NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 Structure of Atom

Q: Is the chapter important for competitive exams like JEE/ NEET?

Yes, this chapter is important for competitive exams. One should go through its important topics and exercises given in NCERT Exemplar solutions for class 11 Chemistry Chapter 2.

Q: Are there numerically based exercises too?

Yes, there are many numericals given in NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 that can be used by the students to solve and evaluate themselves.

Q: What are some of the important topics in the chapter?

The important topics include quantum numbers, atomic models.

Q: Can the solutions of NCERT be downloaded for future reference?

Yes, one can easily download Class 11 Chemistry NCERT exemplar solutions chapter 2.

Edited By Sumit Saini | Updated on Sep 25, 2021 - 5:41 p.m. IST

NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 Structure of Atom gives an in-depth knowledge of the topics, various theories and models associated with it, like – Rutherford’s Atomic model, Thomson’s Model Bohr’s model, Quantum Mechanical Model of an atom and so on. One can refer to the NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 for detailed explanations and the complete steps required while solving numerical questions related to wavelength, frequency and the energy associated with electromagnetic radiation.

NCERT exemplar class 11 Chemistry solutions chapter 2: MCQ (Type 1)

Question:1

Which of the following conclusions could not be derived from Rutherford's α-particle scattering experiment?
(a) Most of the space in the atom is empty.
(b) The radius of the atom is about 10 ^-10 m while that of nucleus is 10 ^-15 m.
(c) Electrons move in a circular path of fixed energy called orbits.
(d) Electrons and the nucleus are held together by electrostatic forces of attraction.
Answer:

The answer is the option (c)
Electrons move in a circular path of fixed energy called orbits because it does not explain anything about an atom’s stability, it’s electronic structures, their energies, and distribution around the nucleus.

Question:2

Which of the following options does not represent ground state electronic configuration of an atom?
(i) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²
(ii) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s²
(iii) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
(iv) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
Answer:

The correct answer is the option (b) as the correct configuration would be
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

Question:3

The probability density plots of 1s and 2s orbitals are given in the following figures.

The density of dots in a region represents the probability density of finding electrons in the region. On the basis of the above diagram, which of the following statements is incorrect?
(a) 1s and 2s orbitals are spherical in shape.
(b) The probability of finding the electron is maximum near the nucleus.
(c) The probability of finding the electron at a given distance is equal in all directions.
(d) The probability density of electrons for 2s orbital decreases uniformly as distance from the nucleus increases.
Answer:

The answer is the option (d)
The probability density of electrons for 2s orbital decreases uniformly as distance from the nucleus increases. The reason being that the probability density for the 2s orbital first sharply reduces to zero and then starts increasing whereas for 1s orbital the probability density is at its peak at the nucleus and then starts decreasing gradually on moving away from it.

Question:4

Which of the following statements is not correct about the characteristics of cathode rays?
(a) They start from the cathode and move towards the anode.
(b) They travel in straight line in the absence of an external electrical or magnetic field.
(c) Characteristics of cathode rays do not depend upon the material of electrodes in cathode ray tube.
(d) Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube.
Answer:

The answer is the option (d)
Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube because we know that nature of the gas in the cathode rays and characteristics of the cathode rays have no bearing on the material of the electrodes.

Question:5

Which of the following statements about the electron is incorrect?
(a) It is a negatively charged particle.
(b) The mass of electron is equal to the mass of neutron.
(c) It is a basic constituent of all atoms.
(d) It is a constituent of cathode rays.
Answer:

The answer is the option (b)
The mass of electron is equal to the mass of neutron because the mass of a neutron is $1.67\times 10^{-27}$ kg and that of an electron is $9.1\times 10^{-31}$ kg.

Question:6

Which of the following properties of atom could be explained correctly by Thomson Model of atom?
(a) Overall neutrality of atom.
(b) Spectra of hydrogen atom.
(c) Position of electrons, protons, and neutrons in atom.
(d) Stability of atom.
Answer:

The answer is the option (a)
Overall neutrality of atom as the Thomson model has been unable to explain it. Initially this model did explain about the overall neutrality of an atom but later tests stated otherwise.

Question:7

Two atoms are said to be isobars if
(a) They have same atomic number but different mass number.
(b) They have same number of electrons but different number of neutrons.
(c) They are same number of neutrons but different number of electrons.
(d) Sum of the number of protons and neutrons is same but the number of protons is different.
Answer:

The answer is the option (d)
Sum of the number of protons and neutrons is same, but the number of protons is different because isobars have a different atomic number.

Question:8

The number of radial nodes for 3p orbital is
(a) 3 (b) 4 (c) 2 (d) 1

Answer:

The answer is the option (d)
1 because for 3p orbital, n = 3-1-1 = 1 which is the number of radial nodes.

