Atomic Structure -

1. Two elements X and Y have the following electronic configurations:
X: 1s² 2s²2p⁶ 3s² 3p¹
Y: 1s² 2s² 2p⁴
Which compound are these elements most likely to form?
A. X₂Y
B. X₃Y₂
C. X₂Y₃
D. XY₃

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Correct Option: C.
Option A Explained
One X atoms loses three electrons, so two X atoms six electrons. One Y accepts two eelctrons, so the number of electrons lost by two X atoms is not equal to the number of electrons gained by one Y atom. This means that the formula X₂Y is wrong.
Option B Explained
Thsi formula is wrong because three X atoms lose nine electrons while two Y atoms accepts four electrons. The electrons lost are not equal to the electrons gained.
Option C Explained
This is correct formula because two X atoms loses six electrons while three Y atoms accepts six electrons. The number of electrons lost are equal to the number of electrons gained so the ration 2:3 is correct.
Option D Explained
One X loses three electrons while three Y require six electrons, so the number of electrons lost and gained are not equal.

2. Which particles are directly involved in ordinary chemical reactions?
A. Protons
B. Neutrons
C. Electrons
D. Electron, protons, and neutrons

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Correct Option: C
Option A Explained: Incorrect. Protons remain inside the nucleus and do not participate in chemical reactions.
Option B Explained: Incorrect. Neutrons also remain inside the nucleus and take no part in reactions.
Option C Explained: Correct. Chemical reactions involve the gain, loss, or sharing of electrons only.
Option D Explained: Incorrect. Only electrons, not protons or neutrons, participate in ordinary chemical reactions.

3. In cathode-ray experiments, the same type of negatively charged particle was obtained regardless of the gas used in the discharge tube or the material of the electrodes. What is the most important conclusion from this observation?
A. Electrons are present only in gases
B. Electrons are fundamental constituents of all atoms
C. Electrons are produced when a high voltage is applied
D. Different elements contain different types of electrons

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Correct Option: B
Option A Explained: Incorrect. Electrons were obtained from different gases and electrode materials, showing that they are not limited to gases.
Option B Explained: Correct. The same electron was obtained from different substances, showing that electrons are common or univeral particle of all atoms. Ther is no atom in this universe having no electrons.
Option C Explained: Incorrect. High voltage helps remove and accelerate electrons; it does not create electrons.
Option D Explained: Incorrect. Electrons have the same basic properties regardless of the element from which they originate.

4. The charge of the electron was accurately measured using the oil-drop experiment by:
A. J. J. Thomson
B. Ernest Rutherford
C. Robert A. Millikan
D. Wilhelm C. Roentgen

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Correct Option: C
Option A Explained: Incorrect. Thomson discovered the electron and measured its charge-to-mass ratio, not its absolute charge.
Option B Explained: Incorrect. Rutherford is known for discovering the atomic nucleus.
Option C Explained: Correct. Millikan determined the fundamental charge of the electron using the oil-drop experiment.
Option D Explained: Incorrect. Roentgen is credited with the discovery of X-rays.

5. Which of the following electromagnetic radiations has the longest wavelength?
A. Infrared rays
B. Ultraviolet rays
C. Gamma rays
D. X-rays

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Correct Option: A
Option A Explained: Correct. Infrared rays have longer wavelengths than ultraviolet rays, X-rays, and gamma rays.
Option B Explained: Incorrect. Ultraviolet rays have shorter wavelengths than infrared and visible light.
Option C Explained: Incorrect. Gamma rays have the shortest wavelengths of all electromagnetic radiation.
Option D Explained: Incorrect. X-rays have shorter wavelengths than both ultraviolet and infrared radiation.

6. A continuous absorption spectrum is produced mainly by:
A. Excited atoms
B. Excited molecules
C. Ground-state molecules
D. Ground-state atoms

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Correct Option: C
Option A Explained: Incorrect. Excited atoms usually produce line emission spectra, not continuous absorption spectra.
Option B Explained: Incorrect. Excited molecules mainly emit spectra rather than absorb continuously.
Option C Explained: Correct. Ground-state molecules absorb a broad, continuous range of energies due to their rotational and vibrational transitions.
Option D Explained: Incorrect. Ground-state atoms generally produce line absorption spectra, not continuous ones.

7. Who proposed the equation describing the energy emitted when an electron moves from a higher energy level (n₂) to a lower energy level (n₁)?
A. Albert Einstein
B. Niels Bohr
C. Ernest Rutherford
D. Werner Heisenberg

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Correct Option: B
Option A Explained: Incorrect. Einstein explained the photoelectric effect but did not propose this atomic energy equation.
Option B Explained: Correct. Bohr introduced quantized energy levels and the relation ΔE = hν for electron transitions. He added that each electron has a specific amount (quantized) of energy in a given shell.
Option C Explained: Incorrect. Rutherford proposed the nuclear model of the atom, not this energy equation.
Option D Explained: Incorrect. Heisenberg is known for the uncertainty principle.

8.Which statement about the hydrogen atom according to Bohr’s model is correct?
A. The radius of each orbit is an integral multiple of the Bohr radius (0.053 nm)
B. The angular momentum of an electron in the nth orbit is n × h/2π
C. The energy of the nth orbit is n times the ground-state energy
D. Electron continuously emit energy during its revolution around the nucleus

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Correct Option: B
Option A Explained: Incorrect. The radius is proportional to n², not simply an integral multiple of n.
Option B Explained: Correct. Bohr’s quantization rule states that angular momentum of electron equals nh/2π. It measn that each electron in a given shell has a specific amount of energy.
Option C Explained: Incorrect. The energy of an orbit is proportional to −1/n², not directly proportional to n.
Option D Explained: Incorrect. This was the postulate of Rutherford atomic model, not Bohr.

9. How does the energy difference between successive energy levels in a hydrogen atom change as the principal quantum number increases?
A. It increases
B. It decreases
C. It first increases and then decreases
D. It first decreases and then increases

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Correct Option: B
Option A Explained: Incorrect. Higher energy levels become closer together, not farther apart.
Option B Explained: Correct. The spacing between energy levels decreases as the electron moves farther from the nucleus. The reason is that the energy of electrons increases as we move away from nucleus, but the increase becomes lesser.
Option C Explained: Incorrect. There is no increase followed by a decrease in this spacing.
Option D Explained: Incorrect. This trend does not occur in the hydrogen atom.

10. Which quantum number tells us about orientation of orbitals?
A. Principal quantum number
B. Spin quantum number
C. Azimuthal quantum number
D. Magnetic quantum number

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Correct Option: D
Option A Explained: Incorrect. The principal quantum number (n) tells us about the main energy level or shell of an electron and how far it is from the nucleus. It does not describe which direction in space an orbital points.
Option B Explained: Incorrect. The spin quantum number tells us the spin direction of an electron within an orbital, either clockwise or anticlockwise. It has nothing to do with orbital orientation in space.
Option C Explained: Incorrect. The azimuthal quantum number tells us about the shape of an orbital, such as spherical for s or dumbbell for p. It does not tell us its direction in space.
Option D Explained: Correct. The magnetic quantum number tells us how an orbital is oriented in three-dimensional space relative to the x, y, and z axes. For example, it distinguishes px, py, and pz orbitals from each other.

11. The relative energies of 4s, 4p and 3d orbitals are in the order:
A. 3d < 4p < 4s
B. 4p < 4s < 3d
C. 4s < 3d < 4p
D. 4p < 3d < 4s

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Correct Option: C
Option A Explained: Incorrect. This places 3d as the lowest energy orbital, but 4s actually has lower energy than 3d and is filled before it according to the Aufbau principle.
Option B Explained: Incorrect. This places 4p as the lowest, but 4p is filled after both 4s and 3d, so it cannot have the lowest energy among these three.
Option C Explained: Correct. Following the (n+l) rule and the Aufbau principle, 4s fills first, then 3d, and then 4p, so the increasing order of energy is 4s < 3d < 4p.
Option D Explained: Incorrect. This order places 4s as the highest energy orbital, but 4s is actually filled before both 3d and 4p because it has the lowest energy of the three.

