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 511-540 of 592 questions

Preparing practice and saved progress…

511 Medium
511 of 592
Analyzing +4 | -1
Assertion :- In hydrogen energy of 4s is more than 3d.

Reason :- An orbital with lower value of (n+ℓ)(n+\ell) has smaller energy than the orbital with higher value of (n+ℓ).(n+\ell).
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B is correct because both the assertion and reason are true, but the reason is not the correct explanation for the assertion. The energy of orbitals in multi-electron atoms depends on both n and l, but the (n+l) rule is an approximation and not always accurate.
512 Easy
512 of 592
Understanding +4 | -1
STATEMENT-1 : γ\gamma-rays have very high penetrating power.

STATEMENT-2 : γ\gamma-rays are electromagnetic radiations of high energy.
View answer and explanation
Correct option: A
If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1
Option A (If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1) is correct because gamma rays have high penetrating power due to their high energy electromagnetic radiation nature.
513 Medium
513 of 592
Application +4 | -1
The number of wave made by an electron moving in an orbit having maximum magnetic quantum number +3 is :
View answer and explanation
Correct option: A
4
Option A (4) is correct because the number of waves made by an electron is equal to the principal quantum number, which is one more than the maximum magnetic quantum number.
514 Medium
514 of 592
Application +4 | -1
Photodissociation of water H2O(l)+hv→H2( g){H}_2 {O}({l})+h v \rightarrow {H}_2({~g}) +1/2O2( g)+1 / 2 {O}_2({~g}) has been suggested as a source of hydrogen. The heat absorbed in this reaction is 289.5 kJ/mole289.5 \mathrm{~kJ} / \mathrm{mole} of water decomposed. The maximum wavelength that would provide the necessary energy assuming that one photon causes the dissociation of one water molecule is ( 1eV=96.5 kJ/mol1 \mathrm{eV}=96.5 \mathrm{~kJ} / \mathrm{mol} )
View answer and explanation
Correct option: A
413.33 nm413.33 \mathrm{~nm}
Option A (413.33 nm413.33 \mathrm{~nm}) is correct because it is calculated using the energy of the photon required for photodissociation of water.
515 Medium
515 of 592
Understanding +4 | -1
What is the correct orbital designation for the electron with the quantum numbers, n=4,l=3,m=−2,s=1/2?n = 4,l = 3,m = - 2,s = 1/2?
View answer and explanation
Correct option: B
4f
Option B (4f) is correct because with n=4 and l=3, the electron is in the 4f orbital.
516 Medium
516 of 592
Application +4 | -1
What is the energy (kJ/mol)({kJ} / {mol}) associated with the de-excitation of an electron from n=6{n}=6 to n=2{n}=2 in He+{He}^{+}ion?
View answer and explanation
Correct option: C
1.16×1031.16 \times 10^3
Option C (1.16×1031.16 \times 10^3) is correct because it is the energy calculated using the Rydberg formula for the de-excitation of an electron in a He+{He}^{+} ion.
517 Medium
517 of 592
Remembering +4 | -1
The number of radial nodes in 3 s3 {~s} and 2p2 {p} orbitals are respectively
View answer and explanation
Correct option: A
2,0
Option A (2,0) is correct because the number of radial nodes is calculated using the formula n - l - 1, where n is the principal quantum number and l is the azimuthal quantum number.
518 Easy
518 of 592
Application +4 | -1
The radius of first Bohr's orbit for hydrogen is 0.53A∘0.53 \mathop {\text{A}}\limits^ \circ. The radius of third Bohr's orbit would be
View answer and explanation
Correct option: D
4.77A∘4.77\mathop {\text{A}}\limits^ \circ
Option D (4.77A∘4.77\mathop {\text{A}}\limits^ \circ) is correct because the radius of the nth Bohr orbit is proportional to n2n^2, so the third orbit's radius is 9 times the first.
519 Medium
519 of 592
Understanding +4 | -1
Number of electrons in −CONH2- \mathrm{CON}{\mathrm{H}_2} are
View answer and explanation
Correct option: A
24
Option A (24) is correct because the total number of electrons in −CONH2-CONH_2 is the sum of the valence electrons of each atom plus the electrons from the bonds.
520 Easy
520 of 592
Remembering +4 | -1
The energy of a photon is inversely proportional to its
View answer and explanation
Correct option: A
Wave length
Option A (Wave length) is correct because the energy of a photon is inversely proportional to its wavelength, as described by the equation E = hc/λ.
521 Medium
521 of 592
Understanding +4 | -1
Which of the following sets of quantum number are correct

