This chapter explores the discovery of sub-atomic particles (electrons, protons, neutrons), the evolution of atomic models (Thomson’s, Rutherford’s, Bohr’s), electronic configuration, valency, atomic number, mass number, and isotopes and isobars. Understanding atomic structure is fundamental to chemistry and explains the behaviour of elements.
Introduction
In Chapter 3, we learnt that atoms and molecules are the fundamental building blocks of matter. Different kinds of matter exist due to different atoms constituting them. Now the questions arise: What makes the atom of one element different from another? Are atoms really indivisible, as proposed by Dalton, or are there smaller constituents inside the atom?
By the end of the 19th century, scientists faced the challenge of revealing the structure of the atom and explaining its important properties. The elucidation of atomic structure is based on a series of experiments. One of the first indications that atoms are not indivisible comes from studying static electricity and how electricity is conducted by different substances.
4.1 Charged Particles in Matter
When we comb dry hair, the comb attracts small pieces of paper. When we rub a glass rod with silk cloth and bring it near an inflated balloon, the balloon gets attracted or repelled. These activities show that on rubbing two objects together, they become electrically charged.
Question: Where does this charge come from?
Answer: This can be explained by knowing that an atom is divisible and consists of charged particles.
Discovery of Sub-Atomic Particles
Many scientists contributed to revealing the presence of charged particles in an atom.
By 1900:
- The atom was known to contain at least one sub-atomic particle – the electron, identified by J.J. Thomson.
E. Goldstein (1886):
- Discovered the presence of new radiations in a gas discharge called canal rays.
- These rays were positively charged radiations.
- This discovery ultimately led to the identification of another sub-atomic particle – the proton.
Properties of Electrons and Protons
Electron (e⁻):
- Charge: Negative (−1)
- Mass: Negligible (about 1/2000th of a proton)
- Can be easily removed from the atom.
Proton (p⁺):
- Charge: Positive (+1)
- Mass: 1 unit (taken as standard)
- Mass is approximately 2000 times that of an electron.
- Present in the interior of the atom (not easily removed).
It seemed that an atom was composed of protons and electrons, mutually balancing their charges. Now the big question was: What sort of structure do these particles form inside the atom?
Questions (4.1) and Solutions
Q1. What are canal rays?
Solution:
- Canal rays are streams of positively charged particles (rays) discovered by E. Goldstein in 1886.
- They were observed in a gas discharge tube.
- These rays are made up of positively charged ions of the gas present in the discharge tube.
- The discovery of canal rays led to the identification of protons.
Q2. If an atom contains one electron and one proton, will it carry any charge or not?
Solution:
- Electron has a charge of −1.
- Proton has a charge of +1.
- Total charge = (−1) + (+1) = 0.
- Therefore, the atom will be electrically neutral (no charge).
4.2 The Structure of an Atom
Dalton’s atomic theory (Chapter 3) suggested that the atom was indivisible and indestructible. But the discovery of electrons and protons inside the atom led to the failure of this aspect of Dalton’s theory. It became necessary to know how electrons and protons are arranged within an atom. Many scientists proposed various atomic models.
4.2.1 THOMSON’S MODEL OF AN ATOM
J.J. Thomson was the first scientist to propose a model for the structure of an atom.
Thomson’s Proposal
Thomson proposed that the atom is like a Christmas pudding or a watermelon:
- The positive charge is spread throughout the atom like the red edible part of a watermelon.
- Electrons are embedded in this positively charged sphere like the seeds in a watermelon.
Postulates of 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. So, the atom as a whole is electrically neutral.
Limitations
Although Thomson’s model explained that atoms are electrically neutral, the results of experiments carried out by other scientists (particularly Rutherford’s alpha-particle scattering experiment) could not be explained by this model.
4.2.2 RUTHERFORD’S MODEL OF AN ATOM
Ernest Rutherford was interested in knowing how the electrons are arranged within an atom. He designed an experiment to find out.
Rutherford’s Alpha-Particle Scattering Experiment
Setup:
- Fast-moving alpha (α)-particles were made to fall on a thin gold foil.
- Gold foil was chosen because it could be made very thin (about 1000 atoms thick).
- α-particles are doubly-charged helium ions (He²⁺) with mass 4 u.
