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

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
Answer: B
The electron configuration of an atom shows the number of electrons in each sublevel in each energy level of the ground-state atom. The first sublevel filled will be the $1s$ sublevel, then the $2s$ sublevel, the $2p$ sublevel, the $3s,3p,4s,3d,$ and so on. Atomic number of magnesium is 12 and Electron configuration is $1s^{2}2s^{2}2p^{6}3s^2.$ So, according to Bohr's formula of distribution of electrons $(2n^2),$,K shell $(n=1)$ can accommodate 2 electrons, L shell $(n=2)$ can accommodate 8 electrons and the rest 2 electrons are accommodated in M shell. So, its electronic configuration is $2,8,2$.

Question:2

Rutherford’s ‘alpha $(\alpha )$ particles scattering experiment’ resulted in to discovery of
(a) Electron
(b) Proton
(c) Nucleus in the atom
(d) Atomic mass
Answer: C
Rutherford's alpha particle scattering experiment changed the way we think of atoms.
Before the experiment the best model of the atom was known as the Thomson or "plum pudding" model. Rutherford directed beams of alpha particles (which are the nuclei of helium atoms and hence positively charged) at thin gold foil to test this model and noted how the alpha particles scattered from the foil. In this model the positive material is concentrated in a small but massive (lot of mass - not size) region called the nucleus. Rutherford’s ‘alpha (a) particles scattering experiment’ resulted in to discovery of Nucleus in the atom.

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?
(a) ${}_{15}^{31}X$
(b) ${}_{16}^{31}X$
(c) ${}_{15}^{16}X$
(d) ${}_{16}^{15}X$
Answer: A
The atomic number of an atom is equal to the number of protons in the nucleus of an atom or the number of electrons in an electrically neutral atom.
Atomic number = Number of protons
It is represented with the letter ‘Z.’ The number of protons and neutrons combined to give us the mass number of an atom. It is represented using the letter ‘A.’
${}_{15}^{31}X$ is the correct representation of the element

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)
Answer: D
The Dalton’s atomic theory explains the laws of chemical combination (the Law of Constant Composition and the Law of Multiple Proportions). Dalton was the first person to recognize a workable distinction between the fundamental particle of an element (atom) and that of a compound (molecule). Dalton’s gave us ‘atoms cannot be created or destroyed.’
His theory gave us nothing on radioactivity.

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 $\alpha$–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
Rutherford’s model described the atom as a tiny, dense, positively charged core called a nucleus, in which nearly all the mass is concentrated, around which the light, negative constituents, called electrons, circulate at some distance, much like planets revolving around the Sun. In the experiment some alpha particles were deflected slightly, suggesting interactions with other positively charged particles within the atom. Still other alpha particles were scattered at large angles, while a very few even bounced back toward the source. Only a positively charged and relatively heavy target particle, such as the proposed nucleus, could account for such strong repulsion.

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)
Answer: D
The atomic number of an atom is equal to the number of protons in the nucleus of an atom or the number of electrons in an electrically neutral atom.
Atomic number = Number of protons, It is represented with the letter ‘Z.’
The number of protons and neutrons combined to give us the mass number of an atom. It is represented using the letter ‘A.’

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 stabilize the atom
(a) (i), (ii) and (iii)
(b) (i) and (iii)
(c) (i) and (iv)
(d) (i), (iii) and (iv)
Answer: A
According to the postulates of Thomson’s atomic model, an atom resembles a sphere of positive charge with electrons (negatively charged particles) present inside the sphere. The positive and negative charge is equal in magnitude and therefore an atom has no charge as a whole and is electrically neutral. Therefore, (i), (ii) and (iii) are correct in Thomson’s model of atom.

Question:8

Rutherford’s $\alpha$–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)
Answer: B
The Conclusion of Rutherford's scattering experiment are most of the space inside the atom is empty because most of the α-particles passed through the gold foil without getting deflected. Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space. There is a positively charged center in an atom called the nucleus. Nearly all the mass of an atom resides in the nucleus.

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
Answer: B
Mass number (A) of the atom = 27
Number of neutrons in the atom = 14
Number of protons in the atom = 27-14 =13
Number of electrons in the atom = 13
Therefore, this element is aluminum (Al).
Number of electrons in the Al atom = 13
Number of electrons in the $A{l}^{3+}$ ion = 13 - 3 = 10
As it is formed from the neutral atom by loss of 3 electrons.
Number of electrons in ion with 3 positive charges = 13 - 3 = 10

Question:10

Identify the $M{g}^{2+}$ ion from the figure where, ''n'' and ''p'' represent the number of neutrons and protons respectively.

