NCERT Exemplar Class 9 Science Chapter 4 Structure of the Atoms

NCERT Exemplar Class 9 Science Chapter 4 Structure of the Atoms are part of NCERT Exemplar Class 9 Science. Here we have given NCERT Exemplar Class 9 Science Solutions Chapter 4 Structure of the Atoms.

NCERT Exemplar Class 9 Science Solutions Chapter 4 Structure of the Atoms

Multiple Choice Questions

Question 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
Solution:

Question 2.
Rutherford’s ‘alpha (α) particles scattering experiment’resulted into discovery of
(a) electron
(b) proton
(c) nucleus in the atom
(d) atomic mass.
Solution:
(c) Rutherford’s alpha (α) particles scattering experiment resulted into discovery of nucleus in the atom. A large number of particles went straight through the atom while a very small number of particles were deflected back showing the presence of positively charged nucleus.

Question 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?

Solution:
(a): Number of electrons = 15,
Number of neutrons = 16
Hence atomic number of element X is 15 and atomic mass is 31. Hence the element is represented as ₁₅³¹X.

Question 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).
Solution:
(d) Dalton’s atomic theory successfully explained law of conservation of mass, law of constant composition and law of multiple proportion.

Question 5.
Which of the following statements about Rutherford’s model of atom are correct?
(i) Considered the nucleus as positively charged
(ii) Established that the α-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)
Solution:
(a) Rutherford’s model of atom explained presence of nucleus in the centre and electrons around the nucleus revolving in round orbitals like the solar system.

Question 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)
Solution:
(d) Mass number = p + n
Atomic number = p = e

Question 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 in the positively charged sphere.
(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)
Solution:
(a) Thomson’s model could be compared with a raisin pudding model according to which the mass of atom is uniformly distributed over the atom in the form of positive charge and electrons are uniformly distributed over the atom. Electrons do not attract each other.

Question 8.
Rutherford’s a-particle 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 exists 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)
Solution:
(b) Two important observations of Rutherford’s a-particles scattering experiment were – (i) The mass and positive charge of the atom is concentrated in the nucleus.
(ii) Most of the space in the atom is empty.

Question 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
Solution:
(b) Mass number of the atom = 27
Number of neutrons = 14
Number of protons = 27 – 14 = 13
Number of electrons in the atom = 13
Number of electrons in ion with 3 positive charges = 13 – 3 = 10

Question 10.
Identify the Mg2⁺ ion from the Fig. 4.1 where, n and p represent the number of neutrons and protons respectively.
Solution:
(d) In magnesium atom :²⁴₁₂Mg
Number of n = 12, p = 12, e = 12
Number of electrons in Mg2⁺ = 10

Question 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.
Solution:
(c) Two isotopes have same number of electrons and protons but different number of neutrons.

Question 12.
Elements with valency 1 are
(a) always metals
(b) always metalloids
(c) either metals or non-metals
(d) always non-metals.
Solution:
(c) Element with valency 1 can be both metals or non-metals.

Question 13.
The first model of an atom was given by
(a) N. Bohr
(b) E. Goldstein
(c) Rutherford
(d) J.J. Thomson.
Solution:
(d) J.J. Thomson proposed first model of an atom.

Question 14.
An atom with 3 protons and 4 neutrons will have a valency of
(a) 3
(b) 7
(c) 1
(d) 4.
Solution:
(c) For the atom n = 4, p = 3, hence e = 3 Distribution of electrons = 2,1

Question 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.
Solution:
(a) Aluminium has 13 electrons. Its electron distribution is 2, 8, 3.

Question 16.
Which of the following in Fig. 4.2 do not represent Bohr’s model of an atom correctly?
(a) (i) and (ii).2
(b) (ii) and (iii)
(c)(ii) and (iv)
(d) (i) and (iv)
Solution:
(c) Fig. (ii) contains 4 electrons in K shell and fig. (iv) contains 9 electrons in L shell which are not in accordance with Bohr’s model.

Question 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.
Solution:
(a) An atom has equal number of electrons and protons since it is’neutral.

Question 18.
Atomic models have been improved over the years. Arrange the following atomic models in the order of their chronological order
(i) Rutherford’s atomic model
(ii) Thomson’s atomic model
(iii) Bohr’s atomic model
(a) (i), (ii) and (iii)
(b) (ii), (iii) and (i)
(c) (ii), (i) and (iii)
(d) (iii), (ii) and (i).
Solution:
(c) Thomson’s atomic model, followed by Rutherford’s model which is followed by Bohr’s model.