- Since α-particles are much heavier than protons, Rutherford did not expect large deflections.
Expected Result:
- α-particles would be slightly deflected by the sub-atomic particles in gold atoms.
Observations (Totally Unexpected Results)
- Most of the fast-moving α-particles passed straight through the gold foil without deflection.
- Some α-particles were deflected by small angles.
- Surprisingly, one out of every 12,000 particles appeared to rebound (deflected by 180°).
In Rutherford’s words: “This result was almost as incredible as if you fire a 15-inch shell at a piece of tissue paper and it comes back and hits you.”
Analogy to Understand the Experiment
Imagine a blindfolded child throwing stones:
- At a solid wall → every stone hits and makes a sound.
- At a barbed-wire fence → most stones pass through the gaps; only a few hit the wires.
Similarly, Rutherford concluded:
- Most of the space inside the atom is empty because most α-particles passed through without deflection.
- Very few particles were deflected, indicating that the positive charge occupies very little space.
- A very small fraction of α-particles were deflected by 180°, indicating that all the positive charge and mass are concentrated in a very small volume called the nucleus.
- Rutherford calculated that the radius of the nucleus is about 10⁵ times less than the radius of the atom.
Rutherford’s Nuclear Model of the Atom
Based on his experiment, Rutherford proposed the nuclear model with the following features:
- There is a positively charged centre in an atom called the nucleus. Nearly all the mass of an atom resides in the nucleus.
- The electrons revolve around the nucleus in circular paths.
- The size of the nucleus is very small as compared to the size of the atom.
Drawbacks of Rutherford’s Model
Problem:
- According to electromagnetic theory, any charged particle moving in a circular orbit undergoes acceleration.
- During acceleration, charged particles radiate energy.
- Thus, the revolving electron would lose energy and eventually fall into the nucleus.
- If this were true, the atom should be highly unstable and matter would not exist in the form we know.
- But we know that atoms are quite stable.
Rutherford’s model could not explain the stability of the atom.
4.2.3 BOHR’S MODEL OF ATOM
To overcome the objections raised against Rutherford’s model, Neils Bohr put forward the following postulates:
Postulates of Bohr’s Model
- Only certain special orbits known as discrete orbits (or stationary orbits) of electrons are allowed inside the atom.
- While revolving in these discrete orbits, the electrons do not radiate energy.
These orbits or shells are called energy levels. They are represented by:
- Letters: K, L, M, N, …
- Numbers: n = 1, 2, 3, 4, …
4.2.4 NEUTRONS
In 1932, J. Chadwick discovered another sub-atomic particle which had:
- No charge (neutral)
- Mass nearly equal to that of a proton
It was named the neutron (n).
Properties:
- Neutrons are present in the nucleus of all atoms, except hydrogen.
- The mass of an atom is given by the sum of the masses of protons and neutrons present in the nucleus.
Questions (4.2) and Solutions
Q1. On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.
Solution:
- According to Thomson’s model, an atom consists of a positively charged sphere with electrons embedded in it.
- The magnitude of positive charge is equal to the magnitude of negative charge (from electrons).
- Since positive and negative charges are equal and opposite, they cancel each other.
- Therefore, the atom as a whole is electrically neutral.
Q2. On the basis of Rutherford’s model of an atom, which sub-atomic particle is present in the nucleus of an atom?
Solution:
- According to Rutherford’s model, the nucleus contains protons (positively charged particles).
- Later, it was discovered that the nucleus also contains neutrons (neutral particles).
- So, the nucleus contains protons and neutrons (collectively called nucleons).
Q3. Draw a sketch of Bohr’s model of an atom with three shells.
Solution:
- In Bohr’s model, the nucleus is at the centre.
- Electrons revolve around the nucleus in circular orbits (shells).
- The shells are named K, L, M (or n = 1, 2, 3).
Sketch:
text M shell (n=3)
L shell (n=2)
K shell (n=1)
[Nucleus (p⁺, n)]
(Draw concentric circles with the nucleus at the centre and electrons as dots on the circles.)
Q4. 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?
Solution:
- The observations would be similar because all atoms have a similar structure (small, dense nucleus with electrons revolving around it).