12. With increase in the value of principal quantum number “n”, the shape of the s-orbitals remains the same although their sizes:
A. Decrease
B. Remain the same
C. Increase
D. May or may not remain the same

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Correct Option: C
Option A Explained: Incorrect. As the principal quantum number increases, the electron shell gets farther from the nucleus, so the orbital becomes bigger, not smaller.
Option B Explained: Incorrect. The size of an s-orbital is directly related to its principal quantum number, so it cannot stay the same when n changes.
Option C Explained: Correct. A higher principal quantum number means the electron is, on average, farther from the nucleus, so the s-orbital becomes larger in size while keeping its spherical shape.
Option D Explained: Incorrect. The size change with increasing n is a definite and predictable trend, not something uncertain or variable.

13. Number of electrons in the outermost shell of chloride ion (Cl⁻) is:
A. 17
B. 1
C. 7
D. 8

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Correct Option: D
Option A Explained: Incorrect. 17 is the atomic number of chlorine, which equals the number of protons and also the number of electrons in a neutral chlorine atom, not the number of electrons in the outer shell of the ion.
Option B Explained: Incorrect. Chlorine does not have only 1 electron in its outer shell; this number is far too small to represent the outermost shell of any form of chlorine.
Option C Explained: Incorrect. A neutral chlorine atom has 7 electrons in its outermost shell, but the chloride ion has gained one extra electron, so this number no longer applies to Cl⁻.
Option D Explained: Correct. Chloride ion (Cl⁻) has 18 electrons in total, with the configuration 2, 8, 8. This means its outermost shell holds 8 electrons, giving it a stable noble gas-like configuration.

14. Correct order of energy in the given sub-shells is:
A. 5s > 3d > 3p > 4s
B. 3p > 3d > 5s > 4s
C. 5s > 3d > 4s > 3p
D. 3p > 3d > 4s > 5s

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Correct Option: C
Option A Explained: Incorrect. This order wrongly places 3p below 4s, but 3p actually has lower energy than 4s, so 3p should be the lowest of these four, not placed above 4s.
Option B Explained: Incorrect. This order wrongly shows 3p as having the highest energy, but 3p actually has the lowest energy among these four sub-shells because of its smaller principal quantum number.
Option C Explained: Correct. Using the (n+l) rule, the increasing order of energy is 3p < 4s < 3d < 5s, so from highest to lowest energy the order is 5s > 3d > 4s > 3p, which matches this option exactly.
Option D Explained: Incorrect. This order places 5s as having the lowest energy, but 5s actually has the highest energy among these four sub-shells because of its larger principal quantum number.

15. There are four orbitals s, p, d and f. Which increasing order is correct with respect to the increasing energy of the orbitals?
A. 4s < 4p < 4d < 4f
B. 4s < 4d < 4f < 4p
C. 4p < 4s < 4f < 4d
D. 4f < 4s < 4d < 4p

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Correct Option: A
Option A Explained: Correct. For orbitals with the same principal quantum number (here n = 4), energy increases as the azimuthal quantum number increases, so the correct increasing order is s < p < d < f.
Option B Explained: Incorrect. This order incorrectly places 4p after 4d and 4f, but 4p actually has lower energy than both 4d and 4f within the same shell.
Option C Explained: Incorrect. This order wrongly starts with 4p as the lowest energy orbital, but 4s always has the lowest energy among orbitals of the same shell.
Option D Explained: Incorrect. This order wrongly places 4f as the lowest energy orbital, but 4f actually has the highest energy among s, p, d and f orbitals of the same shell.

16. The maximum number of electrons in electronic configuration can be calculated by using the formula:
A. 2l+1
B. 2n²
C. 2n²+2
D. 2n²+1

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Correct Option: B
Option A Explained: Incorrect. The expression 2l+1 gives the number of possible orbitals (orientations) within a sub-shell, not the maximum number of electrons in a whole shell.
Option B Explained: Correct. The maximum number of electrons that a principal shell with quantum number n can hold is given by the formula 2n². For example, the second shell (n=2) can hold 2(2)²=8 electrons.
Option C Explained: Incorrect. Adding an extra 2 to the formula gives a wrong, larger number of electrons than a shell can actually hold.
Option D Explained: Incorrect. Adding an extra 1 to 2n² does not match the actual maximum electron capacity of any real electron shell.

17. Number of electrons in ₃₁Ga³⁺ will be:
A. 28
B. 30
C. 29
D. 34

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Correct Option: A
Option A Explained: Correct. Gallium has atomic number 31, meaning a neutral gallium atom has 31 electrons. Since Ga³⁺ has lost 3 electrons, it is left with 31 − 3 = 28 electrons.
Option B Explained: Incorrect. 30 electrons would mean only 1 electron was lost, but the 3+ charge on the ion means 3 electrons must have been removed.
Option C Explained: Incorrect. 29 electrons would mean only 2 electrons were removed, which does not match the 3+ charge shown on the ion.
Option D Explained: Incorrect. 34 electrons is even more than the neutral atom itself has, which is impossible for a positively charged ion that has lost electrons.

18. Identify the correct shape of a p-orbital.
A. A single sphere with no lobes
B. Four petal-shaped lobes arranged like a four-leaf clover
C. Two lobes of equal size placed on opposite sides of the nucleus along one straight axis (a dumbbell shape)
D. Two lobes along one axis with an extra ring (torus) around the middle

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Correct Option: C
Option A Explained: Incorrect. A single sphere with no lobes describes the shape of an s-orbital, not a p-orbital.
Option B Explained: Incorrect. Four petal-shaped lobes arranged like a clover describe the shape of most d-orbitals, not a p-orbital.
Option C Explained: Correct. A p-orbital has a dumbbell shape made of two equal lobes on either side of the nucleus, arranged along one of the three axes (x, y, or z).
Option D Explained: Incorrect. Two lobes with an extra ring around the middle describe the shape of the dz² orbital, which belongs to the d-orbital family, not a p-orbital.

19. Which is the correct electronic configuration of chromium (₂₄Cr)?
A. 1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d⁴
B. 1s²,2s²,2p⁶,3s²,3p⁶,4s¹,3d⁵
C. 1s²,2s²,2p⁶,3s²,3p⁶,3d⁶
D. 1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d⁶

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Correct Option: B
Option A Explained: Incorrect. This configuration follows the expected simple filling pattern but does not match chromium’s actual electron arrangement, which is a well-known exception to the usual rule.
Option B Explained: Correct. Chromium is a well-known exception to the normal filling order. One electron moves from 4s into 3d so that the 3d sub-shell becomes half-filled (3d⁵), which is more stable, giving the configuration 4s¹3d⁵.
Option C Explained: Incorrect. This configuration has too few electrons in total and completely misses the 4s electron that chromium actually has.
Option D Explained: Incorrect. This configuration adds up to more electrons than chromium actually has and does not reflect its real half-filled stability exception.

20. Maximum number of electrons in a sub-shell is given by:
A. 2(2l-1)
B. 2(2l+1)
C. 2(l+1)
D. 2l+1

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Correct Option: C.
Correct Option: B
Option A Explained: Incorrect. Using minus instead of plus inside the bracket gives a smaller and incorrect number of electrons for any sub-shell.
Option B Explained: Correct. The correct formula for the maximum number of electrons in any sub-shell is 2(2l+1), where l is the azimuthal quantum number of that sub-shell.
Option C Explained: Incorrect. This formula does not correctly account for the actual number of orbitals in a sub-shell, so it gives wrong electron counts.
Option D Explained: Incorrect. The expression 2l+1 only gives the number of orbitals in a sub-shell, not the number of electrons; it must still be multiplied by 2.

21. The order of energy level in ₁₉K is:
A. 4s, 4p
B. 4s, 3d
C. 3p, 4s
D. 3s, 3d

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Correct Option: B
Option A Explained: Incorrect. Potassium’s last electron goes into the 4s orbital, not 4p, so 4p is not relevant to describing its energy level order here.
Option B Explained: Correct. In potassium, the last electron enters the 4s orbital instead of 3d because 4s has lower energy than 3d at this point, showing that 4s is filled before 3d.
Option C Explained: Incorrect. While 3p is filled before 4s, this option does not address the actual comparison between 4s and 3d energies in potassium.
Option D Explained: Incorrect. Potassium’s configuration does not involve filling of 3d before 3s in any way relevant here, so this pairing does not describe its energy order.