Answer: D
The atomic number (Z) of magnesium is 12 and
the mass number (A) is 24 $(p=12andn=A-Z=24-12=12)$.
Electronic configuration of $Mg$ atom is 2, 8, 2.
So, the electronic configuration of $M{g}^{2+}$ ion is 2, 8 (the positive charge is acquired from the loss of 2 electrons).

Question:11

In a sample of ethyl ethanoate $\left({\text{CH}}_3{\text{COOC}}_2{\text{H}}_5\right)$ 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 have 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
Isotopes are atoms of the same element (with same atomic number) with different mass numbers. The difference in mass number is because of the different number of neutrons present in the atoms. The two O-atoms in ${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5$ can have different number of neutrons only if the two O-atoms are isotopes.

Question:12

Elements with valency 1 are:
(a) always metals
(b) always metalloids
(c) either metals or non-metals
(d) always non-metals
Answer: C
Metals and non-metals both can have valency 1. If an element show positive valency (like sodium), it is a metal; otherwise (negative valency like chlorine) it is a non-metal.

Question:13

The first model of an atom was given by
(a) N. Bohr
(b) E. Goldstein
(c) Rutherford
(d) J.J. Thomson
Answer: D
The first model of an atom was given by J.J. Thomson. According to him, an atom consists of a sphere of positive charge with negatively charged electrons embedded in it.

Question:14

An atom with 3 protons and 4 neutrons will have a valency of
(a) 3
(b) 7
(c) 1
(d) 4
Answer: C
Electronic configuration of this element will be 2, 1 (Atomic number $Z=3$) Since number of electrons in the outermost shell is 1, hence the valency of the atom will be 1.
Given that, number of protons in an atom = 3 and number of neutrons = 4
Electronic configuration of ${}_{3}Li=2,1$

Question:15

The electron distribution in an aluminum atom is
(a) 2,8,3
(b) 2,8,2
(c) 8,2,3
(d) 2,3,8
Answer: A
The atomic number of aluminum is 13 and the first shell can have at the most two electrons.
Therefore, Electronic configuration of ${}_{13}Al=2,8,3$

Question:16

Which of the following in figure 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
Answer: C
The first shell cannot have more than 2 electrons and the second shell cannot have more than 8 electrons. Figures (ii) and (iv) not correctly represent the Bohr’s model of an atom.

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
Answer: A
Atom is electrically neutral. It is possible only if an atom has equal number of protons and electrons.

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)
Answer: C
The correct order of the improvements in atomic models is as Thomson’s atomic model (ii), Rutherford’s atomic model (i) and Bohr’s atomic model (iii).

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.
Answer:

Yes, it is true for the common, stable isotope of hydrogen (protium), H (protium) has one electron, one proton and no neutron.

Question:20

Write any two observations which support the fact that atoms are divisible.
Answer:
Discovery of electrons and discovery of protons support the fact that atoms are divisible. during a chemical reaction, there is an either transfer of electrons or sharing of electrons between different atoms which leads to the rearrangement of atoms. Presence of isotopes for the same element is possible due to the difference in the number of neutrons also support the fact that atoms are divisible.

Question:21

Will ${}^{35}Cl$ and ${}^{37}Cl$ have different valencies ? Justify your answer.
Answer:
${}^{35}Cl$ and ${}^{37}Cl$ are isotopes of the same element - chlorine. Isotopes have the same number of electrons and protons. They differ only in the number of neutrons. Hence, their valencies do not differ. They have the same valency.

Question:22

Why did Rutherford select a gold foil in his $\alpha$-ray scattering experiment?
Answer:
A light metal cannot be used because on being hit by fast moving $\alpha$-particle, the atom of light metal will be simply pushed forward and no scattering can occur. Gold is heavy metal and for the scattering experiment, Rutherford wanted a metal sheet which could be as thin as possible. Gold is the most malleable of all known metals. It can easily be converted into very thin sheets. Hence, Rutherford selected a gold foil for his alpha-ray scattering experiment.

Question:23

Find out the valency of the atoms represented by the Figure (a) and (b).

Answer:
Atom (a) has zero valency since it has 8 electrons in the valence shell and has achieved a stable configuration. Atom (b) has a valency of 1 since it has 7 electrons in the valence shell. Atom (b) can accept on more electron in order to achieve a stable (octet) configuration.

Question:24

One electron is present in the outermost 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 outermost shell?
Answer:

The ion formed by loss of one electron will have positive nature and one positive (+1) charge on the cation formed. When the single electron present in the outer most shell of the atom of an element X is removed, the value of the net charge on the ion will be equal to the amount of charge present on one electron.

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).
Answer :
Since the atomic number (Z) of chlorine is 17,
Atomic number of chlorine atom = 17
So, its electronic configuration is
K L M
2 8 7.
The electronic configuration of chlorine will be 2, 8, 7. There are eight electrons in the L shell (second shell).

