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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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301 Hard
301 of 592
Application +4 | -1
An energy of 24.6  eV24.6\,\,eV is required to remove one of the electrons from Helium atom. The energy required to remove both electrons from Helium atom is
View answer and explanation
Correct option: D
79  eV79\,\,eV
Option D (79 eV) is correct because the total energy required is the sum of the first and second ionization energies of Helium.
302 Easy
302 of 592
Remembering +4 | -1
For a p-electron orbital angular momentum is
View answer and explanation
Correct option: A
2ℏ\sqrt 2 \hbar
Option A (2ℏ\sqrt 2 \hbar) is correct because the orbital angular momentum for a p-electron (l=1) is calculated using the formula l(l+1)ℏ\sqrt{l(l+1)}\hbar, which results in 2ℏ\sqrt{2}\hbar.
303 Medium
303 of 592
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.
304 Easy
304 of 592
Understanding +4 | -1
The KE of electron in He+H{e^ + } will be maximum in:
View answer and explanation
Correct option: C
1st orbit
Option C (1st orbit) is correct because kinetic energy is inversely proportional to the radius of the orbit, and the radius is smallest in the first orbit.
305 Easy
305 of 592
Understanding +4 | -1
Which of the following configurations is correct for iron ?
View answer and explanation
Correct option: D
1s2,2s22p6,3s23p6,4s2,3d61 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6, 4 s^2, 3 d^6
Option D (1s2,2s22p6,3s23p6,4s2,3d61 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6, 4 s^2, 3 d^6) is correct because it represents the correct electronic configuration of iron (Fe), which has 26 electrons.
306 Hard
306 of 592
Application +4 | -1
When electromagnetic radiation of wavelength 300 nm300 \mathrm{~nm} falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68×105 J mol−11.68 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}. What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted?
View answer and explanation
Correct option: A
517 nm
Option A (517 nm) is correct because it is the maximum wavelength calculated using the photoelectric effect equation and the given kinetic energy of the emitted electrons.
307 Medium
307 of 592
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.
308 Medium
308 of 592
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.
309 Easy
309 of 592
Application +4 | -1
The orbital angular momentum of an electron in the 3d-orbital is
View answer and explanation
Correct option: B
6h2π\sqrt 6 \frac{h}{{2\pi }}
Option B (6h2π\sqrt 6 \frac{h}{{2\pi }}) is correct because it is calculated using the formula for orbital angular momentum, which depends on the azimuthal quantum number (l).
310 Medium
310 of 592
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.
311 Medium
311 of 592
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.
312 Medium
312 of 592
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.
313 Medium
313 of 592
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.
314 Easy
314 of 592
Analyzing +4 | -1
Which of the following element will have the same number of electrons in ss and as well as pp-type of orbitals?
View answer and explanation
Correct option: B
Mg⁡(Z=12)\operatorname{Mg}(Z=12)
Option B is correct because Magnesium (Mg) has the electronic configuration 1s22s22p63s21s^22s^22p^63s^2, which has an equal number of electrons in s-orbitals (6) and p-orbitals (6).
315 Easy
315 of 592
Understanding +4 | -1
The ratio of charge and mass would be greter for
View answer and explanation
Correct option: B
Electron
Option B (Electron) is correct because the charge-to-mass ratio is significantly higher for an electron due to its very small mass.
316 Easy
316 of 592
Understanding +4 | -1
Which of the following set of quantum number is possible?
View answer and explanation
Correct option: C
n=4,I=3,m=−3,s=12n=4, I=3, m=-3, s=\frac{1}{2}
The correct answer is C (n=4,I=3,m=−3,s=12n=4, I=3, m=-3, s=\frac{1}{2}) because it satisfies all the rules for quantum numbers: n > 0, 0 ≤ l < n, -l ≤ m ≤ +l, and s = ±1/2.
317 Medium
317 of 592
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.
318 Easy
318 of 592
Understanding +4 | -1
Bohr’s model can explain
View answer and explanation
Correct option: B
Spectrum of an atom or ion containing one electron only
Option B (Spectrum of an atom or ion containing one electron only) is correct because Bohr's model is based on the hydrogen atom, which has only one electron, and can be extended to other one-electron species.
319 Easy
319 of 592
Remembering +4 | -1
Rutherford's experiment which established the nuclear model of the atom used a beam of
View answer and explanation
Correct option: D
Helium nuclei, which impinged on a metal foil and got scattered
Option D (Helium nuclei, which impinged on a metal foil and got scattered) is correct because Rutherford's experiment used alpha particles (helium nuclei) to bombard a gold foil.
320 Medium
320 of 592
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.
321 Medium
321 of 592
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.
322 Medium
322 of 592
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.
323 Medium
323 of 592
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.
324 Medium
324 of 592
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.
325 Easy
325 of 592
Remembering +4 | -1
Electron behaves both as a particle and a wave. This was proposed by
View answer and explanation
Correct option: C
de-Broglie
Option C (de-Broglie) is correct because de-Broglie proposed the wave-particle duality of electrons.
326 Easy
326 of 592
Remembering +4 | -1
The planck’s constant has a unit of :
View answer and explanation
Correct option: C
Angular momentum
Option C (Angular momentum) is correct because Planck's constant has units of energy multiplied by time, which is equivalent to the units of angular momentum.
327 Hard
327 of 592
Application +4 | -1
Determine the number of revolutions made by electron in one second in the 2nd 2^{\text {nd }} Bohr orbit of hydrogen atom. (Take r0=0.53×10−10 mr_0=0.53 \times 10^{-10} \mathrm{~m} )
View answer and explanation
Correct option: D
8.23×10148.23 \times 10^{14}
The correct answer is 8.23×10148.23 \times 10^{14} because it is calculated using the electron speed and orbital radius in the 2nd Bohr orbit.
328 Easy
328 of 592
Remembering +4 | -1
The shape of orbital with l=1l = 1 and m=0m = 0 is
View answer and explanation
Correct option: B
Dumbell
Option B (Dumbell) is correct because an orbital with l=1l = 1 corresponds to a p-orbital, which has a dumbbell shape.
329 Easy
329 of 592
Analyzing +4 | -1
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Loss of electron from hydrogen atom results in nucleus of ∼1.5×10−3 pm\sim 1.5 \times 10^{-3} \, \mathrm{pm} size.

Reason R: Proton (H+)\left(\mathrm{H}^{+}\right)always exists in combined form.

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 A and R are correct but R is NOT the
correct explanation of A
The correct answer is D because both the assertion and the reason are correct, but the reason is not the correct explanation of the assertion. The size of the nucleus is indeed around 1.5×10−3 pm1.5 \times 10^{-3} \, \mathrm{pm}, and protons always exist in combined form, but these two facts are unrelated.
330 Hard
330 of 592
Analyzing +4 | -1
Statement-1 : Nuclide 30Al13{ }^{30} {Al}_{13} is less stable than 40Ca20{ }^{40} {Ca}_{20}

Statement-2 : Nuclide having odd number of protons and neutrons are generally unstable.
View answer and explanation
Correct option: A
Statement- 1 is True, Statement- 2 is True, Statement- 2 is a correct explanation for Statement - 3
Option A (Statement- 1 is True, Statement- 2 is True, Statement- 2 is a correct explanation for Statement - 3) is correct because nuclides with even numbers of protons and neutrons are generally more stable.
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