NCERT Exemplar Solution Class 9 Science Chapter 4
NCERT Exemplar Solution Class 9 Science Chapter 4 Structure Of The Atoms all Questions Answer Solution. NCERT Exemplar Class 9 Science Chapter 4 Solution PDF.
NCERT Exemplar Solution Class 9 Science Chapter 4: Overview
NCERT Exemplar Solution Class 9 Science Chapter 4 |
|
Board | NCERT |
Topic | Exemplar Problem Solution |
Class | 9 |
Subject | Science |
Chapter | 4 |
Chapter Name | Structure Of The Atoms |
NCERT Exemplar Solution Class 9 Science Chapter 4 Structure Of The Atoms
1) Which of the following correctly represent the electronic distribution in the Mg atom?
a) 3, 8, 1
b) 2, 8, 2
c) 1, 8, 3
d) 8, 2, 2
Answer:- b) 2,8,2
Mg has Atomic number 12.
These will be distributed in K, L, M shells as 2, 8, 2 .
2) Rutherford’s alpha particles scattering experiment resulted in to discovery of
a) Electron
b) Proton
c) Nucleus in the atom
d) Atomic mass
Answer:- c) Nucleus in the atom
Nucleus is present in the atom was result of scattering experiment. Some Alpha rays returned back which showed that a very small amount of positive charge is gathered at the middle which is called as nucleus.
3) The number of electrons in an element X is 15 and the number of neutrons is 16. Which of the following is the correct representation of the element?
31X15
31X16
16X15
15X16
Answer:- a) 31X15
Here 31 is atomic mass number which is addition of electrons or protons and neutrons. And 16 is atomic number which the total number of electrons present in the atom. And it is correctly represented as a) 31X15
4) Dalton’s atomic theory successfully explained.
i) Law of conservation of mass
ii) Law of constant composition
iii) Law of radioactivity
iv) Law of multiple proportion
a) I, II, and III
b) I, III and IV
c) II, III, and IV
d) I, II, and IV
Answer:- d) I, II, and IV
It does not explain the law of radioactivity. But Atomic theory of Dalton explained all the three laws – law of conservation of mass, law of constant composition, law of multiple proportion.
5) Which of the following statements about Rutherford’s model of atom are correct?
i) Considered the nucleus as positively charge.
ii) Established that the Alpha- particles are four times as heavy as a hydrogen atom.
iii) Can be compared to solar system
iv) Was in agreement with Thomson’s model.
a) I and III
b) II and III
c) I and IV
d) Only I
Answer:- a) I and III
Rutherford’s alpha ray scattering experiment resulted that nucleus is a positively charged. These positively charged nucleus deflect the rays. And it can be compared to solar system also.
6) Which of the following are true for an element?
I) Atomic number = number of protons + number of electrons.
II) Mass number= number of protons + number of neutrons.
III) Atomic mass= number of protons= number of neutrons
IV) Atomic number= number of protons = number of electrons
a) I and II
b) I and III
c) II and III
d) II and IV
Answer:- d) II and IV
Atom has Atomic mass and atomic number.
Atomic mass is Addition of neutrons and protons present in atom. And atomic number is total number of protons and Electrons which is same.
7) In the Thomson’s model of atom, which of the following statements are correct?
I) The mass of the atom is assumed to be uniformly distributed over the atom.
II) The Positive charge is assumed to be uniformly distributed over the atom
III) The electrons are uniformly distributed over the atom
IV) The electrons attract each other to stabilise the atom.
a) I, II and III
b) I and III
c) I and IV
d) I, III, and IV
Answer:- a) I, II and III
Thomson’s Atomic model is also known as plum pudding model. Which states that mass of atom are uniformly distributed, electrons are embedded in it and positive charge is distributed uniformly.
8) Rutherford’s alpha particles scattering experiment showed that
I) Electrons have negative charge
II) The mass and positive charge of the atom is concentrated in the nucleus
III) Neutron exist in the nucleus
IV) Most of the space in atom is empty
Which of the above statements are correct?
a) I and III
b) II and IV
c) I and IV
d) III and IV
Answer:- b) II and IV
Rutherford’s scattering experiment showed that positive charge is centred in the nucleus. The nucleus is very small as compared to atom. Because some alpha particles were returned from the atom. And The most of the spaces are empty in the atom. Because the Alpha rays were passed through the atom without interaction.