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?
Answer:
Element X has 6 electrons in the outermost shell. In order to acquire noble gas configuration, element X requires 2 electrons.
$X+2e^{-}\to X^{2-}$
when an atom has 6 electrons in its outermost shell and accepts 2 more electrons, the charge on the ion formed would be – 2.

Question:27

What information do you get from the figure about the atomic number, mass number and valency of atoms X, Y and Z? Give your answer in a tabular form.

Answer:
The atomic number, mass number and valency of the atoms X, Y and Z are given on the table below:-

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.
Answer:
The statement is not correct. According to this statement
$p>n>e$
But actually, the number of protons in an atom can never be greater than the number of neutrons. The number of protons will be equal to or less than the number of neutrons. The number of electrons and protons are always equal in a neutral atom.

Question:29

Calculate the number of neutrons present in the nucleus of an element X which is represented ${}_{15}^{31}\text{X}$
Answer:
Mass number (A) = No. of protons (Z) + No. of neutrons
But the mass number is given as 31 and the number of protons is 15.
No. of protons (Z) + No. of neutrons = 31
Number of neutrons = 31 – number of protons
Number of neutrons = 31 - 15 = 16
16 neutrons are present in the nucleus of the element X.

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

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?
Answer:
Elements with the same mass number (A) but different atomic numbers (Z) are called isobars. Calcium and Argon are isobars. Hence ${}_{20}^{40}\text{Ca}$ and is ${}_{18}^{40}\text{Ar}$ are isobars

Question:32

Complete the Table 4.1 on the basis of information available in the symbols given below.
(a) ${}_{17}^{35}\text{Cl}$
(b) ${}_6^{12}\text{C}$
(C) ${}_{35}^{81}\text{Br}$
Answer:
The table can be completed as follows:

$n_p$of Cl = Atomic number = 17

$n_p$ of C = Atomic number = 6

$n_p$ of Br = Atomic number = 35

$n_n$ of Cl = Atomic number $=A-Z=35-17=18$

$n_n$ of C = Mass number - Atomic number $=A-Z=12-6=6$

$n_n$ of Br = Mass number - Atomic number $=A-Z=81-35=46$

Question:33

Helium atom has 2 electrons in its valence shell but its valency is not 2. Explain.
Answer:
Helium atom has only one shell (K shell) which can have maximum 2 electrons. Helium atom has 2 electrons in its outermost shell and its duplet is complete. Hence the valency of helium atom is zero. It is called noble gas or inert gas.

Question:34

Fill in the blanks in the following statements.
(a) Rutherford’s $\alpha$-particle scattering experiment lead 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…………..
Answer:
(a) nucleus
Rutherford’s a-particle scattering experiment lead to the discovery of the nucleus.
(b) atomic number, mass number
Isotopes have same atomic number but different mass number
Example:
${}_6^{12}\text{C}{,}_6^{13}\text{C}{,}_6^{14}\text{C}$
(c) 0 and 1
Neon: Configuration: 2, 8
Valency: 0
Chlorine: Configuration: 2,8,7
It needs 1 electron to form stable noble gas configuration.
Valency: 1
(d) silicon – Atomic number 14 – configuration - 2, 8, 4
sulphur – Atomic number 16 - configuration -2, 8, 6

Question:35

An element X has a mass number 4 and atomic number 2. Write the valency of this element?
Answer:
The element X has its atomic number as 2. Thus, an atom of X contains two electrons. Thus, the elements has only K shell containing 2 electrons i.e., a complete duplet
Hence, its valency = 0.

Question:36

Why do helium, neon and argon have a zero valency?
Answer:
Noble gases have full filled octet, i.e., the subshell are completely filled. They have the general electronic configuration as $n{s}^{2}n{p}^6$.
Valency is a combining capacity of an element. As there are no free electrons in the valence shell as well as the elements are already in a stable state, so the valency of the noble gases is zero.

Question:37

The ratio of the radii of hydrogen atom and its nucleus is $\sim {10}^5$. Assuming the atom and the nucleus to be spherical.
(a) What will be the ratio of their sizes ?
(b) If atom is represented by planet earth ${}^{\prime }{R}_e^{\prime }=6.4\times {10}^{6}m$, estimate the size of the nucleus.
Answer:

Given is the ratio of the radii of hydrogen atom and its nucleus is $\sim {10}^5$.
And assuming the atom and the nucleus to be spherical.
(i) Sphere of volume $=\frac{4}{3}\pi r^3$
Let ''R'' be the radius of the atom and ''r'' be the radius of nucleus,
Therefore, $R={10}^{5}r$
Volume of atom $=\frac{4}{3}\pi R^3=\frac{4}{3}\pi {\left({10}^{5}r\right)}^3$
Volume of nucleus$=\frac{4}{3}\pi r^3$
Ratio of size of the atom to that of nucleus$={10}^{15}$
(ii) If the atom is represented by planet earth (Radius R_e)
Radius of nucleus $=\frac{\text{Radius of Earth}}{{10}^5}=\frac{6.4\times {10}^6}{{10}^5}m=64m$

