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JEE • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Mixed difficulty

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331 Easy
331 of 592
Application +4 | -1
If there are five radial nodes, when what can be the correct representation of the orbital for n=8n = 8
View answer and explanation
Correct option: A
8d
Answer A (8d) is correct because the number of radial nodes is given by n - l - 1, and solving for l gives the d orbital.
332 Easy
332 of 592
Application +4 | -1
The number of unpaired electrons in the Fe2+{Fe}^{2+} ion is
View answer and explanation
Correct option: B
4
Option B (4) is correct because the electronic configuration of Fe2+ has 4 unpaired electrons in the 3d orbitals.
333 Medium
333 of 592
Understanding +4 | -1
Angular momentum of electron in 3rd stationary orbit of hydrogen atom is
View answer and explanation
Correct option: B
3h2π\frac{{3h}}{{2\pi }}
Option B (3h2π\frac{{3h}}{{2\pi }}) is correct because the angular momentum of an electron in the nth orbit is given by nh2π\frac{{nh}}{{2\pi }}, and for the 3rd orbit, n=3.
334 Easy
334 of 592
Understanding +4 | -1
If the uncertainty in the position of electron is zero, the uncertainty in its momentum would be
View answer and explanation
Correct option: D
Infinite
Option D (Infinite) is correct because, according to the Heisenberg uncertainty principle, if the uncertainty in position is zero, the uncertainty in momentum must be infinite.
335 Medium
335 of 592
Remembering +4 | -1
What is the packet of energy called?
View answer and explanation
Correct option: B
photon
Option B (photon) is correct because a photon is defined as a packet of energy of electromagnetic radiation.
336 Medium
336 of 592
Analyzing +4 | -1
The last electron placed in the third (n=3) quantum shell for:
View answer and explanation
Correct option: C
Cu
Copper (Cu) has an electronic configuration of [Ar] 3d10 4s1, where the last electron fills the 3d subshell (n=3). Therefore, the correct answer is C.
337 Medium
337 of 592
Analyzing +4 | -1
Consider the following pairs of electrons
(A)  (a) n=3,  l=1, ml=1,  ms=+12           (b) n=3,  l=2, ml=1,  ms=+12 (B)(a)n=3,  l=2, ml=−2,  ms=−12       (b)n=3,  l=2, ml=−1,  ms=−12(C) (a)n=4,  l=2, ml=2,  ms=+12        (b)n=3,  l=2, ml=2,  ms=+12\begin{array}{cccccccccccccccccccc} {\mathrm{(A)}\,\,\mathrm{(a)}\,\mathrm{n = 3,}\,\,\mathrm{l = 1,}\,{\mathrm{m}_\mathrm{l}}\mathrm{ = 1,}\,\,{\mathrm{m}_\mathrm{s}}\mathrm{ = + }\frac{\mathrm{1}}{\mathrm{2}}}\\ {\,\,\,\,\,\,\,\,\,\,\,\mathrm{(b)}\,\mathrm{n = 3,}\,\,\mathrm{l = 2,}\,{\mathrm{m}_\mathrm{l}}\mathrm{ = 1,}\,\,{\mathrm{m}_\mathrm{s}}\mathrm{ = + }\frac{\mathrm{1}}{\mathrm{2}}}\\ {\,\mathrm{(B) (a) n = 3,}\,\,\mathrm{l = 2,}\,{\mathrm{m}_\mathrm{l}}\mathrm{ = } - \mathrm{2,}\,\,{\mathrm{m}_\mathrm{s}}\mathrm{ = } - \frac{\mathrm{1}}{\mathrm{2}}}\\ {\,\,\,\,\,\,\,\mathrm{(b) n = 3,}\,\,\mathrm{l = 2,}\,{\mathrm{m}_\mathrm{l}}\mathrm{ = } - 1\mathrm{,}\,\,{\mathrm{m}_\mathrm{s}}\mathrm{ = } - \frac{\mathrm{1}}{\mathrm{2}}}\\ {\mathrm{(C)}\,\mathrm{(a) n = 4,}\,\,\mathrm{l = 2,}\,{\mathrm{m}_\mathrm{l}}\mathrm{ = 2,}\,\,{\mathrm{m}_\mathrm{s}}\mathrm{ = + }\frac{\mathrm{1}}{\mathrm{2}}}\\ {\,\,\,\,\,\,\,\,\mathrm{(b) n = 3,}\,\,\mathrm{l = 2,}\,{\mathrm{m}_\mathrm{l}}\mathrm{ = 2,}\,\,{\mathrm{m}_\mathrm{s}}\mathrm{ = + }\frac{\mathrm{1}}{\mathrm{2}}} \end{array}
The pairs of electrons present in degenerate orbitals is/are:
View answer and explanation
Correct option: B
Only B
Option B (Only B) is correct because degenerate orbitals have the same n and l values but different m values, and only pair B satisfies this condition.
338 Medium
338 of 592
Understanding +4 | -1
In the Balmer series of H\mathrm{H}-spectrum, successive lines tend towards a limiting value of
View answer and explanation
Correct option: A
RˉHc4\frac{\bar{R}_{\mathrm{H}} c}{4}
Option A is correct because in the Balmer series, as n approaches infinity, the wavelength converges to a limit, and the frequency (and thus the energy) approaches RˉHc4\frac{{\bar{{R}}}_{{\mathrm{{H}}}} c}{{4}}$.
339 Easy
339 of 592
Remembering +4 | -1
The unit of wavelength (nm) is equal to:
View answer and explanation
Correct option: A
10 Å
Option A (10 Å) is correct because 1 nanometer (nm) is equal to 10 Angstroms (Å).
340 Medium
340 of 592
Understanding +4 | -1
The electronic configuration of P in H3PO4{\mathrm{H}_3}\mathrm{P}{\mathrm{O}_4} is
View answer and explanation
Correct option: C
1s2,2s2,2p61{s^2},2{s^2},2{p^6}
The correct answer is C (1s2,2s2,2p61{s^2},2{s^2},2{p^6}) because in H3PO4{\mathrm{H}_3}\mathrm{P}{\mathrm{O}_4}, phosphorus is in the +5 oxidation state (P5+{\mathrm{P}^{5 + }}), having lost its 3s and 3p electrons.
341 Medium
341 of 592
Analyzing +4 | -1
Statement - I: If a nucleus may undergo positron decay as well as K-capture, the amount of
energy released in positron decay is more than that in K-capture.

