NCERT Science Chapter 4 | Structure of the Atom | In-Text Questions & Answers | Exercises Questions & Answers | Notes & pdf
Structure of the Atom – Backdrop – In 1986, scientist E. Goldstein discovered the presence of positively charged radiations in a gas discharge. These are termed canal rays. It resulted to the discovery of a sub-atomic particle i.e. proton. Another sub-atomic particle i.e. electron is identified by scientist J. J. Thompson in 1897.
Q1. Discuss structure of the atom.
Dalton’s atomic theory postulates that atom of an element is indivisible and indestructible. Later scientists establish through various experiment that atom is also divisible and it contains three sub-atomic particles ⇒ 1) Proton, 2) Electron and 3) Neutron.
A. The centre of atom is called nucleus. It particularly contains proton and neutron. Nucleus also is known as nucleon.
B. Electrons exist in shell or orbit around nucleus and they are continuously revolving.
However, hydrogen (H) atom does not contain neutron.
Q2. State the properties of electron.
Electron is a fundamental particle of the atom and has negative charge. Number of electrons present in an atom is equal to the numbers of proton present in that atom. So, total negative charge is equal to total positive charge and they both nullify each other to make an atom electrically neutral and stable. Mass of electron is almost 1836 times lighter than proton. As a result, mass of electron is neglected in calculating mass of an atom of an element. Till now, there is no particle found containing so much less electrical charge as electron. That’s why quantity of charge of electron is used as unit. Generally it is written as -1eo. In conclusion, we say that electron has one unit negative charge and zero mass due to negligible mass.
Q3. State the properties of proton.
Proton is another fundamental particle of the atom. It has positive charge whereas electron has negative charge. Quantity of the charge of the proton and quantity of the charge of the electron are equal but opposite in nature. As atom has equal number of protons and electrons, so total positive charge is equal to total negative charge and they both nullify each other to make atom electrically neutral and stable. Mass of the proton is almost 1836 times heavier than electron. So, mass of an atom is actually summation of total mass of the protons present in an atom of an element. Generally it is written as 1p1. Explicitly symbol on1 indicates that proton has one (1) unit positive charge and 1 u mass. Hydrogen(H) atom contains one proton and one electron and due to this, mass of single atom of Hydrogen (H) is equal to the mass of single proton. As a result, proton sometimes is written as 1H1.
Q4. State the properties of neutron.
Neutron is another fundamental particle of atom. However all elements except Hydrogen (H) have neutron in their atom. Neutron has no charge and neutral in nature. Mass of neutron is almost equal to proton. Generally it is written as on1. Explicitly the on1 symbol indicates that neutron has zero charge and 1 u mass.
Q5. Compare proton, electron and neutron in structure of the atom.
| Answer: | |||
|---|---|---|---|
|
Properties |
Proton |
Electron |
Neutron |
|
1.67 x 10-27 Kg |
9.1 x 10-31 Kg |
1.675 x 10-27 Kg |
|
|
Radius |
1.2 x 10-15 Mtr |
2.8 x 10-15 Mtr |
1.2 x 10-15 Mtr |
|
Nature of Charge |
Positive ( +ve ) |
Negative ( -ve ) |
Neutral |
|
Quantity of Charge |
1.6 x 10-19 Coulomb |
1.6 x 10-19 Coulomb |
Zero |
Q6. What is Thompson’s model of structure of the atom.
In early 20th century, through various experiments scientist E. Goldstein, J.J. Thompson and others discovered and identified sub-atomic charged particle proton and electron. Scientist J.J. Thompson put structure of the atom as follows –
Firstly, atom consists positively charged sphere where negatively charged electrons are embedded in it.
Secondly, quantity of positive charge and negative charge are equal. Both nullify each other to make the atom stable and electrically neutral.
Q7. What is Rutherford's model of structure of the atom and it's shortcomings?
Rutherford’s α-particle scattering experiment⇒
E. Rutherford did α-particle scattering experiment to know how the electrons are arranged inside an atom. E. Rutherford selected a thin layer of gold foil (about 1000 atoms thick) & fast moving α-particles are passed through it. The α-particles are basically doubly charged helium ions. Therefore, it has a mass of 4 u. So, that fast moving α-particles are associated with considerable amount of energy. From the experiment, E. Rutherford found & concluded that –
♦ Most of the α-particles passed straight through the gold foil. It indicates that most of the space inside the atom is empty.
