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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 1-30 of 297 questions

1 Medium

1 of 297

Application +4 | -1

The time taken by the electron in one complete revolution in the

n^{\text {th }}

Bohr's orbit of the hydrogen atom is

View answer and explanation

Correct option: B

Directly proportionalto

{n}^3

Option B (Directly proportionalto

{n}^3

) is correct because the time period is directly proportional to the cube of the principal quantum number.

2 Medium

2 of 297

Remembering +4 | -1

The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency '

\mathrm{A}

\times 10^{12}

hertz and that has a radiant intensity in that direction of

\frac{1}{{ }^{\prime} \mathrm{B}^{\prime}}

watt per steradian. 'A' and 'B' are respectively

View answer and explanation

Correct option: B

540 and 683

Option B (540 and 683) is correct because the candela is defined with a frequency of 540 × 10¹² Hz and a radiant intensity of 1/683 watt per steradian.

3 Medium

3 of 297

Remembering +4 | -1

Hamiltonian operator

\hat{H}

is the sum of energy operator. These are

View answer and explanation

Correct option: C

kinetic and potential

Option C (kinetic and potential) is correct because the Hamiltonian operator represents the total energy of a system, which is the sum of kinetic and potential energies.

4 Medium

4 of 297

Application +4 | -1

Radiation corresponding to the transition

\mathrm{n}=4

\mathrm{n}=2

in hydrogen atoms falls on a certain metal (work function

=2.0 \mathrm{eV}

). The maximum kinetic energy of the photoelectrons will be:

View answer and explanation

Correct option: A

0.55 \mathrm{~eV}

Option A (

0.55 \mathrm{~eV}

) is correct because the maximum kinetic energy of the photoelectrons is the energy of the incident radiation minus the work function of the metal.

5 Medium

5 of 297

Remembering +4 | -1

3p_z

orbital has

(a) One radial node

(b) Two radial nodes

(c) One angular node

(d) Two angular nodes

View answer and explanation

Correct option: B

a and c only

Option B (a and c only) is correct because a 3pz orbital has one radial node and one angular node.

6 Medium

6 of 297

Application +4 | -1

The radius of second Bohr’s orbit of hydrogen atom is

View answer and explanation

Correct option: C

0.2116 nm

Option C (0.2116 nm) is correct because it is the result of applying the formula for the radius of the nth Bohr orbit with n=2 and Z=1.

7 Medium

7 of 297

Application +4 | -1

The ratio of

\left( {{E_2} - {E_1}} \right)\,to\,\left( {{E_4} - {E_3}} \right)

for the Hydrogen atom is equal to

View answer and explanation

Correct option: B

Option D (12) is closest to the correct answer of approximately 15.4, which is the ratio of the energy differences for the given transitions in a hydrogen atom.

8 Medium

8 of 297

Remembering +4 | -1

The region in the electromagnetic spectrum where the Balmer series lines appear is

View answer and explanation

Correct option: D

visible

Option D (visible) is correct because the Balmer series lines appear in the visible region of the electromagnetic spectrum.

9 Medium

9 of 297

Remembering +4 | -1

Bohr frequency condition is given by

View answer and explanation

Correct option: A

v = \frac{{\Delta E}}{h}

Option A (

v = \frac{{\Delta E}}{h}

) is correct because it represents the Bohr frequency condition, relating the frequency of emitted radiation to the energy difference between energy levels.

10 Medium

10 of 297

Analyzing +4 | -1

Which is not isoelectronic with the other three?

View answer and explanation

Correct option: A

\mathrm{CO}

Option A (CO) is correct because it has 14 electrons, while the other three species have 16 electrons.

11 Medium

11 of 297

Understanding +4 | -1

Which of the following sets of quantum numbers is permissible for an electron in an atom?

View answer and explanation

Correct option: C

{n}=2, \ell=1, {m}=0, {s}=+1 / 2

Option C (

{n}=2, \ell=1, {m}=0, {s}=+1 / 2

) is correct because it follows the rules for permissible quantum numbers: n can be any positive integer, l ranges from 0 to n-1, m ranges from -l to +l, and s is +1/2 or -1/2.

12 Medium

12 of 297

Understanding +4 | -1

Which of the following orbital is represented by the complete wave function

\psi_{410}

View answer and explanation

Correct option: C

4 p

Option C (4p) is correct because the quantum numbers n=4 and l=1 correspond to the 4p orbital.

13 Medium

13 of 297

Analyzing +4 | -1

Of the species,

NO,\,\,N{O^ + },\,\,N{O^{2 + }}\,\,\mathrm{and}\,\,N{O^ - }

the one with minimum bond strength is

View answer and explanation

Correct option: D

N{O^ - }

Option D (

NO^−

) is correct because it has the lowest bond order (2) compared to the other species, resulting in the minimum bond strength.

14 Medium

14 of 297

Understanding +4 | -1

The maximum tendency to form the gaseous unipositive ion is for the element with configuration

View answer and explanation

Correct option: B

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}

Option B (

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}

) is correct because elements with electronic configurations close to noble gases readily lose electrons to achieve a stable configuration, and this configuration represents Aluminum which easily forms Al+.

15 Medium

15 of 297

Understanding +4 | -1

4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is

View answer and explanation

Correct option: B

5f > 6p > 5p > 4d

Option B (

5f > 6p > 5p > 4d

) is correct because the energy of an orbital is determined by the (n+l) rule, and when (n+l) is the same, the orbital with higher n has higher energy.