Statement - II: The loss in K-capture is equal to the difference in atomic masses of parent
nucleus and the daughter nucleus but in positron decay, additional mass loss equal to the
mass of two electrons also occur.
View answer and explanation
Correct option: A
If both Statement-I and Statement-II are true, and Statement-II is the correct
explanation of Statement-I
Option A is correct because both statements are true, and Statement II correctly explains why the energy released in positron decay is more than in K-capture.
342 Medium
342 of 592
Application +4 | -1
Total number of electrons having n+l=3 in Cr(z=24)n + l = 3\,in\,Cr\left( {z = 24} \right) atom in its ground state is
View answer and explanation
Correct option: A
8
Option A (8) is correct because the electron configuration of Cr is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵, and the combinations with n + l = 3 are 2p (6 electrons) and 3s (2 electrons), totaling 8 electrons.
343 Medium
343 of 592
Application +4 | -1
A laser used to read CDs emits red light of wavelength 700 nm. How many photons does it emit each second if its power is 1.0 W?
View answer and explanation
Correct option: D
3.52×1018 s−13.52 \times 10^{18} \mathrm{~s}^{-1}
Option D (3.52×1018 s−13.52 \times 10^{18} \mathrm{~s}^{-1}) is correct because it is the number of photons emitted per second calculated using the given power and wavelength of the laser.
344 Medium
344 of 592
Application +4 | -1
Uncertainty in position of a 0.25g particle is 10−5m.{10^{ - 5}}m. The uncertainty in its velocity will be
View answer and explanation
Correct option: B
2.1×10−262.1 \times {10^{ - 26}}
Option B (2.1×10−262.1 \times {10^{ - 26}}) is correct because it is the uncertainty in velocity calculated using the Heisenberg Uncertainty Principle.
345 Hard
345 of 592
Application +4 | -1
The total spin resulting from a d7{d^7} configuration is:
View answer and explanation
Correct option: D
±3/2\pm 3/2
Option D (±3/2\pm 3/2) is correct because the total spin is calculated by multiplying \±1/2\pm 1/2 by the number of unpaired electrons in the d7 configuration.
346 Easy
346 of 592
Application +4 | -1
An atom has the electronic configuration of 1s2,2s22p6,3s23p63d10,4s24p51 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^{10}, 4 s^2 4 p^5. Its atomic mass is 80 . Its atomic number and the number of neutrons in its nucleus shall be:
View answer and explanation
Correct option: A
35 and 45
Option A (35 and 45) is correct because the electronic configuration indicates 35 electrons, which is the atomic number, and the difference between the atomic mass (80) and atomic number (35) gives the number of neutrons (45).
347 Medium
347 of 592
Analyzing +4 | -1
n,  ln,\,\,l and mm values of orbitals A and B are given below.
nlmA320B310\begin{array}{cccccccccccccccccccc} {}&n&l&m \\ {\text{A}}&3&2&0 \\ {\text{B}}&3&1&0 \end{array}
Which statement about A and B orbitals is incorrect?

