experiment, Rutherford concluded the nuclear model of an atom with features like:  There is a positively charged centre in an atom (later on it was called the nucleus of the atom). Nucleus stores almost all the mass of an atom.  The electrons revolve around the nucleus in circular paths called orbits.  The nucleus is very small in size in comparison with the size of the atom. From Rutherford’s experiment it was concluded that the atom consists of a heavy but very small positive nucleus surrounded by negative electrons. But then what stops the electrons falling into the nucleus? One idea could be that the atom is like a mini solar system with electrons rotating in orbits around the nucleus, like the planets orbit the Sun. One problem with this model is that if an electron were to move in this way it would create a changing electric and magnetic field resulting in emission of EM radiation. The revolution of the electron in a circular orbit is not expected to be stable. Any particle in a circular orbit would undergo acceleration and in this state charged particles would radiate energy. If this were so, the atom should be highly unstable and matter would not exist in the current form. But the atoms are very much stable. Bohr’s Model of Atom: Further Neils Bohr came out with following theory about the atomic model for overcoming the drawback in Rutherford’s model of atom.  Only certain special orbits known as discrete orbits of electrons are allowed inside the atom.  While revolving in discrete orbits the electrons do not radiate energy. He named these orbits as energy levels and represented them by the letters K, L, M, N,… or the numbers, n=1, 2, 3, 4 etc….. N Shell (n = 4) M Shell (n =3) L Shell (n = 2) K Shell (n = 1) Nucleus Fig - 4 Distribution of Electrons in different Shells (Orbits): Bohr and Bury suggested the following rules for the number of electrons to be revolving in different energy levels or shells:  The maximum number of electrons present in a shell is given by the formula 2n², where ‘n’ is the orbit number or energy level index, 1,2,3,…. Hence the maximum number of electrons in different shells can be as following: 1st orbit or K-shell will be = 2 × 1² = 2, 2nd orbit or L-shell will be = 2 × 2² = 8,
  • 5. 5 3rd orbit or M-shell will be = 2 × 3² = 18, 4th orbit or N-shell will be = 2 × 4² = 32, and so on.  The maximum number of electrons that can be accommodated in the outermost orbit is 8.  Electrons are not accommodated in a given shell, unless the inner shells are filled. That is, the shells are filled in a step-wise manner. Below Chart shows the number of electrons in each orbit/Shell for few elements. Elements Symbols K Shell L Shell M Shell N Shell Total Electrons Valency Hydrogen H 1 1 1 Helium He 2 2 0 Lithium Li 2 1 3 1 Beryllium Be 2 2 4 2 Boron B 2 3 5 3 Carbon C 2 4 6 4 Nitrogen N 2 5 7 3 Oxygen O 2 6 8 2 Fluorine F 2 7 9 1 Neon Ne 2 8 10 0 Sodium Na 2 8 1 11 1 Magnesium Mg 2 8 2 12 2 Aluminum Al 2 8 3 13 3 Silicon Si 2 8 4 14 4 Phosphorus P 2 8 5 15 3, 5 Sulfur S 2 8 6 16 2 Chlorine Cl 2 8 7 17 1 Argon Ar 2 8 8 18 0 Calcium Ca 2 8 8 2 20 2 Titanium Ti 2 8 10 2 22 4 Nickel Ni 2 8 16 2 28 2 Table - 1 Valency: The electrons present in the outermost shell of an atom are known as the valence electrons. From the Bohr-Bury scheme, we also know that the outermost shell of an atom can accommodate a maximum of 8 electrons. It was observed that the atoms of elements completely filled with 8 electrons in the outermost shell show little chemical activity. In other words, their combining capacity or valency is zero. Of these inert elements, the helium atom has two electrons in its outermost shell and all other elements have atoms with eight electrons in the outermost shell. The combining capacity of the