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JEE • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Medium

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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 151-180 of 297 questions

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151 Medium
151 of 297
Understanding +4 | -1
The ratio of the radius of the orbit for the electron orbiting the hydrogen nucleus to that of an electron orbiting a deuterium nucleus is:
View answer and explanation
Correct option: A
1∶1
The correct answer is A (1∶1) because the radius of the orbit depends on the atomic number (Z) and the principal quantum number (n), and since both hydrogen and deuterium have Z=1, the ratio is 1:1.
152 Medium
152 of 297
Application +4 | -1
Calculate the number of electrons which will together weigh one gram.
View answer and explanation
Correct option: A
1.098×10271.098 \times 10^{27}
The correct answer is A (1.098×10271.098 \times 10^{27}) because it correctly calculates the number of electrons required to weigh one gram using the mass of a single electron.
153 Medium
153 of 297
Analyzing +4 | -1
According to Bohr's atomic theory, which of the following is correct ?
View answer and explanation
Correct option: C
Frequency of revolution of electron in an orbit ∝Z2n3\propto \frac{Z^2}{n^3}
Option C is correct because the stability of the conjugate base determines the base strength; alkyl carbanions are more basic than vinyl carbanions, which are more basic than acetylide carbanions.
154 Medium
154 of 297
Analyzing +4 | -1
For an atom or ion having single electron, compare the energies of the following orbitals.

S1=\mathrm{S}_1=A spherically symmetrical orbital having two spherical nodes.

S2=\mathrm{S}_2= An orbital which is double dumb-bell and has no radial node.

S3=\mathrm{S}_3= An orbital with orbital angular momentum zero and three radial nodes.

S4=\mathrm{S}_4= An orbital having one planar and one radial node.
View answer and explanation
Correct option: B
S1=S2=S4<S3\mathrm{S}_1=\mathrm{S}_2=\mathrm{S}_4 < \mathrm{S}_3
Option A is correct because for a single-electron system, energy depends only on the principal quantum number (n), so orbitals with the same n have the same energy.
155 Medium
155 of 297
Understanding +4 | -1
Given below are two statements:
Statement I Rutherford’s gold foil experiment cannot explain the line spectrum of hydrogen atom.
Statement II Bohr’s model of hydrogen atom contradicts Heisenberg’s uncertainly principle.
In the light of the above statements, choose the most appropriate answer from the options given below:
View answer and explanation
Correct option: D
Both Statement I and statement II are true
Option D is correct because both statements are true. Rutherford's model couldn't explain line spectra, and Bohr's model contradicts the uncertainty principle.
156 Medium
156 of 297
Analyzing +4 | -1
What is the difference between two species if one has atomic mass =14=14 and atomic number =7=7 whereas the other has atomic mass =14=14 and atomic number =6=6 ?
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because a difference in atomic number implies a difference in the number of protons and electrons, and consequently, the number of neutrons.
157 Medium
157 of 297
Application +4 | -1
When electrons are de-exciting to the ground state from nnth orbit of hydrogen atoms, 15 spectral lines are formed. The shortest wavelength among these will be
View answer and explanation
Correct option: D
3635R\frac{36}{35 R}
Option D (3635R\frac{{36}}{{35 R}}) is correct because it is obtained by calculating the wavelength for the transition from n=6 to n=1 using the Rydberg formula.
158 Medium
158 of 297
Understanding +4 | -1
Arrange the following type of radiations in increasing order of frequency.

A. Microwaves

B. Amber light from traffic signal

C. Radiation from FM radio

D. Cosmic rays

E. X – rays
View answer and explanation
Correct option: B
C < A < B < E < D
The correct answer is C (C < A < B < D < E) because it represents the correct increasing order of frequency for the given types of electromagnetic radiation.
159 Medium
159 of 297
Analyzing +4 | -1
For which of the following element, all of its existing ion Mx+\mathrm{M}^{{x}+} will be diamagnetic?
View answer and explanation
Correct option: D
Na\mathrm{Na}
Option D is correct because Na+ has a noble gas configuration with no unpaired electrons, making it diamagnetic.
160 Medium
160 of 297
Analyzing +4 | -1
Assertion :- Mass numbers of most of the elements are fractional.

Reason :- Mass numbers are obtained by comparing with the mass number of carbon taken as
12.
View answer and explanation
Correct option: D
If both Assertion & Reason are False
Option D is correct because both the Assertion and Reason are false. Mass numbers are integers, and atomic masses are obtained by comparing with the mass of carbon-12.
161 Medium
161 of 297
Application +4 | -1
The radii of two of the first four Bohr&#39;s orbits of the hydrogen atom are in the ratio 1∶4. The energy difference between them may be:
View answer and explanation
Correct option: B
either 2.55 eV or 10.2 eV
The maximum number of emission lines observed is 4 because there is only one hydrogen atom, so only direct transitions to lower levels are possible.
162 Medium
162 of 297
Understanding +4 | -1
Maximum number of electrons which can be accommodated in a g-subshell is:
View answer and explanation
Correct option: B
18
Option B (18) is correct because a g-subshell has l=4, and the number of electrons it can accommodate is 2(2l+1) = 2(2*4+1) = 18.
163 Medium
163 of 297
Understanding +4 | -1
(I)   n=3,   l=2,  m1 =−2\mathrm{(I)}\,\,\,n = 3,\,\,\,l = 2,\,\,{m_1}\, = - 2
(II)   n=3,   l=1,   m1=0\mathrm{(II)}\,\,\,n = 3,\,\,\,l = 1,\,\,\,{m_1} = 0
(III)   n=3,  l=0,  m1=−1\mathrm{(III)}\,\,\,n = 3,\,\,l = 0,\,\,{m_1} = - 1
(IV)   n=3,  l=2,  m1=0\mathrm{(IV)}\,\,\,n = 3,\,\,l = 2,\,\,{m_1} = 0
(V)   n=3,  l=3,  m1=−2\mathrm{(V)}\,\,\,n = 3,\,\,l = 3,\,\,{m_1} = - 2
Of these question state designation which does not describe an allowed state for an electron in an atom?
View answer and explanation
Correct option: B
III and V
Option B (III and V) is correct because in option III, mlm_l cannot be -1 when l=0, and in option V, l cannot be 3 when n=3.
164 Medium
164 of 297
Understanding +4 | -1
Which of the following is not possible for 4p4 p or 3d3 d electrons?
View answer and explanation
Correct option: C
n=3,l=3,m=+3,s=+1/2n=3, l=3, m=+3, s=+1 / 2
Option C (n=3,l=3,m=+3,s=+1/2n=3, l=3, m=+3, s=+1 / 2) is correct because for n=3, the possible values of l are 0, 1, and 2. l=3 is not possible.
165 Medium
165 of 297
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.
166 Medium
166 of 297
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.
167 Medium
167 of 297
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.
168 Medium
168 of 297
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.
169 Medium
169 of 297
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}}$.
170 Medium
170 of 297
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.
171 Medium
171 of 297
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.
172 Medium
172 of 297
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.
173 Medium
173 of 297
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.
174 Medium
174 of 297
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.
175 Medium
175 of 297
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).
176 Medium
176 of 297
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.
177 Medium
177 of 297
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.
178 Medium
178 of 297
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.
179 Medium
179 of 297
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.
180 Medium
180 of 297
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.
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