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JEE • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 421-450 of 592 questions

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421 Medium
421 of 592
Remembering +4 | -1
Which experiment is responsible for finding out the charge on an electron?
View answer and explanation
Correct option: A
Millikan’s oil drop experiment
Option A (Millikan’s oil drop experiment) is correct because Millikan's oil drop experiment was designed to measure the charge of a single electron.
422 Hard
422 of 592
Analyzing +4 | -1
A single electron species in energy level (orbit number nn ) with energy XX was provided with excess of energy so that it jumps to higher energy level with energy YY. If it can emit radiations of six different wavelengths on de-excitement between these two energy levels, then the correct relation is
View answer and explanation
Correct option: C
XY=1+3n\sqrt{\frac{X}{Y}}=1+\frac{3}{n}
Option C (XY=1+3n\sqrt{\frac{X}{Y}}=1+\frac{3}{n}) is correct because it relates the energies of two levels with n and n+3, considering the emission of six different wavelengths.
423 Medium
423 of 592
Application +4 | -1
The energy of a photon is 3×10−123 \times {10^{ - 12}} ergs. What is its wavelength in nm? (h=6.62×10−27ergs,  c=3×1010cm/s)\left( {h = 6.62 \times {{10}^{ - 27}}\mathrm{ergs},\;c = 3 \times {{10}^{10}}\mathrm{cm}/\mathrm{s}} \right)
View answer and explanation
Correct option: A
662
The correct answer is A (662) because the wavelength is calculated using the energy of the photon and the Planck-Einstein relation, resulting in 662 nm.
424 Easy
424 of 592
Remembering +4 | -1
Possible number of orientations of a subshell is:
View answer and explanation
Correct option: C
2l+12l + 1
Option C (2l+12l + 1) is correct because the number of possible orientations of a subshell is determined by the magnetic quantum number (ml), which ranges from -l to +l, including 0, giving a total of 2l + 1 orientations.
425 Medium
425 of 592
Application +4 | -1
What is the maximum number of electrons in an atom that can have the following quantum numbers n=4,  m1=+1?n = 4,\;{m_1} = + 1?
View answer and explanation
Correct option: D
6
Option D (6) is correct because with n=4 and ml=+1, we are looking at the p-orbital, which can hold a maximum of 6 electrons.
426 Medium
426 of 592
Application +4 | -1
The radius of Bohr's first orbit in H-atom is 0.053 nm. The radius of second orbit in He+\mathrm{H}{\mathrm{e}^ + } would be:
View answer and explanation
Correct option: C
0.1060 nm
Option C (0.1060 nm) is correct because the radius of the nth Bohr orbit is proportional to n²/Z, where n is the principal quantum number and Z is the atomic number.
427 Medium
427 of 592
Understanding +4 | -1
Assertion: Dalton's Atomic Theory was later modified when the existence of subatomic particles was discovered.

Reason: The discovery of subatomic particles showed that atoms are not indivisible and can be further broken down into smaller particles.
View answer and explanation
Correct option: B
Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion
Option B (Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion) is correct because while both statements are true, the reason doesn't fully explain why Dalton's theory needed modification.
428 Easy
428 of 592
Remembering +4 | -1
The speed of the cathode rays is:
View answer and explanation
Correct option: B
Less than light
Option B (Less than light) is correct because cathode rays are streams of electrons, which have mass and cannot travel at the speed of light.
429 Medium
429 of 592
Remembering +4 | -1
The radius of the stationary state which is also called Bohr radius is given by the expression rn=n2a0\mathrm{r_n=n^2 a_0} where the value of a0\mathrm{a}_0 is
View answer and explanation
Correct option: A
52.9 pm52.9 \mathrm{~pm}
Option A (52.9 pm52.9 \mathrm{~pm}) is correct because the Bohr radius, denoted as a0a_0, has a value of 52.9 picometers.
430 Medium
430 of 592
Application +4 | -1
Which hydrogen like species will have same radius as that of Bohr orbit hydrogen atom?
View answer and explanation
Correct option: B
n=2,  Be3+n = 2,\mathrm{\;B}{\mathrm{e}^{3 + }}
Option B (n=2,  Be3+n = 2,\mathrm{\;B}{\mathrm{e}^{3 + }}) is correct because it satisfies the condition where the radius of the hydrogen-like species is equal to the Bohr radius of the hydrogen atom.