- Most α-particles would pass through.
- Some would be deflected by small angles.
- Very few would be deflected by large angles or rebound.
- The exact number of deflections might vary depending on the atomic number (number of protons) of the metal used.
Questions (4.2.4) and Solutions
Q1. Name the three sub-atomic particles of an atom.
Solution:
The three sub-atomic particles are:
- Electron (e⁻) – negatively charged
- Proton (p⁺) – positively charged
- Neutron (n) – neutral (no charge)
Q2. Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?
Solution:
- Atomic mass = Number of protons + Number of neutrons
- 4 = 2 + Number of neutrons
- Number of neutrons = 4 − 2 = 2 neutrons.
4.3 How are Electrons Distributed in Different Orbits (Shells)?
The distribution of electrons into different orbits was suggested by Bohr and Bury.
Rules for Electronic Configuration
Rule 1: The maximum number of electrons in a shell is given by the formula 2n², where n is the shell number (or energy level).
Maximum electrons in each shell:
- K shell (n=1): 2 × 1² = 2 electrons
- L shell (n=2): 2 × 2² = 8 electrons
- M shell (n=3): 2 × 3² = 18 electrons
- N shell (n=4): 2 × 4² = 32 electrons
Rule 2: The outermost shell can hold a maximum of 8 electrons (except K shell, which can hold only 2).
Rule 3: Electrons are not accommodated in a given shell unless the inner shells are filled. That is, shells are filled in a step-wise manner (starting from the innermost shell).
Table 4.1: Composition of Atoms of the First Eighteen Elements
| Element | Symbol | Atomic Number | Protons | Neutrons | Electrons | K | L | M | N | Valency |
|---|---|---|---|---|---|---|---|---|---|---|
| Hydrogen | H | 1 | 1 | 0 | 1 | 1 | – | – | – | 1 |
| Helium | He | 2 | 2 | 2 | 2 | 2 | – | – | – | 0 |
| Lithium | Li | 3 | 3 | 4 | 3 | 2 | 1 | – | – | 1 |
| Beryllium | Be | 4 | 4 | 5 | 4 | 2 | 2 | – | – | 2 |
| Boron | B | 5 | 5 | 6 | 5 | 2 | 3 | – | – | 3 |
| Carbon | C | 6 | 6 | 6 | 6 | 2 | 4 | – | – | 4 |
| Nitrogen | N | 7 | 7 | 7 | 7 | 2 | 5 | – | – | 3 |
| Oxygen | O | 8 | 8 | 8 | 8 | 2 | 6 | – | – | 2 |
| Fluorine | F | 9 | 9 | 10 | 9 | 2 | 7 | – | – | 1 |
| Neon | Ne | 10 | 10 | 10 | 10 | 2 | 8 | – | – | 0 |
| Sodium | Na | 11 | 11 | 12 | 11 | 2 | 8 | 1 | – | 1 |
| Magnesium | Mg | 12 | 12 | 12 | 12 | 2 | 8 | 2 | – | 2 |
| Aluminium | Al | 13 | 13 | 14 | 13 | 2 | 8 | 3 | – | 3 |
| Silicon | Si | 14 | 14 | 14 | 14 | 2 | 8 | 4 | – | 4 |
| Phosphorus | P | 15 | 15 | 16 | 15 | 2 | 8 | 5 | – | 3, 5 |
| Sulphur | S | 16 | 16 | 16 | 16 | 2 | 8 | 6 | – | 2 |
| Chlorine | Cl | 17 | 17 | 18 | 17 | 2 | 8 | 7 | – | 1 |
| Argon | Ar | 18 | 18 | 22 | 18 | 2 | 8 | 8 | – | 0 |
Questions (4.3) and Solutions
Q1. Write the distribution of electrons in carbon and sodium atoms.
Solution:
Carbon (C):
- Atomic number = 6
- Number of electrons = 6
- Electronic configuration: K = 2, L = 4 (or 2, 4)
Sodium (Na):
- Atomic number = 11
- Number of electrons = 11
- Electronic configuration: K = 2, L = 8, M = 1 (or 2, 8, 1)
Q2. If K and L shells of an atom are full, then what would be the total number of electrons in the atom?
