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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 241-270 of 297 questions

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241 Medium
241 of 297
Understanding +4 | -1
Which of the following statements concerning the quantum numbers are correct?

(i) Angular quantum number determines the threedimensional shape of the orbital.

(ii) The principal quantum number determines the orientation and energy of the orbital.

(iii) Magnetic quantum number determines the size of the orbital.

(iv) Spin quantum number of an electron determines the orientation of the spin of electron relative to the chosen axis.

The correct set of option is
View answer and explanation
Correct option: B
(i) and (iv)
Option B is correct because the angular quantum number determines the shape of the orbital, and the spin quantum number determines the orientation of the electron's spin.
242 Medium
242 of 297
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.
243 Medium
243 of 297
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.
244 Medium
244 of 297
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.
245 Medium
245 of 297
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.
246 Medium
246 of 297
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.
247 Medium
247 of 297
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.
248 Medium
248 of 297
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.
249 Medium
249 of 297
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.
250 Medium
250 of 297
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.
251 Medium
251 of 297
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.
252 Medium
252 of 297
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.
253 Medium
253 of 297
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.
254 Medium
254 of 297
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.
255 Medium
255 of 297
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.
256 Medium
256 of 297
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.
257 Medium
257 of 297
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.
258 Medium
258 of 297
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.
259 Medium
259 of 297
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.
260 Medium
260 of 297
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.
261 Medium
261 of 297
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.
262 Medium
262 of 297
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.
263 Medium
263 of 297
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.
264 Medium
264 of 297
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.
265 Medium
265 of 297
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).
266 Medium
266 of 297
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.
267 Medium
267 of 297
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.
268 Medium
268 of 297
Remembering +4 | -1
Which of the following statement is/are correct about Hertz (Hz)(\mathrm{Hz})?
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because all the given statements about Hertz (Hz) are accurate and well-established.
269 Medium
269 of 297
Remembering +4 | -1
Cathode rays have same charge to mass ratio as
View answer and explanation
Correct option: A
β\beta-rays
Option A (β\beta-rays) is correct because cathode rays are streams of electrons, and beta particles are also electrons emitted from the nucleus.
270 Medium
270 of 297
Application +4 | -1
The ratio of wavelength values of series limit lines (n2=∞)(n_2 = \infty) of Balmer series and Paschen series are
View answer and explanation
Correct option: A
0.172916667
Option A (04:09) is correct because the ratio of the wavelengths is found by using the Rydberg formula for the series limit of the Balmer and Paschen series.
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