9) The ion of an element has 3 positive charges. Mass number of the atom is 27 and the number of neutrons is 14. What is the number of electrons in the ion?
a) 13
b) 10
c) 14
d) 16
Answer:- b) 10
Mass number = 27
No. Of Neutrons = 14
No. Of electrons = mass number – No. Neutrons
= 27 – 14
= 13 electrons
And there are 3 positive charges means the element donate 3 Electrons.
Therefore,
13 – 3 = 10 electrons
10) Identify the Mg 2+ ion from the fig. 4.1 where, n and p represent the number of neutrons and protons respectively.
(A)
(B)
(C)
(d)
Answer:- d)
The Mass number of Mg atom = 24
No. Of neutrons in Mg atom = 12
No. Of protons in Mg atom = 12
But as the Mg atom is donor of 2 Electrons and formed Mg 2+ ion
Thus there should be 10 electrons in the atom. Therefore d) is correct.
(11) In a sample of ethyl ethanoate (CH3COOC2H5) the two oxygen atoms have the same number of electrons but different number of neutrons. Which of the following is the correct reason for it?
(a) One of the oxygen atoms has gained electrons
(b) One of the oxygen atoms has gained two neutrons
(c) The two oxygen atoms are isotopes
(d) The two oxygen atoms are isobars.
Answer:- c) the two oxygen atoms are isotopes.
Ethanoate (CH3COOC2H5) have two different number of neutrons and same number of electrons then both the oxygen atoms should be isotopes of each other.
Because isotopes have same atomic number but different atomic mass number i.e. same number of electrons and different neutrons.
12) Elements with valency 1 are
a) Always metals
b) Always metalloids
c) Either metals or non- metals
d) Always non- metals.
Answer:- c) either metals or non- metals
Elements with -1 Valency are Non- metal. While the elements with Valency +1 are metals.
13) The first model of an atom was given by
a) N. Bohr
b) E. Goldstein
c) Rutherford
d) J.J. Thomson
Answer:- d) J.J Thomson
The model of Atom was first invented by J.J. Thomson .
14) An atom with 3 protons and 4 neutrons will have a valency of
a) 3
b) 7
c) 1
d) 4
Answer:- c) 1
3 protons means there should be 3 Electrons in the atom
K shell will contain 2 electron to complete the shell
L will contain remaining 1 Electron thus Valency is 1.
15) The electron distribution in an aluminium atom is
a) 2,8,3
b) 2,8,2
c) 8,2,3
d) 2,3,8
Answer:- a) 2,8,3
Aluminium has the Atomic number 13.
K shell will be completed in 2 electrons,
L shell will completed in 8 Electrons
And remaining 3 will be in the M shell. Thus the electronic configuration of the Al is 2,8,3.
16) Which of the following in fig. 4.2 do not represent Bohr’s model of an atom correctly?
a) I and II
b) II and III
c) II and IV
d) I and IV
(a)
b)
c)
d)
Answer:- c) II and IV
Bohr’s model of an atom shows k shell i.e. first shell can contains only 2 electron to fulfill the shell. And L shell can contain maximum 8 Electrons in it. Therefore these two are wrong.
17) Which of the following statement is always correct?
a) An atom has equal number of electrons and protons.
b) An atom has equal number of electrons and neutrons.
c) An atom has equal number of protons and neutrons
d) An atom has equal number of electrons, protons and neutrons.
Answer:-a) an atom has equal number of electrons and protons.
Every atom has equal number of protons and Electrons to keep the atom electrically neutral.
18) Atomic models have been improved over the years. Arrange the following atomic models in the their chronological order
a) Rutherford’s atomic model
b) Thomson’s atomic model
c) Bohr’s atomic model
a) I, II and III
b) II, III and I
c) II, I, and III
d) III, II, and I
Answer:- c) II, I and III
The correct arrangement of atomic models according to years is –
Thomson’s Atomic model – 1904
Rutherford’s Atomic model – 1911
Bohr’s atomic model –1913
Short Answer Questions
19) Is it possible for the atom of an element to have one electron, one proton and no neutron. If so, name the element.