Solution:
- K shell can hold maximum 2 electrons.
- L shell can hold maximum 8 electrons.
- Total electrons = 2 + 8 = 10 electrons.
- This corresponds to Neon (Ne) atom.
4.4 Valency
The electrons present in the outermost shell of an atom are known as valence electrons.
From the Bohr-Bury scheme, we know that the outermost shell can accommodate a maximum of 8 electrons (called an octet). Atoms of elements with 8 electrons in the outermost shell (or 2 for helium) show little chemical activity – their valency is zero. These are called inert or noble gases.
Definition of Valency
The combining capacity of the atoms of elements (their tendency to react and form molecules) is explained as an attempt to attain a fully-filled outermost shell (octet). This is done by:
- Sharing electrons
- Gaining electrons
- Losing electrons
Valency is the number of electrons gained, lost, or shared to achieve an octet in the outermost shell.
How to Determine Valency
Case 1: If the outermost shell has 1 to 4 electrons, valency = number of electrons in the outermost shell.
Examples:
- Hydrogen (H): 1 electron in outermost shell → valency = 1
- Lithium (Li): 1 electron → valency = 1
- Sodium (Na): 1 electron → valency = 1
- Magnesium (Mg): 2 electrons → valency = 2
- Aluminium (Al): 3 electrons → valency = 3
- Carbon (C): 4 electrons → valency = 4
Case 2: If the outermost shell has 5 to 7 electrons, valency = 8 − number of electrons in the outermost shell.
Examples:
- Nitrogen (N): 5 electrons → valency = 8 − 5 = 3
- Oxygen (O): 6 electrons → valency = 8 − 6 = 2
- Fluorine (F): 7 electrons → valency = 8 − 7 = 1
- Chlorine (Cl): 7 electrons → valency = 8 − 7 = 1
Case 3: If the outermost shell has 8 electrons (or 2 for helium), valency = 0 (inert/noble gases).
Examples:
- Helium (He): 2 electrons → valency = 0
- Neon (Ne): 8 electrons → valency = 0
- Argon (Ar): 8 electrons → valency = 0
Question (4.4) and Solution
Q1. How will you find the valency of chlorine, sulphur and magnesium?
Solution:
Chlorine (Cl):
- Atomic number = 17
- Electronic configuration: 2, 8, 7
- Outermost shell has 7 electrons.
- Valency = 8 − 7 = 1
Sulphur (S):
- Atomic number = 16
- Electronic configuration: 2, 8, 6
- Outermost shell has 6 electrons.
- Valency = 8 − 6 = 2
Magnesium (Mg):
- Atomic number = 12
- Electronic configuration: 2, 8, 2
- Outermost shell has 2 electrons.
- Valency = 2
4.5 Atomic Number and Mass Number
4.5.1 ATOMIC NUMBER
The atomic number (Z) is defined as the total number of protons present in the nucleus of an atom.
- All atoms of an element have the same atomic number.
- In fact, elements are defined by the number of protons they possess.
- For hydrogen, Z = 1 (1 proton in nucleus).
- For carbon, Z = 6 (6 protons in nucleus).
4.5.2 MASS NUMBER
The mass number (A) is defined as the sum of the total number of protons and neutrons present in the nucleus of an atom.
- Protons and neutrons are also called nucleons.
- The mass of an atom resides in its nucleus.
Formula:
textMass Number (A) = Number of Protons + Number of Neutrons
Examples:
- Carbon: 6 protons + 6 neutrons = mass number 12 u
- Aluminium: 13 protons + 14 neutrons = mass number 27 u
Notation for an Atom
An atom is represented as:
text A
Z X
Where:
- X = Symbol of element
- A = Mass number
- Z = Atomic number
Example: Nitrogen is written as ¹⁴₇N (mass number 14, atomic number 7).
Questions (4.5) and Solutions
Q1. 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:
- Atomic number = Number of protons = 8
(ii) Charge on the atom:
- Number of protons = 8 (positive charge = +8)
- Number of electrons = 8 (negative charge = −8)
- Total charge = (+8) + (−8) = 0
- The atom is electrically neutral (no charge).
Q2. With the help of Table 4.1, find out the mass number of oxygen and sulphur atom.
