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JEE • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Mixed difficulty

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Practice format and chapter quiz

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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211 Medium
211 of 592
Remembering +4 | -1
If r1{r_1} is the radius of the first orbit, the radius of nth{n^{th}} orbit of H-atom is given by
View answer and explanation
Correct option: A
r1n2{r_1}{n^2}
Option A (r1n2{r_1}{n^2}) is correct because the radius of the nth orbit of a hydrogen atom is given by rn=r1n2{r_n} = {r_1}{n^2}, where r1{r_1} is the radius of the first orbit.
212 Medium
212 of 592
Understanding +4 | -1
Correct energy value order is
View answer and explanation
Correct option: D
ns,(n−1)d,np,(n−1)fns,\left( {n - 1} \right)d,np,\left( {n - 1} \right)f
Option D (ns,(n−1)d,np,(n−1)fns,( {n - 1} )d,np,( {n - 1} )f) is correct because subshells with lower (n+l) values have lower energy, determining the order.
213 Medium
213 of 592
Understanding +4 | -1
As the orbit number increases, the difference in two consecutive energy levels
View answer and explanation
Correct option: C
decreases
Option C (decreases) is correct because the energy difference between consecutive energy levels decreases as the orbit number increases, as shown by the formula E=−1312n2z2 kJ/molE = \frac{{-1312}}{{n^2}} z^2 \mathrm{~kJ} / \mathrm{mol}.
214 Hard
214 of 592
Application +4 | -1
The de-Broglie wavelength of an electron in the ground state of hydrogen atom is :

(K.E.=13.6eV;1eV=1.602×10−19 J)\left({K} . {E} .=13.6 {eV} ; 1 {eV}=1.602 \times 10^{-19} {~J}\right)
View answer and explanation
Correct option: C
0.3328 nm0.3328 {~nm}
Option C (0.3328 nm) is correct because it is calculated using the de-Broglie wavelength formula with the given kinetic energy.
215 Medium
215 of 592
Application +4 | -1
The K\mathrm{K} and L\mathrm{L} shell of an element are completely filled and there are 16 electrons in M-shell and 2 electrons in N\mathrm{N}-shell. The atomic number of the element is
View answer and explanation
Correct option: B
28
Option B (28) is correct because the atomic number is the sum of all electrons in the filled shells.
216 Easy
216 of 592
Understanding +4 | -1
STATEMENT-1 : Half-filled and fully-filled degenerate orbitals are more stable.