a) n=1, l=1, m=+2n = 1,\,l = 1,\,m = + 2 b) n=2, l=1, m=+1n = 2,\,l = 1,\,m = + 1

c) n=3, l=2, m=−2n = 3,\,l = 2,\,m = - 2 d) n=3, l=4, m=−2n = 3,\,l = 4,\,m = - 2
View answer and explanation
Correct option: B
b,c
Option B (b,c) is correct because the allowed values for the quantum numbers are: 0≤l≤n−10 \le l \le n-1 and −l≤m≤+l-l \le m \le +l.
522 Easy
522 of 592
Application +4 | -1
What will be the difference in mass number if the number of neutrons halved and the number of electrons double in 6C12{{}_6{C^{12}}}
View answer and explanation
Correct option: A
3
Option A (3) is correct because halving the neutrons and doubling the electrons results in a mass number difference of 3.
523 Easy
523 of 592
Remembering +4 | -1
According to Dalton, the smallest particle of matter which is indivisible, is called:
View answer and explanation
Correct option: B
Atom
Option B (Atom) is correct because, according to Dalton's atomic theory, the atom is the smallest indivisible particle of matter.
524 Easy
524 of 592
Understanding +4 | -1
What is the trend of energy of Bohr’s orbits?
View answer and explanation
Correct option: A
Energy of the orbit increases as we move away from the nucleus
Option A is correct because the energy of an electron in Bohr's orbit increases as the electron moves farther away from the nucleus (higher n value).
525 Easy
525 of 592
Remembering +4 | -1
What is the ration of mass of an electron to the mass of a proton?
View answer and explanation
Correct option: C
1:18371:1837
Option C (1:18371:1837) is correct because the mass of an electron is approximately 1/1837th the mass of a proton.
526 Medium
526 of 592
Remembering +4 | -1
The possibility of finding an electron in an orbital was conceived by:
View answer and explanation
Correct option: D
Schrödinger
Option D (Schrödinger) is correct because Erwin Schrödinger conceived the concept of atomic orbitals, which are regions of space where there is a high probability of finding an electron.
527 Medium
527 of 592
Application +4 | -1
Which transition in the hydrogen atomic spectrum will have the same wavelength as the Balmer transition (i.e. n=4\mathrm{n}=4 to n=2\mathrm{n}=2 ) of He+\mathrm{He}^{+}spectrum
View answer and explanation
Correct option: D
n=2\mathrm{n}=2 to n=1\mathrm{n}=1
The correct answer is n=2\mathrm{n}=2 to n=1\mathrm{n}=1 because the wavelength of the Balmer transition (n=4 to n=2) of He+\mathrm{He}^{+} is equal to the wavelength of the transition from n=2 to n=1 in hydrogen.
528 Hard
528 of 592
Application +4 | -1
Given that in the H-atom the transition energy for n=1 to n=2n = 1\,to\,n = 2, 10.2eV, the energy for the same transition in Be3+\mathrm{B}{\mathrm{e}^{3 + }} is:
View answer and explanation
Correct option: C
163.2 eV
Option C (163.2 eV) is correct because the energy transition is proportional to Z2Z^2, where Z is the atomic number.
529 Easy
529 of 592
Understanding +4 | -1
Which of the following wave properties is proportional to
energy for electromagnetic radiation?
View answer and explanation
Correct option: B
Wave number
Option B (Wave number) is correct because energy is directly proportional to the wave number of electromagnetic radiation, as indicated by the formula E ∝ ν̄.
530 Easy
530 of 592
Application +4 | -1
According to (n+l) rule after completing 'np' level the electron enters to:
View answer and explanation
Correct option: B
(n+1)s\left( {n + 1} \right)s
Option B ((n + 1)s) is correct because according to the (n+l) rule, after filling the np level, the next electron occupies the (n+1)s level due to its lower energy.
531 Easy
531 of 592
Application +4 | -1
An electron has a velocity of 300 m/sec300 \mathrm{~m} / \mathrm{sec} that is accurate to 0.01%0.01 \% with what accuracy can be locate the position of this electron :
View answer and explanation
Correct option: D
0.00190.0019 meter
Option D (0.00190.0019 meter) is correct because it is calculated using the Heisenberg uncertainty principle, relating the uncertainty in position and momentum.
532 Easy
532 of 592
Remembering +4 | -1
The equation, λ=hmv\lambda = \frac{h}{{mv}} was deduced by
View answer and explanation
Correct option: B
de-Broglie
Option B (de-Broglie) is correct because de-Broglie deduced the equation λ=hmv\lambda = \frac{h}{{mv}} relating wavelength to momentum.
533 Medium
533 of 592
Understanding +4 | -1
Which of the following conclusions could not be derived from Rutherford’s α\alpha - particle scattering experiment
View answer and explanation
Correct option: C
Electrons move in a circular path of fixed energy called orbits
Option C (Electrons move in a circular path of fixed energy called orbits) is correct because Rutherford's experiment established the nuclear model but did not propose the concept of electrons moving in fixed energy orbits, which was introduced by Bohr.
534 Medium
534 of 592
Application +4 | -1
The number of electrons in an atom with atomic number 105 having (n+l)=8\left( {n + l} \right) = 8 are:
View answer and explanation
Correct option: B
17
Option B (17) is correct because the element with atomic number 105 has 17 electrons with (n + l) = 8 based on its electronic configuration.
535 Easy
535 of 592
Analyzing +4 | -1
Statement (S): Humphrey series discovered in H\mathrm{H} atomic spectra has lowest energy radiations.