22. Which of the following elements in its atomic state has electrons fully occupying the first two spherically symmetrical orbitals?
A. Oxygen
B. Helium
C. Beryllium
D. Carbon

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Correct Option: C
Option A Explained: Incorrect. Oxygen’s electrons fill beyond just the first two s-orbitals; it also has electrons in the 2p orbitals, which are not spherically symmetrical.
Option B Explained: Incorrect. Helium only completely fills the very first spherically symmetrical orbital (1s); it does not have enough electrons to reach the second one (2s).
Option C Explained: Correct. Beryllium has the configuration 1s²2s², meaning both of its first two spherically symmetrical (s-type) orbitals are exactly and completely filled with no electrons left over.
Option D Explained: Incorrect. Carbon has electrons beyond the first two s-orbitals, occupying the 2p orbitals as well, which are not spherically symmetrical.

23. What are the values of principal quantum number and azimuthal quantum number for the last electron in the chlorine atom?
A. 1, 6
B. 1, 3
C. 3, 1
D. 6, 1

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Correct Option: C
Option A Explained: Incorrect. A principal quantum number of 1 would mean the electron is in the very first shell, but chlorine’s last electron is actually in the third shell.
Option B Explained: Incorrect. An azimuthal quantum number of 3 would mean an f-type orbital, but chlorine’s last electron sits in a p-type orbital, not an f-type orbital.
Option C Explained: Correct. Chlorine’s electron configuration ends at 3p⁵, so its last electron has a principal quantum number of 3 (third shell) and an azimuthal quantum number of 1 (p sub-shell).
Option D Explained: Incorrect. A principal quantum number of 6 is far too high; chlorine’s outermost electrons are only in the third shell, not the sixth.

24. Which of the following electronic configurations is/are correct?
(i) ₂₉Cu = 1s²2s²2p⁶,3s²3p⁶,4s¹,3d⁹
(ii) ₂₉Cu = [Ar]4s¹3d¹⁰
(iii) ₂₄Cr = [Ar]4s¹3d¹⁰
(iv) ₂₄Cr = [Ar]4s²3d⁴
A. (i) only
B. (ii) only
C. (i) and (ii) only
D. (ii) and (iii) only

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Correct Option: B
Option A Explained: Incorrect. Statement (i) is written incorrectly and does not properly represent copper’s 29 electrons in the right order, so it cannot be a correct configuration.
Option B Explained: Correct. Only statement (ii) is correct. Copper is an exception to the normal filling order; one electron shifts from 4s to 3d so the 3d sub-shell becomes completely filled (3d¹⁰), giving the more stable arrangement [Ar]4s¹3d¹⁰.
Option C Explained: Incorrect. Although (ii) is correct, (i) is written incorrectly and does not correctly show copper’s configuration, so this combination is wrong.
Option D Explained: Incorrect. Statement (iii) wrongly applies copper’s exception to chromium; chromium actually becomes stable with a half-filled 3d⁵, not a fully filled 3d¹⁰, so (iii) is incorrect.

25. Quantum number which describes the orientation of orbitals in three-dimensional space is:
A. Spin quantum number
B. Azimuthal quantum number
C. Magnetic quantum number
D. Principal quantum number

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Correct Option: C
Option A Explained: Incorrect. The spin quantum number describes the spin of an electron within an orbital, not the orbital’s direction in space.
Option B Explained: Incorrect. The azimuthal quantum number tells us the shape of an orbital, such as s, p, or d, but not which direction it points.
Option C Explained: Correct. The magnetic quantum number describes how an orbital is oriented in three-dimensional space, distinguishing, for example, between px, py, and pz orbitals.
Option D Explained: Incorrect. The principal quantum number tells us about the size and energy of the shell an electron belongs to, not the direction the orbital faces.

26. Which of the following is the electronic configuration of Cr?
A. [Ar],3d⁵,4s²
B. [Ar],3d⁴,4s²
C. [Ar],3d⁶,4s⁰
D. [Ar],3d⁵,4s¹

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Correct Option: D
Option A Explained: Incorrect. This configuration wrongly keeps both 4s electrons together with a 3d⁵ arrangement, which does not match chromium’s real, more stable structure.
Option B Explained: Incorrect. This configuration would be expected from simple Aufbau filling, but chromium does not actually follow this pattern because of the extra stability of a half-filled 3d sub-shell.
Option C Explained: Incorrect. This configuration puts all six valence electrons into the 3d sub-shell and none in 4s, which does not match chromium’s actual electron arrangement.
Option D Explained: Correct. Chromium’s real ground state configuration is [Ar]3d⁵4s¹, since a half-filled 3d sub-shell together with one electron in 4s is more stable than other possible arrangements.

27. The relationship between quantum number n and l is:
A. n = l – 1
B. l = n – 2
C. l = n – 1
D. n = l – 2

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Correct Option: C
Option A Explained: Incorrect. This equation gives negative or meaningless values of n for small l, and does not represent the true relationship between these two quantum numbers.
Option B Explained: Incorrect. This formula does not correctly show the maximum possible value of l for a given shell; it underestimates it by one.
Option C Explained: Correct. For any given principal quantum number n, the azimuthal quantum number l can take whole number values from 0 up to a maximum of n − 1, so l = n − 1 correctly shows this upper limit.
Option D Explained: Incorrect. This equation gives an incorrect and much too small value of n compared to l, and does not match the real quantum number rule.

28. Quantum number values for “2p” orbitals are:
A. n=2, l=1
B. n=1, l=2
C. n=1, l=0
D. n=2, l=0

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Correct Option: A
Option A Explained: Correct. In “2p,” the number 2 represents the principal quantum number (n=2, meaning the second shell), and “p” represents an azimuthal quantum number of l=1.
Option B Explained: Incorrect. Here the principal quantum number is wrongly given as 1 and l as 2, but l can never be greater than n−1, so l cannot be 2 when n is only 1.
Option C Explained: Incorrect. This pair describes a “1s” orbital, not “2p,” since l=0 always corresponds to an s-type orbital.
Option D Explained: Incorrect. Here l is wrongly given as 0, which would describe an s-orbital, but a p-orbital must have l=1.

29. Which pair has 1 electron in its outermost ‘s’ orbital?
A. Li & Fe
B. Na & Cr
C. K & Mn
D. H & He

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Correct Option: B
Option A Explained: Incorrect. Lithium does have 1 electron in its outer 2s orbital, but iron’s outer shell has 2 electrons in 4s, not 1.
Option B Explained: Correct. Sodium’s outer shell configuration is 3s¹, and chromium is a special exception with configuration [Ar]4s¹3d⁵, so both have exactly 1 electron in their outermost s-orbital.
Option C Explained: Incorrect. Potassium does have 1 electron in 4s, but manganese’s outer shell has 2 electrons in 4s, not 1.
Option D Explained: Incorrect. Hydrogen does have 1 electron in 1s, but helium’s 1s orbital is completely filled with 2 electrons, not 1.

30. M-shell contains:
A. s
B. s, p, d
C. s, p, d, f
D. s, p

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Correct Option: B
Option A Explained: Incorrect. The M-shell (n=3) is not limited to only the s subshell; it also contains p and d subshells since l can range from 0 to n-1=2.
Option B Explained: Correct. For the M-shell (n=3), the azimuthal quantum number l can take values 0, 1, and 2, corresponding to the s, p, and d subshells respectively.
Option C Explained: Incorrect. The f subshell only appears from n=4 onwards, since the maximum value of l is n-1, so the M-shell (n=3) cannot contain an f subshell.
Option D Explained: Incorrect. This option leaves out the d subshell, which is definitely present in the M-shell since l=2 is allowed when n=3.

31. Total number of electrons that can be accommodated in the f-subshell:
A. 6
B. 14
C. 10
D. 18

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Correct Option: B
Option A Explained: Incorrect. 6 electrons is the maximum capacity of a p-subshell (l=1), not an f-subshell.
Option B Explained: Correct. For an f-subshell, l=3, so using the formula 2(2l+1) = 2(7) = 14, the f-subshell can hold a maximum of 14 electrons.
Option C Explained: Incorrect. 10 electrons is the maximum capacity of a d-subshell (l=2), not an f-subshell.
Option D Explained: Incorrect. 18 electrons does not match the formula for any single subshell’s electron capacity.