Short Answer Type questions

Question 19.
Is it possible for the atom of an element to have one electron, one proton and no neutron. If so, name the element.
Solution:
Yes, hydrogen has one electron, one proton and no neutron. It is represented as ¹₁H.

Question 20.
Write any two observations which support the fact that atoms are divisible.
Solution:
With the discovery of electrons and protons it was established that atom is divisible and is made up of negatively charged electrons and positively charged protons along with some neutral particles called neutrons.

Question 21.
Will ³⁵CIand ³⁷CI have different valencies? Justify your answer.
Solution:
No, both ³⁵CI and ³⁷CI will have same valencies, as ³⁵CI and ³⁷CI are isotopes. The isotopes have same number of electrons and protons. They differ only in the number of neutrons. Their electron distribution will be same.

Question 22.
Why did Rutherford select a gold foil in his α-ray scattering experiment?
Solution:
Gold is a heavy metal with high mass number. A light metal cannot be used because on being hit by fast moving α-particles, the atom of light metal will be simply pushed forward and no scattering can occur. Moreover, gold is highly malleable and can be beaten to get very thin foils.

Question 23.
Find out the valency of the atoms represented by the Fig. 4.3 (a) and (b).

Solution:
(a) It has electronic configuration = 2,8,8. Its outermost shell has complete octet. Hence, its valency = 0.
(b) It has electronic configuration = 2, 7. It can easily gain one electron to complete its outermost octet.
Hence, its valency = – 1.

Question 24.
One electron is present in the outer most shell of the atom of an element X. What would be the nature and value of charge on the ion formed if this electron is removed from the outer most shell?
Solution:
X – le^– → X⁺. The ion formed by loss of one electron will have positive nature and one positive (+1) charge on the cation formed.

Question 25.
Write down the electron distribution of chlorine atom. How many electrons are there in the L shell? (Atomic number of chlorine is 17).
Solution:

L shell will have 8 electrons.

Question 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?
Solution:
X – 2e^– —> X^2-; when an atom has 6 electrons in its outermost shell and it accepts 2 electrons, it gets two negative charges.

Question 27.
What information do you get 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.

Solution:

Element	Atomic number	Mass number	Valency
X	5	11	3
Y	8	18	2
Z	15	31	3,5

Question 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.
Solution:
The given statement is not correct. As number of protons is never greater than number of neutron. Number of neutrons can be equal to or greater than number of protons but number of protons is equal to number of electrons for an atom since it is neutral.

Question 29.
Calculate the number of neutrons present in the nucleus of an element X which is represented ³¹₁₅X
Solution:
In ³¹₁₅X, number of protons = number of electrons = 15
Number of neutrons =Mass number – no. of protons = 31 – 15 = 16

Question 30.
Match the names of the Scientists given in column A with their contributions towards the understanding of the atomic structure as given in column B

Column (A)		Column (B)
(a)	Ernest Rutherford	(i)	Indivisibility of atoms
(b)	JJ.Thomson	(ii)	Stationary orbits
(c)	Dalton	(Hi)	Concept of nucleus
(d)	Neils Bohr	(iv)	Discovery of electrons
(e)	James Chadwick	(v)	Atomic number
(f)	E. Goldstein	(vii)	Neutron
(9)	Mosley	(vii)	Canal rays

Solution:
(a) (iii)
(b) (iv)
(c) (i)
(d) (ii)
(e) (vi)
(f) (vii)
(g) (v)

Question 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?
Solution:
A pair of elements in which the elements have same mass number but different atomic numbers are called isobars. Hence ⁴⁰₂₀Ca and is ⁴⁰₁₈Ar are isobars.

Question 32.
Complete the Table 4.1 on the basis of information available in the symbols given below.
(a) ³⁵₁₇Cl
(b) ¹²₆C
(C)⁸¹₃₅Br

Element	np	n„

Solution:

Element	np	nn
3517Cl	17	18
126C	6	6
8135Br	35	46

Question 33.
Helium atom has 2 electrons in its valence shell but its valency is not 2, Explain.
Solution:
Helium has only one shell (K – shell) and the maximum number of electrons within K shell can be 2. Hence the valency is zero.