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16 Medium

16 of 297

Application +4 | -1

Ratio of energy of photon of wavelength

3000\mathop A\limits^ \circ

and

6000\mathop A\limits^ \circ

View answer and explanation

Correct option: B

2: 1

Option B (2:1) is correct because the energy of a photon is inversely proportional to its wavelength, so the ratio of energies is the inverse ratio of wavelengths.

17 Medium

17 of 297

Application +4 | -1

What is the wavelength in

\mathrm{nm}

of the spectral line associated with a transition from

n=3

n=2

for the

\mathrm{Li}^{2+}

ion?

View answer and explanation

Correct option: C

Option C (73) is correct because it is the wavelength calculated using the Rydberg formula for the given transition in the

\mathrm{Li}^{2+}

ion.

18 Medium

18 of 297

Analyzing +4 | -1

Which of the following electronic transitions requires for the greatest quantity of energy be absorbed by a H-atom?

View answer and explanation

Correct option: D

n=1

n=\infty

Option D (

n=1

n=\infty

) is correct because the energy difference is the greatest when the electron is completely removed from the atom.

19 Medium

19 of 297

Analyzing +4 | -1

Assertion: All p-orbitals have only one planar node.

Reason: The number of radial nodes depends on the principal quantum number only.

View answer and explanation

Correct option: C

If A is true but R is false

The correct answer is C (If A is true but R is false) because all p-orbitals have one planar node, but the number of radial nodes depends on both the principal quantum number and the azimuthal quantum number.

20 Medium

20 of 297

Application +4 | -1

The radius of which of the following orbit is same as that of first Bohr’s orbit of Hydrogen atom

View answer and explanation

Correct option: D

B{e^{ + 3}}\left( {n = 2} \right)

Option D (

Be^{+3}(n=2)

) is correct because the radius of its second orbit is the same as the first Bohr's orbit of Hydrogen atom, calculated using the formula r = 0.529 * (n²/Z).

21 Medium

21 of 297

Understanding +4 | -1

As ‘n’ value increases de-broglie’s wavelength of electron

View answer and explanation

Correct option: B

Increases

Option B is correct because as 'n' (principal quantum number) increases, the de Broglie wavelength of an electron increases, as the momentum decreases with increasing energy levels.

22 Medium

22 of 297

Analyzing +4 | -1

Assertion :- A spectral line will be seen for a

2p_x – 2p_y

transition.

Reason :- Only Balmer lines are observed in the visible region.

View answer and explanation

Correct option: D

If both Assertion & Reason are False

Option D (4,2) is correct because the molecule has 3 chiral centers, leading to a maximum of 8 stereoisomers, but due to symmetry, there are only 4 stereoisomers with 2 meso forms.

23 Medium

23 of 297

Application +4 | -1

Thewavelength of a

150 \mathrm{~g}

rubber ball moving with a velocity of

50 {~ms}^{-1}

is :

View answer and explanation

Correct option: D

8.83 \times 10^{-33} {~cm}

Option D (

8.83 \times 10^{-33} {~cm}

) is correct because it is calculated using the de Broglie wavelength formula with the given mass and velocity.

24 Medium

24 of 297

Analyzing +4 | -1

Which of the following is not iso electronic with

{H}_2 {S}?

View answer and explanation

Correct option: B

Oxide ion

Option B (Oxide ion) is not isoelectronic with

{H}_2 {S}

because oxide ion has 10 electrons, while

{H}_2 {S}

has 18 electrons.

25 Medium

25 of 297

Remembering +4 | -1

The ratio of charge to mass of an electron in coulombs per gram was determined by J.J. Thomson. He determined this ratio by measuring the deflection of cathode rays in electric and magnetic fields. What value did he find for this ratio?

View answer and explanation

Correct option: A

-1.76 \times 10^8

coulombs/g

Option A (

-1.76 \times 10^8

coulombs/g) is correct because it represents the experimentally determined charge-to-mass ratio of an electron.

26 Medium

26 of 297

Understanding +4 | -1

For hydrogen atom, the correct order of energy of orbitals is

View answer and explanation

Correct option: D

4f=4d=4p>3d=3p=3s

Option D (4f=4d=4p>3d=3p=3s) is correct because for a hydrogen atom, energy depends only on the principal quantum number (n).

27 Medium

27 of 297

Understanding +4 | -1

Consider the following statements:

I. The speed of light does not depends upon the nature of the medium through which it passes.

II. The beam of light is deviated or refracted from its original path as it passes from one medium to another.

III. When a ray of white light is passed through a prism, the wave with shorter wavelength bends more than the one with a longer wavelength.

Choose the correct statements and select the correct option.

View answer and explanation

Correct option: B

II and III

Option B (II and III) is correct because the speed of light depends on the medium, light refracts when passing between media, and shorter wavelengths bend more in a prism.

28 Medium

28 of 297

Remembering +4 | -1

The number of radial nodes of 3s and 2p-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.

29 Medium

29 of 297

Application +4 | -1

The mass of a negative meson is 208 time the mass of electron. If the

e / m

ratio of

\alpha

-particle is

x

unit, then the

e / m

ratio of this meson will be

View answer and explanation

Correct option: C

17.65\,x

unit

Option C (

17.65\,x

unit) is correct because the e/m ratio is inversely proportional to the mass, and the mass of the meson is given relative to the electron.

30 Medium

30 of 297

Understanding +4 | -1

An electron jumps from lower orbit to higher orbit, when

View answer and explanation

Correct option: B

Energy is absorbed

Option B (Energy is absorbed) is correct because electrons absorb energy when transitioning from a lower to a higher energy orbit.

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Key Takeaways

✓ This set contains 297 single-correct-answer MCQs on STRUCTURE OF ATOM.
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