a. A has more energy than B

b. A possess electron density along axial direction only

c. B has one angular node and one radial node

d. Probability of finding electron density in B is zero along XZ-plane.
View answer and explanation
Correct option: D
b & d
The correct answer is D (b & d) because a 3d orbital does not necessarily have electron density only along the axial direction, and the probability of finding electron density in a 3p orbital is not zero along the XZ-plane (it's zero along the XY plane for 3pz).
348 Medium
348 of 592
Application +4 | -1
The radiation is emitted when a hydrogen atom goes from a higher energy state to a lower energy state. The wavelength of one line in visible region of atomic spectrum of hydrogen is 6.63×10−7 m6.63 \times 10^{-7} \mathrm{~m}. Energy difference between the two states is
View answer and explanation
Correct option: A
3.0×10−19 J3.0 \times 10^{-19} \mathrm{~J}
Option A (3.0 x 10⁻¹⁹ J) is correct because the energy difference between the two states is calculated using the formula E = hc/λ.
349 Medium
349 of 592
Analyzing +4 | -1
If E1,E2{E_1},{E_2} and E3{E_3} represents kinetic energy of electron, α−\alpha - particle and proton respectively (having same De Broglie wavelength) then
View answer and explanation
Correct option: A
E1>E3>E2{E_1} > {E_3} > {E_2}
Option A (E1>E3>E2{E_1} > {E_3} > {E_2}) is correct because for the same de Broglie wavelength, kinetic energy is inversely proportional to mass, and the electron has the smallest mass, followed by the proton, and then the alpha particle.
350 Medium
350 of 592
Understanding +4 | -1
The electron density of dxy{d_{xy}} orbital in yz plane is
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because the dxy{d_{xy}} orbital has lobes lying between the x and y axes, so the electron density in the yz plane (where x=0) is zero.
351 Easy
351 of 592
Understanding +4 | -1
Statement (S): The location and momentum of an electron in an orbital are complementry to each other.

Explanation (E): The statement is against Heisenberg uncertainty principle.
View answer and explanation
Correct option: A
S is correct but E is wrong
Option A (S is correct but E is wrong) is correct because while the statement reflects the uncertainty principle, the explanation misinterprets its meaning.
352 Medium
352 of 592
Remembering +4 | -1
The number of nodal planes ‘d’ orbital has
View answer and explanation
Correct option: C
Two
Option C (Two) is correct because a 'd' orbital has two nodal planes. The number of nodal planes is equal to the azimuthal quantum number (ll), which is 2 for d-orbitals.
353 Easy
353 of 592
Understanding +4 | -1
The ground state electronic configuration of Fe3+F{e^{3 + }} is
View answer and explanation
Correct option: C
[Ar]3d5\left[ {Ar} \right]3{d^5}
Option C is correct because Fe3+ has lost 3 electrons from Fe, resulting in a 3d5 configuration.
354 Easy
354 of 592
Remembering +4 | -1
Which is not basic postulate of Dalton’s atomic theory?
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Correct option: C
Atoms of all elements are alike, including their masses
The correct answer is C (Atoms of all elements are alike, including their masses) because Dalton's atomic theory states that atoms of the *same* element are alike, not atoms of *all* elements.
355 Easy
355 of 592
Remembering +4 | -1
The electron is
View answer and explanation
Correct option: B
β\beta-ray particle
Option B (β\beta-ray particle) is correct because an electron is a type of beta particle, which is a high-energy, high-speed electron or positron emitted during radioactive decay.
356 Hard
356 of 592
Application +4 | -1
Photons having energy equivalent to binding energy of 4th 4^{\text {th }} state of He+\mathrm{He}^{+}ion is used on the metal surface of work function 1.4eV1.4 \mathrm{eV}. If electrons are further accelerated through
the potential difference of 4 V4 \mathrm{~V} then the minimum value of de-Broglie wavelength associated with the electron is
View answer and explanation
Correct option: C
5 A5 \mathrm{~A}
Option C (5 A) is correct because the de-Broglie wavelength is calculated using the total kinetic energy of the electron.
357 Medium
357 of 592
Application +4 | -1
The wavelength of radiation emitted out in the transition n=4n=4 to n=1n=1 in Li2+\mathrm{Li}^{2+} ion is
View answer and explanation
Correct option: B
16135R\frac{16}{135 R}
Option B (16135R\frac{{16}}{{135 R}}) is correct because it is the reciprocal of the wavenumber calculated using the Rydberg formula for the given transition in Li2+\mathrm{Li}^{2+} ion.
358 Easy
358 of 592
Analyzing +4 | -1
An increasing order (lowest first) for the values of e/m for electron (e), proton (p), neutron (n) and alpha (α) particle is:
View answer and explanation
Correct option: B
n,α,p,e
Option B (n,α,p,e) is correct because the e/m ratio increases in the order: neutron (0), alpha, proton, and electron due to their respective charges and masses.
359 Easy
359 of 592
Understanding +4 | -1
Which of the following electron transition in H-atom will require the energy equivalent to its ionisation energy
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Correct option: D
From n=1n = 1 to n= ∝n = \, \propto
Option D is correct because ionization energy is the energy required to remove an electron from the ground state (n=1) to infinity (n=∞).
360 Hard
360 of 592
Application +4 | -1
What is the mass of a photon of sodium light with a wavelength of 5890 A0?(h=6.63×10−27erg−s){A^0}?(h = 6.63\mathrm{ \times }{10^{ - 27}}erg - s)
View answer and explanation
Correct option: D
3.752×10−33g3.752\mathrm{ \times }{10^{ - 33}}g
Option D (3.752×10−33g3.752\mathrm{ \times }{10^{ - 33}}g) is correct because the mass of a photon is calculated using the de Broglie wavelength equation.
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