♦ Very few particles were deflected by small angles. It indicates that positive charge occupies very little space.
♦ A very small fraction of particles (01 out of 12000) were deflected by 180o. It indicates that positive charge & mass of the atom were concentrated in a very small volume within the atom.
♦ Based on the aforesaid reasoning, E. Rutherford put following model of structure of the atom –
Rutherford’s model of structure of the atom:
Firstly, positive charge is concentrated in very smaller place in atom. That positively charged centre is known as nucleus. Basically size of nucleus is very smaller than atom. The radius of atom is approximately 105 times of nucleus. Mass of atom is also concentrated in nucleus.
Secondly, electrons are revolving around nucleus in circular orbits.
Shortcomings of Rutherford’s model:
As per classical electrodynamics continuous revolving charged particle electron will radiate energy and will fall into nucleus. This will make atom unstable but we know that atom is quite stable.
Q8. What is Bohr's model of structure of the atom?
Scientist Neils Bohr presents a new concept about structure of the atom that overcomes the shortcomings of Rutherford model. Bohr’s theory states that electron is revolving along special discrete orbits around nucleus and electron does not radiate energy while revolving along these orbit.
These orbits or shells also are known as energy levels and are defined as K, L, M, N etc. Maximum number of shell as per size of the atom is seven (07). Shells are marked sequentially by number such as K shell (n=1), L shell (n=2), M shell (n=3), N shell (n=4) etc. Specifically each shell contains fixed number of electrons that follows the rule of 2n2.
| Bohr's model - | ||
|---|---|---|
|
Name of the Shell |
Mark Number of the Shell (n)
|
Maximum Number of Electron in Shell (2n2) |
|
K |
1 |
2 x 12 = 2 |
|
L |
2 |
2 x 22 = 8 |
|
M |
3 |
2 x 32 = 18 |
|
N |
4 |
2 x 42 = 32 |
Q9. State the concept of neutron.
Answer: Hydrogen-(H) atom is the lightest atom containing single proton and single electron. After Hydrogen we get Helium atom whose mass is four times heavier than Hydrogen atom. So it indicates that there should be four nos. proton in the nucleus of Helium atom. But in experiment, two nos. electron in the orbits of Helium atom are found. Therefore, there is no chance of four nos. proton existing in the nucleus of the Helium atom as it will become electrically unstable if nos. of proton and electron are not same. Hence, there must be two nos. of proton in the nucleus of Helium atom. Then from where the extra mass comes in Helium atom?
In 1932, scientist J. Chadwick solved this problem. He discovered another sub-atomic particle neutron in the nucleus of an atom. Basically neutron has no charge and mass of neutron is almost equal to proton. Generally it is written as on1 . Explicitly the symbol on1 expresses that neutron has zero charge and 1 u mass. Helium atom consists two nos. neutron in addition of two nos. proton. For this reason Helium atom is four times heavier than Hydrogen atom.
Q10. How electrons are distributed in shells of an atom?
Firstly ⇒ Memorise the table of electron distribution in K, L, M, N orbits or shells:
| Bohr's model - | ||
|---|---|---|
|
Name of the Shell |
Mark Number of the Shell (n)
|
Maximum Number of Electron in Shell (2n2) |
|
K |
1 |
2 x 12 = 2 |
|
L |
2 |
2 x 22 = 8 |
|
M |
3 |
2 x 32 = 18 |
|
N |
4 |
2 x 42 = 32 |
Secondly ⇒ Always keep in mind octet rule for electron distribution in structure of the atom:–
♦ Maximum eight (08) electrons are accommodated in the outermost shell. Outermost shell containing 08 nos. electrons is called an octet.
♦ Filling of shell occurs step by step i.e. after complete filling of one particular orbit, filling of the next orbit starts. It means that at first filling of K-shell is done. Then L-shell is filled. After that M-shell & N-shell are filled respectively. And it continues in the same fashion.
♦ But there are some violations of aforesaid rules. In case of some elements, without complete filling of one particular orbit, filling of the next orbit starts. For example – K, Ca etc.
Q11. Show the chart of electron distribution of first 36 elements.