Question:9

Number of angular nodes for 4d orbital is __________ .
(a) 4 (b) 3 (c) 2 (d) 1
Answer:

The answer is the option (c)
2 because we know that for dxy orbitals, there are two angular nodes (represented as “I”) that pass through the origin and bisect the xy plane that contains the z-axis.

Question:10

Which of the following is responsible to rule out the existence of definite paths or trajectories of electrons?
(a) Pauli's exclusion principle
(b) Heisenberg's uncertainty principle
(c) Hund's rule of maximum multiplicity
(d) Aufbau principle
Answer:

The answer is the option (b)
Heisenberg's uncertainty principle, which states that it is not possible to simultaneously determine the velocity and position of an electron correctly.

Question:11

Total number of orbitals associated with third shell will be _______.
(a) 2
(b) 4
(c) 9
(d) 3
Answer:

The answer is the option (c)
9, This is so because the total number of orbitals associated with the shell are given by n², wherein “n” is the shell. O, in this case, it would be 3² = 9.

Question:12

Orbital angular momentum depends on
(a) l
(b) n and l
(c) n and m
(d) m and s
Answer:

The answer is the option (a)
I, because “I” defines the 3-dimensional shape of the orbital and is commonly known as the subsidiary quantum number or the orbital angular momentum. For any given value of “n”, the “I” can have n values that range from 0 to n-1.

Question:13

Chlorine exists in two isotopic forms, Cl-37 and Cl-35, but its atomic mass is 35.5. This indicates the ratio of Cl-37 and Cl-35 is approximately
(a) 1 : 2
(b) 1 : 1
(c) 1:3
(d) 3:1
Answer:

The answer is the option (c)
1:3 , because we know that chlorine is made up of two isotopes that have atomic masses of 35 u and 37 u in the ratio 1:3.

Question:14

The pair of ions having same electronic configuration is............
(a) $Cr^{3+},Fe^{3+}$
(b) $Fe^{3+},Mn^{2+}$
(c) $Fe^{3+},CO^{3+}$
(d) $Sc^{3+},Cr^{3+}$
Answer:

The answer is the option (b) $Fe^{3+},Mn^{2+}$ because they have the same number of electrons, hence the same electronic configuration.

Question:15

For the electrons of oxygen atom, which of the following statements is correct?
(a) Z_eff for an electron in a 2s orbital is the same as Z_eff for an electron in a 2p
(b) An electron in the 2s orbital has the same energy as an electron in the 2p
(c) Z_eff for an electron in Is orbital is the same as Z_eff for an electron in a 2s orbital.
(d) The two electrons present in the 2s orbital have spin quantum numbers m_s but of opposite sign.
Answer:

The answer is the option (d)
The two electrons present in the 2s orbital have spin quantum numbers ms but of opposite sign. This is so because in the same orbital plane, only two electrons can exist, and they must have opposite signs.

Question:16

If traveling at same speeds, which of the following matter waves have the shortest wavelength?
(a) Electron
(b) Alpha particle (He²⁺ )
(c) Neutron
(d) Proton

Answer:

The answer is the option (b) Alpha particle (He²⁺ ) because higher is the mass, shorter is the wavelength.

NCERT exemplar class 11 Chemistry solutions chapter 2: MCQ (Type 2)

Question:17

Identify the pairs which are not of isotopes?
i) $_{6}^{12}\textrm{X},\; _{6}^{13}\textrm{Y}$
ii) $_{17}^{35}\textrm{X},\; _{17}^{37}\textrm{Y}$
iii) $_{6}^{14}\textrm{X},\; _{7}^{14}\textrm{Y}$
iv) $_{4}^{8}\textrm{X},\; _{5}^{8}\textrm{Y}$

Answer:

The answer is the options (iii) and (iv) because isotopes must have different atomic mass number but an identical atomic number.

Question:18

Out of the following pairs of electrons, identify the pairs of electrons present in degenerate orbitals :
i) (a) $n=3,\; l=2,\; m_{l}=-2,m_{s}=-\frac{1}{2}$
(b) $n=3,\; l=2,\; m_{l}=-1,m_{s}=-\frac{1}{2}$
ii) (a) $n=3,\; l=1,\; m_{l}=1,m_{s}=+\frac{1}{2}$
(b) $n=3,\; l=2,\; m_{l}=1,m_{s}=+\frac{1}{2}$
iii) (a) $n=4,\; l=1,\; m_{l}=1,m_{s}=+\frac{1}{2}$
(b) $n=3,\; l=2,\; m_{l}=1,m_{s}=+\frac{1}{2}$
iv) (a) $n=3,\; l=2,\; m_{l}=+2,m_{s}=-\frac{1}{2}$
(b) $n=3,\; l=2,\; m_{l}=+2,m_{s}=+\frac{1}{2}$

Answer:

The answers are the options (i) and (iv) because the degenerate orbitals have the same n and l values. Herein, the other two options have different values of n and l; therefore they cannot possess similar energy.

