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JEE • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Mixed difficulty

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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 1-30 of 592 questions
1 Easy
1 of 592
Application +4 | -1
Which of the following is paramagnetic?
View answer and explanation
Correct option: A
Ni++\mathrm{Ni}^{++}
Option A (Ni++\mathrm{Ni}^{++}) is correct because it has unpaired electrons in its d-orbital configuration, making it paramagnetic.
2 Medium
2 of 592
Application +4 | -1
The time taken by the electron in one complete revolution in the nth n^{\text {th }} Bohr's orbit of the hydrogen atom is
View answer and explanation
Correct option: B
Directly proportionalto n3{n}^3
Option B (Directly proportionalto n3{n}^3) is correct because the time period is directly proportional to the cube of the principal quantum number.
3 Medium
3 of 592
Remembering +4 | -1
The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency ' A\mathrm{A} ' ×1012\times 10^{12} hertz and that has a radiant intensity in that direction of 1′B′\frac{1}{{ }^{\prime} \mathrm{B}^{\prime}} watt per steradian. 'A' and 'B' are respectively
View answer and explanation
Correct option: B
540 and 683
Option B (540 and 683) is correct because the candela is defined with a frequency of 540 × 1012 Hz and a radiant intensity of 1/683 watt per steradian.
4 Easy
4 of 592
Remembering +4 | -1
Quantum number which defines the orientation of orbital present in a subshell is
View answer and explanation
Correct option: C
Magnetic quantum number (m)
Option C (Magnetic quantum number (m)) is correct because the magnetic quantum number defines the orientation of an orbital in space.
5 Easy
5 of 592
Understanding +4 | -1
Maximum number of electrons associated with ℓ=1,  m=0\ell = 1,\,\,m = 0 in magnesium atom is
View answer and explanation
Correct option: B
2
The correct answer is B (2) because only the two electrons in the 2p orbital with m=0 satisfy the given conditions in Magnesium's electronic configuration.
6 Hard
6 of 592
Application +4 | -1
The nucleus and an atom can be assumed to be spherical. The radius of the nucleus of mass no. A is given by 1.25×10−13×A1/3cm.1.25 \times {10^{ - 13}} \times {A^{1/3}}\mathrm{cm}.. The atomic radius of atom is 1 Å. If the mass no. is 64, the fraction of the atomic volume that is occupied by nucleus is:
View answer and explanation
Correct option: D
1.25×10−131.25 \times {10^{ - 13}}
Option D (1.25×10−131.25 \times {10^{ - 13}}) is correct because it represents the ratio of the volume of the nucleus to the volume of the atom, calculated using their respective radii and the formula for the volume of a sphere.
7 Easy
7 of 592
Understanding +4 | -1
The nucleus of the element having atomic number 25 and atomic weight 55 will contain
View answer and explanation
Correct option: A
25 protons and 30 neutrons
Option A (25 protons and 30 neutrons) is correct because the number of protons equals the atomic number, and the number of neutrons is the difference between the atomic weight and the atomic number.
8 Medium
8 of 592
Remembering +4 | -1
Hamiltonian operator H^\hat{H} is the sum of energy operator. These are
View answer and explanation
Correct option: C
kinetic and potential
Option C (kinetic and potential) is correct because the Hamiltonian operator represents the total energy of a system, which is the sum of kinetic and potential energies.
9 Medium
9 of 592
Application +4 | -1
Radiation corresponding to the transition n=4\mathrm{n}=4 to n=2\mathrm{n}=2 in hydrogen atoms falls on a certain metal (work function =2.0eV=2.0 \mathrm{eV}). The maximum kinetic energy of the photoelectrons will be:
View answer and explanation
Correct option: A
0.55 eV0.55 \mathrm{~eV}
Option A (0.55 eV0.55 \mathrm{~eV}) is correct because the maximum kinetic energy of the photoelectrons is the energy of the incident radiation minus the work function of the metal.
10 Hard
10 of 592
Understanding +4 | -1
In case of photoelectric effect, slope of V0\mathrm{V}_0 vs v\mathrm{v} curve is (where V0\mathrm{V}_0-stopping potential, v=\mathrm{v}= subjected frequency)
View answer and explanation
Correct option: B
he\frac{\mathrm{h}}{\mathrm{e}}
Option B (he\frac{\mathrm{h}}{\mathrm{e}}) is correct because the slope of the stopping potential vs. frequency curve in the photoelectric effect is given by Planck's constant divided by the elementary charge.
11 Medium
11 of 592
Remembering +4 | -1
3pz3p_z orbital has