Answer:-
Yes, it is possible for the atom of Hydrogen. Hydrogen atom has only 1 proton and only one 1 electron and there is no neutron in it. it is represented as 1H1.
20) Write any two observations which support the fact that atoms are divisible.
Answer:- The facts which supports the atoms are divisible are –
a) Atom is divisible into electron and proton, these is are proved with their discovery. The discovery of electron and proton shows it can cut into parts.
b) Transfer of electron to form the ionic compounds shows that it can be divisible.
c) One single elements has difference in their proton and Electrons also neutrons, these is proved by the element has difference isotopes, and isobars.
21) Will 35Cl and 37Cl have different valencies ? Justify your answer.
Answer:- No. It does not have different valencies. Chlorine element is same in both the form 35 Cl and 37 Cl. These are isotopes of each other. Isotopes have same Atomic number but difference in their atomic mass number. But the valency of an atom is dependent over the atomic number of element. Here Atomic number of Chlorine is same in both the form thus valency is also same.
22) Why did Rutherford select a gold foil in his alpha- ray scattering experiment?
Answer:-
Rutherford selected gold metal for alpha ray scattering experiment because, to perform the experiment a thin sheet of an metal was needed, and gold is highly malleable metal. It can be converted into thinnest sheet . Therefore Rutherford selected Gold metal for scattering experiment.
23) Find out the valency of the atoms represented by the fig. 4.3 (a) and (b).
Answer:-
The valency of atom represented in
b) Atom shown in fig. a) has 0 valency, because the outer valence shell has total 8 electrons which complete the shell and makes the stable configuration. Thus the valency is 0
c) Atom shown in fig. b) has valency +1 because it’s outer valence shell has 7 atoms, it has one atom less in its outer shell, therefore it can take 1 electron from another atom to complete the octet. Thus the valency of that atom is +1.
24) One electron is present in the outer most shell of the atom of an element X. What would be the and value of charge on the ion formed if this electron is removed from the outer most shell?
Answer:- if one electron is removed from the outermost shell of an atom then the atom of an element becomes the donor of the electron which gives +1 charge on it. And the formation of cation takes place over the atom.
25) Write down the electron distribution of Chlorine atom. How many Electrons are there in the L shell ? (Atomic number of Chlorine is 17)
Answer:- The electronic distribution of elements according to the shell is takes place as –
1st shell = K = max. 2 Electrons
2nd shell = L= max. 8 Electrons
3rdshell = M= max. 18 Electrons
4th shell = N = max. 32 Electrons.
Now, The distribution the electrons of Chlorine Will takes place as (2,8,7)
K = 2 e- , L = 8 e- , M = 7 e-
Thus, L shell of Chlorine will contain 8 electrons.
26) In the atom of an element X, 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed?
Answer:-
If the atom of an element X Have an 6 electrons in its outermost shell and if it acquires to attain noble gas configuration i.e a stable configuration then the atom of an element will have to accept two electrons from another atom and thus the charge will be -2 over it.
27) What information do you from the fig. 4.4 about the atomic number, mass number and valency of atoms X, Y, and Z ? Give your answer in a tabular form.
Answer:-
Atomic No. | Mass No. | Valency | |
X | 5 | 11 | 3 |
Y | 8 | 18 | 2 |
Z | 15 | 31 | 3,5 |
28) In response to a question, a student stated that in an atom, the number of protons is greater than the number of neutrons, which in turn is greater than the number of electrons. Do you agree with the statement? Justify your answer.
Answer:- No. I am completely disagree with the statement. Because the statement is wrong. In any atom of an element the number of protons and the number of electrons are always equal. And the number of protons can be equal to the neutrons or less than the total number of neutrons. Thus the given statement is wrong.
29) Calculate the number of neutrons present in the nucleus of an element X which is represented as 31X 15.
