CBSE Class 11 Chemistry Structure Of Atom Worksheet Set 01

Structure Of Atom MCQ Questions with Answers Class 11 Chemistry

Question- Which of the following does not contain number of neutrons equal to that of 40Ar 18 ?

(a) 41K
     19

(b) 43
     21Sc

(c) 40Sc
     21

(d) Ca 42
      20

Answer-(c)

Question-The radius of 1st Bohr’s orbit for hydrogen atom is ‘r’. The radius of second Bohr’s orbit is

(a) 4r

(b) r³

(c) 4r²

(d) r^1/3

Answer-(a)

Question-The number of electrons and neutrons of an element is 18 and 20 respectively. Its mass number is

(a) 2

(b) 17

(c) 37

(d) 38

Answer-(d)

Question- The ideal body, which emits and absorbs radiations of all frequencies, is called a black body and the radiation emitted by such a body is called

(a) white body radiation

(b) black body radiation

(c) black body emission

(d) None of these

Answer-(b)

Question-Nucleons are

(a) only neutrons

(b) neutrons + protons

(c) neutrons + protons + electrons

(d) neutrons + electrons

Answer-(b)

Question-Number of protons, neutrons and electrons in the element 89^{X 231} is

(a) 89, 89, 242

(b) 89, 142, 89

(c) 89, 71, 89

(d) 89, 231, 89

Answer-(b)

Question-Atoms with same mass number but different atomic numbers are called

(a) isotopes

(b) isobars

(c) isochores

(d) None of these

Answer-(b)

Question-An electron from one Bohr stationary orbit can go to next higher orbit

(a) by emission of electromagnetic radiation

(b) by absorption of any electromagnetic radiation

(c) by absorption of electromagnetic radiation of particular frequency

(d) without emission or absorption of electromagnetic radiation

Answer-(c)

Question-An element has atomic number 11 and mass number 24.What does the nucleus contain?

(a) 11 protons, 13 neutrons

(b) 11 protons, 13 neutrons, 13 electrons

(c) 13 protons, 11 neutrons

(d) 13 protons, 11 electrons

Answer-(a)

Question-What is the difference between two species if one has atomic mass = 14 and atomic number = 7 whereas the other has atomic mass = 14 and atomic number = 6 ?

(a) Neutrons

(b) Protons

(c) Electrons

(d) All of these

Answer-(d)

Question- When a metal surface is exposed to solar radiations

(a) The emitted electrons have energy less than a maximum value of energy depending upon frequency of incident radiations

(b) The emitted electrons have energy less than maximum value of energy depending upon intensity of incident radiation

(c) The emitted electrons have zero energy

(d) The emitted electrons have energy equal to energy of photons of incident light

Answer-(a)

Question-If the wavelength of the electromagnetic radiation is increased to thrice the digital value, then what will be the percent change in the value of frequency of the electromagnetic radiation.

(a) Increases by 33%

(b) Decreases by 33%

(c) Increases by 66%

(d) Decreases by 66%

Answer-(d)

Question- Which one of the following pairs is not correctly matched ?

(a) Rutherford-Proton

(b) J.J. Thomson-Electron

(c) J.H. Chadwick-Neutron

(d) Bohr-Isotopes

Answer-(d)

Question- From the data given below A, B, C and D respectively are,
(A) 10 e^–, atomic no. 11 (B) 10 e^–, atomic no. 6
(C) 10 e^–, atomic no. 10 (D) 10 e^–, atomic no. 9

(a) Na⁺, C^4–, Ne, F^–

(b) C^4–, Ne, Na^–, F^–

(c) F^–, Na⁺, Ne, C^4–

(d) F–, Na⁺, C^4–, Ne

Answer-(a)

Question-Which of the following is related with both wave nature and particle nature ?

(a) Interference

(b) E = mc²

(c) Diffraction

(d) E = hv

Answer-(d)

Question- ‘A’ represents mass no. and Z represents atomic no. then α - decay is characterized by

(a) Z increases by 2, A decreases by 4

(b) Z decreases by 2, A increases by 4

(c) Z decreases by 2, A decreases by 4

(d) Z increases by 2, A increases by 4.

Answer-(c)

Question- In the photo-electron emission, the energy of the emitted electron is

(a) greater than the incident photon

(b) same as than of the incident photon

(c) smaller than the incident photon

(d) proportional to the intensity of incident photon

Answer-(c)

Question- In Bohr’s model, atomic radius of the first orbit is y, the radius of the 3rd orbit, is

(a) y / 3

(b) y

(c) 3y

(d) 9y

Answer-(d)

Question-Which of the following statements do not form a part of Bohr’s model of hydrogen atom ?

