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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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121 Medium
121 of 297
Application +4 | -1
If the de-Broglie wavelength of a particle of mass mm is 100 times its velocity, then its value in terms of its mass ( m{m} ) and Planck's constant (h)(h) is
View answer and explanation
Correct option: B
10hm10 \sqrt{\frac{{h}}{{m}}}
Option B (10hm10 \sqrt{\frac{{h}}{{m}}}) is correct because it is derived from the de Broglie wavelength formula, λ=hmv\lambda = \frac{h}{mv}, where the velocity is given as λ100\frac{\lambda}{100}.
122 Medium
122 of 297
Application +4 | -1
The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species.
View answer and explanation
Correct option: A
1632 S2−{ }_{16}^{32} \mathrm{~S}^{2-}
Option A (1632 S2−{}_{16}^{32} \mathrm{~S}^{2-}) is correct because the species has 16 protons, 16 neutrons, and 18 electrons, indicating sulfur with a mass number of 32 and a 2- charge.
123 Medium
123 of 297
Remembering +4 | -1
The total number of nodes in a subshell are given by
View answer and explanation
Correct option: C
n−l−1{n}-l-1
Option C (n−l−1{n}-l-1) is correct because the total number of nodes in a subshell is the sum of radial and angular nodes, where radial nodes are n−l−1{n}-l-1.
124 Medium
124 of 297
Analyzing +4 | -1
Which of the following is correctly matched?
View answer and explanation
Correct option: A
Momentum of H atom when electrons return from n=2 to n=1: 3Rh4\frac{{3Rh}}{4}
Option A is correctly matched because the momentum of the H atom equals the momentum of the emitted photon when the electron transitions from n=2 to n=1, which is calculated to be 3Rh4\frac{{3Rh}}{{4}}.
125 Medium
125 of 297
Understanding +4 | -1
For a d – electron, the orbital angular momentum is
View answer and explanation
Correct option: A
6 ℏ\sqrt 6 \,\hbar
Option A (6 ℏ\sqrt 6 \,\hbar) is correct because the orbital angular momentum for a d-electron (l=2) is given by l(l+1)ℏ=2(2+1)ℏ=6ℏ\sqrt{l(l+1)}\hbar = \sqrt{2(2+1)}\hbar = \sqrt{6}\hbar.
126 Medium
126 of 297
Analyzing +4 | -1
Which of the following is not permissible arrangement of electrons in an atom
View answer and explanation
Correct option: A
n=3,  l=2,  m=−3,  s=−12n = 3,\,\,l = 2,\,\,m = - 3,\,\,s = - \frac{1}{2}
Option A is correct because for n=3 and l=2, m can only range from -2 to +2, therefore m=-3 is not permissible.
127 Medium
127 of 297
Analyzing +4 | -1
Statement I: Fe3+\mathrm{Fe}^{3+} ion is more stable than Fe2+\mathrm{Fe}^{2+} ion in the gaseous state.
Statement II: Fe3+\mathrm{Fe}^{3+} ion has more number of unpaired electrons than Fe2+\mathrm{Fe}^{2+} ion.
View answer and explanation
Correct option: B
Both Statement I and statement II are true
The correct answer is B (Both Statement I and statement II are true) because Fe3+ has a half-filled d orbital configuration, making it more stable, and it has more unpaired electrons than Fe2+.
128 Medium
128 of 297
Understanding +4 | -1
Which of the following orbital have nodal plane in XY-plane?
View answer and explanation
Correct option: D
dyzd_{yz}
Option D (dyzd_{yz}) is correct because the dyzd_{yz} orbital lies in the YZ plane, making the XY plane a nodal plane where the probability of finding an electron is zero.
129 Medium
129 of 297
Application +4 | -1
In hydrogen atomic spectrum, a series limit is found at 12186.3 cm−1c{m^{ - 1}}. Then, it belongs to
View answer and explanation
Correct option: C
Paschen series
Option C (Paschen series) is correct because the series limit corresponds to n1=3n_1 = 3, which defines the Paschen series.
130 Medium
130 of 297
Application +4 | -1
The most probable radius (in pm) for finding the electron inHe +{ }^{+}is:
View answer and explanation
Correct option: C
26.526.5
Option C (26.5) is correct because the most probable radius is calculated using the formula a0/Z, where a0 is the Bohr radius (52.9 pm) and Z is the atomic number.
131 Medium
131 of 297
Understanding +4 | -1
Which of the following has the highest value of specific charge?
View answer and explanation
Correct option: D
Electron
Option D (Electron) is correct because the electron has the highest charge-to-mass ratio due to its small mass.
132 Medium
132 of 297
Remembering +4 | -1
The momentum of a photon is p,p, the energy associated with photon is given by:
View answer and explanation
Correct option: B
Ep\frac{E}{p}
Option B (Ep\frac{E}{p}) is correct because the energy of a photon is related to its momentum by the equation E = pc, which gives c = E/p.
133 Medium
133 of 297
Analyzing +4 | -1
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason
R.
Assertion A: Energy of 2s orbital of hydrogen atom is greater than that of 2s orbital of lithium.
Reason R : Energies of the orbitals in the same subshell decrease with increase in the atomic number.
In the light of the above statements, choose the correct answer from the options given below.
View answer and explanation
Correct option: A
Both A and R are true and R is the correct explanation of A.
Option A is correct because both the assertion and reason are true, and the reason correctly explains the assertion. The energy of a 2s orbital decreases with increasing atomic number due to the increased nuclear charge.
134 Medium
134 of 297
Application +4 | -1
An electron revolves round Li2+\mathrm{Li}^{2+} nucleus at a distance of 1.587 A∘1.587\,\mathop {\text{A}}\limits^ \circ. The speed of electron should be
View answer and explanation
Correct option: A
2.188×106 m/s2.188 \times 10^6 \mathrm{~m} / \mathrm{s}
Option A (2.188×106m/s2.188 \times 10^6 m/s) is correct because it calculates the speed of the electron using the Bohr model and the given radius for Li²⁺.
135 Medium
135 of 297
Application +4 | -1
Number waves in a Bohr orbit of H – atom is 3. Its potential energy would be
View answer and explanation
Correct option: B
−3.02eV- 3.02\mathrm{ }eV
Option B (−3.02eV- 3.02\mathrm{ }eV) is correct because the potential energy is calculated from the total energy of the electron in the given Bohr orbit.
136 Medium
136 of 297
Understanding +4 | -1
Select the correct statements from the following.