Explanation (E): The series belongs to n1=6.n_1=6.
View answer and explanation
Correct option: C
Both S and E are correct and E is correct explanation of S
Option C is correct because the Humphrey series corresponds to transitions to n1=6, and transitions to higher energy levels result in lower energy radiations.
536 Easy
536 of 592
Analyzing +4 | -1
S-I: Fe+3\mathrm{Fe}^{+3} ion is less stable than Fe+2.\mathrm{Fe}^{+2}.

S-II : Fe+3\mathrm{Fe}^{+3} ion has less number of unpaired electrons than Fe+2.\mathrm{Fe}^{+2}.
View answer and explanation
Correct option: B
Both S-I and S-II are incorrect
The correct answer is B (Both S-I and S-II are incorrect) because Fe+3 is more stable than Fe+2 due to its half-filled d-orbital configuration, and it has more unpaired electrons.
537 Medium
537 of 592
Analyzing +4 | -1
In which one of the following pairs the two species are both isoelectronic and isotopic?
(Atomic numbers : Ca=20,Ar=18, K=19,Mg=12{Ca}=20, {Ar}=18, {~K}=19, {Mg}=12, Fe=26,Na=11{Fe}=26, {Na}=11 )
View answer and explanation
Correct option: B
39 K+{ }^{39} {~K}^{+}and 40 K+{ }^{40} {~K}^{+}
Option B is correct because 39 K+{ }^{39} {~K}^{+} and 40 K+{ }^{40} {~K}^{+} are both isoelectronic (same number of electrons) and isotopic (same atomic number, different mass number).
538 Medium
538 of 592
Analyzing +4 | -1
Assertion: The speed of light depends upon the nature of the medium through which it passes.

Reason: When a ray of white light is passed through a prism, the wave with shorter wavelength bends less than the one with a longer wavelength.
View answer and explanation
Correct option: B
If both A and R are true but R is not the correct explanation of A
The correct answer is B (If both A and R are true but R is not the correct explanation of A) because both the assertion and reason are true, but the reason does not explain why the speed of light depends on the medium.
539 Easy
539 of 592
Analyzing +4 | -1
S-I: Line spectrum of Li+2\mathrm{Li}^{+2} and He+\mathrm{He}^{+}are identical.

S-II: Isoelectronic species produce identical spectrum.
View answer and explanation
Correct option: A
Both S-I and S-II are correct
Option A is correct because both statements are correct. Both Li+2 and He+ have one electron and thus identical spectra. Isoelectronic species with the same number of electrons produce identical spectra.
540 Medium
540 of 592
Application +4 | -1
If a0{a}_0 is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength (λ)(\lambda) of the electron present in the second orbit of hydrogen atom? [n:[ n : any integer]
View answer and explanation
Correct option: B
8πa0n\frac{8 \pi a_0}{n}
Option B (8πa0n\frac{8 \pi a_0}{n}) is correct because the de-Broglie wavelength is calculated using the circumference of the orbit and Bohr's radius.
Revision notes and key takeaways

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

Showing 511-540 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

Answer feedback and difficulty

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.