Following is the electron distribution table of the element nos. 01 to 20 ⇒
| Element No. | Name of the Element | Nos. of Proton | Nos. of Electron | Electron Distribution in Shell | Nos. of Neutron |
|---|---|---|---|---|---|
|
1. |
Hydrogen (H) |
1 |
1 |
K - 1 |
|
|
2. |
Helium (He) |
2 |
2 |
K - 2 |
2 |
|
3. |
Lithium (Li) |
3 |
3 |
K - 2, L - 1 |
4 |
|
4. |
Beryllium (Be) |
4 |
4 |
K - 2, L - 2 |
5 |
|
5. |
Boron (B) |
5 |
5 |
K - 2, L - 3 |
6 |
|
6. |
Carbon (C) |
6 |
6 |
K - 2, L - 4 |
6 |
|
7. |
Nitrogen (N) |
7 |
7 |
K - 2, L - 5 |
7 |
|
8. |
Oxygen (O) |
8 |
8 |
K - 2, L - 6 |
8 |
|
9. |
Fluorine (F) |
9 |
9 |
K - 2, L - 7 |
10 |
|
10. |
Neon (Ne) |
10 |
10 |
K - 2, L - 8 |
10 |
|
11. |
Sodium (Na) |
11 |
11 |
K - 2, L - 8, M - 1 |
12 |
|
12. |
Magnesium (Mg) |
12 |
12 |
K - 2, L - 8, M - 2 |
12 |
|
13. |
Aluminium (Al) |
13 |
13 |
K - 2, L - 8, M - 3 |
14 |
|
14. |
Silicon (Si) |
14 |
14 |
K - 2, L - 8, M - 4 |
14 |
|
15. |
Phosphorus (P) |
15 |
15 |
K - 2, L - 8, M - 5 |
16 |
|
16. |
Sulphur (S) |
16 |
16 |
K - 2, L - 8, M - 6 |
16 |
|
17. |
Chlorine (Cl) |
17 |
17 |
K - 2, L - 8, M - 7 |
18 |
|
18. |
Argon (Ar) |
18 |
18 |
K - 2, L - 8, M - 8 |
22 |
|
19. |
Potassium (K) |
19 |
19 |
K - 2, L - 8, M - 8, N - 1 |
20 |
|
20. |
Calcium (Ca) |
20 |
20 |
K - 2, L - 8, M - 8, N - 2 |
20 |
Following is the electron distribution table of the element nos. 19 to 36 ⇒
| Element No. | Name of the Element | Nos. of Proton | Nos. of Electron | Electron Distribution in Shell | Nos. of Neutron |
|---|---|---|---|---|---|
|
21. |
Scandium (Sc) |
21 |
21 |
K - 1, L - 8, M - 9, N - 2 |
24 |
|
22. |
Titanium (Ti) |
22 |
22 |
K - 2, L - 8, M - 10, N - 2 |
26 |
|
23. |
Vanadium (V) |
23 |
23 |
K - 2, L - 8, M - 11, N - 2 |
28 |
|
24. |
Chromium (Cr) |
24 |
24 |
K - 2, L - 8, M - 13, N - 1 |
28 |
|
25. |
Manganese (Mn) |
25 |
25 |
K - 2, L - 8, M - 13, N - 2 |
30 |
|
26. |
Iron (Fe) |
26 |
26 |
K - 2, L - 8, M - 14, N - 2 |
30 |
|
27. |
Cobalt (Co) |
27 |
27 |
K - 2, L - 8, M - 15, N - 2 |
32 |
|
28. |
Nickel (Ni) |
28 |
28 |
K - 2, L - 8, M - 16, N - 2 |
31 |
|
29. |
Copper (Cu) |
29 |
29 |
K - 2, L - 8, M - 18, N - 1 |
35 |
|
30. |
Zinc (Zn) |
30 |
30 |
K - 2, L - 8, M - 18, N - 2 |
35 |
|
31. |
Gallium (Ga) |
31 |
31 |
K - 2, L - 8, M - 18, N - 3 |
39 |
|
32. |
Germanium (Ge) |
32 |
32 |
K - 2, L - 8, M - 18, N - 4 |
41 |
|
33. |
Arsenic (As) |
33 |
33 |
K - 2, L - 8, M - 18, N - 5 |
42 |
|
34. |
Selenium (Se) |
34 |
34 |
K - 2, L - 8, M - 18, N - 6 |
45 |
|
35. |
Bromine (Br) |
35 |
35 |
K - 2, L - 8, M - 18, N - 7 |
45 |
|
36. |
Krypton (Kr) |
36 |
36 |
K - 2, L - 8, M - 18, N - 8 |
48 |
Q12. What is valence electron?