Question 34.
Fill in the blanks in the following statements.
(a) Rutherford’s a-partide scattering experiment led to the discovery of the_____.
(b) Isotopes have same ______ but different ______.
(c) Neon and chlorine have atomic numbers 10 and 17 respectively. Their valencies will be______ and _______ respectively.
(d) The electronic configuration of silicon is _____ and that of sulphur is______.
Solution:
(a) nucleus
(b) atomic number, mass number
(c) 0 and 1
(d) silicon – 2, 8, 4
sulphur – 2, 8, 6

Question 35.
An element X has a mass number 4 and atomic number 2. Write the valency of this element?
Solution:
The element X has its atomic number as 2. Thus, an atom of X contains two electrons in it. These 2 electrons fill the X-shell completely.
Thus the valency of X is zero.

Long Answer Type Questions

Question 36.
Why do Helium, Neon and Argon have a zero valency?
Solution:
Helium has only K shell and it is completely filled with 2 electrons. Argon and neon have 8 electrons in their outermost shell which is the maximum number of electrons that can be accommodated in the outermost shell, hence their valency is zero.

Question 37.
The ratio of the radii of hydrogen atom and its 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 ‘R_e‘ = 6.4 x 10⁶m, estimate the size of the nucleus.
Solution:


Question 38.
Enlist the conclusions drawn by Rutherford from his α-ray scattering experiment.
Solution:
Rutherford concluded from a-particle scattering experiment that

1. Most of the space in the atom is empty because most of the α-particles passed through the gold foil without getting deflected.
2. A small fraction of a-particles were deflected indicating that the positive charge of the atom occupies very little space, (viii) A very small fraction of a-particles were deflected by 180°, indicating that all the positive charge and mass of the atom were concentrated in a very small volume within the atom.
   From the data Rutherford also concluded that the radius of the nucleus is about 105 times less than Mg2+ the radius of the atom

Question 39.
In what way is the Rutherford’s atomic model different from that of Thomson’s atomic model?
Solution:
Thomson’s model of the atom proposed the resin pudding structure in which electrons are embedded like resins in a positively charged sphere of pudding or cake. Rutherford’s model could be compared with solar system. According to his model the positive charge is concentrated in a very small space in the centre which was called nucleus. Electrons revolve around the nucleus in well defined orbits.

Question 40.
What were the drawbacks of Rutherford’s model of an atom?
Solution:
Drawbacks of Rutherford’s model of an atom:

• Any particle in a circular motion would undergo acceleration and would radiate energy. Thus revolving electron would lose energy and finally fall into the nucleus. Thus atom should be unstable and should not exist in the stable form.
• The model did not give any arrangement of electrons around the nucleus.

Question 41.
What are the postulates of Bohr’s model of an atom?
Solution:
According to Neils Bohr’s model of the structure of the atom

1. Electrons revolve around the nucleus in fixed paths called orbits.
2. Only certain special orbits are allowed inside the atom.
3. While revolving in discrete orbits the electrons do not radiate energy. They absorb energy to go to the higher level and emit energy to go tot he lower level. These orbits are called energy levels which are represented as K, L, M, N or 1, 2, 3, 4

Question 42.
Show diagramatically the electron distributions in a sodium atom and a sodium ion and also give their atomic number.
Solution:
₁₁Na -2,8,1 (11 electrons)
Na – 1e^–—> Na⁺ (10 electrons) Electronic configuration 2, 8.

The atomic number of an element is equal to the number of protons in its atom. Since, sodium atom and sodium ion contain the same number of protons, therefore, the atomic number of both is 11.

Question 43.
In the Gold foil experiment of Geiger and Marsden, that paved the way for Rutherford’s model of an atom, ~ 1.00% of the α-particles were found to deflect at angles > 50°. If one mole of a-particles were bombarded on the gold foil, compute the number of a-particles that would deflect at angles less than 50°.
Solution:
% of a-particles deflected more than 50° = 1%
% of a-particles deflected less than 50°
= 100 -1 = 99%
Number of a-particles bombarded = 1 mole = 6.022 x 10²³particles
Number of particles that deflected at an angle less than 50°
= 99/100x 6.022 x 10²³100
\(\begin{aligned} &=\frac{99}{100} \times 6.022 \times 10^{23} \\ &=\frac{596.178}{100} \times 10^{23} \\=& 5.96 \times 10^{23} \end{aligned}\)

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