Question:19

Which of the following sets of quantum numbers are correct?
n l m₁
(i) 1 1 +2
(ii) 2 1 +1
(iii) 3 2 -2
(iv) 3 4 -2

Answer:

The answer is the options (ii) and (iii)
As, in the case n = 1, then I ≠ 1, therefore option (i) is incorrect.
In case n = 3, I ≠ 4, therefore option (iv) is also incorrect.
In case, n = 2, I = 0, 1. When l = 1, then m = -1, 0, +1. Therefore option (ii) is correct.
In case n = 3, l= 0, 1, 2. When l = 2, m = -2, -1, 0, +1, +2. Therefore, option (iii) is also correct.

Question:20

In which of the following pairs, the ions are iso-electronic?
(i) $Na^{+}, Mg^{2+}$
(ii) $Al^{3+}, O^{-}$
(iii) $Na^{+}, O^{2-}$
(iv) $N^{3-}, Cl^{-}$
Answer:

The answer is the options (i) and (iii) because in order for the ions to be iso-electronic, they must have the same number of electrons.

Question:21

Which of the following statements concerning the quantum numbers are correct?
(i) Angular quantum number determines the three dimensional shape of the orbital.
(ii) The principal quantum number determines the orientation and energy of the orbital.
(iii) Magnetic quantum number determines the size of the orbital.
(iv) Spin quantum number of an electron determines the orientation of the spin of electron relative to the chosen axis.
Answer:

The answer is the options (i) and (iv) as only these two statements are correct. In contrast, the principal quantum number determines the orbit's size while the orientation of the electron cloud within a sub-shell is determined by the magnetic quantum number. Hence, statements (ii) and (iii) are incorrect.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 2: Short answer type

Question:22

Arrange s, p and d sub-shells of a shell in the increasing order of effective nuclear charge (Zeff) experienced by the electron present in them.
Answer:

The arrangement of sub-shells in order of effective nuclear charge is: d < p < s. The net positive charge experienced by an outer electron is known as an effective nuclear charge (Zeff). This charge decreases with an increase of azimuthal quantum number (l), that is, the s orbital electron will be more tightly bound to the nucleus than p orbital electron which in turn will be better tightly bound than the d orbital electron.

Question:23

Show the distribution of electrons in oxygen atom (atomic number 8) using orbital diagram.
Answer:

The orbital diagram of oxygen will be -

Question:24

Nickel atom can lose two electrons to form Ni²⁺ ion. The atomic number of nickel is 28. From which orbital will nickel lose two electrons.
Answer:

$Ni=[Ar]\; 3d^{8}4s^{2}$
So from 4s orbital, Ni will lose electron

Question:25

Which of the following orbitals are degenerate?
$3d_{xy},4d_{xy},3d_{{z}^{2}},3d_{yz},4d_{yz},4d_{{z}^{2}}$

Answer:

Orbitals with the same shells and sub-shells are known as degenerate orbitals. From the given options, the options that represent degenerate orbitals due to the same main shell, i.e. n=3 are $3d_{xy},3d_{z^{2}},3d_{yz}$
Moreover, $4d_{xy},4d_{yz},4d_{z^{2}}$ represent degenerate orbitals due to the same value of n = 4.

Question:26

Calculate the total number of angular nodes and radial nodes present in 3p orbital.

Answer:

Number of angular nodes= l = 1
Number of radial nodes = n-l-1 = 3-1-1 =1

Question:27

The arrangement of orbitals on the basis of energy is based upon their (n+l) value. Lower the value of (n+l), lower is the energy. For orbitals having same values of (n+l), the orbital with lower value of n will have lower energy.
Based upon the above information, arrange the following orbitals in the increasing order of energy.
(a) 1s, 2s, 3s, 2p
(b) 4s, 3s, 3p, 4d
(c) 5p, 4d, 5d, 4f, 6s
(d) 5f, 6d, 7s, 7p
Based upon the above information, solve the questions given below:
(a) Which of the following orbitals has the lowest energy?
4d, 4f, 5s, 5p
(b) Which of the following orbitals has the highest energy?
5p, 5d, 5f, 6s, 6p
Answer:

I. (a) 1s<2s<2p<3s
(b) 3s<3p<4s<4d
(c) 4d<5p<6s<4f<5d
(d) 7s < 5f < 6d < 7p
II. (a) 5s
(b) 5f

Question:28

Which of the following will not show deflection from the path on passing through an electric field? Proton, cathode rays, electron, neutron.
Answer:

Neutron will not show deflection from the path on passing through an electric field, as the particles are neutral in nature, thereby being unaffected by any electrical field.

