Skip to main content
Back to STRUCTURE OF ATOM

Your Progress

0 / 592 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 481-510 of 592 questions

Preparing practice and saved progress…

481 Easy
481 of 592
Application +4 | -1
The momentum of a photon of wavelength 6626 nm\mathrm{nm} will be
View answer and explanation
Correct option: A
10−28 kg ms−110^{-28} \mathrm{~kg} \mathrm{~ms}^{-1}
Option A (10−28 kg ms−110^{-28} \mathrm{~kg} \mathrm{~ms}^{-1}) is correct because the momentum of a photon is calculated using the formula p = h/λ, where h is Planck's constant and λ is the wavelength.
482 Easy
482 of 592
Analyzing +4 | -1
The electronic configuration of an element is 1s2,2s2 2p6,3s2 3p6 3d5,4s1. 1{s^2},2{s^2}\,2{p^6},3{s^2}\,3{p^6}\,3{d^5},4{s^1}.\, This represent its:
View answer and explanation
Correct option: D
Both 2 and 3
Option D is correct because the given electronic configuration represents the ground state of Chromium (Cr) and can also represent the cationic form of Manganese (Mn+).
483 Easy
483 of 592
Remembering +4 | -1
The number of electrons accommodated in an orbit with principle quantum number 2, is
View answer and explanation
Correct option: D
8
Option D (8) is correct because the number of electrons accommodated in an orbit is given by 2n22n^2, where n is the principal quantum number. For n=2, the number of electrons is 2(22)=82(2^2) = 8.
484 Medium
484 of 592
Application +4 | -1
For which of the following, the radius will be same as for hydrogen atom having n=1?
View answer and explanation
Correct option: C
Be3+,n=2\mathrm{B}{\mathrm{e}^{3 + }},n = 2
The correct answer is Be3+,n=2\mathrm{B}{\mathrm{e}^{3 + }},n = 2 because the radius of a hydrogen-like atom is proportional to n²/Z, and for Be³⁺ with n=2, the radius is the same as hydrogen with n=1.
485 Easy
485 of 592
Application +4 | -1
Which has the maximum value of magnetic moment?
View answer and explanation
Correct option: D
Fe3+\mathrm{Fe}^{3+}
Option D (Fe3+\mathrm{Fe}^{3+}) is correct because it has the maximum number of unpaired electrons, leading to the highest magnetic moment.
486 Easy
486 of 592
Understanding +4 | -1
An isostere is
View answer and explanation
Correct option: A
NO2−{NO}_2^{-}and O3{O}_3
Option A (NO2−{NO}_2^{-}and O3{O}_3) is correct because they have the same number of atoms and the same number of electrons.
487 Easy
487 of 592
Remembering +4 | -1
Zeeman effect explains splitting of lines in
View answer and explanation
Correct option: A
Magnetic Field
Option A (Magnetic Field) is correct because the Zeeman effect describes the splitting of spectral lines when an atom is placed in a magnetic field.
488 Medium
488 of 592
Understanding +4 | -1
An electron that has quantum number n=3 and m=2:
View answer and explanation
Correct option: D
Must have l=2l = 2
Option D (Must have l=2l = 2) is correct because if m=2, then l must be greater than or equal to 2. Since n=3, l can be 0, 1, or 2. Therefore, l must be 2.
489 Medium
489 of 592
Application +4 | -1
Bohr radius for the hydrogen atom (n = 1) is approximately 0.53  Ao.0.53\,\,\mathop \mathrm{A}\limits^\mathrm{o} . The radius for the first excited state ( n = 2) orbit is (in Ao)\mathop \mathrm{A}\limits^\mathrm{o} )
View answer and explanation
Correct option: D
2.12
The correct answer is D (2.12) because the radius of an orbit is proportional to n²/Z, where n is the principal quantum number and Z is the atomic number.
490 Easy
490 of 592
Remembering +4 | -1
Degeneracy of s, p, d, f and g sub-orbits respectively are
View answer and explanation
Correct option: B
1,3, 5, 7, 9
Option B (1,3, 5, 7, 9) is correct because the degeneracy of s, p, d, f, and g sub-orbits corresponds to the number of orbitals, which is given by 2l+1.
491 Medium
491 of 592
Understanding +4 | -1
In the Bohr’s orbit, what is the ratio of kinetic energy and total energy of the electron ?
View answer and explanation
Correct option: A
-1
Option A (-1) is correct because the ratio of kinetic energy to total energy in Bohr's orbit is always -1.
492 Easy
492 of 592
Understanding +4 | -1
The de-Broglie equation applies
View answer and explanation
Correct option: C
All the material objects in motion
Option C (All the material objects in motion) is correct because the de Broglie equation applies to all matter, relating wavelength to momentum.