(a) Energy of the electrons in the orbits are quantized

(b) The electron(s) in the orbit nearest to the nucleus has the lowest energy

(c) Electrons revolve in different orbits around the nucleus

(d) The position and velocity of the electrons in the orbit cannot be determined simultaneously

Answer-(d)

Question- Which one of the following is not the characteristic of Planck’s quantum theory of radiation ?

(a) The energy is not absorbed or emitted in whole number or multiple of quantum

(b) Radiation is associated with energy

(c) Radiation energy is not emitted or absorbed continuously but in the form of small packets called quanta

(d) This magnitude of energy associated with a quantum is proportional to the frequency.

Answer-(a)

Question- The third line of the Balmer series, in the emission spectrum of the hydrogen atom, is due to the transition from the

(a) fourth Bohr orbit to the first Bohr orbit

(b) fifth Bohr orbit to the second Bohr orbit

(c) sixth Bohr orbit to the third Bohr orbit

(d) seventh Bohr orbit to the third Bohr orbit

Answer-(b)

Q1 - What were the discrepancies observed in Bohr's model?
Answer: The major discrepancies and limitations observed in Bohr's model of the atom are as follows:
1. It failed to account for the finer details (doublet structure) of the hydrogen atom spectrum observed using spectroscopic techniques.
2. It was unable to explain the splitting of spectral lines in the presence of a magnetic field (Zeeman effect) or an electric field (Stark effect).
3. It could not explain the spectra of multi-electron atoms (atoms containing more than one electron).
4. It contradicted Heisenberg's Uncertainty Principle by considering electrons to occupy well-defined circular orbits with simultaneously fixed positions and momenta.
5. It failed to explain the ability of atoms to form molecules via chemical bonds (shape and geometry of molecules).
In simple words: Bohr's model only worked well for single-electron systems like hydrogen and could not explain advanced spectral phenomena or wave-particle duality.

Exam Tip: Mention the Zeeman effect, Stark effect, and failure for multi-electron species to secure full credit.

Q2 - How many protons and neutron are in the following nuclei?
\({}_{15}^{31}\text{P},\) {}_{17}^{37}\text{Cl},\ {}_{20}^{40}\text{Ca} \)
Answer: The atomic number (\(Z\)) gives the number of protons, and the mass number (\(A\)) minus the atomic number gives the number of neutrons (\(A - Z\)).
1. For \({}_{15}^{31}\text{P}\):
Number of protons = \(Z = 15\)
Number of neutrons = \(A - Z = 31 - 15 = 16\)
2. For \({}_{17}^{37}\text{Cl}\):
Number of protons = \(Z = 17\)
Number of neutrons = \(A - Z = 37 - 17 = 20\)
3. For \({}_{20}^{40}\text{Ca}\):
Number of protons = \(Z = 20\)
Number of neutrons = \(A - Z = 40 - 20 = 20\)
In simple words: Protons are equal to the lower subscript (atomic number), and neutrons are the difference between the upper superscript (mass number) and the lower subscript.

Exam Tip: Always double-check subtraction for mass number minus atomic number to avoid silly calculation errors.

Q3 - Which of the following orbitals are not possible -
2d, 4f, 6d, 3g
Answer: The orbitals that are not possible are 2d and 3g.
Reasoning:
- For a given principal quantum number \(n\), the azimuthal quantum number \(l\) can take values from \(0\) to \(n - 1\).
- For \(n = 2\), the possible values of \(l\) are \(0\) (\(s\)-orbital) and \(1\) (\(p\)-orbital). A \(d\)-orbital corresponds to \(l = 2\), which is not possible when \(n = 2\). Therefore, 2d is not possible.
- For \(n = 3\), the possible values of \(l\) are \(0, 1, 2\) (\(s, p, d\)). A \(g\)-orbital corresponds to \(l = 4\) (\(s=0, p=1, d=2, f=3, g=4\)), which requires at least \(n = 5\). Therefore, 3g is not possible.
In simple words: Azimuthal quantum number l must always be strictly less than principal quantum number n (\(l < n\)).

Q4 - If the largest value of m for an electron is +2, then the electron may be present in what type of sub shell?
Answer: The electron may be present in a d-subshell (or any higher subshell like f, g, etc., though d is the lowest subshell where \(m = +2\) is possible).
Reasoning:
- The magnetic quantum number \(m\) ranges from \(-l\) to \(+l\) through zero.
- If the maximum value of \(m\) is \(+2\), then the azimuthal quantum number \(l\) must be at least \(2\).
- An azimuthal quantum number of \(l = 2\) corresponds to a \(d\)-subshell.
In simple words: Since the maximum magnetic quantum number is +2, the orbital angular momentum quantum number must be 2, which defines a d-subshell.