Previously we learnt that outermost shell or orbit can accommodate maximum eight (08) nos. electron. Elements like Neon, Argon, Krypton etc. possess octet and have eight nos. electron in the outermost shell. These elements show zero chemical activity and are called inert element.
However, only exception is Helium. Helium has only two (02) instead of eight (08) nos. electron in outermost shell. But it also shows characteristics like inert elements.
Basically chemical activities of an element depend on the nos. of electron existing in the outermost shell. And valence electron of an element is the number of electrons existing in the outermost shell. Boron’s atomic structure is K=2, L=3. Hence, outermost shell L of Boron 03 nos. electron. Therefore, Boron’s valence electron is 3. Similarly valence electron of Carbon is 4 and Oxygen is 6.
Q13. How do we derive valency from electron distribution in shells?
Basically valency is the combining capacity of an element to take part in chemical reaction. Also we get that atoms always tend to possess an octet. The nos. of electrons to be added or released from the outermost shell to turn that shell an octet is the valency of an element.
Following rules to follow to derive valency –
A. Firstly, if the outermost shell contains 1 or 2 or 3 nos. electron, then to possesses an octet it is easier to release electrons than adding.
B. Secondly, if the outermost shell contains 5 or 6 or 7 nos. electron, then to possesses an octet it is easier to add electrons than releasing.
C. Thirdly, if the outermost shell contains 4 nos. electron, then it is similar to release electrons or to add electrons for possessing an octet.
However, all aforesaid rules are just simplistic. There are exceptions too as many elements have multiple valency and inert elements have zero valency due to possessing an octet.
Following are some examples of derivation of valency –
Sulphur’s electron distribution in structure of the atom is 2, 8, 6. In this case outermost shell possesses 6 nos. electron. Therefore, to possess an octet, Sulphur needs to add 02 nos. electron in the outermost shell or to release 06 nos. electron from the outermost shell. Generally it is easier to add 02 nos. electron rather than releasing 06 nos. electron. Hence, valency of Sulphur is 2 .
Magnesium’s electron distribution in structure of the atom is 2, 8, 2. In this case outermost shell possesses 2 nos. electron. Therefore, to possess an octet, Magnesium needs to add 06 nos. electron in the outermost shell or to release 02 nos. electron from the outermost shell. Generally it is easier to release 02 nos. electron rather than adding 06 nos. electron. Hence, valency of Magnesium is 2.
Silicon’s electron distribution in the structure of atom is 2, 8, 4. In this case outermost shell possesses 4 nos. electron. Therefore, to possess an octet, Silicon needs to add 04 nos. electron in the outermost shell or to release 04 nos. electron from the outermost shell. Both are same. Hence, valency of Silicon is 4.
Q14. Show the chart of derivation of valency of first 20 elements from electron distribution.
Following is the table of deriving valency of first 20 elements from electron distribution chart ⇒
| Name of the Element | Nos. of Electron | Electron Distribution in Shell | Valency |
|---|---|---|---|
|
Hydrogen (H) |
1 |
K - 1 |
1 |
|
Helium (He) |
2 |
K - 2 |
|
|
Lithium (Li) |
3 |
K - 2, L - 1 |
1 |
|
Beryllium (Be) |
4 |
K - 2, L - 2 |
2 |
|
Boron (B) |
5 |
K - 2, L - 3 |
3 |
|
Carbon (C) |
6 |
K - 2, L - 4 |
4 |
|
Nitrogen (N) |
7 |
K - 2, L - 5 |
3 |
|
Oxygen (O) |
8 |
K - 2, L - 6 |
2 |
|
Fluorine (F) |
9 |
K - 2, L - 7 |
1 |
|
Neon (Ne) |
10 |
K - 2, L - 8 |
|
|
Sodium (Na) |
11 |
K - 2, L - 8, M - 1 |
1 |
|
Magnesium (Mg) |
12 |
K - 2, L - 8, M - 2 |
2 |
|
Aluminium (Al) |
13 |
K - 2, L - 8, M - 3 |
3 |
|
Silicon (Si) |
14 |
K - 2, L - 8, M - 4 |
4 |
|
Phosphorus (P) |
15 |
K - 2, L - 8, M - 5 |
3,5 |
|
Sulphur (S) |
16 |
K - 2, L - 8, M - 6 |
2 |
|
Chlorine (Cl) |
17 |
K - 2, L - 8, M - 7 |
1 |
|
Argon (Ar) |
18 |
K - 2, L - 8, M - 8 |
|
|
Potassium (K) |
19 |
K - 2, L - 8, M - 8, N - 1 |
1 |
|
Calcium (Ca) |
20 |
K - 2, L - 8, M - 8, N - 2 |
2 |
Q15. What is atomic number?