STATEMENT-2 : Extra stability is due to the symmetrical distribution of electrons and exchange energy.
View answer and explanation
Correct option: A
If both the statement are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-4
Option A is correct because both statements are true, and Statement-2 correctly explains why half-filled and fully-filled orbitals are more stable.
217 Medium
217 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
The correct answer is B (18) because a g-subshell has l=4, and the number of electrons is calculated using the formula 2(2l+1).
218 Medium
218 of 592
Application +4 | -1
Yellow light emitted from a sodium lamp has a wavelength (λ)(\lambda) of 580 nm580 \mathrm{~nm}. Calculate the frequency (v) and wave number ( vˉ)\bar{v}) of the yellow light.
View answer and explanation
Correct option: A
5.17×1014 s−1,1.724×104 cm−15.17 \times 10^{14} \mathrm{~s}^{-1}, 1.724 \times 10^4 \mathrm{~cm}^{-1}
Option A (5.17×1014 s−1,1.724×104 cm−15.17 \times 10^{14} \mathrm{~s}^{-1}, 1.724 \times 10^4 \mathrm{~cm}^{-1}) is correct because it accurately calculates the frequency and wave number using the given wavelength and the speed of light.
219 Easy
219 of 592
Remembering +4 | -1
Maximum number of electrons in a subshell is given by
View answer and explanation
Correct option: B
2(2l+1)2\left( {2l + 1} \right)
Option B (2(2l+1)2\left( {2l + 1} \right)) is correct because it represents the total number of electrons that can occupy a subshell with a given azimuthal quantum number l.
220 Medium
220 of 592
Analyzing +4 | -1
Highest energy will be absorbed to eject out the electron in the
configuration.
View answer and explanation
Correct option: B
1s22s22p31{s^2}2{s^2}2{p^3}
The correct answer is B (1s22s22p31s^22s^22p^3) because the half-filled p-orbital configuration provides extra stability, requiring more energy to remove an electron.
221 Hard
221 of 592
Analyzing +4 | -1
When photon of energy 4.25 eV4.25 \mathrm{~eV} strikes the surface, the ejected electron has maximum kinetic energy TAT_A expressed in eV\mathrm{eV} and de Broglie wavelength λA.\lambda_A. The maximum kinetic energy of photoelectrons liberated by another metal BB by photons of 4.70 eV4.70 \mathrm{~eV} is TBT_B (TB=TA−1.5 eV).\left(T_B=T_A-1.5 \mathrm{~eV}\right). If de Broglie wavelength of the electron is λB(λB=2λA),\lambda_B\left(\lambda_B=2 \lambda_A\right), then which is not correct?
View answer and explanation
Correct option: C
Work function of BB is 1.20 eV1.20 \mathrm{~eV}
Option D (A is correct but R is not correct) is correct because the basicity decreases due to the lanthanide contraction, but the reason given about covalent character is incorrect.
222 Easy
222 of 592
Analyzing +4 | -1
Ions which have the same electronic configuration are those of
View answer and explanation
Correct option: C
Potassium and calcium
Option C is correct. K⁺ and Ca²⁺ both have the same electronic configuration: 1s²2s²2p⁶3s²3p⁶, which is the electronic configuration of Argon.
223 Easy
223 of 592
Remembering +4 | -1
For a Bohr atom angular momentum M of the electron is (n=0,1,2,…)
View answer and explanation
Correct option: D
nh2π\frac{{nh}}{{2\mathrm{\pi }}}
Option D (nh2π\frac{{nh}}{{2\mathrm{\pi }}}) is correct because the angular momentum of an electron in a Bohr atom is quantized and given by this formula.
224 Easy
224 of 592
Remembering +4 | -1
Mass of a proton in Kg\mathrm{Kg} is
View answer and explanation
Correct option: D
1.66×10−271.66 \times 10^{-27}
Option D (1.66×10−271.66 \times 10^{-27}) is correct because the mass of a proton is approximately 1.66×10−271.66 \times 10^{-27} kg.
225 Medium
225 of 592
Application +4 | -1
How fast is an electron moving if it has a wavelength equal to the distance it travels in one second? 
View answer and explanation
Correct option: A
hm\sqrt{\frac{h}{m}}
Option A (hm\sqrt{\frac{h}{m}}) is correct because it is derived by equating the wavelength to the velocity using the de Broglie equation.
226 Medium
226 of 592
Analyzing +4 | -1
The wave function of an electron in the lowest energy state of a hydrogen atom is proportional to e−r/a0e^{-r / a_0}, where rr is the distance of the electron from the nucleus. Probability is maximum when electron is at
View answer and explanation
Correct option: A
the nucleus
Option A (the nucleus) is correct because the probability density is proportional to the square of the wave function, which is maximum at the nucleus (r=0).
227 Hard
227 of 592
Analyzing +4 | -1
The frequency v of certain line of the Lyman series of the atomic spectrum of hydrogen satisfies the following conditions:
(i) It is the sum of the frequencies of another Lyman line and a Balmer line.
(ii) It is the sum of the frequencies of a certain line, a Lyman line, and a Paschen line.
(iii) It is the sum of the frequencies of a Lyman and a Paschen line but no Bracket line.
To what transition does v correspond?
View answer and explanation
Correct option: D
n2=4  to  n1=1{n_2} = 4\;\mathrm{to\;}{n_1} = 1
Option D (n2 = 4 to n1 = 1) is correct because it satisfies all the given conditions for the frequency ν of the Lyman series.
228 Easy
228 of 592
Application +4 | -1
If threshold wavelength (λ0)\left( {{\lambda ^0}} \right) for ejection of electron from metal is 330 nm. Then work function for the photo electric emission is
View answer and explanation
Correct option: C
6×10−19J6 \times {10^{ - 19}}J
Option C (6×10−19J6 \times {10^{ - 19}}J) is correct because it is the work function calculated using the threshold wavelength and Planck's constant.
229 Hard
229 of 592
Application +4 | -1
Suppose the mass of electron is decreased by 25%25 \%. How will it affect the Rydberg constant?
View answer and explanation
Correct option: C
It reduces to 75%75 \% of its original value.
The Rydberg constant is directly proportional to the mass of the electron. Therefore, if the mass decreases by 25%, the Rydberg constant also decreases to 75% of its original value, making option C correct.
230 Medium
230 of 592
Understanding +4 | -1
The study of photoelectric effect is useful in understanding :
View answer and explanation
Correct option: A
Conservation of energy
Option A (Conservation of energy) is correct because the photoelectric effect demonstrates the conversion of light energy into kinetic energy of electrons, thus illustrating energy conservation.
231 Easy
231 of 592
Application +4 | -1
Hydrogen atoms are excited to n=4n = 4 state, In the spectrum of emitted radiation, number of lines in the
UV and visible regions are resp.
View answer and explanation
Correct option: D
3,23,2
Option D (3,23,2) is correct because when hydrogen atoms are excited to n=4, there are 3 Lyman series lines (UV region) and 2 Balmer series lines (visible region).
232 Easy
232 of 592
Understanding +4 | -1
Assertion: Isobars have the same atomic mass number but different atomic numbers.