atoms of elements, that is, their tendency to react and form molecules with atoms of the same or different elements was thus explained as an attempt to attain a fully-filled outermost shell. An outermost-shell, which had eight electrons was said to possess an octet. Atoms would thus react, so as to achieve an octet in the outermost shell. This is done by sharing, gaining or losing electrons. The number of electrons gained, lost or shared so as to make the octet of electrons in the outermost shell, gives us directly the combining capacity of the element. For example, hydrogen/ lithium/sodium atoms contain one electron each in their outermost shell; therefore each one of them can lose one electron. So, they are said to have valency of one. The valencies of magnesium and aluminum are 2and 3 respectively, because magnesium has two electrons in its outermost shell and aluminum has three electrons in its outermost shell. If the number of electrons in the outermost shell of an atom is close to its full capacity, then valency is determined in a different way. For example, the fluorine atom has 7 electrons in the outermost shell, and its
  • 6. 6 valency could be 7. But it is easier for fluorine to gain one electron instead of losing seven electrons. Hence, its valency is determined by subtracting seven electrons from the octet and this gives us a valency of one for fluorine. Therefore, an atom of each element has a definite combining capacity, called its valency. ATOMIC NUMBER: We know that protons are present in the nucleus of an atom. It is the number of protons of an atom, which determines its atomic number. It is denoted by ‘Z’. All atoms of an element have the same atomic number Z. In fact, elements are defined by the number of protons they possess. For hydrogen, Z = 1, because in hydrogen atom, only one proton is present in the nucleus. Similarly, for carbon, Z = 6. Therefore, the atomic number is defined as the total number of protons present in the nucleus of an atom. Elements Symbols Atomic Number No of Protons No of Neutrons Mass Number No of Electrons Hydrogen H 1 1 0 1 1 Helium He 2 2 2 4 2 Lithium Li 3 3 4 7 3 Beryllium Be 4 4 5 9 4 Boron B 5 5 6 11 5 Carbon C 6 6 6 12 6 Nitrogen N 7 7 7 14 7 Oxygen O 8 8 8 16 8 Fluorine F 9 9 10 19 9 Neon Ne 10 10 10 20 10 Sodium Na 11 11 12 23 11 Magnesium Mg 12 12 12 24 12 Aluminum Al 13 13 14 27 13 Silicon Si 14 14 14 28 14 Phosphorus P 15 15 16 31 15 Sulfur S 16 16 16 32 16 Chlorine Cl 17 17 18 35 17 Argon Ar 18 18 22 40 18 Table - 2 MASS NUMBER: After studying the properties of the sub atomic particles of an atom, we can conclude that mass of an atom is practically due to protons and neutrons alone. These are present in the nucleus of an atom. Hence Protons and neutrons are also called nucleons. Therefore, the mass of an atom resides in its nucleus. For example, mass of carbon is 12 u because it has 6 protons and 6 neutrons, 6 u + 6 u = 12 u. Similarly, the mass of aluminum is 27 u (13 protons+14 neutrons). The mass number is defined as the sum of the total number of protons and neutrons present in the nucleus of an atom. It is denoted by ‘A’. In the notation for an atom, the atomic number, mass number and symbol of the element are to be written as: Mass Number 14 Symbol of element for example, nitrogen is written as: N Atomic Number 7 Isotopes: In nature, a number of atoms of some elements have been identified, which have the same atomic number but different mass numbers. For example, take the case of hydrogen atom; it has three atomic species, namely protium - mass number 1 and Atomic number 1, deuterium - mass number 2 and Atomic number 1 and tritium - mass number 3 and Atomic number 1. The atomic number of each one is 1, but the mass number is 1, 2 and 3, respectively.