Basically atomic number of an element is the number of proton present in the nucleus of an atom. Single proton exists in the nucleus of Hydrogen atom. Hence, atomic number of Hydrogen is 1. Similarly, atomic number of Helium is 2 and atomic number of Lithium and Oxygen are 3 and 8 respectively. In physics, all the elements are particularly listed according to their atomic number i.e. number of proton present in the nucleus. In the first place we find Hydrogen as first element due to having single proton in the nucleus. Symbol A denotes atomic number of an element.
Q16. What is mass number?
Basically mass number of an element is the sum of number of protons and neutrons present in the nucleus of an atom. Oxygen has 08 nos. proton and 08 nos. neutron present in the nucleus of it’s atom. Therefore, mass number of oxygen is 16 i.e. 8+8=16. Symbol Z denotes mass number of an element.
Q17. How do you represent an element using atomic number and mass number?
Generally an element is written by it’s atomic number and mass number as follows –
A
[ Symbol of Element ]
Z
♦ Z ⇒ Atomic number = number of protons present in nucleus which is equal to the number of electron present in atom. ♠ A ⇒ Mass number = number of protons and neutrons present in nucleus in single atom. ♣ [A – Z] = number of neutrons present in nucleus in single atom. For example ⇒ 168O, 42He, 126C, 115B, 3517Cl, 11H etc.
Q18. What is isotope?
We know that atoms of an element have same number of proton in it’s nucleus. But it is also found that atoms of same element may contain different number of neutrons. These atoms of same element containing different number of neutrons are called isotopes. For example ⇒
♦ Hydrogen : 11H (Protium), 21H (Deuterium), 31H(Tritium). ♠ Helium : 32He, 42He.
♣ Nitrogen : 147N, 157N. ♦ Chlorine : 3517Cl, 3717Cl. ♠ Lithium : 63Li, 73Li. ♣ Carbon : 126C, 136C, 146C.
♦ Oxygen : 168O, 178O, 188O. ♠ Uranium : 23592U, 23892U, 23992U.
Q19. What is isobar?
Different elements have different atomic number as atomic number of every element is fixed. But there are atoms of different elements having same mass number i.e. sum of total nos. of proton and neutron is equal. The atoms of different elements having same mass number are called isobars. For example ⇒ 4018Ar & 4020Ca, 11550Sn & 11549In etc.
Q20. What is average atomic mass?
Generally some elements exist in nature in a combination of it’s isotopes. Neon has two isotopes – one has atomic mass 20 and other has atomic mass 22. Ratio of existence in nature of these isotopes is 9:1. So if we analyse Neon gas, we will get 90% atoms having atomic mass 20 and 10% atoms having atomic mass 22. As a result, we calculate average atomic mass. Therefore, average atomic mass of Neon ⇒ (90×20 + 10×22)/100 = 20.2.
Similarly, Chlorine has two isotopes – one has atomic mass 35 and other has atomic mass 37. Ratio of existence in nature of these isotopes is 3:1. So if we analyse Chlorine gas, we will get 75% atoms having atomic mass 35 and 25% atoms having atomic mass 37. Therefore, average atomic mass of Chlorine ⇒ (75×35 + 25×37)/100 = 35.5.
In-Text Questions & Answers | NCERT Class 9 Science Chapter 4 | Structure of the Atom |
In-Text Questions & Answers | NCERT Class 9 Science Chapter 4 | Structure of the Atom | Page Nos. 47
1. What are canal rays?
It is discovered by Goldstein in 1886. Canal rays are positively charged radiations that consist positively charged particles i.e. protons. Also we may term it as anode ray or positive ray.
2. If an atom contains one electron & one proton, will it carry any charge or not?
It will neither carry any positive nor negative charge. Because both proton & electron carry same amount of charge i.e. 1.6 x 10-19 Coulomb. But nature of their charge is opposite to each other. Proton carry positive charge while electron carry negative charge. So the atom will be neutral.