32. The n+l value for 5d is:
A. 5
B. 6
C. 7
D. 8

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Correct Option: C
Option A Explained: Incorrect. This value is too low; it does not correctly add the principal quantum number (5) and the azimuthal quantum number of d (2).
Option B Explained: Incorrect. This value is one less than the correct sum of n and l for the 5d orbital.
Option C Explained: Correct. For the 5d orbital, n=5 and l=2 (since d corresponds to l=2), so n+l = 5+2 = 7.
Option D Explained: Incorrect. This value is one more than the correct sum; it does not match n=5 and l=2 for the 5d orbital.

33. Each electron in an atom must have its own unique set of quantum numbers is a statement of:
A. Aufbau principle
B. Hund’s rule
C. Pauli exclusion principle
D. Heisenberg Uncertainty Principle

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Correct Option: C
Option A Explained: Incorrect. The Aufbau principle describes the order in which electrons fill orbitals from lowest to highest energy; it does not state that each electron needs a unique set of quantum numbers.
Option B Explained: Incorrect. Hund’s rule describes how electrons are distributed among orbitals of equal energy, preferring to stay unpaired before pairing; it is not about the uniqueness of quantum numbers.
Option C Explained: Correct. This statement is the exact definition of the Pauli exclusion principle: no two electrons in the same atom can have an identical set of all four quantum numbers.
Option D Explained: Incorrect. Heisenberg uncertainty principle tells that the position and momentum of an electron moving in an orbit cannot be determined simultaneously. It tells nothing about quantum numbers of electrons.

34. According to which scientist, the probability of finding an electron at a certain position is possible?
A. Bohr
B. Hund
C. de-Broglie
D. Schrodinger

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Correct Option: D
Option A Explained: Incorrect. Bohr’s model placed electrons in fixed circular orbits with definite paths and did not describe probability of location.
Option B Explained: Incorrect. Hund is known for his rule about electron filling in equal-energy orbitals, not for probability-based models of electron position.
Option C Explained: Incorrect. de-Broglie proposed that particles like electrons have wave-like properties, which was an important step, but he did not develop the mathematical probability model itself.
Option D Explained: Correct. Schrodinger developed the wave equation that gives the probability of finding an electron in a particular region of space, forming the basis of the modern quantum mechanical model of the atom.

35. Which gas in the discharge tube produces the lightest canal ray particles?
A. Ar
B. H2
C. He
D. Ne

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Correct Option: B
Option A Explained: Incorrect. Argon atoms are much heavier than hydrogen molecules, so canal ray particles from argon would not be the lightest.
Option B Explained: Correct. Hydrogen is the lightest gas among these options, so the positive ions (canal rays) produced from hydrogen gas in the discharge tube are the lightest possible particles.
Option C Explained: Incorrect. Helium atoms are heavier than hydrogen molecules, so canal rays from helium are not the lightest.
Option D Explained: Incorrect. Neon atoms are much heavier than hydrogen molecules, so canal rays from neon are far from being the lightest.

36. Which element has the ground state electronic configuration of 1s²,2s²,2p⁶,3s²,3p⁶?
A. Ar
B. Na
C. Cl
D. S

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Correct Option: A
Option A Explained: Correct. Counting the electrons in this configuration (2+2+6+2+6=18) gives atomic number 18, which corresponds to argon, a noble gas with a completely filled outer shell.
Option B Explained: Incorrect. Sodium has only 11 electrons, so its configuration would be much shorter and would not include a filled 3p subshell.
Option C Explained: Incorrect. Chlorine has 17 electrons, one less than this configuration, so its 3p subshell would only be partially filled (3p⁵), not completely filled (3p⁶).
Option D Explained: Incorrect. Sulfur has 16 electrons, so its 3p subshell would only hold 4 electrons (3p⁴), not the full 6 shown in this configuration.

37. What is the proton (Atomic Number) of an element that has four unpaired electrons in its ground state?
A. 6
B. 22
C. 14
D. 26

Show Answer

Correct Option: D
Option A Explained: Incorrect. Carbon (atomic number 6) has the configuration 1s²2s²2p², which gives only 2 unpaired electrons in the 2p orbitals, not 4.
Option B Explained: Incorrect. Titanium (atomic number 22) has the configuration [Ar]3d²4s², which gives only 2 unpaired electrons in the 3d orbitals, not 4.
Option C Explained: Incorrect. Silicon (atomic number 14) has the configuration [Ne]3s²3p², which gives only 2 unpaired electrons, not 4.
Option D Explained: Correct. Iron (atomic number 26) has the configuration [Ar]3d⁶4s². Following Hund’s rule, the six 3d electrons fill as five single electrons first and then one pairs up, leaving exactly four unpaired electrons.

38. Principal quantum number is represented by the symbol:
A. m
B. n
C. s
D. l

Show Answer

Correct Option: B
Option A Explained: Incorrect. The letter “m” is used to represent the magnetic quantum number, not the principal quantum number.
Option B Explained: Correct. The principal quantum number, which tells us about the main energy shell of an electron, is always represented by the symbol “n”.
Option C Explained: Incorrect. The letter “s” is used to represent the spin quantum number, not the principal quantum number.
Option D Explained: Incorrect. The letter “l” is used to represent the azimuthal (angular momentum) quantum number, not the principal quantum number.

39. Shape of the sub-shell is explained by which quantum number:
A. Principal quantum number
B. Azimuthal quantum number
C. Magnetic quantum number
D. Spin quantum number

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Correct Option: B
Option A Explained: Incorrect. The principal quantum number tells us about the size and energy of the shell, not the shape of a subshell.
Option B Explained: Correct. The azimuthal quantum number determines the shape of a subshell, for example spherical for s, dumbbell for p, and cloverleaf-like for d.
Option C Explained: Incorrect. The magnetic quantum number tells us about the orientation of an orbital in space, not its overall shape.
Option D Explained: Incorrect. The spin quantum number describes the spin of the electron itself, and has no connection to the shape of a subshell.

40. The electronic configuration for degenerate orbitals is explained by:
A. Aufbau’s Principle
B. (n+l) rule
C. Hund’s rule
D. Pauli’s exclusion principle

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Correct Option: C
Option A Explained: Incorrect. The Aufbau principle explains the overall order in which subshells are filled with increasing energy, not specifically how electrons are distributed among orbitals of the same energy.
Option B Explained: Incorrect. The (n+l) rule helps decide the relative energy order of different subshells; it does not describe how electrons fill orbitals of equal energy within the same subshell.
Option C Explained: Correct. Hund’s rule states that electrons fill degenerate (equal-energy) orbitals singly first, with parallel spins, before any pairing occurs, which explains their distribution in such orbitals.
Option D Explained: Incorrect. The Pauli exclusion principle limits each orbital to a maximum of two electrons with opposite spins; it does not describe the specific order of filling degenerate orbitals.

41. Maximum number of electrons which can be placed in one orbital is:
A. 1
B. 2
C. 3
D. 5

Show Answer

Correct Option: B
Option A Explained: Incorrect. A single orbital is not limited to just 1 electron; it can actually hold up to 2 electrons as long as their spins are opposite.
Option B Explained: Correct. According to the Pauli exclusion principle, each orbital can hold a maximum of 2 electrons, and these two electrons must have opposite spins.
Option C Explained: Incorrect. No orbital in an atom can ever hold 3 electrons, since this would require two electrons to share the same set of all four quantum numbers.
Option D Explained: Incorrect. 5 electrons in a single orbital is far too many, and would completely violate the Pauli exclusion principle.

42. Greater the wavelength associated with the photon:
A. Greater is its energy
B. Smaller is its energy
C. Its energy will be variable
D. Its energy will remain constant

Show Answer

Correct Option: B
Option A Explained: Incorrect. Energy and wavelength of a photon are inversely related, so a greater wavelength cannot mean greater energy.
Option B Explained: Correct. Since the energy of a photon is given by E = hc/λ, a greater wavelength (λ) directly corresponds to a smaller amount of energy.
Option C Explained: Incorrect. The relationship between energy and wavelength of a photon is fixed and predictable, not randomly variable.
Option D Explained: Incorrect. Energy does not stay constant as wavelength changes; it decreases in a definite, predictable way as wavelength increases.