Question:29

An atom having atomic mass number 13 has 7 neutrons. What is the atomic number of the atom?
Answer:

A=n+p
P=A-n=13-7=6
No. of protons = no. of electrons = atomic number
Therefore, the atomic number of the atom is 6.

Question:30

Wavelengths of different radiations are given below:
λ (A) = 300 nm
λ (B) = 300 μ m
λ (C) = 3 nm
λ (D) = 30 angstrom
Arrange these radiations in the increasing order of their energies.
Answer:

λ(B) < λ(A) < λ(c) = λ (D)

Question:31

The electronic configuration of valence shell of Cu is $3d^{10}4s^{1}$ and not $3d^{9}4s^{2}$ . How is this configuration explained?
Answer:

When the two sub-shells differ slightly in their energies, an electron shifts from a sub-shell of lower energy (4s) to a sub-shell of a higher energy (3d). It has been found that there is extra stability associated with this electronic configuration as stability in 3d¹⁰4s¹, d orbital is completely filled and s is half filled. So it is more stable configuration.

Question:32

The Balmer series in the hydrogen spectrum corresponds to the transition from n₁ = 2 to n₂= 3, 4. This series lies in the visible region. Calculate the wave number of line associated with the transition in Balmer series when the electron moves to n= 4 orbit. (R_H= 109677 cm^–1).
Answer:

Here n₁ = 2, n₂= 4
$^{-}v=109677\left ( \frac{1}{2^{2}}-\frac{1}{4^{2}} \right )$
for n₁= 2, n₂= 4 transition in Balmer series.
$=109667\left ( \frac{1}{4}-\frac{1}{16}\right )$
$=20564.44\; cm^{-1}$

Question:33

According to de Broglie, matter should exhibit dual behaviour, that is both particle and wave like properties. However, a cricket ball of mass 100 g does not move like a wave when it is thrown by a bowler at a speed of 100 km/h. Calculate the wavelength of the ball and explain why it does not show wave nature.
Answer:

According to be Broglie,
The wavelength = λ = h / mv
m = 100 g = 0.1 kg.
v = 100 km/hr = (100 $\times$ 1000)/(60 $\times$ 60) = 1000/36 m/s
h =6.626×10^–34 Js
Therefore, in the present case.
$\lambda =\frac{6.626\times 10-34J.s}{0.1kg\times \frac{1000}{36}m/s}$
$=238.5\times 10^{-36}m^{-1}$
Since the wavelength is negligible to be detected, thus, it does not show wave nature.

Question:34

What is the experimental evidence in support of the idea that electronic energies in an atom are quantized?

Answer:

The bright line spectrum shows that the energy levels in an atom are quantized. Classical mechanics is not able to explain the working of electrons, atoms, and molecules, etc. This is because classical mechanics ignores the concept of dual nature of matter. Quantum mechanics considers these theories.

Question:35

Out of electron and proton which one will have, a higher velocity to produce matter waves of the same wavelength? Explain it.
Answer:

Electron will have a higher velocity.
$\lambda =\frac{h}{p}=\frac{h}{mv}$
Lighter the mass higher will be the velocity of the particle.

Question:36

A hypothetical electromagnetic wave is shown in Fig. 2.2. Find out the wavelength of the radiation.

Answer:

λ = 4 X 2.16 pm
= 8.64 pm

Question:37

Chlorophyll present in green leaves of plants absorbs light at $4.620 \times 10^{14}Hz$ . Calculate the wavelength of radiation in nanometer. Which part of the electromagnetic spectrum does it belong to?
Answer:

We know that
$\lambda =\frac{c}{v}$
For this problem,
$C = 3 \times 10^{8} ms^{-1}$
$V= 4.620 \times 10^{14} Hz.$
$\lambda =3\times 10^{8}ms^{-1}/4.620\times 10^{14}Hz$
$=0.6494\times 10^{-6}m$
Therefore, we can say that it belongs to the visible light of the spectrum

Question:38

What is the difference between the terms orbit and orbital?
Answer:

Orbit	Orbital
Orbits are circular and disc-like.	Different orbitals have different shapes, i.e. s orbitals are spherically symmetrical, p-orbitals are dumb-bell shaped and so on.
It represents the planar motion of an electron around the nucleus.	It represents the three-dimensional motion of an electron around the nucleus.
Orbit is a well-defined circular path around the nucleus in which the electrons revolve.	An orbital is a three-dimensional space around the nucleus within which the probability of finding an electron is maximum.