Solution:
Oxygen (O):
- Protons = 8
- Neutrons = 8
- Mass number = 8 + 8 = 16
Sulphur (S):
- Protons = 16
- Neutrons = 16
- Mass number = 16 + 16 = 32
4.6 Isotopes
In nature, atoms of some elements have the same atomic number but different mass numbers.
Example: Hydrogen
- Protium (¹₁H): 1 proton, 0 neutrons, mass number = 1
- Deuterium (²₁H or D): 1 proton, 1 neutron, mass number = 2
- Tritium (³₁H or T): 1 proton, 2 neutrons, mass number = 3
All three have the same atomic number (1) but different mass numbers.
Other examples:
- Carbon: ¹²₆C and ¹⁴₆C
- Chlorine: ³⁵₁₇Cl and ³⁷₁₇Cl
Definition of Isotopes
Isotopes are atoms of the same element having the same atomic number but different mass numbers.
Properties:
- Chemical properties of isotopes are similar (same number of electrons).
- Physical properties are different (different masses).
Average Atomic Mass
Many elements exist as a mixture of isotopes. The average atomic mass of an element is calculated based on the percentage abundance of each isotope.
Example: Chlorine
- Chlorine occurs in nature in two isotopic forms:
- ³⁵Cl (mass = 35 u) – 75% abundance
- ³⁷Cl (mass = 37 u) – 25% abundance
Average atomic mass:
text= (35 × 0.75) + (37 × 0.25)
= 26.25 + 9.25
= 35.5 u
This does not mean any single chlorine atom has mass 35.5 u. It means the average mass of a large number of chlorine atoms is 35.5 u.
Applications of Isotopes
- An isotope of uranium is used as fuel in nuclear reactors.
- An isotope of cobalt is used in the treatment of cancer.
- An isotope of iodine is used in the treatment of goitre.
4.6.1 ISOBARS
Isobars are atoms of different elements with different atomic numbers but the same mass number.
Example:
- Calcium (²⁰Ca): Atomic number = 20, Mass number = 40
- Argon (¹⁸Ar): Atomic number = 18, Mass number = 40
Both have the same mass number (40) but different atomic numbers.
Questions (4.6) and Solutions
Q1. For the symbols H, D and T, tabulate three sub-atomic particles found in each of them.
Solution:
| Symbol | Name | Protons | Neutrons | Electrons |
|---|---|---|---|---|
| ¹₁H | Protium | 1 | 0 | 1 |
| ²₁H (D) | Deuterium | 1 | 1 | 1 |
| ³₁H (T) | Tritium | 1 | 2 | 1 |
Q2. Write the electronic configuration of any one pair of isotopes and isobars.
Solution:
Isotopes (Carbon-12 and Carbon-14):
- ¹²₆C: Atomic number = 6, Electronic configuration: 2, 4
- ¹⁴₆C: Atomic number = 6, Electronic configuration: 2, 4
(Same electronic configuration because same number of electrons)
Isobars (Calcium-40 and Argon-40):
- ⁴⁰₂₀Ca: Atomic number = 20, Electronic configuration: 2, 8, 8, 2
- ⁴⁰₁₈Ar: Atomic number = 18, Electronic configuration: 2, 8, 8
(Different electronic configurations because different number of electrons)
End-of-Chapter Exercises – Questions and Solutions
Q1. Compare the properties of electrons, protons and neutrons.
Solution:
| Property | Electron (e⁻) | Proton (p⁺) | Neutron (n) |
|---|---|---|---|
| Charge | Negative (−1) | Positive (+1) | Neutral (0) |
| Mass | Negligible (1/2000 u) | 1 u | 1 u |
| Location | Outside nucleus (in shells) | Inside nucleus | Inside nucleus |
| Discovered by | J.J. Thomson | E. Goldstein | J. Chadwick (1932) |
Q2. What are the limitations of J.J. Thomson’s model of the atom?
Solution:
- Thomson’s model could not explain the results of Rutherford’s alpha-particle scattering experiment.
- It could not explain why most α-particles passed through the gold foil without deflection.
- It could not explain why some α-particles were deflected by large angles or rebounded.
- The model did not have a nucleus concept.
Q3. What are the limitations of Rutherford’s model of the atom?