43. Photon of which of the following series will have the largest wavelength?
A. Bracket series
B. Pfund series
C. Balmer series
D. Paschen series

Show Answer

Correct Option: B
Option A Explained: Incorrect. The Bracket series involves transitions ending at n=4, which have larger energy gaps and shorter wavelengths than transitions ending at n=5.
Option B Explained: Correct. The Pfund series involves transitions ending at the highest energy level (n=5) among these choices, giving the smallest energy gaps and therefore the largest (longest) wavelength photons.
Option C Explained: Incorrect. The Balmer series involves transitions ending at n=2, which are the largest energy gaps among these series and therefore produce the shortest wavelength photons, not the largest.
Option D Explained: Incorrect. The Paschen series involves transitions ending at n=3, which have larger energy gaps and shorter wavelengths than the Pfund series.

44. The equation v̄ = R[1/2² − 1/n²] where n > 2 is the equation of H-spectrum for:
A. Paschen series
B. Pfund series
C. Bracket series
D. Lyman series
E. Balmer series

Show Answer

Correct Option: E
Option A Explained: Incorrect. The Paschen series equation uses 1/3² instead of 1/2² as the fixed term, since its transitions end at n=3, not n=2.
Option B Explained: Incorrect. The Pfund series equation uses 1/5² as the fixed term, since its transitions end at n=5, not n=2.
Option C Explained: Incorrect. The Bracket series equation uses 1/4² as the fixed term, since its transitions end at n=4, not n=2.
Option D Explained: Incorrect. The Lyman series equation uses 1/1² as the fixed term, since its transitions end at n=1, not n=2.
Option E Explained: Correct. Since the fixed term here is 1/2² and n is greater than 2, this equation exactly represents the Balmer series, where all transitions end at the second energy level (n=2).

45. Which of the following sub-shells are allowed by Azimuthal quantum number in the N-shell of an element?
A. s- sub-shell
B. s, p, d & f sub-shells
C. s & p sub-shells
D. s, p, & d sub-shells

Show Answer

Correct Option: B
Option A Explained: Incorrect. The N-shell (n=4) is not limited to only the s orbital; higher orbitals are also allowed within it.
Option B Explained: Correct. For the N-shell (n=4), the azimuthal quantum number l can take the values 0, 1, 2, and 3, which correspond to s, p, d, and f orbitals respectively, since the maximum value of l is n-1=3.
Option C Explained: Incorrect. This option leaves out the d and f orbitals, which are indeed allowed in the N-shell.
Option D Explained: Incorrect. Remember that the “n” value of a shell or orbit shows the number of sub-shells present in it.

46. Which concept was introduced by the scientist Bohr?
A. Energy levels
B. Orbitals
C. Sub-orbits
D. Sub-energy levels

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Correct Option: A
Option A Explained: Correct. Bohr introduced the idea that electrons move only in certain fixed circular paths called energy levels, each having a definite, quantized amount of energy.
Option B Explained: Incorrect. The concept of orbitals as three-dimensional probability regions came later, from the quantum mechanical model developed by Schrodinger, not Bohr.
Option C Explained: Incorrect. “Sub-orbits” is not the term associated with Bohr’s model; his model was based on whole energy levels, not further subdivided orbits.
Option D Explained: Incorrect. Sub-energy levels (subshells) came from later, more detailed quantum mechanical models, not from Bohr’s original simple energy-level model.

47. What is the electronic configuration of C?
A. 1s²,2s²,2p²
B. 1s²,2s²,2p⁴
C. 1s²,2s²,2p⁶,3s²
D. 1s²,2s²,2p⁶,3s²,3p³

Show Answer

Correct Option: A
Option A Explained: Correct. Carbon has atomic number 6, so its electrons fill as 1s²2s²2p², matching this configuration exactly.
Option B Explained: Incorrect. This configuration (2p⁴) has a total of 8 electrons, which corresponds to oxygen, not carbon.
Option C Explained: Incorrect. This configuration has 12 electrons in total, which corresponds to magnesium, not carbon.
Option D Explained: Incorrect. This configuration has 15 electrons in total, which corresponds to phosphorus, not carbon.

48. Why do electrons have opposite spins when they are in the same orbital?
A. This condition reduces friction
B. This condition creates more energy
C. This condition results in zero magnetism and removes the charge of the electron
D. This condition results in less repulsion and opposite magnetic fields

Show Answer

Correct Option: D
Option A Explained: Incorrect. Electrons do not experience physical friction, so this is not a real or scientific reason for their spin arrangement.
Option B Explained: Incorrect. Opposite spins actually help make the arrangement more stable and lower in energy, not higher in energy.
Option C Explained: Incorrect. Electrons always keep their negative charge regardless of spin direction; spin pairing does not remove their charge.
Option D Explained: Correct. When two electrons in the same orbital have opposite spins, their magnetic fields oppose each other, which lowers the overall repulsion between them and makes the paired arrangement more stable.

49. With the increase in the value of principal quantum number “n” the size of the s-orbitals:
A. Decrease
B. Increase
C. May or may not remain the same
D. Remain the same

Show Answer

Correct Option: B
Option A Explained: Incorrect. As the principal quantum number becomes larger, the electron is found farther from the nucleus on average, so the orbital cannot become smaller.
Option B Explained: Correct. A larger value of the principal quantum number places the electron farther from the nucleus, causing the s-orbital to become larger in size.
Option C Explained: Incorrect. The change in orbital size with increasing n is a definite, predictable trend, not something uncertain.
Option D Explained: Incorrect. The size of an s-orbital is directly linked to the value of n, so it cannot remain unchanged as n increases.

50. Maximum electrons that can be placed in a p subshell are:
A. 2
B. 10
C. 6
D. 14

Show Answer

Correct Option: C
Option A Explained: Incorrect. 2 electrons is the maximum capacity of a single orbital, not a whole p subshell, which contains three orbitals.
Option B Explained: Incorrect. 10 electrons is the maximum capacity of a d subshell, not a p subshell.
Option C Explained: Correct. A p subshell has three orbitals (px, py, pz), and since each orbital holds a maximum of 2 electrons, the p subshell can hold a maximum of 3×2=6 electrons.
Option D Explained: Incorrect. 14 electrons is the maximum capacity of an f subshell, not a p subshell.

51. Apply the (n+l) rule and calculate which of the following orbitals has the maximum energy.
A. 4s
B. 2d
C. 2p
D. 3s

Show Answer

Correct Option: A
Option A Explained: Correct. For 4s, n+l = 4+0 = 4, which is the highest sum among the valid orbitals listed, so it has the maximum energy according to the (n+l) rule.
Option B Explained: Incorrect. The orbital “2d” does not actually exist, since for n=2 the azimuthal quantum number l can only be 0 or 1, and d requires l=2.
Option C Explained: Incorrect. For 2p, n+l = 2+1 = 3, which is lower than the value for 4s, so 2p has lower energy.
Option D Explained: Incorrect. For 3s, n+l = 3+0 = 3, which is lower than the value for 4s, so 3s has lower energy.

52. The mass of an electron is:
A. 1836 times more than the mass of a proton
B. 1836 times less than the mass of a proton
C. 1836 times more than the mass of a neutron
D. 1836 times more than the mass of a hydrogen atom

Show Answer

Correct Option: B
Option A Explained: Incorrect. This reverses the true relationship; the electron is far lighter than the proton, not heavier.
Option B Explained: Correct. The mass of an electron is about 1836 times smaller than the mass of a proton, which is why electrons contribute very little to an atom’s total mass.
Option C Explained: Incorrect. Since a neutron’s mass is very close to a proton’s mass, the electron is also about 1836 times lighter than a neutron, not heavier.
Option D Explained: Incorrect. A hydrogen atom’s mass is essentially the mass of its one proton, so the electron is about 1836 times lighter than a hydrogen atom too, not heavier.

53. The shape of an orbital is determined by which quantum number?
A. n
B. m
C. l
D. s

Show Answer

Correct Option: C
Option A Explained: Incorrect. The principal quantum number “n” tells us the size and energy of the shell, not the shape of the orbital.
Option B Explained: Incorrect. The magnetic quantum number “m” tells us the orientation of the orbital in space, not its basic shape.
Option C Explained: Correct. The azimuthal quantum number “l” determines the shape of an orbital, such as spherical for s, dumbbell for p, and more complex shapes for d and f.
Option D Explained: Incorrect. The spin quantum number “s” describes the spin of the electron itself, and has nothing to do with orbital shape.