Question:39

Table-tennis ball has a mass 10 g and a speed of 90 m/s. If speed can be measured with an accuracy of 4% what will be the uncertainty in speed and position?
Answer:

As per the Heisenberg uncertainty principal, it is not possible to determine the position and velocity of a particle at the same time.
$\Delta x.\Delta p\geq \frac{h}{4\pi }$
As per the information given in the question.
Mass of the ball = 4g
Speed = 90 m/s
Uncertainty in the speed of ball = $\Delta v=\frac{4}{100}\times 90=3.6\; m/s$
Uncertainty in position = $\Delta x=\frac{h}{4\pi m\Delta v}$
$6.626\times 10^{-34}/4\times 3.14\times 10\times 10^{-3}\times 3.6$
$=1.46\times 10^{-33}m$

Question:40

The effect of uncertainty principle is significant only for motion of microscopic particles within an accuracy of 4% what will be the uncertainty in speed and position?
Answer:

If we consider mass as 10^-6 kg (1 milligram), then we know that
$\Delta v.\Delta x=h/4\pi m$
$=\Delta v.\Delta x=h/4\times 3.14\times 10^{-6}$
$=6.626\times 10^{-34}/4\times 3.14\times 10^{-6}$
$=0.52\times 10^{-28}m^{2}s^{-1}$
As the value is negligible and is highly insignificant for the uncertainty principle to be applicable to the particle.

Question:41

Hydrogen atom has only one electron, so mutual repulsion between electrons is absent. However, in multielectron atoms mutual repulsion between the electrons is significant. How does this affect the energy of an electron in the orbitals of the same principal quantum number in multielectron atoms?
Answer:

The energy of an electron in a hydrogen atom is determined by the principal quantum number. Thus, the energy of the orbitals increases as: 1s< 2s= 2p< 3s= 3p= 3d< 4s= 4p < 4d= 4f.
However, the energy of an electron in a multielectron atom is dependent on not just the principal quantum number (shell) as well as also on the azimuthal quantum number (subshell) which is quite unlike that of an hydrogen atom. This implies that, for a given principal quantum number, s, p, d, f, all will have different energies.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 2: Matching Type

Question:42

Match the following species with their corresponding ground state electronic configuration.

Atom / Ion	Electric Configuration
$Cu$	$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10}$
$Cu^{2+}$	$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10}4s^{2}$
$Zn^{2+}$	$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10}4s^{1}$
$Cr^{3+}$	$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{9}$
	$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{3}$

Answer:

The answer is as follows: -

- (c)
– (d)
– (a)
– (e)

Question:43

Match the quantum numbers with the information provided by these.

Quantum Number	Information provided
Principal Quantum Number	Orientation of the orbital
Azimuthal Quantum Number	Energy and size of orbital
Magnetic Quantum Number	Spin of electron
Spin Quantum Number	Shape of the orbital

Answer:

– (b)
– (d)
– (a)
– (c)

Question:44

Match the following rules with their statements.

Rules	Statements
(i) Hund's Rule	(a) No two electrons in an atom can have the same set of four quantum numbers.
(ii) Aufbau Principle	(b) Half - filled and completely filled orbitals have extra stability.
(iii) Pauli Exclusion Principle	(c) The pairing of electrons in the orbitals belonging to the same subshell does not take place until each orbital is singly occupied.
(iv) Heisenberg's Uncertainty Principle	(d) It is impossible to determine the exact position and exact momentum of a subatomic particle simultaneously.
	(e) In the ground state of atoms, orbitals are filled in the order of their increasing energies.

Answer:

– (c)
– (e)
– (a)
– (d)

Question:45

Match the following :

Column I	Column II
X-rays	V = 10⁰ - 10⁴ Hz
UV	V = 10¹⁰ Hz
Long Radio Waves	V = 10¹⁶ Hz
Microwave	V = 10¹⁸ Hz

Answer:

(i). - (d)
(ii) - (c)
(iii) – (a)
(iv) – (b)

Question:46

Match the following:

Column I	Column II
Photon	(a) Value is 4 for N shell
Electron	(b) Probability density
$\psi ^{2}$	(c) Always positive value
Principal quantum number n	(d) Exhibits both momentum and wavelength

Answer:

1. - (d)
2. – (d)
3. – (b), (c)
4. – (a), (c)

Question:47

Match species given in Column I with the electronic configuration given in Column II.

Column I	Column II
(i) $Cr$	(a) $[Ar]3d^{8}4s^{0}$
(ii) $Fe^{2+}$	(b) $[Ar]3d^{10}4s^{1}$
(iii) $Ni^{2+}$	(c) $[Ar]3d^{6}4s^{0}$
(iv) $Cu$	(d) $[Ar]3d^{5}4s^{1}$
	(e) $[Ar]3d^{6}4s^{2}$

Answer:

(i) - (d)
(ii) – (c)
(iii) – (a)
(iv) – (b)

NCERT Exemplar Class 11 Chemistry Solutions Chapter 2: Assertion and Reason Type

Question:48

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): All isotopes of a given element show the same type of chemical behaviour.
Reason (R): The chemical properties of an atom are controlled by the number of electrons in the atom.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.