(a) One radial node

(b) Two radial nodes

(c) One angular node

(d) Two angular nodes
View answer and explanation
Correct option: B
a and c only
Option B (a and c only) is correct because a 3pz orbital has one radial node and one angular node.
12 Medium
12 of 592
Application +4 | -1
The radius of second Bohr’s orbit of hydrogen atom is
View answer and explanation
Correct option: C
0.2116 nm
Option C (0.2116 nm) is correct because it is the result of applying the formula for the radius of the nth Bohr orbit with n=2 and Z=1.
13 Medium
13 of 592
Application +4 | -1
The ratio of (E2−E1) to (E4−E3)\left( {{E_2} - {E_1}} \right)\,to\,\left( {{E_4} - {E_3}} \right) for the Hydrogen atom is equal to
View answer and explanation
Correct option: B
15
Option D (12) is closest to the correct answer of approximately 15.4, which is the ratio of the energy differences for the given transitions in a hydrogen atom.
14 Medium
14 of 592
Remembering +4 | -1
The region in the electromagnetic spectrum where the Balmer series lines appear is
View answer and explanation
Correct option: D
visible
Option D (visible) is correct because the Balmer series lines appear in the visible region of the electromagnetic spectrum.
15 Easy
15 of 592
Understanding +4 | -1
Which is not permissible subshell?
View answer and explanation
Correct option: A
2d
Option A (2d) is correct because the d subshell starts from the third energy level (n=3), so 2d is not a permissible subshell.

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16 Easy
16 of 592
Analyzing +4 | -1
Statement (S): H-atom has only one electron in its orbit, but several spectral lines are noticed

Explanation (E): The H\mathrm{H}-spectra is observed in H2\mathrm{H}_2 gas
View answer and explanation
Correct option: C
Both S and E are correct and E is correct explanation of S
Option A is incorrect. The explanation is wrong, it should be that the H-spectra is observed in a sample of H atoms, not H2 gas. The spectral lines arise from electronic transitions within the H atom.
17 Medium
17 of 592
Remembering +4 | -1
Bohr frequency condition is given by
View answer and explanation
Correct option: A
v=ΔEhv = \frac{{\Delta E}}{h}
Option A (v=ΔEhv = \frac{{\Delta E}}{h}) is correct because it represents the Bohr frequency condition, relating the frequency of emitted radiation to the energy difference between energy levels.
18 Hard
18 of 592
Application +4 | -1
Uncertainty in measuring the speed of a particle is numerically equal to the uncertainty in measuring its position. The value of these uncertainties will be
View answer and explanation
Correct option: D
Both (1) and (3)
Option D (Both (1) and (3)) is correct because the uncertainty principle states that the product of uncertainties in position and momentum must be greater than or equal to h4π\frac{h}{4 \pi}, implying the uncertainty can be equal to or greater than h4πm\sqrt{\frac{h}{4 \pi m}}.
19 Easy
19 of 592
Application +4 | -1
When the value of azimuthal quantum number is 3, the maximum and minimum values of spin multiplicity are
View answer and explanation
Correct option: B
8,1
Option B (8,1) is correct because when the azimuthal quantum number is 3 (f orbital), the maximum spin multiplicity is 8 (7 unpaired electrons) and the minimum is 1 (all paired).
20 Easy
20 of 592
Remembering +4 | -1
The orientation of an atomic orbital is governed by
View answer and explanation
Correct option: C
m
Option C (m) is correct because the orientation of an atomic orbital in space is governed by the magnetic quantum number (m).
21 Medium
21 of 592
Analyzing +4 | -1
Which is not isoelectronic with the other three?
View answer and explanation
Correct option: A
CO\mathrm{CO}
Option A (CO) is correct because it has 14 electrons, while the other three species have 16 electrons.
22 Easy
22 of 592
Remembering +4 | -1
The energy of different orbitals in an atom or ion having only one electron depends on
View answer and explanation
Correct option: A
nn only
Option A (nn only) is correct because for a one-electron system, the energy of an orbital depends only on the principal quantum number (nn).
23 Medium
23 of 592
Understanding +4 | -1
Which of the following sets of quantum numbers is permissible for an electron in an atom?
View answer and explanation
Correct option: C
n=2,ℓ=1,m=0,s=+1/2{n}=2, \ell=1, {m}=0, {s}=+1 / 2
Option C (n=2,ℓ=1,m=0,s=+1/2{n}=2, \ell=1, {m}=0, {s}=+1 / 2) is correct because it follows the rules for permissible quantum numbers: n can be any positive integer, l ranges from 0 to n-1, m ranges from -l to +l, and s is +1/2 or -1/2.
24 Easy
24 of 592
Understanding +4 | -1
Statement (S): Number of waves in an orbit of atom is equal to number of that orbit.