Solution:
- Rutherford’s model could not explain the stability of the atom.
- According to electromagnetic theory, a charged particle moving in a circular orbit radiates energy.
- The revolving electron should continuously lose energy and spiral into the nucleus.
- But atoms are stable in nature, which this model could not explain.
Q4. Describe Bohr’s model of the atom.
Solution:
Bohr’s model proposed:
- Electrons revolve around the nucleus in fixed circular orbits called energy levels or shells.
- These orbits are designated as K, L, M, N or n = 1, 2, 3, 4.
- While revolving in these orbits, electrons do not radiate energy.
- Each orbit has a fixed energy level.
- Electrons can jump from one orbit to another by absorbing or emitting energy.
- The nucleus contains protons and neutrons (nucleons).
Q5. Compare all the proposed models of an atom given in this chapter.
Solution:
| Model | Features | Limitations |
|---|---|---|
| Thomson’s Model | Positive charge spread throughout atom; electrons embedded like seeds in watermelon. | Could not explain α-particle scattering results. |
| Rutherford’s Model | Nucleus at centre with protons; electrons revolve around nucleus. | Could not explain stability of atom. |
| Bohr’s Model | Electrons in fixed orbits; do not radiate energy while in orbit. | Most accepted; explains stability. |
Q6. Summarise the rules for writing distribution of electrons in various shells for the first eighteen elements.
Solution:
The rules are:
- Maximum electrons in a shell = 2n², where n is the shell number.
- K (n=1): 2 electrons
- L (n=2): 8 electrons
- M (n=3): 18 electrons
- N (n=4): 32 electrons
- The outermost shell can hold a maximum of 8 electrons.
- Shells are filled in a step-wise manner starting from the innermost shell (K shell first, then L, then M, etc.).
Q7. Define valency by taking examples of silicon and oxygen.
Solution:
Valency is the combining capacity of an atom – the number of electrons an atom can lose, gain, or share to achieve a stable electronic configuration (octet).
Silicon (Si):
- Atomic number = 14
- Electronic configuration: 2, 8, 4
- Outermost shell has 4 electrons.
- Silicon can share 4 electrons to complete octet.
- Valency of silicon = 4
Oxygen (O):
- Atomic number = 8
- Electronic configuration: 2, 6
- Outermost shell has 6 electrons.
- Oxygen needs 2 more electrons to complete octet.
- Valency of oxygen = 2
Q8. Explain with examples: (i) Atomic number, (ii) Mass number, (iii) Isotopes, (iv) Isobars. Give any two uses of isotopes.
Solution:
(i) Atomic Number (Z):
- Definition: Total number of protons in the nucleus of an atom.
- Example: Hydrogen has atomic number 1 (1 proton); Carbon has atomic number 6 (6 protons).
(ii) Mass Number (A):
- Definition: Sum of protons and neutrons in the nucleus.
- Example: Carbon-12 has 6 protons + 6 neutrons = mass number 12.
(iii) Isotopes:
- Definition: Atoms of the same element with same atomic number but different mass numbers.
- Example: Hydrogen isotopes – Protium (¹H), Deuterium (²H), Tritium (³H).
(iv) Isobars:
- Definition: Atoms of different elements with different atomic numbers but same mass number.
- Example: Calcium-40 (²⁰Ca) and Argon-40 (¹⁸Ar).
Two uses of isotopes:
- Uranium isotope is used as fuel in nuclear reactors.
- Cobalt isotope is used in treatment of cancer.
Q9. Na⁺ has completely filled K and L shells. Explain.
Solution:
- Sodium (Na) atom:
- Atomic number = 11
- Electronic configuration: 2, 8, 1
- It has 1 electron in the outermost M shell.
- Sodium ion (Na⁺):
- Na loses 1 electron from M shell to form Na⁺.
- Electronic configuration of Na⁺: 2, 8
- K shell has 2 electrons (completely filled).
- L shell has 8 electrons (completely filled).
- M shell is now empty.
- Therefore, Na⁺ has completely filled K and L shells.
Q10. If bromine atom is available in the form of two isotopes ⁷⁹Br (49.7%) and ⁸¹Br (50.3%), calculate the average atomic mass of bromine atom.