Answer:-We have given
The element 31X15
Atomic mass of an element is nearby equal to the number of and the number of neutrons present in it. Here it is given as 31.
And,
The atomic number which is shows the total number of protons or Electrons present in the atom of an element. Which is given here 15.
So, to calculate the number of neutrons
Mass number = number of protons + number of neutrons
No. Of neutrons = mass number – No. Of protons
No. Of neutrons = 31- 15
No. Of neutrons = 16
Therefore, the total number of neutrons present in the element X are 16.
30) Match the names of the Neutrons of the scientists given in column A with their contributions towards the understanding of the atomic structure as given in column B.
(A) |
(B) |
a) Earnest Rutherford | I)Indivisibility of atoms |
b) J.J Thomson | II)Stationary orbits |
c) Dalton | III Concept of nucleus |
d) Neil’s bohr | IV Discovery of electrons |
e) James Chadwick | V Atomic number |
f) E. Goldstein | VI neutron |
g) Mosley | VII canal rays |
Answer:-
Name of the scientists | Discovery of atomic structure by them
(Answers) |
a) Earnest Rutherford
|
III) Concept of Nucleus |
b) J.J. Thomson | IV) Discovery of electrons
|
c) Dalton | I) Indivisibility of atoms |
d) Neil’s Bohr | II)Stationery orbits |
e) James Chadwick | VI) neutron |
f) E. Goldstein | VII) canal rays |
g) Mosley | V)Atomic number |
31) The atomic number of calcium and argon are 20 and 18 respectively, but the mass number of both these elements is 40. What is the name given to such a pair of elements?
Answer:-
The name given to such pairs of an elements is Isobars .
Isobars have same mass numbers but different atomic numbers.
Here Calcium and argon have different Atomic numbers 20 and 18 but both have same mass 40. Thus these are Isobars of each others.
32) Complete the table 4.1 on the basis of information available in the symbols given below.
35 Cl 17
12 C 6
81 Br 35
Answer:-
Element | np | nn |
Cl | 17 | 18 |
C | 6 | 6 |
Br | 35 | 46 |
33) Helium atom has 2 electrons in its valance shell but it’s valency is not 2, explain.
Answer:-
The helium atom has Atomic number 2. It Means that, It has only two electrons in an atom. the outermost shell of helium which is K has the maximum capacity of 2 Electrons. Which is filled up with two electrons present near the Helium atom. Thus,He atom does not have any valence shell unfilled,If the octet is completely filled then the valency becomes zero. the electrons of helium fulfil the outermost shell thus the valency of helium is 0 .
34) Fill in the blanks in the following statements.
a) Rutherford’s Alpha- particles scattering experiment led to the discovery of the _________.
Answer:- Rutherford’s Alpha – particles scattering experiment led to the discovery of the atomic nucleus.
b) Isotopes have same ________ but different ____________.
Answer:- Isotopes have same atomic number but different mass number.
c) Neon and Chlorine have atomic numbers 10 and 17 respectively. Their valencies will be __________ and __________ respectively.
Answer:- Neon and chlorine have atomic Numbers 10 and 17 respectively. Their valencies will be 0 and 1 respectively.
d) The electronic configuration of silicon is _________ and that of sulphur is _________.
Answer:- The electronic configuration of silicon is 2,8,4 and that of sulphur is 2,8,6 .
35) An element X has a mass number 4 and atomic number 2. Write the valency of this element?
Answer:-
The given element X has mass number 4 and atomic number 2.
According to these information these element is Helium. 4He2
The outermost shell of the Helium atom which is K shell is completely filled thus it does not have any shell remained incomplete. Hence the valency is zero of the Helium.
Long Answer Questions
36) Why do Helium, Neon and argon have a zero valency?
Answer:- These all the elements are noble gas elements Or 18th group elements.
These elements have their outermost shell fully filled, and there is no any space remained vacant. So that the Valency of these elements becomes zero.
The maximum number of electrons present in outermost shell of Helium is 2. Neon has 8 electrons in its outermost shell, and argon has 18 Electrons in its outermost shell. Which completes it’s octet fully. And thus these all have zero valency.