Explanation (E): Number of waves in an orbit is derived by 2πrnλ\frac{2 \pi r_n}{\lambda}
View answer and explanation
Correct option: C
Both S and E are correct and E is correct explanation of S
Option C is correct because the number of waves in an orbit is equal to the orbit number, and this is derived from the relationship between the circumference of the orbit and the wavelength of the electron.
25 Easy
25 of 592
Remembering +4 | -1
Normally, the time taken in the transition is :
View answer and explanation
Correct option: D
10−8sec{10^{ - 8}}\mathrm{sec}
Option D (10−8sec{10^{ - 8}}\mathrm{sec}) is correct because electronic transitions in atoms typically occur on the order of nanoseconds.
26 Medium
26 of 592
Understanding +4 | -1
Which of the following orbital is represented by the complete wave function ψ410\psi_{410} ?
View answer and explanation
Correct option: C
4p4 p
Option C (4p) is correct because the quantum numbers n=4 and l=1 correspond to the 4p orbital.
27 Medium
27 of 592
Analyzing +4 | -1
Of the species, NO,  NO+,  NO2+  and  NO−NO,\,\,N{O^ + },\,\,N{O^{2 + }}\,\,\mathrm{and}\,\,N{O^ - } the one with minimum bond strength is
View answer and explanation
Correct option: D
NO−N{O^ - }
Option D (NO−NO^−) is correct because it has the lowest bond order (2) compared to the other species, resulting in the minimum bond strength.
28 Medium
28 of 592
Understanding +4 | -1
The maximum tendency to form the gaseous unipositive ion is for the element with configuration
View answer and explanation
Correct option: B
1s22s22p63s23p11 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}
Option B (1s22s22p63s23p11 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}) is correct because elements with electronic configurations close to noble gases readily lose electrons to achieve a stable configuration, and this configuration represents Aluminum which easily forms Al+.
29 Medium
29 of 592
Understanding +4 | -1
4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is
View answer and explanation
Correct option: B
5f>6p>5p>4d5f > 6p > 5p > 4d
Option B (5f>6p>5p>4d5f > 6p > 5p > 4d) is correct because the energy of an orbital is determined by the (n+l) rule, and when (n+l) is the same, the orbital with higher n has higher energy.
30 Medium
30 of 592
Application +4 | -1
Ratio of energy of photon of wavelength 3000A∘3000\mathop A\limits^ \circ and 6000A∘6000\mathop A\limits^ \circ is
View answer and explanation
Correct option: B
2:12: 1
Option B (2:1) is correct because the energy of a photon is inversely proportional to its wavelength, so the ratio of energies is the inverse ratio of wavelengths.
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