  • 7. 7 On the basis of these examples, isotopes are defined as the atoms of the same element, having the same atomic number but different mass numbers. Therefore, we can say that there are three isotopes of hydrogen atom, namely protium, deuterium and tritium. Many elements consist of a mixture of isotopes. Each isotope of an element is a pure substance. The chemical properties of isotopes are similar but their physical properties are different. Chlorine occurs in nature in two isotopic forms, with masses 35 u and 37 u in the ratio of 3:1. Obviously, the question arises: what should we take as the mass of chlorine atom? Let us find out. The average atomic mass of chlorine atom, on the basis of above data, will be [(35x75/100) + (37x25/100)] = [(105/4) + (37/4)] = 142/4 = 35.5 u The mass of an atom of any natural element is taken as the average mass of all the naturally occurring atoms of that element. If an element has no isotopes, then the mass of its atom would be the same as the sum of protons and neutrons in it. But if an element occurs in isotopic forms, then we have to know the percentage of each isotopic form and then the average mass is calculated. This does not mean that any one atom of chlorine has a fractional mass of 35.5 u. It means that if you take a certain amount of chlorine, it will contain both isotopes of chlorine and the average mass is 35.5 u. Applications Since the chemical properties of all the isotopes of an element are the same, normally we are not concerned about taking a mixture. But some isotopes have special properties which find them useful in various fields. Some of them are: (i) An isotope of uranium is used as a fuel in nuclear reactors. (ii) An isotope of cobalt is used in the treatment of cancer. (iii) An isotope of iodine is used in the treatment of goitre. ISOBARS: Let us consider two elements — calcium, atomic number 20, and argon, atomic number 18. The number of protons in these atoms is different, but the mass number of both these elements is 40. That is, the total number of nucleons is the same in the atoms of this pair of elements. Atoms of different elements with different atomic numbers, which have the same mass number, are known as isobars. Important: Mass of Electron = me = 9.11 x 10ˉ³¹ Kg = approximately 1/2000 mass of Proton (mp) Mass of Proton = mp = 1.672 x 10ˉ²⁷ Kg Mass of Neutron = mn = 1.674 x 10ˉ²⁷ Kg Charge on Electron = -1.6 x 10 x 10ˉ¹⁹ Coulombs (-1e) Charge on Proton = 1.6 x 10 x 10ˉ¹⁹ Coulombs (+1e) (1 coulomb is equivalent to charge provided by approximately 6.242 x 10¹⁸ protons) Charge of Neutron = 0 Questions and Answers: 1. What are canal rays The positively charged radiations in a gas discharge were discovered and called canal rays by E Goldstein. 2. If an atom contains one electron and one proton, will it carry any charge or not? No 3. On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole. Thomson proposed that: (i) An atom may consist of a positively charged sphere and the electrons are embedded in it. (ii) The negative and positive charges are equal in magnitude. So, the atom as a whole is electrically neutral. 4. On the basis of Rutherford’s model, which subatomic particle is present in the nucleus of an atom? Proton as positively charged particle. 5. Draw a sketch of Bohr’s model of an atom with three shells. See Fig 4 and draw
  • 8. 8 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? If a metal foil used which has a heavier nucleus than gold then scattering will be more, but with lighter Nucleus than gold metal foils will scatter the α-particle lesser. 7. Name the three sub-atomic particles of an atom. Electrons, protons and neutrons. 8. Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have? 2 (Two) 9. Write the distribution of electrons in carbon and sodium atoms. i) Carbon – No. of electrons 8: K shell – 2, L Shell – 4 (1s² 2s²2p²) ii) Sodium – No. of Electrons 11: K shell – 2, L shell – 8, M shell – 1 (1s² 2s²2p⁶ 3s¹) 10. If K and L shells of an atom are full, then what would be the total number of electrons in the atom? K shell 2 + L shell 8 No. of electrons in the atom = 10 11. How will you find the valency of Chlorine, Sulphur and Magnesium? Chlorine has 17 electrons: K shell = 2, L shell = 8, M shell = 7. It has 1 electron short to from octet, so Valency of chlorine is 1 as it can accept 1 electron Sulpher has 16 electrons: K shell = 2, L shell = 8, M shell = 6. It has 2 electrons short to from octet, so Valency of Sulpher is 2 as it can accept 2 electrons. Magnesium 12 electrons: K shell = 2, L shell = 8, M shell = 2. It has 2 electron in the outermost orbit so Valency of chlorine is 1 as it can loose 1 electron. 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? Atomic No. is 8 (ii) What is the charge on the atom? Charge on atom is 0 as atoms are neutral. 13. What is the mass number of Oxygen and Sulphur atoms? Oxygen has 8 Protons and 8 Neutrons; therefore Mass number of Oxygen is 16. Sulpher has 16 Protons and 16 Neutrons therefore Mass number of Sulpher is 32. 