A. Atoms of all elements are composed of two fundamental particles.

B. The mass of the electron is 9.10939×10−31 kg9.10939 \times 10^{-31} \mathrm{~kg}.

C. All the isotopes of a given element show same chemical properties.

D. Protons and electrons are collectively known as nucleons.

E. Dalton's atomic theory, regarded the atom as an ultimate particles of matter.

Choose the correct answer from the options given below.
View answer and explanation
Correct option: B
B, C and E only
Option A (4) is correct because each boron atom in diborane (B2H6B_2H_6) is sp3sp^3 hybridized, involving four orbitals.
137 Medium
137 of 297
Application +4 | -1
The de Broglie wavelength associated with a ball of mass 200 g200 {~g} and moving at a speed of 5 m/h5 {~m} / {h}, is of the order of (h=6.625×10−34Js\left({h}=6.625 \times 10^{-34} {Js}\right. )
View answer and explanation
Correct option: C
10−25 m10^{-25} {~m}
Option C (10−25 m10^{-25} {~m}) is correct because it calculates the de Broglie wavelength using the formula λ=hmv\lambda = \frac{{h}}{{mv}} with appropriate unit conversions.
138 Medium
138 of 297
Remembering +4 | -1
The radii of an atom and atomic nucleus is of the order of
View answer and explanation
Correct option: C
10−10 m10^{-10} \mathrm{~m} and 10−15 m10^{-15} \mathrm{~m}
Option C (10−10 m10^{-10} \mathrm{~m} and 10−15 m10^{-15} \mathrm{~m}) is correct because it represents the approximate order of magnitude for the radii of an atom and its nucleus, respectively.
139 Medium
139 of 297
Application +4 | -1
The wavelength of a spectral line emitted by hydrogen atom in the Lyman series is 1615Rcm\frac{{16}}{{15R}}\mathrm{cm}. What is the value of n2{n_2}?(R=Rydberg constant)
View answer and explanation
Correct option: C
4
Option C (4) is correct because it is derived from the Rydberg formula for the Lyman series.
140 Medium
140 of 297
Application +4 | -1
What is the ratio of maximum λ\lambda of Lyman and Balmer series
View answer and explanation
Correct option: A
46508
Option A (May-27) is correct and should be 27/5, because it is calculated by finding the ratio of the maximum wavelengths of the Lyman and Balmer series using the Rydberg formula.
141 Medium
141 of 297
Analyzing +4 | -1
In Bohr's model of the hydrogen atom, let r,vr, v and EE represent the orbit radius, speed of an electron and the total energy of the electron, respectively. Which of following relation is proportional to the orbit number nn ?
View answer and explanation
Correct option: A
v⋅rv \cdot r
The correct answer is C (KI) because it forms a yellow-brown precipitate with copper sulfate that turns white with excess sodium thiosulfate, and a yellow precipitate with silver ions insoluble in ammonium hydroxide.
142 Medium
142 of 297
Remembering +4 | -1
The ratio of the radii of the first three Bohr orbits is :
View answer and explanation
Correct option: C
1:4:91: 4: 9
Option C (1: 4: 9) is correct because the radius of a Bohr orbit is proportional to n2n^2, where n is the principal quantum number.
143 Medium
143 of 297
Application +4 | -1
The quantum number of four electrons are given below:
I. n=4,  l=2,  ml=−2,  ms=−12n = 4,\,\,l = 2,\,\,{m_l} = - 2,\,\,{m_s} = - \frac{1}{2}
II. n=3,  l=2,  ml=1,  ms=+12n = 3,\,\,l = 2,\,\,{m_l} = 1,\,\,{m_s} = + \frac{1}{2}
III. n=4,  l=1,  ml=0,  ms=+12n = 4,\,\,l = 1,\,\,{m_l} = 0,\,\,{m_s} = + \frac{1}{2}
IV. n=3,  l=1,  ml=1,  ms=−12n = 3,\,\,l = 1,\,\,{m_l} = 1,\,\,{m_s} = - \frac{1}{2}
The correct order of their increasing energies will be
View answer and explanation
Correct option: C
IV<II<III<I \mathrm{IV < II < III < I}\,