431 Hard
431 of 592
Application +4 | -1
The ratio of the difference in energy between the first and the second Bohr orbit to that between the second and the third Bohr orbit is
View answer and explanation
Correct option: D
275\frac{{27}}{5}
Option D (275\frac{{27}}{5}) is correct because it is the ratio calculated by using the energy difference formula for Bohr orbits.
432 Medium
432 of 592
Remembering +4 | -1
The expression for Bohr's frequency rule is
View answer and explanation
Correct option: A
v=ΔEh=E2−E1hv=\frac{\Delta E}{h}=\frac{E_2-E_1}{h}
Option A (v=ΔEh=E2−E1hv=\frac{\Delta E}{h}=\frac{E_2-E_1}{h}) is correct because it represents Bohr's frequency rule, which relates the frequency of emitted or absorbed radiation to the energy difference between two energy levels.
433 Medium
433 of 592
Application +4 | -1
It the radius of first Bohr orbit be ao{a_\mathrm{o}}, then the radius of the third orbit would be
View answer and explanation
Correct option: C
9ao9{a_\mathrm{o}}
Option C (9ao9{a_\mathrm{o}}) is correct because the radius of the nth Bohr orbit is proportional to n squared.
434 Easy
434 of 592
Understanding +4 | -1
Two electron in an atm of an element cannot have:
View answer and explanation
Correct option: D
An identical set of quantum numbers
Option D (An identical set of quantum numbers) is correct because the Pauli Exclusion Principle states that no two electrons in an atom can have the same set of four quantum numbers.
435 Medium
435 of 592
Understanding +4 | -1
S-I: Wavelength of limiting line of lyman series is less than wavelength of limiting line of Balmer series.

S-II: Rydberg constant value is same for all elements.
View answer and explanation
Correct option: C
S-I is correct and S-II is incorrect
The correct answer is C because the limiting line of the Lyman series has a shorter wavelength than the Balmer series, and the Rydberg constant depends on the element.
436 Easy
436 of 592
Understanding +4 | -1
The maximum number of atomic orbitals associated with a principal quantum number 5 is:
View answer and explanation
Correct option: D
25
Option D (25) is correct because the maximum number of atomic orbitals associated with a principal quantum number n is n², and for n=5, it is 5² = 25.
437 Medium
437 of 592
Remembering +4 | -1
The maximum number of electrons accommodated in all 5f-orbitals
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Correct option: C
14
Option C (14) is correct because each orbital can accommodate a maximum of 2 electrons, and there are 7 f-orbitals (5f), so 7 * 2 = 14 electrons.
438 Easy
438 of 592
Remembering +4 | -1
Which of the given radiations is having high penetrating power and not affected by electrical and magnetic field?
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Correct option: C
Gamma rays
Option C (Gamma rays) is correct because gamma rays have high penetrating power and are not affected by electric or magnetic fields due to being uncharged photons.
439 Hard
439 of 592
Application +4 | -1
O2\mathrm{O}_2 undergoes photochemical dissociation into one normal oxygen atom and one oxygen atom, 1.2 eV\mathrm{eV} more energetic than normal. The dissociation of O2\mathrm{O}_2 into two normal atoms of oxygen requires 482.5 kJ/mol482.5 \mathrm{~kJ} / \mathrm{mol}. The maximum wavelength effective for photochemical dissociation of O2\mathrm{O}_2 is (1eV=(1 \mathrm{eV}= 96.5 kJ/mol96.5 \mathrm{~kJ} / \mathrm{mol} )
View answer and explanation
Correct option: D
200 nm200 \mathrm{~nm}
Option D (200 nm) is correct because it is calculated by finding the total energy required for dissociation and then using that energy to determine the maximum wavelength.
440 Medium
440 of 592
Application +4 | -1
In H\mathrm{H}-atom, the wave number ratio is 108:7108: 7 is for
View answer and explanation
Correct option: C
first Lyman and first Paschen transition.
Option C (first Lyman and first Paschen transition) is correct because the given wave number ratio matches the ratio calculated for the first transitions of the Lyman and Paschen series in the hydrogen atom spectrum.