14. For the symbol H, D and T tabulate three sub-atomic particles found in each of them. For H electrons 1, Protons 1, Neutrons 0 Total Sub atomic particles 2 For D electrons 1, Protons 1, Neutrons 1 Total Sub atomic particles 3 For T electrons 1, Protons 1, Neutrons 2 Total Sub atomic particles 4 15. Write the electronic configuration of any one pair of isotopes and isobars. 12 6C No of Protons 6 + No. of Neutrons 6 Atomic mass no. of a Carbon isotope 12. 14 6C No of Protons 6 + No. of Neutrons 8 Atomic mass no. of another Carbon isotope 14. Argon has 18 protons and 22 neutrons and mass no. is 40. Calcium has 20 Protons + 20 neutrons and Mass no. is 40 this is an example of Isobars. Worked Exercises: 1. Compare the properties of electrons, protons and neutrons. Electron – has negative charge of -1.6 x 10ˉ¹⁹ Coulombs, mass of 9.11 x 10ˉ³¹ Kg = approximately 1/2000 mass of Proton, revolves around nucleus in definite orbits. Protons have equal and opposite charge of +1.6x10ˉ¹⁹ coulombs, having mass of 1.672 x 10ˉ²⁷ Kg, Remains stable inside the atom nucleus. Neutrons are also placed in the atom nucleus along with protons, has 0 charge. Protons and neutrons together called nucleons. 2. What are the limitations of J.J. Thomson’s model of the atom? 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.
  • 9. 9 3. What are the limitations of Rutherford’s model of the atom? The limitations of Rutherford's model of the atom are It failed to explain the stability of an atom. It doesn't explain the spectrum of hydrogen and other atoms. 4. Describe Bohr’s model of the atom. The atom consists of a small positively charged nucleus at its center. 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 the atom are contained in the nucleus. Only certain orbits known as discrete orbits of electrons are allowed inside the atom. While revolving in these discrete orbits electrons do not radiate energy. These orbits or cells are represented by the letters K, L, M, N etc. or the numbers, n = 1, 2, 3, 4, . . As shown in Fig - 4. 5. Compare all the proposed models of an atom given in this chapter. Thomson’s model: An atom consists of a 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. Rutherford’s model: An atom consists of a positively charged center 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 compared to the size of the atom. Bohr’s model: Bohr agreed with almost all points as said by Rutherford except regarding the revolution of electrons for which he added that there are only certain orbits known as discrete orbits inside the atom in which electrons revolve around the nucleus. While revolving in its discrete orbits the Electrons do not radiate energy. 6. Summarize the rules for writing of distribution of electrons in various shells for the first 18 elements. The rules for writing of the distribution of electrons in various shells for the first eighteen elements are given below. If n gives the number of orbit or energy level, then 2n² gives the maximum number of electrons possible in a given orbit or energy level. Thus, First orbit or K-shell will have 2 electrons, Second orbit or L-shell will have 8 electrons, Third orbit or M-shell will have 18 electrons. If it is the outermost orbit, then it should have not more than 8 electrons. There should be step-wise filling of electrons in different orbits, i.e., electrons are not accompanied in a given orbit if the earlier orbits or shells are incompletely filled. 7. Define valency by taking examples of silicon and oxygen. The valency of an element is the combining capacity of that element. The valency of an element is determined by the number of valence electrons present in the atom of that element. Valency of Silicon: It has electronic configuration: 2, 8, 4 Thus, the valency of silicon is 4 as these electrons can be shared with others to complete octet. Valency of Oxygen: It has electronic configuration: 26 Thus, the valency of oxygen is 2 as it will gain 2 electrons to complete its octet. 8. Explain with examples (i) Atomic number, (ii) Mass number, (iii) Isotopes and iv) Isobars. Give any two uses of isotopes. (i) 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. So, the atomic number of nitrogen is 7. (ii) (ii) Mass number: The mass number of 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 5 protons and 6 neutrons. So, the mass number of boron is 5 + 6 = 11. (iii) Isotopes: These are atoms of the same element having the same atomic number, but different mass numbers. For example, chlorine has two isotopes with atomic number 17 but mass numbers 35 and 37 represented by 35 37 Cl and Cl 17 17 (iv) Isobars: These are atoms having the same mass number, but different atomic numbers i.e., isobars are atoms of different elements having the same mass number. For example, Ne has atomic number 10 and sodium has atomic number 11 but both of them have mass numbers as 22 represented by 22 22 Ne and Na 10 11 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.