In-Text Questions & Answers | NCERT Class 9 Science Chapter 4 | Structure of the Atom | Page Nos. 49
1. On the basis of Thompson’s model of an atom, explain how the atom is neutral as a whole.
According to the scientist J.J. Thompson, atom consists positively charged sphere whereas negatively charged electrons are embedded in it. Quantity of positive charge and negative charge are equal. Both nullify each other to make the atom stable and electrically neutral.
2. On the basis of Rutherford’s model of an atom, which sub-atomic particle is present in the nucleus of an atom?
According to the scientist E. Rutherford, positive charge is concentrated in very smaller place in an atom known as nucleus. That positive charge is due to the presence of positively charged particles i.e. protons.
3. Draw a sketch of Bohr’s model of an atom with three shells.
Draw the sketch from Q8.
4. 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?
Nothing will change in observation. Because gold is taken as it is malleable & can be fold easily. A thin foil can be made easily from gold which is difficult in case of other metal.
In-Text Questions & Answers | NCERT Class 9 Science Chapter 4 | Structure of the Atom | Page Nos. 49
1. Name the three sub-atomic particles of an atom.
Proton, Electron & Neutron are the three sub-atomic particle of an atom.
2. Helium atom has an atomic mass of 4 u & two protons in its nucleus. How many neutrons does it have?
Mass of the proton is 1 u whereas mass of the electron is negligible. Also mass of the neutron is 1 u. In this case total mass is 4 u. Also given, number of protons is 2. So the mass of protons is 2 u. Therefore, number of neutrons is (4 u -2 u)/1 u = 2.
In-Text Questions & Answers | NCERT Class 9 Science Chapter 4 | Structure of the Atom | Page Nos. 50
1. Write the distribution of electrons in carbon & sodium atoms.
Read the chart in Q11 in detail to know electron distribution of first 36 elements. Electron distribution of carbon ⇒ K – 2, L – 4. Electron distribution of sodium ⇒ K – 2, L – 8, M – 1.
2. If K & L shells of an atom are full, then what would be the total number of electrons in the atom?
K – shell contains maximum 2 nos. electrons & L – shell contains maximum 8 nos. electrons. So there will be total 10 nos. electrons if K & L shells are full.
In-Text Questions & Answers | NCERT Class 9 Science Chapter 4 | Structure of the Atom | Page Nos. 52
1. How will you find the valency of chlorine, sulphur & magnesium?
Chlorine’s electron distribution in structure of the atom is 2, 8, 7. In this case outermost shell possesses 7 nos. electron. Therefore, to possess an octet, Chlorine’s needs to add 01 no. electron in the outermost shell or to release 07 nos. electron from the outermost shell. Generally it is easier to add 01 no. electron rather than releasing 07 nos. electron. Hence, valency of Chlorine is 1.
Sulphur’s electron distribution in structure of the atom is 2, 8, 6. In this case outermost shell possesses 6 nos. electron. Therefore, to possess an octet, Sulphur needs to add 02 nos. electron in the outermost shell or to release 06 nos. electron from the outermost shell. Generally it is easier to add 02 nos. electron rather than releasing 06 nos. electron. Hence, valency of Sulphur is 2.
Magnesium’s electron distribution in structure of the atom is 2, 8, 2. In this case outermost shell possesses 2 nos. electron. Therefore, to possess an octet, Magnesium needs to add 06 nos. electron in the outermost shell or to release 02 nos. electron from the outermost shell. Generally it is easier to release 02 nos. electron rather than adding 06 nos. electron. Hence, valency of Magnesium is 2.
In-Text Questions & Answers | NCERT Class 9 Science Chapter 4 | Structure of the Atom | Page Nos. 52
1. If number of electrons in an atom is 8 & number of protons is also 8, then i) what is the atomic number of the atom? & ii) what is the charge on the atom?
i) Read Q15 in detail. Atomic number is equal to the number of protons. So the atomic number is 8. ii) As the atom contains equal number of electrons & protons, so positive charge & negative charge are equal. Therefore, the atom is neutral.
2. Find the mass number of oxygen & sulphur atom from the chart of Q14.
From the chart of Q14, we get that number of protons & neutrons of oxygen are 8 & 8 respectively. So mass number of oxygen is (8+8) = 16.
Similarly, we get that number of protons & neutrons of sulphur are 16 & 16 respectively. So mass number of oxygen is (16+16) = 32.
