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JEE • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Medium

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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 271-297 of 297 questions

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271 Medium
271 of 297
Application +4 | -1
The energy of an electron in first Bohr orbit of H-atom is -13.6 eV. The possible energy value of electron in the excited state of Li2+\mathrm{L}{\mathrm{i}^{2 + }} is
View answer and explanation
Correct option: C
−30.6 eV- 30.6\,eV
Option C (-30.6 eV) is correct because it's calculated using the formula for energy levels in hydrogen-like atoms, considering the excited state (n=2) and the atomic number of Li2+ (Z=3).
272 Medium
272 of 297
Application +4 | -1
Electromagnetic radiations of wavelength 240 nm\mathrm{nm} are just sufficient to ionize sodium atom. The ionization energy of sodium (in kJ/mol\mathrm{kJ} / \mathrm{mol} ) is
View answer and explanation
Correct option: B
498.58
Option B (498.58) is correct because the ionization energy is calculated using the given wavelength and the formula I.E. = hc/λ.
273 Medium
273 of 297
Application +4 | -1
The principal quantum number of H-atom orbital, if the electron energy is -3.4 eV, will be
View answer and explanation
Correct option: B
2
Option B (2) is correct because the energy of an electron in a hydrogen atom is given by -13.6/n² eV, and solving for n when the energy is -3.4 eV gives n=2.
274 Medium
274 of 297
Application +4 | -1
A typical golf ball weighs 40.0 g40.0 \mathrm{~g}. If it is moving with a velocity of 20.0 ms−120.0 \mathrm{~ms}^{-1}, its de-Broglie wavelength is
View answer and explanation
Correct option: C
8.28×10−25 nm8.28 \times 10^{-25} \mathrm{~nm}
Option B (8.28×10−32 nm8.28 \times 10^{-32} \mathrm{~nm}) is correct because it is calculated using the de Broglie wavelength formula, λ=hmv\lambda = \frac{h}{mv}.
275 Medium
275 of 297
Application +4 | -1
Bond dissociation on energy of Br2\mathrm{Br}_2 is 200 kJ/mole200 \mathrm{~kJ} / \mathrm{mole}. The longest wavelength of photon that can break this bond would be (NA×hc=0.12 J)\left(N_{\mathrm{A}} \times h c=0.12 \mathrm{~J}\right)
View answer and explanation
Correct option: C
6.0×10−7 m6.0 \times 10^{-7} \mathrm{~m}
Option C (6.0×10−7 m6.0 \times 10^{-7} \mathrm{~m}) is correct because it is the longest wavelength of a photon that can break the Br2Br_2 bond, calculated using the energy-wavelength relationship.
276 Medium
276 of 297
Application +4 | -1
According to Heisenberg's uncertainty principle, the product of uncertainties in position and velocities for an electron of mass 9.1×10−31 kg9.1 \times 10^{-31} {~kg} is
View answer and explanation
Correct option: C
5.8×10−5 m2 s−15.8 \times 10^{-5} {~m}^2 {~s}^{-1}
Option C (5.8×10−5 m2 s−15.8 \times 10^{-5} \mathrm{~m}^2 \mathrm{~s}^{-1}) is correct because it is calculated using Heisenberg's uncertainty principle.
277 Medium
277 of 297
Application +4 | -1
Quantum numbers for some electrons are given below

A : n=4,I=1n=4, I=1\qquad B : n=4,I=0n=4, I=0\qquad C: n=3,I=2n=3, I=2\qquad D: n=3,I=1n=3, I=1

The correct increasing order of energy of electrons
View answer and explanation
Correct option: C
D << B << C << A
Option C (D << B << C << A) is correct because it represents the increasing order of energy based on the (n+l) rule and the n value when (n+l) is the same.
278 Medium
278 of 297
Application +4 | -1
The radius of an orbit of H – atom is 4.77×10−10m4.77 \times {10^{ - 10}}m. Then the angular momentum of an electron in this orbit is
View answer and explanation
Correct option: B
1.5h/π1.5h/\pi
Option B (1.5h/π1.5h/\pi) is correct because the radius is used to determine the orbit number (n=3), and then the angular momentum is calculated using the formula L=nh/2πL = nh/2\pi.
279 Medium
279 of 297
Application +4 | -1
A photon of light of wavelength 6000 Å has energy EE. What will be the wavelength of photon of a light which has energy of photon 4E4 E ?
View answer and explanation
Correct option: A
1500 Å
Option A (1500 Å) is correct because energy and wavelength are inversely proportional, so quadrupling the energy reduces the wavelength to one-fourth.
280 Medium
280 of 297
Understanding +4 | -1
ASSERTION (A) : Two electrons in an atom can have the same values of four quantum numbers.