54. When the 6d orbital is filled, the entering electron goes into:
A. 7f
B. 7d
C. 7p
D. 7s

Show Answer

Correct Option: C.
Correct Option: C
Option A Explained: Incorrect. The 7f orbital is filled much later in the sequence, after several other orbitals including 7p, so it is not the immediate next orbital after 6d.
Option B Explained: Incorrect. The 7d orbital comes much later in the filling order, after 7s, 5f, and 7p, so it is not filled right after 6d.
Option C Explained: Correct. Following the standard (n+l) filling order, after the 6d orbital is completely filled, the next electron enters the 7p orbital.
Option D Explained: Incorrect. The 7s orbital is actually filled before 5f and 6d, not after 6d, so it cannot be the next orbital in this sequence.

55. If the value of the azimuthal quantum number is 1, then the value of m will range from:
A. -1 to 1
B. -1 to 2
C. -1 to 3
D. -1 to 4

Show Answer

Correct Option: A
Option A Explained: Correct. When l=1, the magnetic quantum number m can take integer values from -l to +l, which means m can be -1, 0, or +1, i.e., it ranges from -1 to 1.
Option B Explained: Incorrect. This range incorrectly extends beyond +l; for l=1, m cannot go as high as +2.
Option C Explained: Incorrect. This range is far too wide for l=1; such a wide range of m values would only be possible for a much larger value of l.
Option D Explained: Incorrect. This range is much too wide for l=1 and does not follow the rule that m can only range from -l to +l.

56. The dumbbell-like orbitals can move in how many directions?
A. 7
B. 5
C. 1
D. 3

Show Answer

Correct Option: D
Option A Explained: Incorrect. 7 directions would correspond to the number of orientations of an f-type orbital, not a p-type (dumbbell-shaped) orbital.
Option B Explained: Incorrect. 5 directions would correspond to the number of orientations of a d-type orbital, not a p-type orbital.
Option C Explained: Incorrect. A single direction would only apply to a spherical s-type orbital, which has no directional preference at all.
Option D Explained: Correct. Dumbbell-shaped orbitals are p-orbitals, and there are three of them (px, py, pz), each oriented along one of the three perpendicular axes in space.

57. The number of orientations of a sub-shell for which n=4 and l=3 will be:
A. 5
B. 3
C. 7
D. 1

Show Answer

Correct Option: C
Option A Explained: Incorrect. 5 orientations would apply to a d-type subshell (l=2), not an f-type subshell where l=3.
Option B Explained: Incorrect. 3 orientations would apply to a p-type subshell (l=1), not an f-type subshell where l=3.
Option C Explained: Correct. The number of orientations of any subshell equals 2l+1. Since l=3 here, the number of orientations is 2(3)+1=7.
Option D Explained: Incorrect. A single orientation would only apply to an s-type subshell (l=0), not an f-type subshell.

58. Photon of the lowest wavelength is related to:
A. Balmer series
B. Pfund series
C. Bracket series
D. Paschen series

Show Answer

Correct Option: A
Option A Explained: Correct. The Balmer series involves transitions ending at the lowest energy level among these four series (n=2), giving the largest energy gaps and therefore the shortest (lowest) wavelength photons.
Option B Explained: Incorrect. The Pfund series involves transitions ending at a much higher energy level (n=5), giving small energy gaps and long wavelength photons, not the lowest wavelength.
Option C Explained: Incorrect. The Bracket series involves transitions ending at n=4, giving smaller energy gaps and longer wavelengths than the Balmer series.
Option D Explained: Incorrect. The Paschen series involves transitions ending at n=3, giving smaller energy gaps and longer wavelengths than the Balmer series.

59. The maximum e/m ratio for positive rays is obtained when the discharge tube contains:
A. He
B. N2
C. Ne
D. H2

Show Answer

Correct Option: D
Option A Explained: Incorrect. Helium atoms are heavier than hydrogen molecules, so positive ions from helium would have a smaller e/m ratio than those from hydrogen.
Option B Explained: Incorrect. Nitrogen molecules are much heavier than hydrogen molecules, so positive ions from nitrogen would have a much smaller e/m ratio.
Option C Explained: Incorrect. Neon atoms are much heavier than hydrogen molecules, so positive ions from neon would have a smaller e/m ratio.
Option D Explained: Correct. Since hydrogen is the lightest gas among these options, its positive ions have the smallest mass for a given charge, giving them the maximum possible e/m (charge-to-mass) ratio.

60. The correct sequence of electronic configuration filling is:
A. 4p, 5s, 4d, 5p, 6s, 4f, 5d
B. 4p, 4s, 4d, 5p, 6s, 4f, 6d
C. 4p, 4s, 4d, 5p, 5s, 4f, 5d
D. 4p, 3s, 4d, 5p, 6s, 4f, 5d

Show Answer

Correct Option: A
Option A Explained: Correct. This sequence exactly follows the accepted (n+l) filling order of subshells: after 4p comes 5s, then 4d, then 5p, then 6s, then 4f, and then 5d, matching the real order used to build up electron configurations.
Option B Explained: Incorrect. This sequence wrongly places 4s after 4p, but 4s always fills before 4p in any real atom, since it has a lower n+l value.
Option C Explained: Incorrect. This sequence wrongly places 5s after 5p, but 5s should be filled before 5p, not after it.
Option D Explained: Incorrect. This sequence wrongly places 3s after 4p, but 3s is one of the very first subshells filled, long before 4p.

61. Radius of the first orbit of hydrogen is 0.53Å. Which orbit has a radius of 4.77Å?
A. First
B. Second
C. Third
D. Fourth

Show Answer

Correct Option: C
Option A Explained: Incorrect. The first orbit already has the given radius of 0.53Å, not the larger radius of 4.77Å.
Option B Explained: Incorrect. Using the relation radius = n² × 0.53Å, the second orbit would have a radius of 4×0.53=2.12Å, not 4.77Å.
Option C Explained: Correct. Using the relation radius = n² × 0.53Å, setting n=3 gives 9×0.53=4.77Å, which exactly matches the given radius.
Option D Explained: Incorrect. Using the relation radius = n² × 0.53Å, the fourth orbit would have a radius of 16×0.53=8.48Å, which is much larger than 4.77Å.

62. Which of the following has only one orientation in space, in the magnetic field?
A. p orbital
B. s orbital
C. d orbital
D. f orbital

Show Answer

Correct Option: B
Option A Explained: Incorrect. A p orbital (l=1) has three possible orientations in space (m=-1,0,+1), not just one.
Option B Explained: Correct. An s orbital has l=0, so its magnetic quantum number m can only be 0, meaning it has exactly one possible orientation in space; this matches its spherical, direction-independent shape.
Option C Explained: Incorrect. A d orbital (l=2) has five possible orientations in space (m=-2,-1,0,+1,+2), not just one.
Option D Explained: Incorrect. An f orbital (l=3) has seven possible orientations in space (m=-3 to +3), not just one.

63. What’s the mass of one proton?
A. Equal to 1836 electrons
B. Equal to a positron
C. Equal to an electron
D. Equal to 1836 neutrons

Show Answer

Correct Option: A
Option A Explained: Correct. A proton is about 1836 times heavier than an electron, so its mass is often expressed as being equal to the mass of 1836 electrons.
Option B Explained: Incorrect. A positron has the same tiny mass as an electron, so a proton, which is nearly 1836 times heavier, cannot have a mass equal to a positron.
Option C Explained: Incorrect. A proton’s mass is far greater than an electron’s mass, so the two cannot be described as equal.
Option D Explained: Incorrect. A neutron’s mass is very close to a proton’s mass, so a proton cannot be 1836 times as heavy as a neutron.

64. What is the quantum number for the unpaired electron of the chlorine atom (n, l, m)?
A. 2, 1, 0
B. 2, 1, 1
C. 3, 1, 1
D. 3, 0, 0

Show Answer

Correct Option: C
Option A Explained: Incorrect. This option has n=2, but chlorine’s unpaired electron is actually in the third shell (n=3), not the second.
Option B Explained: Incorrect. This option also uses n=2, which is wrong since chlorine’s last unpaired electron belongs to the third shell.
Option C Explained: Correct. Chlorine’s configuration ends in 3p⁵, and its unpaired electron sits in a 3p orbital, giving quantum numbers n=3, l=1, and by convention m=1 for the singly occupied orbital.
Option D Explained: Incorrect. This option uses l=0, which describes an s-type orbital, but chlorine’s unpaired electron is actually found in a p-type orbital.