In-Text Questions & Answers | NCERT Class 9 Science Chapter 4 | Structure of the Atom | Page Nos. 53
1. For the symbol H, D & T tabulate three sub-atomic particles found in each of them.
| Answer: | |||
|---|---|---|---|
|
Symbol ⇒ |
H |
D |
T |
|
Proton |
1 |
01 |
1 |
|
Electron |
1 |
01 |
1 |
|
Neutron |
|
01 |
2 |
2. Write the electronic configuration of any one pair of isotopes & isobars.
Read Q18 in detail to know what is isotope. Isotopes of any element have same atomic number i.e. same nos. of proton or electron. For example – Isotopes of Chlorine are 35Cl17 & 37Cl17 . Both have same nos. of electron i.e. 17 & same electron configuration as 2, 8, 7. Similarly carbon has three isotopes i.e. 12C6, 14C6, 16C6. The electron distribution of carbon isotopes is – 2,4.
Read Q19 in detail to know what is isobars. Isobars have same mass number but different atomic number. 40Ca20 & 40Ar18 are isobars. The electron distribution of 40Ca20 is – 2, 8, 8, 2 whereas 40Ar18 is 2, 8, 8.
Exercises Questions & Answers | NCERT Class 9 Science Science Chapter 4 | Structure of the Atom | Page No. 54 - 56
1. Compare the properties of electrons, protons & neutrons.
Write down the answer of Q5 in detail.
2. What are the limitations of J. J. Thompson’s model of the atom?
Scientist J.J. Thompson opines that atom consists positively charged sphere where negatively charged electrons are embedded in it. But actually positively charge particles (proton) exists at the centre of atom i.e. nucleus & negatively charge particles (electron) revolve around the nucleus.
3. What are the limitations of Rutherford’s model of the atom?
Any accelerating particle in a circular orbit radiates energy. So the orbital revolution of the electron will be unsteady. The rotating electron will lose energy & will finally fall into the nucleus. As a result the atom will be highly unstable which actually does not exist. This is the limitations of J. J. Thompson’s model of the atom.
4. Describe Bohr’s model of the atom.
Write down the answer of Q8 in detail. Also write the electron distribution chart.
5. Compare all the proposed models of the atom given in this chapter.
| Answer: | Thompson's Model | Rutherford's Model | Bohr's Model |
|---|---|---|---|
|
1. |
Atom consists positively charged sphere where negatively charged electrons are embedded in it. Quantity of positive charge and negative charge are equal. Both nullify each other to make the atom stable and electrically neutral. |
Positive charge is concentrated in very smaller place in atom known as nucleus. Basically size of nucleus is very smaller than atom. Mass of atom is also concentrated in nucleus. Electrons are revolving around nucleus in circular orbits. |
Electron is revolving along special discrete orbits around nucleus and electron does not radiate energy while revolving along these orbit. We also term these orbits as energy levels and are defined as K, L, M, N etc. |
6. Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.
a) The electron is distributed by the formula 2n2 where n=shell number. In case of n=1 (K-shell), nos. of electron are accommodated = 2×12 =2. Similarly, for L-shell (n=2), nos. of electron is 2×22=8. Thus it continues.
b) Maximum eight (08) electrons are accommodated in the outermost shell. Outermost shell containing 08 nos. electrons is called an octet.
c) Filling of shell occurs step by step i.e. after complete filling of one particular orbit, filling of the next orbit starts. It means that at first filling of K-shell is done. Then L-shell is filled. After that M-shell & N-shell are filled respectively. And it continues in the same fashion.
d) But there are some violations of aforesaid rules. In case of some elements, without complete filling of one particular orbit, filling of the next orbit starts. For example – K, Ca etc.
7. Define valency by taking examples of silicon & oxygen.
Read the chart of Q14 in detail. Silicon contains 14 nos. electron. Its electron distribution is as K-2, L-8, M-4. As the outermost shell contains 4 nos. electron, then it is similar to release electrons or adding electrons to possesses an octet. So the valency of silicon is 4.
Oxygen contains 08 nos. electron. Its electron distribution is as K-2, L-6. As the outermost shell contains 6 nos. electron, then it is easy to add 02 nos. electrons to possesses an octet. So the valency of oxygen is 2.
8. Explain with examples a) Atomic number, b) Mass number, c) Isotopes, & d) Isobars. Give any two uses of isotopes.
a) Atomic number – Q15, b) Mass number – Q16, c) Isotopes – Q18, d) Isobars – Q19.
