NCERT Exemplar Class 11 Chemistry Solutions Chapter 2 Structure of Atom

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 more 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

A 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 an 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 subshells 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. The lower the value of (n+l), lower is the energy. For orbitals having the same values of (n+l), the orbital with the 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:

Neutrons 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 subshells differ slightly in their energies, an electron shifts from a subshell of lower energy (4s) to a sub-shell of 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 a 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
${}^{-}v=109677(\frac{1}{2^2}-\frac{1}{4^2})$
for n₁ = 2, n₂ = 4 transition in Balmer series.
$=109667(\frac{1}{4}-\frac{1}{16})$
$=20564.44c{m}^{-1}$

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 too small 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:

Electrons 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,
$C=3\times {10}^{8}m{s}^{-1}$
$V=4.620\times {10}^{14}Hz.$
$\lambda =3\times {10}^{8}m{s}^{-1}/4.620\times {10}^{14}Hz$
$=0.6494\times {10}^{-6}m$
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:

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

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
$\mathrm{\Delta }v.\mathrm{\Delta }x=h/4\pi m$
$=\mathrm{\Delta }v.\mathrm{\Delta }x=h/4\times 3.14\times {10}^{-6}$
$=6.626\times {10}^{-34}/4\times 3.14\times {10}^{-6}$
$=0.52\times {10}^{-28}{m}^2{s}^{-1}$
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 not just on the principal quantum number (shell) as well as also on the azimuthal quantum number (subshell), which is quite unlike that of a 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.

Answer:

The answer is as follows: -

1. - (c)

2. – (d)

3. – (a)

4. – (e)

Question:43

Match the quantum numbers with the information provided by these.

Answer:

1. – (b)

2. – (d)

3. – (a)

4. – (c)

Question:44

Match the following rules with their statements.

Answer:

1. – (c)

2. – (e)

3. – (a)

4. – (d)

Question:45

Match the following :

Answer:

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

Question:46

Match the following:

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.

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): The black body is an ideal body that emits and absorbs radiation of all frequencies.
Reason (R): The frequency of radiation emitted by a body goes from a lower frequency to a higher frequency with an increase in temperature.

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the explanation of A.

3. A is true and R is false.

4. Both A and R are false.

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 radiation 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.

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are tru,e and R is not the correct explanation of A.

3. A is true and R is false.

4. Both A and R are false.

Answer:

The answer is option (iii) A is true, and R is false.
Explanation: The effect of the 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 the photoelectric effect? State the result of the photoelectric effect experiment that could not be explained on the basis of the laws of classical physics. Explain this effect on the basis of the quantum theory of electromagnetic radiation.
Answer:

Photoelectric effect states that when specific metals are exposed to a light beam, the metal ejects 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 the 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. The greater is the energy of the photon, the greater will also be the kinetic energy of the ejected electron.

Question:52

Threshold frequency, $v_0$ is the minimum frequency that 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, $hv=h{v}_0+\frac{1}{2}m{v}^2$
Therefore, $\frac{1}{2}m{v}^2=h(v-v_0)$
Where h represents the plank's constant, i.e. $6.626\times {10}^{-34}JS$
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,
$v_0=v-\left(\frac{1}{2}m{v}^2/h\right)=1.0\times {10}^{15}{s}^{-1}-(1.988\times {10}^{-19}/6.626\times {10}^{-34})$
$=1.0\times {10}^{15}{s}^{-1}-0.300030\times {10}^{15}$
$=6.9997\times {10}^{14}{s}^{-1}$ (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 $600nm=6\times {10}^{-7}m$ 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
$\overline{v}=109677\frac{1}{n_i^2}-\frac{1}{n_f^2}$
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 a description of each step and each term.
Answer:

When the hydrogen gas, an electric discharge, the molecules produce the excited hydrogen atoms by dissociating. Electromagnetic radiation is emitted by these excited atoms of discrete frequencies. It can be represented by the following formula :
$\overline{v}=109677\frac{1}{n_i^2}-\frac{1}{n_f^2}$
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$h/2\pi (h=Planc{k}^{\prime }sconstant).$"
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):
$E_1=\frac{-m{e}^4}{8{ϵ}_0^2{h}^2{n}_1^2},E_2=\frac{-m{e}^4}{8{ϵ}_0^2{h}^2{n}_2^2}$
Thus, $E_2-E_1=h\vartheta$$=\frac{m{e}^4}{8{ϵ}_0^2{h}^2}[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$\vartheta =\frac{m{e}^4}{8{ϵ}_0^2{h}^3}[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$ and
$\overline{v}=\frac{\vartheta }{c}=\frac{m{e}^4}{8{ϵ}_0^2{h}^3}[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
Now, on substituting the values form, e, ${ϵ}_0$ and h
$\overline{v}=109677\frac{1}{n_2^2}c{m}^{-1}$

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 Rydberg's Equation for hydrogen as well as an hydrogen-like atom:

$\frac{1}{\lambda }=R_H(\frac{1}{n_1^2}-\frac{1}{n_f^2})$

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

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

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

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

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

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

1. Dual behaviour of matter,

2. 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 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.