Exam Tip: Maximum value of m equals +l for any given subshell.

Q5 - Point out the differences and similarities in the orbitals represented by following sets of quantum numbers-
(i) \(n = 3,\) l = 2,\ m = +2 \)
(ii) \(n = 3,\) l = 2,\ m = -1 \)
Answer:
Similarities:
1. Both orbitals belong to the same principal shell because both have \(n = 3\).
2. Both orbitals belong to the same subshell because both have \(l = 2\) (\(3d\) subshell), meaning they have the same shape and energy (in the absence of a magnetic field).
Differences:
1. They have different magnetic quantum numbers (\(m = +2\) for the first and \(m = -1\) for the second), which means they differ in their spatial orientation in space.
In simple words: Both are 3d orbitals (same size, shape, and energy), but they point in different directions in space due to different magnetic quantum numbers.

Exam Tip: n determines size/energy, l determines shape/subshell, and m determines spatial orientation.

Q6 - The work function of a metal is 4.2 eV. If radiations of 2000 A0 fall on the metal, then find the kinetic energy of fastest photon electron.
Answer:
Given:
Work function (\(W\) or \(h\nu_0\)) = \(4.2\text{ eV}\)
Wavelength of incident radiation (\(\lambda\)) = \(2000\) \AA = 2000 \times 10^{-10}\text{ m} = 2 \times 10^{-7}\text{ m} \)
Planck's constant (\(h\)) = \(6.626 \times 10^{-34}\text{ J s}\)
Speed of light (\(c\)) = \(3 \times 10^8\text{ m/s}\)

Energy of incident photon (\(E\)) is calculated as:
\(E = \frac{hc}{\lambda}\)
\(E = \frac{6.626 \times 10^{-34}\text{ J s} \times 3 \times 10^8\text{ m/s}}{2 \times 10^{-7}\text{ m}}\)
\(E = \frac{1.9878 \times 10^{-25}}{2 \times 10^{-7}} = 9.939 \times 10^{-19}\text{ J}\)

Convert energy from Joules to eV by dividing by \(1.6 \times 10^{-19}\text{ J/eV}\):
\(E_{\text{eV}} = \frac{9.939 \times 10^{-19}}{1.6 \times 10^{-19}} = 6.21\text{ eV}\)

According to Einstein's photoelectric equation, the kinetic energy (\(K.E.\)) of the fastest photoelectron is:
\(K.E. = E - W\)
\(K.E. = 6.21\text{ eV} - 4.2\text{ eV} = 2.01\text{ eV}\)

In simple words: Subtract the work function energy from the total energy of the incident photon to find the kinetic energy of the emitted electron.

Exam Tip: Always ensure energy units are consistent (either both in Joules or both in eV) before subtracting.

Q7 - What is the value of Plank's constant in S.I. Units?
Answer: The value of Planck's constant (\(h\)) in S.I. units (\(\text{J s}\) or \(\text{kg m}^2\text{ s}^{-1}\)) is:
\(h = 6.626 \times 10^{-34}\text{ J s}\) (or \(6.626 \times 10^{-34}\text{ kg m}^2\text{ s}^{-1}\))
In simple words: Planck's constant is approximately \(6.626 \times 10^{-34}\) Joule-seconds.

Exam Tip: Memorize this constant with correct units as it is frequently used in atomic structure and quantum chemistry numericals.

Q8 - When a ball is hit with a hockey stick by a player, it does not make a wave. Why?
Answer: According to de Broglie's wavelength equation (\(\lambda = \frac{h}{mv}\)), the wavelength associated with any moving particle is inversely proportional to its mass (\(m\)). A macroscopic object like a cricket or hockey ball has a very large mass compared to subatomic particles like electrons. Consequently, the associated de Broglie wavelength is extremely small and beyond any experimental measurement or observable macroscopic scale. Therefore, wave characteristics are not observed for macroscopic bodies.
In simple words: Because macroscopic objects have large masses, their de Broglie wavelengths are infinitesimally small and undetectable.

Exam Tip: Mention de Broglie's relation \(\lambda = \frac{h}{mv}\) and state that mass is inversely proportional to wavelength.

Q9 - Which series are produced when electrons from the outer orbits jumps to 3rd orbit?
Answer: The spectral series produced when electrons from outer orbits jump to the 3rd orbit (\(n_1 = 3\) from \(n_2 = 4, 5, 6, \dots\)) is known as the Paschen series.
In simple words: Electron transitions ending at the third energy level produce the Paschen series in the infrared region.

Exam Tip: Recall the hydrogen spectral series order: Lyman (1), Balmer (2), Paschen (3), Brackett (4), Pfund (5).