Solution:
textAverage atomic mass = (79 × 0.497) + (81 × 0.503)
= 39.263 + 40.743
= 79.006 u
≈ 80 u
Answer: Average atomic mass of bromine = 80 u.
Q11. The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes ¹⁶X and ¹⁸X in the sample?
Solution:
Let the percentage of ¹⁶X = x%
Then percentage of ¹⁸X = (100 − x)%
textAverage atomic mass = (16 × x/100) + (18 × (100−x)/100)
16.2 = (16x + 1800 − 18x) / 100
1620 = 1800 − 2x
2x = 1800 − 1620
2x = 180
x = 90%
Answer:
- Percentage of ¹⁶X = 90%
- Percentage of ¹⁸X = 10%
Q12. If Z = 3, what would be the valency of the element? Also, name the element.
Solution:
- Atomic number (Z) = 3
- Number of electrons = 3
- Electronic configuration: 2, 1
- Outermost shell has 1 electron.
- Valency = 1
- Element with Z = 3 is Lithium (Li).
Q13. Composition of the nuclei of two atomic species X and Y are given as under:
| Species | Protons | Neutrons |
|---|---|---|
| X | 6 | 6 |
| Y | 6 | 8 |
Give the mass numbers of X and Y. What is the relation between the two species?
Solution:
For X:
- Protons = 6, Neutrons = 6
- Mass number = 6 + 6 = 12
For Y:
- Protons = 6, Neutrons = 8
- Mass number = 6 + 8 = 14
Relation:
- Both X and Y have the same number of protons (6), so they have the same atomic number.
- But they have different mass numbers (12 and 14).
- Therefore, X and Y are isotopes of the same element (Carbon).
Q14. For the following statements, write T for True and F for False.
(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
Answer: False (F)
(Thomson proposed electrons embedded in a positive sphere; nucleus was discovered by Rutherford.)
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
Answer: False (F)
(Neutron is an independent sub-atomic particle, not a combination of electron and proton.)
(c) The mass of an electron is about 1/2000 times that of proton.
Answer: True (T)
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Answer: False (F)
(Tincture of iodine is a solution of iodine in alcohol; isotopes of iodine are used in treatment of goitre, not for making tincture.)
Q15. Rutherford’s alpha-particle scattering experiment was responsible for the discovery of:
(a) Atomic Nucleus ✓
(b) Electron
(c) Proton
(d) Neutron
Answer: (a) Atomic Nucleus
Q16. Isotopes of an element have:
(a) the same physical properties
(b) different chemical properties
(c) different number of neutrons ✓
(d) different atomic numbers
Answer: (c) different number of neutrons
Q17. Number of valence electrons in Cl⁻ ion are:
(a) 16
(b) 8 ✓
(c) 17
(d) 18
Solution:
- Chlorine atom (Cl) has 17 electrons: 2, 8, 7
- Chlorine ion (Cl⁻) gains 1 electron: 2, 8, 8
- Valence electrons (outermost shell) = 8
Answer: (b) 8
Q18. Which one of the following is a correct electronic configuration of sodium?
(a) 2, 8
(b) 8, 2, 1
(c) 2, 1, 8
(d) 2, 8, 1 ✓
Answer: (d) 2, 8, 1
Q19. Complete the following table:
| Atomic Number | Mass Number | Neutrons | Protons | Electrons | Name |
|---|---|---|---|---|---|
| 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 ion (H⁺) |
Solutions:
Row 1: Atomic number = 9 (Fluorine); Mass number = 9 + 10 = 19; Electrons = 9
Row 2: Neutrons = 32 − 16 = 16; Protons = 16; Electrons = 16
Row 3: Atomic number = 12 (Magnesium); Neutrons = 24 − 12 = 12; Electrons = 12
Row 4: Atomic number = 1 (Deuterium/Hydrogen-2); Neutrons = 2 − 1 = 1; Electrons = 1
Row 5: Atomic number = 1 (Hydrogen ion); Neutrons = 1 − 1 = 0; Electrons = 0 (lost 1 electron)
Download Free Mind Map from the link below
This mind map contains all important topics of this chapter
Visit our Class 9 Science page for free mind maps of all Chapters