  • 10. 10 9. Na+ has completely filled K and L shells. Explain. The atomic number of sodium is 11. So, neutral sodium atom has 11 electrons and its electronic configuration is 2, 8, and 1. But Na+ has 10 electrons. Out of 10, K-shell contains 2 and L-shell 8 electrons respectively. Thus, Na+ has completely filled K and L shells. 10. If bromine atom is available in the form of, say, two isotopes 79 81 Br (.7%) and Br (50.3%), calculate the average atomic mass of bromine atom. 35 35 It is given that two isotopes of bromine are 79 / 35Br (49.7%) and 81 / 35Br (50.3%). Then, the average atomic mass of bromine atom is calculated as below: (79x 49.7/100) + (81x50.3/100) = (3926.3/100) + (4074.3/100) = 8000.6/100 = 80.006 u. 11. The average atomic mass of a sample of an element X is 16.2 u. 16 18 What are the percentages of isotopes 8X and 8X in the sample? Given that the average atomic mass of the sample of element X is 16.2 u. Let the percentage of isotope 18 / 8 X is y%. So the percentage of isotope 16 / 8 X will be (100 - y)%. Hence 18 x y/100 + 16 x (100 – y)/100 =16.2 or 18y/100 + (1600 – 16y)/100 = 16.2 Or (18y + 1600 -16y)/100 = 16.2 or 2y = 1620 – 1600 or y = 20/2 =10 Therefore, the percentage of isotope 18 / 8 X is 10%. And, the percentage of isotope 16 / 8 X is (100 - 10) % = 90%. 12. If Z = 3, what would be the valency of the element? Also, name the element. If Z = 3, that means the atomic number of the element is 3. Its electronic configuration is 2 and 1. Therefore the valency of the element is 1 (since the outermost shell has only one electron). The element with Z = 3 is lithium. 13. Composition of the nuclei of two atomic species X and Y are given as X Y Protons = 6 6, Neutrons = 6 8, Give the mass numbers of X and Y. Establish the relation between them Mass number of X = Number of protons + Number of neutrons = 6 + 6 = 12 Mass number of Y = Number of protons + Number of neutrons = 6 + 8 = 14. These atomic species X and Y have the same atomic number, but different mass numbers. So they are isotopes. 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. - F (b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral. – F (c) The mass of an electron is about 1/2000 times that of proton. – T (d) An isotope of iodine is used for making tincture iodine, which is used as a medicine. – F 15. Rutherford’s alpha-particle scattering experiment was responsible for the discovery of (a) Atomic Nucleus, (b) Electron, (c) Proton (d) Neutron - √ Atomic nucleus. (b) 16. Isotopes of an element have (a) the same physical properties, (b) different chemical properties, (c) Different number of neutrons, (d) different atomic numbers. - √ different numbers of neutrons. 17. Complete the following table. Sl. No. Atomic Number Mass Number Number of Protons Number of Neutron Number of Electron Name of Atomic Species 1 9 18 9 9 9 Fluorine 2 16 32 16 16 16 Sulpher 3 12 24 12 12 12 Magnesium 4 1 2 1 1 1 Deuterium 5 1 1 0 1 0 Hydrogen ion Compiled for with relevant questions and For Classes/grades 9 - 10 (MYP 4 – 5) Base on Chapter Structure of Atom