The correct order is IV < II < III < I because the energy of an electron is determined by the (n+l) rule, and if (n+l) is the same, the electron with lower n has lower energy.
144 Medium
144 of 297
Application +4 | -1
A dye absorb light of λ=4530Ao\lambda=4530 \mathop {\text{A}}\limits^{\text{o}} and then fluorescence of 5080Ao5080 \mathop {\text{A}}\limits^{\text{o}} Assuming that under given conditions 47%C47 \% \mathrm{C} absorbed energy is re-emitted out as fluorescence, calculate ratio of quanta emitted out to the number of quanta absorb
View answer and explanation
Correct option: A
0.5270.527
Option A (0.527) is correct because it represents the ratio of emitted to absorbed quanta, calculated using the wavelengths and the given energy conversion efficiency.
145 Medium
145 of 297
Application +4 | -1
If nucleus and atom are considered as perfect spheres with diameters 4×10−15 m4 \times 10^{-15} \mathrm{~m} and 2×10−102 \times 10^{-10} m\mathrm{m}, respectively, then the ratio of the volumes of nucleus and atom should be
View answer and explanation
Correct option: B
8×10−15:18 \times 10^{-15}: 1
Option B (8×10−15:18 \times 10^{-15}: 1) is correct because the ratio of volumes is proportional to the cube of the ratio of their diameters.
146 Medium
146 of 297
Understanding +4 | -1
Statement I : The amount of light emitted (intensity of radiation) from a black body and its spectral distribution depends only on its temperature
Statement II : When a ray of white light is passed through a prism, the wave with shorter wavelength bends more than the one with a longer wavelength
View answer and explanation
Correct option: A
Both Statement I and Statement II are correct
Option A is correct because both statements are fundamental principles in physics: black body radiation depends on temperature, and shorter wavelengths bend more in a prism.
147 Medium
147 of 297
Remembering +4 | -1
The total energy in 1st 1^{\text {st }} orbit of hydrogen atom is given by
View answer and explanation
Correct option: D
−2π2k2e4mh2-\frac{2 \pi^2 k^2 e^4 m}{h^2}
Option D (−2π2k2e4mh2-\frac{2 \pi^2 k^2 e^4 m}{h^2}) is correct because it represents the total energy of an electron in the first orbit of a hydrogen atom, derived from Bohr's model.
148 Medium
148 of 297
Remembering +4 | -1
In cathode ray discharge tube experiment, when sufficiently high voltage is applied across the electrodes, current starts flowing through a stream of particles from negative electrode to positive electrodes. These particles are called
View answer and explanation
Correct option: D
Both (2) or (3)
Option D (Both (2) or (3)) is correct because the particles emitted from the negative electrode in a cathode ray tube are called cathode rays or cathode ray particles.
149 Medium
149 of 297
Application +4 | -1
What is the de-Broglie wavelength (in nm{nm} ) of an electron in the ground state of hydrogen atom is ?

 [K.E. =13.6 eV;1 eV=1.602×10−19 J ] \text { [K.E. }=13.6 \,{eV} ; 1 \,{eV}=1.602 \times 10^{-19} {~J} \text { ] }
View answer and explanation
Correct option: A
0.33
Option A (0.33) is correct because it is the de-Broglie wavelength calculated using the kinetic energy of the electron.
150 Medium
150 of 297
Understanding +4 | -1
Which of the following does not characterise X-rays?
View answer and explanation
Correct option: C
Deflected by electric and magnetic fields
Option C (Deflected by electric and magnetic fields) is correct because X-rays are electromagnetic radiation and are not deflected by electric or magnetic fields.
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