441 Easy
441 of 592
Remembering +4 | -1
The energy of an electron in nth {n}^{\text {th }} orbit of hydrogen atom is
View answer and explanation
Correct option: C
13.6n2eV\frac{13.6}{n^2} \mathrm{eV}
Option C (13.6n2eV\frac{{13.6}}{{n^2}} \mathrm{{eV}}) is correct because the energy of an electron in the nth orbit of a hydrogen atom is given by the formula E=−13.6/n2E = -13.6/n^2 eV.
442 Medium
442 of 592
Analyzing +4 | -1
Assertion: Ratio of anode rays is different for different gases.
Reason: Proton is the fundamental particle present in gases.
View answer and explanation
Correct option: B
Both assertion and reason are true but reason is not the correct explanation of assertion
Option B (Both assertion and reason are true but reason is not the correct explanation of assertion) is correct because both statements are true, but the reason doesn't explain the assertion.
443 Easy
443 of 592
Understanding +4 | -1
The correct designation of an electron with n=4,l=3,m=2,n = 4,l = 3,m = 2, and s=1/2s = 1/2 is:
View answer and explanation
Correct option: B
4f
Option B (4f) is correct because n=4 and l=3 corresponds to the 4f subshell.
444 Medium
444 of 592
Application +4 | -1
Angular momentum of 2nd Bohr orbit of H – atom is xx . Then find out angular momentum in first excited state of Li2+L{i^{2 + }} is
View answer and explanation
Correct option: D
xx
The correct answer is D (xx) because the angular momentum depends on the principal quantum number (n) and is independent of the atomic number (Z). Both H in the 2nd orbit and Li²⁺ in the first excited state have n=2.
445 Medium
445 of 592
Understanding +4 | -1
If an electron has spin quantum number of +12+ \frac{1}{2} and a magnetic quantum number of -1, it cannot be represented in an
View answer and explanation
Correct option: A
s-orbital
Option A (s-orbital) is correct because an s-orbital has an angular momentum quantum number (l) of 0, and therefore a magnetic quantum number (m) of 0, so m = -1 is not possible for an s-orbital.
446 Medium
446 of 592
Application +4 | -1
If the mass of an electron is 9.11×10−31 kg,9.11 \times {10^{ - 31}}\,{\text{kg}}, and uncertainty in position is 10−8m,{10^{ - 8}}{\text{m}}, the uncertainty in velocity would be:
View answer and explanation
Correct option: B
5.8×103ms−15.8 \times {10^3}{\text{m}}{{\text{s}}^{ - 1}}
Option B (5.8×103 ms−15.8 \times {10^3}\,{\text{m}}{{\text{s}}^{ - 1}}) is correct because it is calculated using the Heisenberg uncertainty principle.
447 Easy
447 of 592
Understanding +4 | -1
S-I: There is one nodal regions in 3s-orbital.

S-II: There is one nodal plane in 3s-orbital.
View answer and explanation
Correct option: B
Both S-I and S-II are incorrect
Option B (Both S-I and S-II are incorrect) is correct because a 3s-orbital has two spherical nodes (nodal regions) and no nodal planes.
448 Medium
448 of 592
Understanding +4 | -1
For 19F−1,16O−2,20Ne,{ }^{19} {F}^{-1},{ }^{16} {O}^{-2}, { }^{20} {Ne}, choose the correct statement.
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because all three species, 19F-1, 16O-2, and 20Ne, have 10 electrons each, making them isoelectronic.
449 Easy
449 of 592
Analyzing +4 | -1
Which set has the same number of unpaired electrons in their ground state
View answer and explanation
Correct option: D
N,P,V
The correct answer is N, P, V because Nitrogen, Phosphorus, and Vanadium each have 3 unpaired electrons in their ground state electron configurations.
450 Medium
450 of 592
Application +4 | -1
The mass of a photon with wavelength 3.6 Å  is
View answer and explanation
Correct option: C
6.135×10−33kg6.135 \times {10^{ - 33}}\mathrm{kg}
Option C (6.135×10−33kg6.135 \times {10^{ - 33}}\mathrm{kg}) is correct because it is the mass calculated using the energy-mass equivalence and the photon's wavelength.
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