Uses of isotopes ⇒ The nuclear reactor uses an isotope of uranium as fuel. The cancer treatment uses an isotope of cobalt.
9. Na+ has completely filled K & L shells. Explain in detail.
Na (atomic number – 11) has electron distribution as 2, 8, 1. Na+ is formed when Na release one electron from it’s outer most shell. So Na+ contains 10 nos. electron. Therefore, electron distribution of Na+ will be as 2, 8. That is why Na+ has completely filled K & L shells.
10. If bromine atom is available in the form of, say, two isotopes 7935Br(49.7%) and 8135Br(50.3%), calculate the average atomic mass of bromine atom.
Firstly, learn average atomic mass in Q20. In this case, average atomic mass of bromine =
[ (79 x 49.70) + (81 x 50.30) ] / 100 = 8000.60 / 100 = 80.006.
11. The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes 168X and 188X in the sample?
Given that, average atomic mass given = 16.2. Now take percentage of 168X in the sample= y. Hence percentage of 188X in the sample is = (100-y).
Therefore, average atomic mass ⇒ 16.2 = [ yx16 + (100-y)x18 ]/100
⇒ 2y = 180,
⇒ y = 90. So the percentage of 168X = 90% & the percentage of 188X = 100-90 =10% .
12. If Z=3, what would be the valency of the element? Also name the element?
If Z=3=atomic number, then the element has 03 nos. electron. So the electron distribution is K – 2, L – 1. Valency is 1 because outermost shell possess 01 no. electron. Read Q14 in detail to learn the valency of first twenty elements.
13. Composition of the nuclei of two atomic species X and Y are given as X (Protons – 6, Neutrons -6) and Y (Protons – 6, Neutrons -8). Find the mass number of X and Y. What is the relation between X and Y?
At first ⇒ mass number of X = 6 + 6 = 12, atomic number of X = 6.
After that ⇒ mass number of Y = 6 + 8 = 14, atomic number of Y = 6.
In conclusion, both X and Y are isotopes as their atomic numbers are same.
14. Write T for true & F for false statements –
a) J.J. Thompson proposed that the nucleus of an atom contains only nucleons.
b) A neutron is formed by an electron & 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 iodine, which is used as a medicine.
a) False, b) False, c) True, d) True.
15. Rutherford’s alpha-particle scattering experiment was responsible for the discovery of –
a) Atomic Nucleus, b) Electron, c) Proton, d) Neutron.
a) Atomic Nucleus.
16. Isotope of an element have ⇒ a) the same physical properties, b) different chemical properties, c) different number of neutrons, d) different atomic numbers.
c) different number of neutrons.
17. Number of valence electrons in Cl– ion are ⇒ a) 16, b) 8, c) 17, d) 18.
b) 8. Atomic number of Chlorine (Cl) is 17. Its electron configuration is 2,8,7. When one electron is added, Cl turns into Cl–. Thus total number of electron becomes 18. As a result, electron configuration becomes 2,8,8. So the number of valence electrons in Cl– ion is 8.
18. Which is the correct electronic configuration of sodium ⇒
a) 2,8
b) 8,2,1
c) 2,1,8
d) 2,8,1
d) 2,8,1.
19. Complete the following table –
| Atomic No. | Mass No. | Nos. of Neutrons | No. of Protons | Nos. of Electrons | Name of the Atomic Species |
|---|---|---|---|---|---|
|
9 |
- |
10 |
- |
- |
- |
|
16 |
32 |
- |
- |
- |
Sulphur |
|
- |
24 |
- |
12 |
- |
- |
|
- |
2 |
- |
1 |
- |
- |
|
- |
1 |
|
1 |
|
- |
| Answer: | |||||
|---|---|---|---|---|---|
|
Atomic No. |
Mass No. |
Nos. of Neutrons |
No. of Protons |
Nos. of Electrons |
Name of the Atomic Species |
|
9 |
19 |
10 |
9 |
9 |
Flourine |
|
16 |
32 |
16 |
016 |
16 |
Sulphur |
|
12 |
24 |
12 |
012 |
12 |
Magnesium |
|
1 |
2 |
2 |
1 |
1 |
Deuterium |
|
1 |
1 |
|
1 |
|
Hydrogen ion |
Structure of the Atom – To Be Continued – Structure of the Atom