Answer:

The answer is the option (i) Both A and R are true and R is the correct explanation of A.
Explanation: All isotopes are having same atomic number that is the same number of electrons and they bear similar chemical properties.

Question:49

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Black body is an ideal body that emits and absorbs radiations of all frequencies.
Reason (R): The frequency of radiation emitted by a body goes from a lower frequency to higher frequency with an increase in temperature.

Answer:

The answer is the option (ii) Both A and R are true, but R is not the explanation of A.
Explanation: A black body is the ideal body, that can emit and absorb the radiations of all frequencies, while the radiation emitted by such a body is known as the black body radiation. The emitted radiation’s specific frequency distribution from a black body is dependent on its temperature. At any given temperature, there is an increase in the intensity of the radiation emitted when the wavelength reaches a maximum value at a given wavelength and then starts reducing with a further reduction of wavelength.

Question:50

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): It is impossible to determine the exact position and exact momentum of an electron simultaneously.
Reason (R): The path of an electron in an atom is clearly defined.

Answer:

The answer is the option (iii) A is true, and R is false.
Explanation: The effect of Heisenberg Uncertainty Principle is considerable specifically for the microscopic objects’ motion and is in fact negligible for the macroscopic objects.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 2: Long Answer Type

Question:51

What is photoelectric effect? State the result of photoelectric effect experiment that could not be explained on the basis of laws of classical physics. Explain this effect on the basis of quantum theory of electromagnetic radiations.
Answer:

Photoelectric effect states that when specific metals are exposed to a light beam, the metal ejects the electrons. This phenomenon is known as the photoelectric effect. The electrons ejected are called photoelectrons.
The results of this experiment were:
i) The electrons are ejected from the metal surface only as the beam of light strikes the surface.
ii) The number of electrons is proportional to the intensity of radiation.
iii) For each metal; there is a minimum frequency, below which photoelectric effect is not observed.
iv) K.E. of electrons is proportional to the frequency of light.
When a photon of adequate energy strikes an electron in the atom of the metal, it transfers its energy to the electron, and the electron is expelled without any delay. Greater is the energy of the photon, greater will also be the kinetic energy of the ejected electron.

Question:52

Threshold frequency, $v_{0}$ is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency $1.0\times 10^{15}\; s^{-1}$ was allowed to hit a metal surface, an electron having $1.988\times 10^{-19}J$ of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted if a photon with a wavelength equal to 600 nm hits the metal surface.
Answer:

As per the quantum theory, $hv=hv_{0}+\frac{1}{2}m\; v^{2}$
Therefore, $\frac{1}{2}m\; v^{2}=h(v-v_{0})$
Where h represents the plank's constant, i.e. $6.626\times 10^{-34}J\; S$
As per the information provided in the question, we must calculate $v_{0}$ which represents the threshold frequency of photons.
v = frequency of incident photons. As per the information provided, $v=1.0\times 10^{15}s^{-1}$
$\frac{1}{2}m\; v^{2}=$ electrons kinetic energy = $1.988\times 10^{-19}J$
Therefore,
$v_{0}=v-\left ( \frac{1}{2}m\; v^{2}/h \right )=1.0\times 10^{15}s^{-1}-\left ( 1.988\times 10^{-19}/6.626\times 10^{-34} \right )$
$= 1.0\times 10^{15} s^{-1} - 0.300030 \times 10^{15}$
$=6.9997\times 10^{14}s^{-1}$ (which is, in fact, the threshold frequency)
The relation between frequency and wavelength can be represented by the equation,
$v=c/\lambda ,$ , where λ represents the wavelength and c represents the speed of light, i.e. $3\times 10^{8}m/s$
Here, in this case of the threshold frequency $v_{0}=c/\lambda _{0}$
Thus, the photon's maximum wavelength can be represented as,
$\lambda _{0}=\frac{c}{v_{0}}=3\times 10^{8}m/s/6.9997\times 10^{14}s^{-1}$
$=4.36\times 10^{-7}m$
Now, we know that $600\; nm=6\times 10^{-7}m$ which is certainly greater than the calculated value of the $\lambda _{0}$ . Thus, the electron will not be emitted in case a photon which has a wavelength that is equal to 600 nm hits the surface of the metal.