65. Which statement about a 4d orbital is correct?
A. It is at a higher energy level than a 4p orbital but has the same shape
B. It is occupied by one electron in an isolated Zn atom
C. It can hold a maximum of 10 electrons
D. It has the highest energy of orbitals with principal quantum number four

Show Answer

Correct Option: C
Option A Explained: Incorrect. While it is true that 4d has higher energy than 4p, its shape is actually different from p orbitals; d orbitals generally have a cloverleaf-like shape, not a dumbbell shape.
Option B Explained: Incorrect. In an isolated zinc atom, the ground state configuration is [Ar]3d¹⁰4s², meaning the 4d orbital is completely empty, not occupied by even one electron.
Option C Explained: Correct. Like every d-type subshell, the 4d subshell has five orbitals, and since each orbital can hold up to 2 electrons, it can hold a maximum of 10 electrons in total.
Option D Explained: Incorrect. Among orbitals with principal quantum number four, the 4f orbital actually has higher energy than 4d, so 4d does not have the highest energy in that group.

66. Which of the following is not true about canal rays?
A. 1.6726×10⁻²⁷ kg
B. 9.54×10⁷ C/kg
C. Slow deflection under electric & magnetic fields
D. Do not cause mechanical motion

Show Answer

Correct Option: D
Option A Explained: Incorrect, meaning this is a true statement. This value correctly represents the approximate mass of a proton, which matches the lightest canal ray particles produced from hydrogen gas.
Option B Explained: Incorrect, meaning this is a true statement. This value is close to the actual charge-to-mass ratio of a proton, which is a valid property of canal ray particles.
Option C Explained: Incorrect, meaning this is a true statement. Because canal ray particles are much heavier than electrons, they deflect only slightly and slowly under electric and magnetic fields compared to cathode rays.
Option D Explained: Correct, meaning this is the false statement. Canal rays, like cathode rays, are actually capable of producing mechanical effects, such as rotating a small paddle wheel placed in their path, so saying they do not cause mechanical motion is incorrect.

67. The amount of energy associated with a quantum of radiation is directly proportional to:
A. Photon
B. Wavelength
C. Frequency
D. Velocity

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Correct Option: C
Option A Explained: Incorrect. A photon is itself the particle or packet of radiation being discussed; it is not a separate quantity that the energy can be “proportional to.”
Option B Explained: Incorrect. Energy is actually inversely proportional to wavelength, meaning a longer wavelength corresponds to lower energy, not higher.
Option C Explained: Correct. According to Planck’s equation E=hν, the energy of a quantum of radiation is directly proportional to its frequency (ν); a higher frequency always means higher energy.
Option D Explained: Incorrect. While all electromagnetic radiation travels at the same constant speed (the speed of light) in a vacuum, this speed does not change with different amounts of energy, so it cannot be called directly proportional to energy.

68. If the value of the azimuthal quantum number is 2, then the total values of the magnetic quantum number will be:
A. 03
B. 05
C. 07
D. 10

Show Answer

Correct Option: B
Option A Explained: Incorrect. This number is too small; for l=2, there are more than 3 possible values of m.
Option B Explained: Correct. The total number of possible values of the magnetic quantum number is given by 2l+1. For l=2, this gives 2(2)+1=5 values (m = -2, -1, 0, +1, +2).
Option C Explained: Incorrect. This number is too high for l=2; a total of 7 values of m would correspond to l=3, not l=2.
Option D Explained: Incorrect. This number does not match the formula 2l+1 for any meaningful value of l, and is far too high for l=2.

69. Total number of directions of f-orbitals in space are:
A. 05
B. 03
C. 07
D. 06

Show Answer

Correct Option: C
Option A Explained: Incorrect. 5 directions would correspond to the orientations of a d-type orbital, not an f-type orbital.
Option B Explained: Incorrect. 3 directions would correspond to the orientations of a p-type orbital, not an f-type orbital.
Option C Explained: Correct. For f-orbitals, l=3, so the number of possible orientations in space is 2l+1=2(3)+1=7.
Option D Explained: Incorrect. 6 directions does not match the formula 2l+1 for any standard type of orbital, so it cannot be correct for f-orbitals.

70. Which of the following quantum numbers is not obtained from the Schrodinger Wave equation?
A. Principal Quantum Number
B. Spin Quantum Number
C. Azimuthal Quantum Number
D. Magnetic Quantum Number

Show Answer

Correct Option: B
Option A Explained: Incorrect. The principal quantum number does arise naturally as a solution when solving the Schrodinger wave equation for the hydrogen atom.
Option B Explained: Correct. The spin quantum number was not obtained from the Schrodinger wave equation; it was introduced separately based on experimental evidence and later justified using relativistic quantum theory.
Option C Explained: Incorrect. The azimuthal quantum number does arise naturally as a solution when solving the Schrodinger wave equation.
Option D Explained: Incorrect. The magnetic quantum number does arise naturally as a solution when solving the Schrodinger wave equation.

71. The p orbital has:
A. 2 lobes
B. 3 lobes
C. 4 lobes
D. 5 lobes

Show Answer

Correct Option: A
Option A Explained: Correct. A p orbital has a dumbbell shape made up of exactly two lobes, one on each side of the nucleus along a particular axis.
Option B Explained: Incorrect. A p orbital does not have three lobes; three lobes is not a typical shape for any single simple atomic orbital.
Option C Explained: Incorrect. Four lobes describe the shape of most d orbitals, not a p orbital.
Option D Explained: Incorrect. Five lobes does not describe any standard simple atomic orbital shape, including p orbitals.

72. . Which of the following electronic configurations is correct for carbon?
A. 1s²2s²2p³
B. 1s²2s²2p⁴
C. 1s²2s²2p²
D. 1s²2s²2p¹

Show Answer

Correct Option: C
Option A Explained: Incorrect. This configuration has 7 electrons total, which corresponds to nitrogen, not carbon.
Option B Explained: Incorrect. This configuration has 8 electrons total, which corresponds to oxygen, not carbon.
Option C Explained: Correct. Carbon has atomic number 6, and its ground state electron configuration is exactly 1s²2s²2p², matching this option.
Option D Explained: Incorrect. This configuration has only 5 electrons total, which corresponds to boron, not carbon.

73. The e/m ratio of a proton is _ that of an electron.
A. 1837 times greater
B. Equal to
C. Greater than
D. Smaller than

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74. How many electrons will be accommodated in a sub-shell with azimuthal quantum number l=2?
A. 2
B. 6
C. 10
D. 12

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Correct Option: C.
Correct Option: C
Option A Explained: Incorrect. 2 electrons is the capacity of a single orbital, not the whole subshell with l=2, which has five orbitals.
Option B Explained: Incorrect. 6 electrons is the maximum capacity of a subshell with l=1 (p subshell), not l=2.
Option C Explained: Correct. For l=2 (a d-type subshell), the maximum number of electrons is given by 2(2l+1) = 2(5) = 10.
Option D Explained: Incorrect. 12 electrons is more than the maximum capacity of any subshell with l=2, and does not match the correct formula.

75. Which of the following elements will show an electronic configuration of the outermost shell like ns², np⁵?
A. C
B. Cl
C. S
D. Si

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Correct Option: B
Option A Explained: Incorrect. Carbon’s outer shell configuration is ns²np², not ns²np⁵, since carbon has only 4 valence electrons.
Option B Explained: Correct. Chlorine belongs to the halogen family, and its outer shell configuration is exactly ns²np⁵, giving it 7 valence electrons.
Option C Explained: Incorrect. Sulfur’s outer shell configuration is ns²np⁴, not ns²np⁵, since sulfur has 6 valence electrons.
Option D Explained: Incorrect. Silicon’s outer shell configuration is ns²np², not ns²np⁵, since silicon has only 4 valence electrons.