REASON (R) : Two electrons in an atom can be present in the same shell, subshell and orbital and have the same spin.
View answer and explanation
Correct option: D
If AA and RR both are wrong.
Option D (If A and R both are wrong) is correct because Assertion (A) is wrong due to Pauli Exclusion Principle, and Reason (R) is also wrong because electrons in the same orbital must have opposite spins.
281 Medium
281 of 297
Remembering +4 | -1
The principal and azimuthal quantum number of electrons in 4f4 f orbitals are
View answer and explanation
Correct option: C
4,3
Option C (4,3) is correct because for a 4f orbital, the principal quantum number (n) is 4 and the azimuthal quantum number (l) is 3.
282 Medium
282 of 297
Application +4 | -1
Based on the equation ΔE=−2.0×10−18 J(1n22−1n12)\Delta E=-2.0 \times 10^{-18} \mathrm{~J}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)
the wavelength of the light that must be absorbed to excite hydrogen electron from level n=1n=1 to level n=2n=2 will be (h=6.625×10−34 J s,c=3×108 m s−1)\left(h=6.625 \times 10^{-34} \mathrm{~J} \mathrm{~s}, c=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)
View answer and explanation
Correct option: A
1.325×10−7 m1.325 \times 10^{-7} \mathrm{~m}
Option A (1.325×10−7 m1.325 \times 10^{-7} \mathrm{~m}) is correct because it is the calculated wavelength of light absorbed during the electron transition from n=1 to n=2 in a hydrogen atom, using the provided energy equation.
283 Medium
283 of 297
Analyzing +4 | -1
Statement-I: Emitted radiation will be 0.408×10−190.408 \times 10^{-19} joule when an electron jump from n=4n=4 to n=2n=2 in H\mathrm{H}-atom.

Statement-II: Balmer series radiations belong to IR region for hydrogen atom only.
View answer and explanation
Correct option: B
Statement I and Statement II both are incorrect
The correct answer is B (Statement I and Statement II both are incorrect) because the emitted radiation energy is 4.08 × 10-19 J, and Balmer series radiations belong to the visible region.
284 Medium
284 of 297
Application +4 | -1
Determine the wavelength of a photon of electromagnetic radiation having energy 2.99×1012 erg mol−1(h=6.625×10−34 Js)2.99 \times 10^{12} \mathrm{~erg} \mathrm{~mol}^{-1} \left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~Js}\right).
View answer and explanation
Correct option: A
4000 A∘4000 ~\mathop {\text{A}}\limits^ \circ
Option A (4000 A∘4000 ~\mathop {\text{A}}\limits^ \circ) is correct because it is the wavelength calculated using the energy of the photon and Avogadro's number.
285 Medium
285 of 297
Understanding +4 | -1
The magnetic quantum number for valency electron of sodium is:
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because the valence electron of sodium is in the 3s orbital, which has l=0, and therefore m=0.
286 Medium
286 of 297
Application +4 | -1
If the magnitude of uncertainties in position (Δx)(\Delta {x}) and momentum (Δp)(\Delta {p}) are equal then which of the following relation is correct?
View answer and explanation
Correct option: A
Δv=1 mh4π\Delta {v}=\frac{1}{{~m}} \sqrt{\frac{{h}}{4 \pi}}
Option A (Δv=1 mh4π\Delta {v}=\frac{1}{{~m}} \sqrt{\frac{{h}}{4 \pi}}) is correct because it correctly applies the Heisenberg uncertainty principle with equal uncertainties in position and momentum.
287 Medium
287 of 297
Application +4 | -1
What is the potential energy of electron in 3
orbit of H-atom ?
View answer and explanation
Correct option: A
−3.02 eV-3.02 \mathrm{~eV}
Option A (−3.02 eV-3.02 \mathrm{~eV}) is correct because the potential energy is twice the total energy, and the total energy of an electron in the 3rd orbit of a hydrogen atom is -1.51 eV, so the potential energy is 2 * -1.51 eV = -3.02 eV.
288 Medium
288 of 297
Understanding +4 | -1
Radial nodes present in 3s3 \mathrm{s} and 2p2 \mathrm{p}-orbitals are respectively
View answer and explanation
Correct option: B
2, 0
Option B (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.
289 Medium
289 of 297
Application +4 | -1
In H-atom, if ‘x’ is the radius of the first orbit, de-Broglie wavelength of an electron in 3rd orbit is
View answer and explanation
Correct option: B
6 πx6\,\pi x
Option B (6πx) is correct because the de-Broglie wavelength is related to the radius of the orbit and the orbit number.
290 Medium
290 of 297
Application +4 | -1
The energy difference will be minimum for which of the following energy levels of H\mathrm{H}-atom?
View answer and explanation
Correct option: B
n=3n=3 and n=4n=4
Option B (n=3 and n=4) is correct because the energy difference between higher energy levels in the H-atom decreases as n increases, making the difference between n=3 and n=4 the smallest among the options.
291 Medium
291 of 297
Analyzing +4 | -1
The value of charge on the oil droplets experimentally observed were −1.6×10−19- 1.6 \times {10^{ - 19}} and −4×10−19- 4 \times {10^{ - 19}} coulomb. The value of the electronic charge, indicated by these results is:
View answer and explanation
Correct option: D
−0.8×10−19- 0.8 \times {10^{ - 19}}
The correct answer is 1.6×10−191.6 \times {10^{ - 19}} because the greatest common divisor of the observed charges is 1.6×10−191.6 \times 10^{-19} C.
292 Medium
292 of 297
Analyzing +4 | -1
Statement (S): The magnetic moment of Mg\mathrm{Mg}-atom is more than K\mathrm{K}-atom as the former has two electrons in outermost shell.