Question:53

When an electric discharge is passed through hydrogen gas, the hydrogen molecules dissociate to produce excited hydrogen atoms. These excited atoms emit electromagnetic radiation of discrete frequencies which can be given by the general formula
$\bar{v}=109677\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}$
What points of Bohr's model of an atom can be used to arrive at this formula? Based on these points derive the above formula giving description of each step and each term.
Answer:

When through the hydrogen gas, an electric discharge is passed, the molecules produce the excited hydrogen atoms by disassociating. Electromagnetic radiation is emitted by these excited atoms of discrete frequencies. It can be represented by the following formula :
$\bar{v}=109677\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}$
2 basic postulates of Bohr's model of an atom can be used to reach this interpretation of the spectrum of atomic hydrogen:
1. The following principle is followed by the criterion for selecting the electron's stationary orbits, "The angular momentum of the electron should be an integral multiple of $h/2\pi (h= Planck's \; constant).$ "
If we assume that at an orbit of radius r, the mass and velocity of the electron is m and v, then its angular momentum can be given by the following formula: -
$mvr=n.h/2\pi ,$ where n is a non-zero positive integer.
2. In accordance with the quantum theory of radiation, when an electron changes its orbit from one orbit to another, there will be a difference of energy between the two energy levels which will be either absorbed or emitted.
Therefore, when from an orbit with energy E2, an electron jumps to an orbit of energy E1 (E2>E1), the difference in energy is given out in the form of quantized radiation. If we assume the frequency of the radiation to be ϑ, then the energy will be given by :
$E_{2}-E_{1}=h\vartheta$
With these two points, Rydberg's constant can be reached –
In a Bohr type system orbit, consider 2 electron orbits that have quantum numbers n1 and n2, in such a way that n2>n1. The corresponding energies are En1 and En2 and En1<En2.
Therefore, during the transfer of electrons from n2 to n1 orbit, the energy emitted will be: -
$E_{2}-E_{1}=h\vartheta$ , wherein, ϑ is the frequency of radiation.
As per the Bohr atomic model for the hydrogen atom (Z=1):
$E_{1}=\frac{-me^{4}}{8\epsilon _{0}^{2}h^{2}n_{1}^{2}},E_{2}=\frac{-me^{4}}{8\epsilon _{0}^{2}h^{2}n_{2}^{2}}$
Thus, $E_{2}-E_{1}=h\vartheta$ $=\frac{me^{4}}{8\epsilon _{0}^{2}h^{2}}\left [ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right ]$
$\vartheta =\frac{me^{4}}{8\epsilon _{0}^{2}h^{3}}\left [ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right ]$ and
$\bar{v} =\frac{\vartheta }{c}=\frac{me^{4}}{8\epsilon _{0}^{2}h^{3}}\left [ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right ]$
Now, on substituting the values form, e, $\epsilon _{0}$ and h
$\bar{v}=109677\frac{1}{n_{2}^{2}} cm^{-1}$

Question:54

Calculate the energy and frequency of the radiation emitted when an electron jumps from n= 3 to n= 2 in a hydrogen atom.
Answer:

When we use the Rydberg's Equation for hydrogen as well as an hydrogen like atom:

$\frac{1}{\lambda }=R_{H}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{f}^{2}} \right )$

Where,

$\lambda$ Represents the wavelength of radiation

$R_{H}$ Is the Rydberg's Constant i.e. $1.097\times 10^{7}m^{-1}$

$n_{f}$ = Higher energy level = 3

$n_{i}$ = Lower energy level = 2

$\frac{1}{\lambda } = 1.097\times 10^7 m^{-1}\times \left ( 1/2^2 -1/3^2\right )$

$\frac{1}{\lambda } = 1.097\times 10^7 m^{-1}\times \frac{5}{36}$

$\lambda = 6.7\times 10^{-7} m$

$\lambda = \frac{c}{\nu }$

where,

$\lambda$ represents the wavelength of the light i.e. $6.7\times 10^{-7}m$

c represents the speed of light i.e. $3\times 10^{8}m/s$

$\nu$ represents the frequency of light i.e. $\frac{c}{\lambda } = \frac{3\times 10^8 m/s}{6.7\times 10^{-7}m} = 0.4\times 10^{15} s^{-1}$

$E = \frac{6.6\times 10^{-34} \times 3\times 10^8}{6.7\times 10^{-7}m}$

Hence, the energy of radiation emitted is 3.028 x 10^-19J and the frequency is 0.4 x 10¹⁵ s^-1

Question:55

Why was a change in the Bohr Model of atom required? Due to which important development (s), concept of movement of an electron in an orbit was replaced by, concept of probability of finding electron in an orbital? What is the name given to the changed model of an atom?

Answer:

In view of the shortcoming of the Bohr's model, attempts were made to develop a more suitable and general model for atoms.
Two important developments which contributed significantly in the formulation of such a model were:

Dual behaviour of matter,
Heisenberg uncertainty principle.