37) The ratio of the radii of hydrogen atom and it’s nucleus is ~ 105. Assuming the atom and the nucleus to be spherical, (i) what will be the ratio of their sizes?
(ii) if atom is represented by planet earth ‘Re‘ = 6.4 × 106 m, estimate the size of the nucleus.
38) Enlist the conclusions drawn by Rutherford from his alpha- ray scattering experiment.
Answer:-
Rutherford performed an experiment over the gold foil which is known as Alpha- ray scattering experiment.
These experiment results the following 3 conclusions.
*Alpha rays passed out straight-
Some Alpha particles directly passed out from the atom without any deflection these shows that the atom has some vacant space inside in it.
* Alpha rays showed slightly deflection-
Some of the alpha particles showed slightly deflection in the angle. These showed that the atom has some positive charges in it.
*Alpha rays returned –
Very less alpha particles returned directly and doesn’t pass out through the atom. These showed that atom has very small concentrated positive charge in middle of the atom.
39) In what way is the Rutherford’s Atomic model different from that of Thomson’s atomic model?
Answer:-
Thomson’s atomic model is also known as plum pudding model.
Thomson’s model states that there is an positively charged sphere which consists of negatively charged Electrons embedded in all over the surface. These positive and negative charges are equal to each other in numbers.
Rutherford proposed an model which is known as Rutherford’s alpha ray scattering experiment.
These experiment states that-
Electrons are revolving around the well organised orbit around the nucleus. These nucleus consists of positive charge at the centre of the atom. And the size of nucleus is very small as compared to atom. He states that all the mass is gathered in the nucleus of an atom.
40) What were the drawbacks of Rutherford’s model of an atom?
Answer:-
The Rutherford’s alpha ray scattering experiment failed to explain the stability of the electrons. the energy loss during the revolutionary motion of an electron. Circular motion will create an acceleration. When an electron move around the nucleus in the orbital path it will loss energy and finally reach to the nucleus and collapse over it. But if the electron collapse over the nucleus then atom and matter will not exist. But as we know atoms are present in stable form. Thus he failed to explain these.
41) What are the postulates of Bohr’s model of an atom?
Answer:-
Postulates of Bohr’s Atomic Model –
- According to the Neil’s Bohr, only some orbits where known as discrete orbits of electrons, which were allowed inside the atom.
- When the Electrons revolve in the discrete orbits these electrons do not radiate energy.
- These orbits are known as Energy levels.
These are represented by an circles.
- Bohr named that orbits or shell from inside to outside as K,L,M,N ……
Or the numbers n= 1,2,3,4…….
42) Show diagrammatically the electron distributions in a sodium atom and a Sodium ion and also give their Atomic number.
Answer:-
At. Number of Na = 11
i.e. 11 electrons and 11 protons
When there is removal or donating of 1 Electron takes place then Sodium atom become Sodium ion and gets +1 charge on it.
Therefore electrons remains in the Sodium ion are 11-1 = 10
And the electronic configuration of the
Na ion is = (2,8)
Atomic number = no. Of electrons = no. Of protons .
Which was 11 initially.
Therefore the number of proton will same in sodium atom and sodium ion i.e. 11.
43) In the gold foil experiment of Gieger and Marsden, that paved the way for Rutherford’s model of an atom, ~ 1.00% of the Alpha- particles were
found to deflect at angles > 50° if one mole of alpha particles were bombarded on the gold foil, compute the number of Alpha- particles that would deflect at angles less than 50°.
Answer:-
Total 1 mole of alpha particles were used for the bombardment in the experiment.
So,
1 mole = 6.022 × 10 23 particles
Now, we have to find out the alpha particles deflected at angle less than 50° .
Percentage Of alpha particles deflected at angle more than 50° = 1 % of total alpha particles bombarded.
Percentage Of alpha particles deflected at angle less than 50° = 99% of total alpha particles bombarded.
Therefore now,
Total number of Alpha- particles deflected at an angle less than 50°
= 5.96× 1023
Thus, the total number of Alpha- particles deflected at an angle less than 50° is 5.96×1023