Explanation (E): The magnetic moment of N\mathrm{N}-atom is more than magnetic moment of O\mathrm{O}-atom and former has more number of unpaired electrons.
View answer and explanation
Correct option: B
S is wrong but E is correct
Option B (S is wrong but E is correct) is correct because the magnetic moment depends on unpaired electrons, and the explanation about N and O is accurate.
293 Medium
293 of 297
Application +4 | -1
If the speed of the electron in the first Bohr orbit of hydrogen atom be ' xx ' then the speed of electron in the third Bohr orbit is
View answer and explanation
Correct option: B
x3\frac{x}{3}
Option B (x3\frac{x}{3}) is correct because the speed of an electron in a Bohr orbit is inversely proportional to the principal quantum number (n).
294 Medium
294 of 297
Understanding +4 | -1
Which of the following set of quantum numbers is impossible for an electron in an atom?
View answer and explanation
Correct option: D
n=4,ℓ=2,m=−3,s=−12{n}=4, \ell=2, {m}=-3, {s}=-\frac{1}{2}
Option D (n=4,ℓ=2,m=−3,s=−12{n}=4, \ell=2, {m}=-3, {s}=-\frac{{1}}{{2}}) is impossible because for ℓ=2\ell=2, the possible values of m are -2, -1, 0, 1, 2. m cannot be -3.
295 Medium
295 of 297
Understanding +4 | -1
When atomic orbitals are filled according to the Aufbau principle. The 6p6 \mathrm{p} orbitals are filled immediately after the:
View answer and explanation
Correct option: B
5d5 \mathrm{d}
Option B (5d) is correct because, according to the Aufbau principle, the 6p orbitals are filled immediately after the 5d orbitals, as the (n+l) value for 5d is slightly lower.
296 Medium
296 of 297
Analyzing +4 | -1
Which of the following electron transition in hydrogen atom will require largest amount of energy?
View answer and explanation
Correct option: A
From n=1 to n=2
Option A (From n=1 to n=2) is correct because the energy required for the transition is inversely proportional to the square of the initial and final energy levels, and this transition has the largest energy difference.
297 Medium
297 of 297
Understanding +4 | -1
The representation of the ground state electronic configuration of He by box-diagram as is wrong because it violates
View answer and explanation
Correct option: C
Pauli exclusion principle
The correct answer is C (Pauli exclusion principle) because the given box diagram violates the Pauli exclusion principle, which states that no two electrons in an atom can have the same set of four quantum numbers.
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