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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 181-210 of 592 questions

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181 Easy
181 of 592
Analyzing +4 | -1
An ion has 18 electrons in the outermost shell, it is
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Correct option: A
Cu+{Cu}^{+}
Option A (Cu+{Cu}^{+}) is correct because Cu+{Cu}^{+} has 28 electrons, and its electronic configuration is 1s22s22p63s23p63d101s^22s^22p^63s^23p^63d^{10}, which has 18 electrons in the outermost shell (3s, 3p, and 3d).
182 Medium
182 of 592
Understanding +4 | -1
Choose the correct energy order for 2s2 s orbital
View answer and explanation
Correct option: A
He(2s)>Li(2s)>Be(2s){He}(2 s)>{Li}(2 s)>{Be}(2 s)
Option A (He(2s)>Li(2s)>Be(2s){He}(2 s)>{Li}(2 s)>{Be}(2 s)) is correct because the energy of an orbital decreases with increasing nuclear charge for the same principal quantum number.
183 Medium
183 of 592
Application +4 | -1
What will be the wave length of an electron moving with 110\frac{1}{10} th of velocity of light?
View answer and explanation
Correct option: A
2.43×10−11 m2.43 \times 10^{-11} \mathrm{~m}
Option A (2.43×10−11 m2.43 \times 10^{-11} \mathrm{~m}) is correct because the wavelength is calculated using the de Broglie wavelength formula, λ=h/mv\lambda = h/mv, with the given velocity.
184 Easy
184 of 592
Understanding +4 | -1
Which one of the following sets of quantum numbers is possible
View answer and explanation
Correct option: C
n=1,l=0,m1=0,ms=+1/2n=1, l=0, m_1=0, m_s=+1 / 2
Option C (n=1,l=0,m1=0,ms=+1/2n=1, l=0, m_1=0, m_s=+1 / 2) is correct because it adheres to all the quantum number rules: ll must be between 0 and n−1n-1, mlm_l must be between −l-l and +l+l, and msm_s must be either +1/2 or -1/2.
185 Medium
185 of 592
Remembering +4 | -1
According to Bohr's theory, the electronic energy of H\mathrm{H}-atom in Bohr's orbit is given by
View answer and explanation
Correct option: B
En=−2.179×10−18×Z2n2J\mathrm{E_n}=-\frac{2.179 \times 10^{-18} \times \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{J}
Option B (En=−2.179×10−18×Z2n2J\mathrm{E_n}=-\frac{2.179 \times 10^{-18} \times \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{J}) is correct because it represents the correct formula for electronic energy of a hydrogen atom in Bohr's orbit.
186 Easy
186 of 592
Understanding +4 | -1
Which of the following particle is not deflected in the magnetic field?
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Correct option: C
Neutron
Option C (Neutron) is correct because neutrons are neutral particles and are not affected by magnetic fields.
187 Medium
187 of 592
Remembering +4 | -1
Amongst the following statements, select the set having statements which was proposed by Dalton.
(1) All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.
(2) When gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same T & P\mathrm{T} \,\&\, \mathrm{P}
(3) Chemical reactions involve reorganization of atoms. These are neither created nor destroyed in a chemical reaction.
(4) Matter consists of indivisible atoms
View answer and explanation
Correct option: B
(1), (3), (4)
Option B ((1), (3), (4)) is correct because these statements align with Dalton's atomic theory.
188 Medium
188 of 592
Understanding +4 | -1
Assertion: The configuration of BeBe-atom cannot be 1s23s21 s^2 3 s^2.
Reason: The electrons enter the various orbitals in the order of increasing energy.
View answer and explanation
Correct option: A
Both assertion and reason are true and reason is the correct explanation of assertion
Option A (Both assertion and reason are true and reason is the correct explanation of assertion) is correct because the Aufbau principle explains why BeBe cannot have the configuration 1s23s21s^23s^2.
189 Medium
189 of 592
Analyzing +4 | -1
Statement (S): Monochromatic X-rays fall on lighter elements such as carbon and show scattering under the name of Compton effect.

Explanation (E): λscattered \lambda_{\text {scattered }} light is always lower than λincident \lambda_{\text {incident }} light.
View answer and explanation
Correct option: A
S is correct but E is wrong
Option A is correct because the statement about Compton effect is correct, but the explanation about the wavelength is wrong. In Compton scattering, the scattered wavelength is always greater than the incident wavelength.