Question:38

Enlist the conclusions drawn by Rutherford from his $\alpha$-ray scattering experiment.
Answer:
The conclusions drawn by Rutherford from his $\alpha$-ray scattering experiment are :
Since most of the $\alpha$-particles passed undeviated through the gold foil, the majority of the space occupied by the matter is vacuum (empty).
Some of the α-particles were deflected by the foil by small angles which indicated that the positive charge of the atom occupies very little space. He named these positively charged particle as “nucleus“.
All the positive charge and mass of the gold atom were concentrated in a very small volume within the atom. The nucleus must also be very heavy.

Question:39

In what way is the Rutherford’s atomic model different from that of Thomson’s atomic model?
Answer:
Thomson model of atom states that electrons are embedded in a positively charged solid material which is spherical in shape. Rutherford model of atom describes that an atom is composed of an atomic nucleus and electrons surrounding the nucleus.
Thomson model of atom does not give any details about the atomic nucleus. Rutherford model of atom explains about the atomic nucleus.
Thomson model of atom states that electrons are embedded in a solid material. Rutherford model of atom states that electrons are located around a central solid material.
Thomson model of atom indicates that the atom is a spherical structure. Rutherford model of atom indicates that an atom has a central solid core surrounded by electrons.

Question:40

What were the drawbacks of Rutherford’s model of an atom?
Answer:
Rutherford model of atom describes that the atom is composed of an atomic nucleus and electrons surrounding the nucleus. This model caused to reject the Thomson model of atom. Rutherford model was proposed by Ernst Rutherford after the discovery of the atomic nucleus. Later, Rutherford model was also rejected because it could not explain why the positively charged nucleus and electrons are not attracted to each other. The orbital revolution of the electron is not expected to be stable. Any particle in a circular orbit would radiate energy. Thus, revolving electrons would lose energy and finally fall into nucleus. If they were so, the atom would be highly unstable and hence matter would not exist in the form that we know. We know that atoms are quite stable

Question:41

What are the postulates of Bohr’s model of an atom?
Answer:
Postulates of Bohr’s Atomic Model are:

• Electrons revolve around the nucleus in a fixed circular path termed “orbits” or “shells” or “energy level.”

• The orbits are termed as “stationary orbit.”

• Every circular orbit will have a certain amount of fixed energy and these circular orbits were termed orbital shells. The electrons will not radiate energy as long as they continue to revolve around the nucleus in the fixed orbital shells.

• 
  

• The different energy levels are denoted by integers such as n=1 or n=2 or n=3 and so on. These are called as quantum numbers. The range of quantum number may vary and begin from the lowest energy level (nucleus side n=1) to highest energy level.

• The different energy levels or orbits are represented in two ways such as 1, 2, 3, 4… or K, L, M, N….. shells. The lowest energy level of the electron is called the ground state

Question:42

Show diagrammatically the electron distributions in a sodium atom and a sodium ion and also give their atomic number.
Answer:
Atomic number of sodium (Z) = 11
Mass number of sodium (A) = 23
∴ Number of protons in the nucleus = 11
Number of neutrons in the nucleus = 23 - 11 = 12
Number of electrons = 11
${}_{11}Na-2,8,1$ (11 electrons)
$Na-1e^{-}\to N{a}^{+}$ (10 electrons) Electronic configuration 2, 8.

Question:43

In the gold foil experiment of Geiger and Marsden, that paved the way for Rutherford’s model of an atom, $\sim$1.00% of the $\alpha$-particles were found to deflect at angles > 50°. If one mole of a-particles were bombarded on the gold foil, compute the number of $\alpha$-particles that would deflect at angles less than 50°.
Answer:
Percentage of $\alpha -$ particles deflected by more than ${50}^o=1{}^o{/}_o$ of total $\alpha -$ particles.
Percentage of $\alpha -$ particles deflected by less than ${50}^o=100{}^o{/}_o-1{}^o{/}_o=99{}^o{/}_o$ of total $\alpha -$ particles.
Number of $\alpha -$ particles bombarded on the gold foil = 1 mole $=6.022\times {10}^{23}$
Number of particles that deflected by an angle less than 50°
$=\frac{99}{100}\times 6.022\times {10}^{23}$
$=\frac{596.178}{100}\times {10}^{23}$
$=5.96\times {10}^{23}$