76. Which of the following sub-shells does not exist?
A. 1p
B. 1s
C. 5d
D. 6f

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Correct Option: A
Option A Explained: Correct. For n=1, the azimuthal quantum number l can only be 0 (since the maximum value of l is n−1=0), so a “1p” subshell, which would need l=1, cannot exist.
Option B Explained: Incorrect. The 1s subshell definitely exists; with n=1 and l=0, it is in fact the very first and simplest subshell in any atom.
Option C Explained: Incorrect. The 5d subshell definitely exists, since for n=5, l can go up to 4, meaning l=2 (d) is a perfectly valid combination.
Option D Explained: Incorrect. The 6f subshell definitely exists, since for n=6, l can go up to 5, meaning l=3 (f) is a perfectly valid combination.

77. The splitting of spectral lines in a magnetic field is called the:
A. Auf-bau principle
B. Pauli’s exclusion principle
C. Stark effect
D. Zeeman effect

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Correct Option: D
Option A Explained: Incorrect. The Aufbau principle describes the order in which electrons fill orbitals; it has nothing to do with spectral line splitting.
Option B Explained: Incorrect. The Pauli exclusion principle describes limits on electrons sharing quantum numbers; it does not explain spectral line splitting in a magnetic field.
Option C Explained: Incorrect. The Stark effect refers to the splitting of spectral lines due to an external electric field, not a magnetic field.
Option D Explained: Correct. The splitting of spectral lines when atoms are placed in a magnetic field is specifically called the Zeeman effect.

78. “In an orbital of an atom, no two electrons can have the same set of four quantum numbers; at least one quantum number must be different.” This statement is:
A. Aufbau principle
B. Pauli’s exclusion principle
C. Hund’s rule
D. Wiswesser rule

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Correct Option: B
Option A Explained: Incorrect. The Aufbau principle is about the order of filling orbitals with increasing energy, not about the uniqueness of quantum numbers.
Option B Explained: Correct. This statement is the exact wording of the Pauli exclusion principle, which says no two electrons in an atom can share an identical set of all four quantum numbers.
Option C Explained: Incorrect. Hund’s rule is about how electrons distribute themselves among orbitals of equal energy, not about quantum number uniqueness.
Option D Explained: Incorrect. There is no standard principle called the “Wiswesser rule” related to quantum numbers in this context; it does not match the description given.

79. The charge of an electron is:
A. 1.602×10⁻¹⁶ C
B. 1.602×10⁻¹⁹ C
C. 1.602×10⁻¹⁷ C
D. 1.602×10⁻¹⁸ C

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Correct Option: B
Option A Explained: Incorrect. This value is off by a factor of 1000 from the correct, experimentally measured charge of an electron.
Option B Explained: Correct. The charge of an electron is a fundamental constant of nature, measured very precisely as 1.602×10⁻¹⁹ coulombs.
Option C Explained: Incorrect. This value is off by a factor of 100 from the correct charge of an electron.
Option D Explained: Incorrect. This value is off by a factor of 10 from the correct charge of an electron.

80. Which statement is not correct for cathode rays?
A. Cathode rays consist of moving material particles
B. Cathode rays have a definite velocity
C. Cathode rays have a definite velocity
D. Cathode rays travel with the velocity of light

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Correct Option: D
Option A Explained: Incorrect, meaning this is a true statement. Cathode rays are actually streams of electrons, which are real material particles with mass, so this statement is true.
Option B Explained: Incorrect, meaning this is a true statement. Cathode ray particles do travel with a definite, measurable velocity that depends on the voltage applied across the discharge tube.
Option C Explained: Incorrect, meaning this is a true statement, repeated here as printed in the original question. Cathode ray particles travel with a definite, measurable velocity, just as described in option B.
Option D Explained: Correct, meaning this is the false statement. Cathode rays travel very fast but always at a speed much less than the speed of light, so claiming they travel exactly at the velocity of light is incorrect.

81. Two elements X and Y have the following electronic configurations:
X: 1s² 2An electron in an atom is completely described by its _ quantum numbers.
A. Eight
B. Six
C. Five
D. Four

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Correct Option: D
Option A Explained: Incorrect. Eight quantum numbers is far more than needed; an electron’s state does not require this many values to describe it.
Option B Explained: Incorrect. Six quantum numbers is more than necessary to fully describe the state of a single electron in an atom.
Option C Explained: Incorrect. Five quantum numbers is still one more than what is actually needed to describe an electron completely.
Option D Explained: Correct. An electron’s complete state in an atom is described using exactly four quantum numbers: principal (n), azimuthal (l), magnetic (m), and spin (s).

82. The only group of lines occurring in the visible region of the hydrogen spectrum is labelled as the:
A. Paschen series
B. Pfund series
C. Balmer series
D. Lyman series

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Correct Option: C
Option A Explained: Incorrect. The Paschen series lies in the infrared region of the spectrum, not the visible region.
Option B Explained: Incorrect. The Pfund series lies in the far infrared region of the spectrum, not the visible region.
Option C Explained: Correct. The Balmer series is the only hydrogen spectral series whose lines fall within the visible range of light that our eyes can detect.
Option D Explained: Incorrect. The Lyman series lies in the ultraviolet region of the spectrum, not the visible region.

83. Select the correct one:
A. Only electrons of atoms get shuffled during a chemical reaction
B. All the sub-atomic particles of an atom take part in a chemical reaction
C. Protons of an atom spearhead the chemical reaction
D. Protons and neutrons of an atom collectively make chemical reactions possible

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Correct Option: A
Option A Explained: Correct. Chemical reactions only involve the rearrangement, sharing, or transfer of electrons between atoms; the nucleus, containing protons and neutrons, remains unaffected.
Option B Explained: Incorrect. Protons and neutrons stay safely inside the nucleus and do not participate in ordinary chemical reactions; only electrons are involved.
Option C Explained: Incorrect. Protons stay fixed inside the nucleus during chemical changes and do not lead or drive chemical reactions.
Option D Explained: Incorrect. Protons and neutrons remain within the nucleus throughout a chemical reaction and play no direct role in making the reaction happen.

84. The value of “l” for a 4p orbital is:
A. 0
B. 1
C. 2
D. 3

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Correct Option: B
Option A Explained: Incorrect. A value of l=0 would describe an s-type orbital, not a p-type orbital.
Option B Explained: Correct. Every p-type orbital, regardless of its principal quantum number, always has an azimuthal quantum number of l=1, so a 4p orbital has l=1.
Option C Explained: Incorrect. A value of l=2 would describe a d-type orbital, not a p-type orbital.
Option D Explained: Incorrect. A value of l=3 would describe an f-type orbital, not a p-type orbital.

85. The element having the electronic configuration of noble gas notation (Kr) 5s²4d⁶ is:
A. ₃₈Sr
B. ₄₄Ru
C. ₄₆Pd
D. ₄₀Zr

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Correct Option: B
Option A Explained: Incorrect. Strontium has atomic number 38, but adding the 36 electrons of krypton to 5s²4d⁶ (8 more electrons) gives atomic number 44, not 38.
Option B Explained: Correct. Krypton has atomic number 36; adding the 2 electrons in 5s and 6 electrons in 4d gives a total of 36+2+6=44 electrons, matching the atomic number of ruthenium.
Option C Explained: Incorrect. Palladium has atomic number 46, which is 2 more than the 44 electrons calculated from this configuration.
Option D Explained: Incorrect. Zirconium has atomic number 40, which is 4 less than the 44 electrons calculated from this configuration.

86. Choose the allowed set of quantum numbers from the following:
A. n=3, l=2, m=0, s=+1/2
B. n=3, l=3, m=0, s=-1/2
C. n=4, l=2, m=4, s=+1/2
D. n=4, l=3, m=4, s=-1/2

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Correct Option: A
Option A Explained: Correct. For n=3, l can be 0, 1, or 2, so l=2 is allowed; for l=2, m can range from -2 to +2, so m=0 is allowed; and s=+1/2 is always an allowed spin value, making this whole set valid.
Option B Explained: Incorrect. For n=3, the maximum allowed value of l is n−1=2, so l=3 is not permitted, making this entire set of quantum numbers invalid.
Option C Explained: Incorrect. For l=2, the magnetic quantum number m can only range from -2 to +2, so a value of m=4 is not permitted, making this set invalid.
Option D Explained: Incorrect. For l=3, the magnetic quantum number m can only range from -3 to +3, so a value of m=4 is not permitted, making this set invalid.

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