190 Medium
190 of 592
Application +4 | -1
Sodium exists in two isotopic forms Na23{Na}^{23} and Na24{Na}^{24} in nature. If Na24{Na}^{24} is present at the abundance of 25%25 \% in the nature. The average atomic mass of the element is :-
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Correct option: C
23.25
Option C (23.25) is correct because it is the average atomic mass calculated considering the isotopic abundance.
191 Easy
191 of 592
Remembering +4 | -1
Which wavelength falls in a X-rays region?
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Correct option: C
1Å
Option C (1Å) is correct because X-rays have wavelengths in the range of approximately 0.01 to 10 Angstroms.
192 Medium
192 of 592
Remembering +4 | -1
Which of the following gave the idea of nucleus of the atom?
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Correct option: C
α\alpha-Rays scattering experiment
Option C (α\alpha-Rays scattering experiment) is correct because Rutherford's alpha-particle scattering experiment led to the discovery of the atomic nucleus.
193 Medium
193 of 592
Application +4 | -1
The momentum of a particle which has a de-Broglie wavelength of 0.1 nm0.1 \mathrm{~nm} is
View answer and explanation
Correct option: D
6.62×10−24 kg ms−16.62 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}
Option D (6.62×10−24 kg ms−16.62 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}) is correct because it is calculated directly from the de-Broglie wavelength formula.
194 Medium
194 of 592
Application +4 | -1
The energy of the electron in the second and the third Bohr's orbits of the hydrogen atom is −5.42×10−12-5.42 \times 10^{-12} erg and −2.41×10−12-2.41 \times 10^{-12} erg respectively. Calculate the wavelength of the emitted radiation when the electron drops from the third to the second orbit.
View answer and explanation
Correct option: A
6.6×103A∘6.6 \times 10^3 \mathop {\text{A}}\limits^ \circ
Option A (6.6×103A∘6.6 \times 10^3 \mathop{\text{A}}\limits^\circ) is correct because it is the calculated wavelength of the emitted radiation when the electron transitions from the third to the second Bohr orbit.
195 Medium
195 of 592
Application +4 | -1
The electronic transition from n=2n=2 to n=1n=1 will produce the shortest wavelength in (where, n=n= principal quantum number)
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Correct option: D
Li2+\mathrm{Li}^{2+}
Option D (Li²⁺) is correct because the shortest wavelength corresponds to the largest energy transition, which is proportional to Z² (atomic number squared).
196 Medium
196 of 592
Application +4 | -1
According to the Bohr theory, which of the following transitions in the Hydrogen atom will give rise to the least energetic photon
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Correct option: C
n = 6 to n = 5
The correct answer is C (n = 6 to n = 5) because the energy difference between these levels is the smallest, resulting in the least energetic photon.
197 Easy
197 of 592
Application +4 | -1
The maximum number of electrons that can be identified with the quantum number ℓ=2\ell=2 for Fe3+\mathrm{Fe}^{3+} ion will be
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Correct option: C
5
Option C (5) is correct because ℓ=2\ell=2 corresponds to the d-orbital, and Fe3+\mathrm{Fe}^{3+} has 5 electrons in its 3d orbital.
198 Medium
198 of 592
Remembering +4 | -1
The number of nodal planes ' 5d5 d ' orbital has is
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Correct option: C
two
Option C (two) is correct because the number of nodal planes in a d-orbital is equal to the azimuthal quantum number (l), which is 2 for a d-orbital.
199 Medium
199 of 592
Application +4 | -1
The wave function ( Ψ\Psi ) of 2 s is given by Ψ2s=122π(1a0)1/2(2−ra0)e−f/2a0\Psi_{2 s}=\frac{1}{2 \sqrt{2 \pi}}\left(\frac{1}{a_0}\right)^{1 / 2}\left(2-\frac{r}{a_0}\right) e^{-f / 2 a_0}

At r=r0r=r_0, radial node is formed. Thus, r0r_0 in terms of aoa_o
View answer and explanation
Correct option: A
r0=2a0r_0=2 a_0
Option A (r0=2a0r_0=2 a_0) is correct because at the radial node, the wave function Ψ2s\Psi_{2s} is equal to zero.