493 Easy
493 of 592
Application +4 | -1
The energy of an electron in Bohr's orbit of hydrogen atom is −13.6 eV-13.6 \mathrm{~eV}. The total electronic energy of a hypothetical He\mathrm{He} atom in which there are no electron-electron repulsions or interactions is
View answer and explanation
Correct option: C
−108.8 eV-108.8 \mathrm{~eV}
Option C (−108.8 eV-108.8 \mathrm{~eV}) is correct because the total electronic energy is calculated by multiplying the energy of one electron by the square of the atomic number and the number of electrons.
494 Hard
494 of 592
Application +4 | -1
Energy for 7.25×10157.25 \times 10^{15} photons of 5.37×1014 s−15.37 \times 10^{14} \mathrm{~s}^{-1} frequency in Einstein unit is
View answer and explanation
Correct option: A
1.20×10−81.20 \times 10^{-8}
Option B (Pent – 3 – en – 1- yne) is correct because it follows the IUPAC naming rules, prioritizing the alkyne functional group and numbering the carbon chain accordingly.
495 Easy
495 of 592
Understanding +4 | -1
Which wave property is directly proportional to energy of electromagnetic radiation:
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because energy of electromagnetic radiation is directly proportional to frequency, wave number and inversely proportional to wavelength. Velocity is constant.
496 Easy
496 of 592
Understanding +4 | -1
The ratio of specific charge of a neutron and an α\alpha-particle is
View answer and explanation
Correct option: B
0
Option B (0) is correct because a neutron has no charge, so its specific charge (charge/mass) is zero, making the ratio zero as well.
497 Medium
497 of 592
Understanding +4 | -1
Number of nodal planes in 4P – subshell will be
View answer and explanation
Correct option: B
1
Option B (1) is correct because the number of nodal planes in a p-orbital is equal to 1.
498 Medium
498 of 592
Application +4 | -1
The first line in the Balmer series in the H\mathrm{H}-atom will have the frequency
View answer and explanation
Correct option: B
4.57×1014 s−14.57 \times 10^{14} \mathrm{~s}^{-1}
Option B (4.57 × 1014 s-1) is correct because it is calculated using the Rydberg formula for the Balmer series with n1=2 and n2=3.
499 Easy
499 of 592
Remembering +4 | -1
When an electron moves from higher orbit to a lower orbit … is produced
View answer and explanation
Correct option: B
Emission spectra
Option B (Emission spectra) is correct because when an electron transitions from a higher energy orbit to a lower energy orbit, it emits energy in the form of photons, resulting in an emission spectrum.
500 Easy
500 of 592
Remembering +4 | -1
Which one of the following orbitals is spherical in shape
View answer and explanation
Correct option: A
4s
Option A (4s) is correct because s-orbitals are always spherical in shape, regardless of the principal quantum number.
501 Medium
501 of 592
Analyzing +4 | -1
There are three elements A, B, C having outermost electronic configuration 3s2,3p4,3p53 s^2, 3 p^4, 3 p^5 respectively. Which statement is incorrect for three elements:
View answer and explanation
Correct option: D
Element B\mathrm{B} and C\mathrm{C} can not form any compound
The rate law is determined by the slow step. By using the equilibrium of the fast step to express the concentration of the intermediate, the rate law can be written as rate = k'[O3][O2]-1. Therefore, the order with respect to O2 is -1, making option A the correct answer.
502 Easy
502 of 592
Remembering +4 | -1
Angular momentum of an electron in an orbital is given by
View answer and explanation
Correct option: B
h2π×l(l+1)\frac{h}{{2\pi }} \times \sqrt {l\left( {l + 1} \right)}
Option B (h2π×l(l+1)\frac{h}{{2\pi }} \times \sqrt {l\left( {l + 1} \right)}) is correct because the angular momentum of an electron in an orbital is given by the formula l(l+1)h2π\sqrt{l(l+1)} \frac{h}{2\pi}.
503 Easy
503 of 592
Application +4 | -1
What are the total numbers of orbitals and electrons for m=0m = 0, in the ground state of ZnZn (atomic number = 30).
View answer and explanation
Correct option: A
7 orbitals, 14 electrons
Option A (7 orbitals, 14 electrons) is correct because in the ground state of Zinc, there are 7 orbitals with m=0, each containing 2 electrons, totaling 14 electrons.
504 Medium
504 of 592
Understanding +4 | -1
Consider the following statements:

I. The energy of electrons in atoms is quantised.

II. The existence of quantised electronic energy levels is a direct result of the wave like properties of electrons.

III. The path of an electron in an atom can never be determined accurately.

The correct statements are
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because all three statements accurately describe the quantum mechanical model of the atom.
505 Medium
505 of 592
Application +4 | -1
The quantum numbers of most energetic electron in Ne atom when it is in first excited state is:
View answer and explanation
Correct option: C
3,  0,  0,  +1/23,\;0,\;0,\; + 1/2
Option C (3,  0,  0,  +1/23,\;0,\;0,\; + 1/2) is correct because the electronic configuration of the first excited state of Ne is 1s22s22p53s11s^22s^22p^53s^1, and the most energetic electron is in the 3s orbital.
506 Easy
506 of 592
Application +4 | -1
The electronic configurations of four elements L,P,Q and R are given below, L=1s2,2s22p4      Q=1s2,2s22p6,3s23p5L = 1{s^2},2{s^2}2{p^4}\;\;\;Q = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^5} , P=1s2,2s22p6,3s1      R=1s2,2s22p6,3s2P = 1{s^2},2{s^2}2{p^6},3{s^1}\;\;\;R = 1{s^2},2{s^2}2{p^6},3{s^2} , The formula of the ionic compounds that can be formed between these elements are:
View answer and explanation
Correct option: C
P2L,  RL,  PQ,  RQ2{P_2}L,\;RL,\;PQ,\;R{Q_2}
Option C (P2L,  RL,  PQ,  RQ2{P_2}L,\;RL,\;PQ,\;R{Q_2}) is correct because it represents the correct ionic formulas based on the valencies of the elements L, P, Q, and R.
507 Medium
507 of 592
Application +4 | -1
The frequency of revolution of electron II excited state He+\text{He}^+ and I excited state of H\text{H}-atom:
View answer and explanation
Correct option: B
3227\frac{32}{27}
Option B (3227\frac{32}{27}) is correct because the frequency of revolution is proportional to Z2n3\frac{Z^2}{n^3}, and the ratio is calculated accordingly.
508 Easy
508 of 592
Application +4 | -1
The energies of the electron in a hydrogen like ion are given by :
En=−(2.18×10−18)Z2(1n2)E_n=-\left(2.18 \times 10^{-18}\right) Z^2\left(\frac{1}{n^2}\right), where nn is principal quantum number and ZZ is the atomic number of the element. The ionisation energy (in kJmol−1\mathrm{kJ} \mathrm{mol}^{-1} ) of the Li2+\mathrm{Li}^{2+} ion is:
View answer and explanation
Correct option: D
11.8×10311.8 \times 10^3
The correct answer is 11.8×10311.8 \times 10^3 because it is calculated using the given formula for ionization energy with Z=3 and n=1, then converting to kJ/mol.
509 Medium
509 of 592
Analyzing +4 | -1
Assertion (A): The number of radial nodes in 3s and 4p orbitals is not equal.

Reason (R): The number of radial nodes in any orbital depends upon the values of nn and ll which are different for 3s and 4p
orbitals.
View answer and explanation
Correct option: D
Both (A) and (R) are false
Option D (Both (A) and (R) are false) is correct because the number of radial nodes depends on n and l, and the number of radial nodes is equal for 3s and 4p orbitals.
510 Easy
510 of 592
Remembering +4 | -1
The shape of pp-orbital is
View answer and explanation
Correct option: C
Dumb-bell
Option C (Dumb-bell) is correct because the p-orbital has a dumbbell shape.
Revision notes and key takeaways

Review STRUCTURE OF ATOM in your textbook or class notes, then use this set to check your understanding.

Showing 481-510 of 592 questions

Your practice

0 / 592 answered · 592 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.