Werner Heisenberg, a German physicist in 1927, stated the uncertainty principle, which is the consequence of duel behaviour of matter and radiation. One of the important implications of the Heisenberg Uncertainty Principle is that it rules out the existence of definite paths or trajectories of electrons and other similar particles.
Quantum mechanics is the theoretical science that deals with the study of the motions of the microscopic objects that have both observable wave-like and particle-like properties. The name of the changed model of the atom is the Quantum Mechanical Model of the atom.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 pdf download can also be used by the students to get in-depth knowledge of topics like quantum numbers and their combinations, writing electronic configurations. These topics are of value from the examination point of view.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 Main Subtopics

NCERT Exemplar Solutions for Class 11 Chemistry Chapter 2 Structure of Atom include the following topics:

2.1 Subatomic Particles

2.1.1 Discovery of Electron

2.1.2 Charge to Mass Ratio of Electron

2.1.3 Charge on Electron

2.1.4 Discovery of Protons and Neutrons

2.2 Atomic Models

2.2.1 Thomson Model of Atom

2.2.2 Rutherford’s Nuclear Model of Atom

2.2.3 Atomic Number and Mass Number

2.2.4 Isobars and Isotopes

2.2.5 Drawbacks of Rutherford’s Model

2.3 Development Leading to Bohr’s Model of Atom

2.3.1 Wave Nature of Electromagnetic Radiation

2.3.2 Planck’s Quantum Theory

2.3.3 Atomic Spectra: Evidence of Quantized Energy Levels

2.4 Bohr’s Model for Hydrogen Atom

2.4.1 Line Spectrum of Hydrogen

2.4.2 Drawbacks of Bohr’s Model

2.5 Towards Quantum Mechanical Model of Atom

2.5.1 Dual Behaviour of Matter

2.5.2 Heisenberg’s Uncertainty Principle

2.6 Quantum Mechanical Model of Atom

2.6.1 Orbitals and Quantum Numbers

2.6.2 Shapes of Atomic Orbitals

2.6.3 Energies of Orbitals

2.6.4 Filling of Orbitals in Atom

2.6.5 Electronic Configuration of atoms

2.6.6 Stability of filled and half-filled subshells

What will students learn from NCERT Exemplar Class 11 Chemistry Solutions Chapter 2?

Through the use of NCERT Exemplar Solutions Class 11 Chemistry Chapter 2 students will be able to understand the concepts of the Quantum Mechanical Model of Atom in which they’ll learn about the shapes and energies of atomic orbitals and how the filling of orbitals takes place in an atom.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 Structure of Atom provides detailed knowledge on topics including the Bohr’s Model for Hydrogen Atom and its drawbacks. The learners will be able to comprehend and solve numerical value questions regarding subatomic particles. The discovery of the subatomic particles and their properties are also discussed.

Chemistry Class 11 Chapter Wise Links

Chapter 1 Some Basic Concepts of Chemistry

Chapter 3 Classification of Elements and Periodicity in Properties

Chapter 4 Chemical Bonding and Molecular Structure

Chapter 5 States of Matter

Chapter 6 Thermodynamics

Chapter 7 Equilibrium

Chapter 8 Redox Reactions

Chapter 9 Hydrogen

Chapter 10 The s-Block Elements

Chapter 11 The p-Block Elements

Chapter 12 Organic Chemistry – Some Basic Principles and Techniques

Chapter 13 Hydrocarbons

Chapter 14 Environmental Chemistry

Important Topics to cover from NCERT Exemplar Class 11 Chemistry Solutions Chapter 2

Some of the important topics from the chapter that must be specifically learned by the students are:

Class 11 Chemistry NCERT Exemplar Solutions Chapter 2 talks about the subatomic particles and also about cathode and anode rays, their origin and properties. Apart from this one can learn about different models of atoms and their drawbacks.
The students will get to learn about electromagnetic radiations and spectrum. It’s properties and reasons for failure have also been discussed with the help of suitable graphs and diagrams. One can also learn about topics like black body radiation and the photoelectric effect.
Topics about the transition of electrons in the Hydrogen atom and its line spectrum that includes Lyman, Balmer, and Paschen series which have been discussed in the NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 and its related questions and solutions are included to offer a better understanding to the students.

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Yes, this chapter is important for competitive exams. One should go through its important topics and exercises given in NCERT Exemplar solutions for class 11 Chemistry Chapter 2.

Question: Are there numerically based exercises too?

Answer:

Yes, there are many numericals given in NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 that can be used by the students to solve and evaluate themselves.

Question: What are some of the important topics in the chapter?

Answer:

The important topics include quantum numbers, atomic models.

Question: Can the solutions of NCERT be downloaded for future reference?

Answer:

Yes, one can easily download Class 11 Chemistry NCERT exemplar solutions chapter 2.

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 Structure of Atom

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NCERT Exemplar Class 11 Solutions

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