200 Easy
200 of 592
Application +4 | -1
The set of quantum numbers n=3,l=0,m=0,s=−1/2n=3, l=0, m=0, s=-1 / 2 belongs to the element
View answer and explanation
Correct option: A
Mg{Mg}
Option A (Mg) is correct because the given quantum numbers (n=3, l=0, m=0, s=-1/2) correspond to the 3s orbital, which is the valence electron configuration of Magnesium.
201 Medium
201 of 592
Application +4 | -1
The λ\mathrm{\lambda } for Hα{H_\alpha } line of Balmer series is  6500 Å. Thus λ\mathrm{\lambda } for Hβ{H_\beta } line of Balmer series is :
View answer and explanation
Correct option: A
4814 Å
Option A (4814 Å) is correct because it is derived from the Rydberg formula for the Balmer series, relating wavelengths to energy level transitions.
202 Medium
202 of 592
Understanding +4 | -1
Consider the element Mg,   AI,   S,  PMg,\,\,\,AI,\,\,\,S,\,\,P and Si,Si, the correct increasing order of their first ionisation enthalpy is
View answer and explanation
Correct option: A
Al < Mg < Si < S < P
Option A (Al < Mg < Si < S < P) is the correct increasing order of first ionization enthalpy, considering exceptions due to stable electronic configurations.
203 Easy
203 of 592
Remembering +4 | -1
Which of the following is Heisenberg uncertainty principle?
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Correct option: A
Δx.Δp≥h4π\Delta x.\Delta p \ge \frac{h}{{4\mathrm{\pi }}}
Option A (Δx.Δp≥h4π\Delta x.\Delta p \ge \frac{h}{{4\mathrm{\pi }}}) is correct because it represents the correct mathematical expression of Heisenberg's uncertainty principle.
204 Easy
204 of 592
Remembering +4 | -1
Which one of the following is considered as the main postulate of Bohr’s model of atom
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Correct option: D
Angular momentum of electron is an integral multiple of h2π\frac{h}{{2\pi }}
Option D is correct because Bohr's postulate states that the angular momentum of an electron is quantized and is an integral multiple of h2π\frac{h}{{2\pi }}.
205 Medium
205 of 592
Application +4 | -1
The ionisation energy of hydrogen atom is 13.6 eV. What will be the ionisation energy of He+\mathrm{H}{\mathrm{e}^ + }?
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Correct option: B
54.4 eV
Option A (31) is correct because the final temperature is calculated by equating the heat lost by iron to the heat gained by water.
206 Medium
206 of 592
Understanding +4 | -1
The total values of magnetic quantum number of an electron when the value of n=2 is:
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Correct option: C
4
Option C (4) is correct because when n=2, l can be 0 or 1. For l=0, m=0 (1 value). For l=1, m=-1, 0, +1 (3 values). Total values = 1 + 3 = 4.
207 Medium
207 of 592
Remembering +4 | -1
The frequency of radiation absorbed or emitted when transition occurs between two stationary states with energies E1E_1 (lower) and E2E_2 (higher) is given by
View answer and explanation
Correct option: D
v=E2−E1hv=\frac{E_2-E_1}{h}
Option D (v=E2−E1hv=\frac{E_2-E_1}{h}) is correct because the frequency of emitted/absorbed radiation is proportional to the energy difference between states, divided by Planck's constant.
208 Easy
208 of 592
Application +4 | -1
The electrons identified by 517 ‘n’ and ‘l’
(i)n=4, l=1, (ii) n=4, l=0, (iii) n=3, l=2 , (iv) n=3, l=1(i)n = 4,\,l = 1,\,(ii)\,n = 4,\,l = 0,\,(iii)\,n = 3,\,l = 2\,,\,(iv)\,n = 3,\,l = 1 , Can be placed in order of increasing from lowest to highest is
View answer and explanation
Correct option: A
iv < ii < iii < i
Option A (iv < ii < iii < i) is correct because it correctly orders the orbitals based on increasing energy using the n+l rule and Hund's rule.
209 Medium
209 of 592
Remembering +4 | -1
α\alpha-particles can be detected by using
View answer and explanation
Correct option: C
Zinc sulphide screen
Option C (Zinc sulphide screen) is correct because alpha particles cause zinc sulphide to fluoresce, allowing their detection.
210 Medium
210 of 592
Remembering +4 | -1
Electrons and nucleus are held together by
View answer and explanation
Correct option: B
Electrostatic forces of attraction
Option B (Electrostatic forces of attraction) is correct because electrons, being negatively charged, are held in orbit around the positively charged nucleus due to electrostatic attraction.
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