Reason: Isobars have different numbers of neutrons.
View answer and explanation
Correct option: A
Both Assertion and Reason are true and Reason is the correct explanation for Assertion
Option A (Both Assertion and Reason are true and Reason is the correct explanation for Assertion) is correct because isobars have the same mass number but different atomic numbers due to having different numbers of protons and neutrons, with the total number of nucleons (protons + neutrons) being the same.
233 Medium
233 of 592
Application +4 | -1
The energy of an electron in the ground state (n=1)(n=1) for He+{He}^{+} ion is −x J,-x \mathrm{~J}, then that for an electron in n=2n=2 state for Be3+{Be}^{3+} ion in J\mathrm{J} is
View answer and explanation
Correct option: D
−x-x
Option A (A=1015\mathrm{A}=10^{15} and Ea=1.9×104 KJ\mathrm{E}_{\mathrm{a}}=1.9 \times 10^4 \mathrm{~KJ}) is correct because it is obtained by comparing the given equation with the Arrhenius equation in logarithmic form.
234 Medium
234 of 592
Application +4 | -1
For the same electronic transition in the following atom or ion, the frequency of the emitted radiation will be maximum for
View answer and explanation
Correct option: D
Li2+\mathrm{Li}^{2+} ion
Option D (Li2+\mathrm{Li}^{2+} ion) is correct because the frequency of emitted radiation is directly proportional to the square of the atomic number (Z), and Li2+ has the highest Z among the given options.
235 Easy
235 of 592
Remembering +4 | -1
The maximum number of electrons possible in a sublevel is equal to
View answer and explanation
Correct option: D
4l+24l + 2
Option D (4l+24l + 2) is correct because the number of orbitals in a sublevel is 2l+1, and each orbital can hold 2 electrons, so the maximum number of electrons is 2(2l+1) = 4l+2.
236 Easy
236 of 592
Understanding +4 | -1
Assertion :- Bohr model is not suitable in case of multi-electron species.

Reason :- It does not tells about electron-electron interaction.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A is correct because Bohr's model, based on hydrogen-like species, fails to account for electron-electron interactions present in multi-electron systems, making it unsuitable for them.
237 Easy
237 of 592
Understanding +4 | -1
Assertion: Dalton's Atomic Theory states that atoms are indivisible and indestructible.
Reason: This statement is based on the observation that in chemical reactions, atoms are rearranged but not destroyed or created.
View answer and explanation
Correct option: A
Both Assertion and Reason are true and Reason is the correct explanation for Assertion
Option A is correct because Dalton's theory indeed states atoms are indivisible and indestructible, and this is based on observations of chemical reactions.
238 Medium
238 of 592
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}.
239 Medium
239 of 592
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.
240 Easy
240 of 592
Remembering +4 | -1
Maximum number of electrons in an orbit is given by:
View answer and explanation
Correct option: B
2n22{n^2}
Option B (2n22{n^2}) is correct because the maximum number of electrons that can occupy an orbit (energy level) is given by the formula 2n22n^2, where n is the principal quantum number.
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