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JEE • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Mixed difficulty

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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 151-180 of 592 questions

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151 Easy
151 of 592
Remembering +4 | -1
The shape of ss-orbital is
View answer and explanation
Correct option: B
Spherical
Option B (Spherical) is correct because the s-orbital is spherically symmetrical around the nucleus.
152 Medium
152 of 592
Remembering +4 | -1
Compared to mass of lightest nucleus the mass of an electron is only:
View answer and explanation
Correct option: C
1/18001/1800
Option C (1/1800) is correct because the mass of an electron is approximately 1/1836 the mass of a hydrogen atom (lightest nucleus).
153 Medium
153 of 592
Understanding +4 | -1
The total number of angular nodes and radial nodes present in 3p orbital
View answer and explanation
Correct option: B
2
Option B (2) is correct because a 3p orbital has one angular node (l=1) and one radial node (n-l-1 = 3-1-1 = 1).
154 Medium
154 of 592
Understanding +4 | -1
The fraction of volume occupied by the nucleus with respect to the total volume of an atom is
View answer and explanation
Correct option: B
10−15{10^{ - 15}}
Option B is correct because the nucleus is extremely small compared to the overall size of the atom, occupying a fraction of approximately 10-15.
155 Medium
155 of 592
Application +4 | -1
The de-Broglie wavelength of the electron in the ground
state of hydrogen atom is (KE = 13.6 eV)
View answer and explanation
Correct option: C
0.3329 nm
Option C (0.3329 nm) is correct because it accurately calculates the de Broglie wavelength using the given kinetic energy of the electron.
156 Medium
156 of 592
Analyzing +4 | -1
A sample of hydrogen atoms containing all the atoms in a particular excited state absorb radiations of a particular wavelength by which the atoms get excited to another excited state. When the atoms finally de-excite to the ground state, they emit the radiations of 10 different wavelengths. Out of these 10 radiations, 7 have wavelengths shorter than the absorbed radiation and 2 have wavelength longer than the absorbed radiation. The orbit number for the initial excited state of atoms is
View answer and explanation
Correct option: A
3
The correct answer is A (3) because the atom absorbs radiation to reach the 5th orbit and emits 2 wavelengths longer than the absorbed one, indicating the initial state was the 3rd orbit.
157 Medium
157 of 592
Application +4 | -1
The wave number of first line of Balmer series of Hydrogen is 15200  cm−115200\,\,c{m^{ - 1}}. The wavenumber of the first line of 15200 cm−115200\,c{m^{ - 1}} ion is
View answer and explanation
Correct option: B
60800 cm−160800\,c{m^{ - 1}}
Option B (60800 cm−160800\,c{m^{ - 1}}) is correct because the wavenumber is proportional to Z2Z^2, where Z is the atomic number.
158 Medium
158 of 592
Application +4 | -1
A dust particle has mass equal to 10−11 g10^{-11} \mathrm{~g}, diameter 10−4 cm10^{-4} \mathrm{~cm} and velocity 10−4 cm s−110^{-4} \mathrm{~cm} \mathrm{~s}^{-1}. The error in measurement of velocity is 0.1%0.1 \%. What will be the uncertainty in its position?
View answer and explanation
Correct option: D
0.527×10−9 cm0.527 \times 10^{-9} \mathrm{~cm}
Answer D (0.527 x 10-9 cm) is correct because it is calculated using the Heisenberg uncertainty principle, considering the error in velocity measurement.
159 Medium
159 of 592
Analyzing +4 | -1
Evaluate the following statements related to group 14 elements for their correctness.

(A)Covalent radius decreases down the group from C\mathrm{C} to Pb\mathrm{Pb} in a regular manner.

(B) Electronegativity decreases from C\mathrm{C} to Pb\mathrm{Pb} down the group gradually.

(C) Maximum covalence of C\mathrm{C} is 4 whereas other elements can expand their covalence due to presence of dd orbitals.

(D) Heavier elements do not form pπ−pπ\mathrm{p} \pi-\mathrm{p} \pi bonds.

(E) Carbon can exhibit negative oxidation states.
Choose the correct answer from the options given below:
View answer and explanation
Correct option: A
(C), (D) and (E) Only
Option A ((C), (D) and (E) Only) is correct because statements C, D, and E are accurate descriptions of group 14 elements' properties.
160 Medium
160 of 592
Understanding +4 | -1
Which one of the following is not the characteristic of Planck's quantum theory of radiation?
View answer and explanation
Correct option: A
The energy is not absorbed or emitted in whole number or multiple of quantum.
Option A is correct because Planck's quantum theory states that energy is absorbed or emitted in whole number multiples of a quantum, not just any amount.
161 Easy
161 of 592
Remembering +4 | -1
The probability of finding the electron in the orbital is
View answer and explanation
Correct option: B
90-95%
Option B (90-95%) is correct because the probability of finding an electron within the defined space of an orbital is conventionally accepted to be within this range.
162 Easy
162 of 592
Analyzing +4 | -1
Assertion: Nuclear binding energy per nucleon is in the order 49Be>37Li>24He{ }_4^9 {Be}>{ }_3^7 {Li}>{ }_2^4 {He}.

Reason: Binding energy per nucleon increases linearly with difference in number of neutrons and protons.
View answer and explanation
Correct option: D
If both the Assertion and Reason are incorrect.
The correct answer is D (If both the Assertion and Reason are incorrect) because the binding energy per nucleon order is reversed, and the reason provided is not the correct explanation.
163 Medium
163 of 592
Application +4 | -1
In the ground state of Cu+\mathrm{C}{\mathrm{u}^ + }, the number of shell occupied, sub-shells occupied, fillied orbitals and unpaired electrons respectively are
View answer and explanation
Correct option: C
3,6,14,0
Option C (3,6,14,0) is correct because Cu+ has 3 shells, 6 subshells, 14 filled orbitals and 0 unpaired electrons.
164 Medium
164 of 592
Analyzing +4 | -1
Statement (S): Wavelength of I line of Humphrey series is more than I line of Lyman series in H-atom.

Explanation (E): ΔE=hcλ.\Delta E=\frac{h c}{\lambda}.
View answer and explanation
Correct option: C
Both S and E are correct and E is correct explanation of S
Option C (Both S and E are correct and E is correct explanation of S) is correct because the energy difference in the Humphrey series is smaller than in the Lyman series, leading to a longer wavelength, and the relationship between energy and wavelength is given by ΔE = hc/λ.
165 Medium
165 of 592
Application +4 | -1
How many moles of electrons weigh one milligram?
View answer and explanation
Correct option: B
1029.1×6.02\frac{10^2}{9.1 \times 6.02}
Option B is correct because it accurately calculates the number of moles of electrons required to weigh one milligram based on the mass of a single electron and Avogadro's number.
166 Easy
166 of 592
Remembering +4 | -1
Mass of neutron is … times the mass of electron
View answer and explanation
Correct option: A
1840
Option A (1840) is correct because the mass of a neutron is approximately 1840 times the mass of an electron.
167 Hard
167 of 592
Remembering +4 | -1
The emission spectrum of hydrogen is found to satisfy the expression for the energy change, ∆E (in joules), such that ΔE=2.18×10−18[1n12−1n22]\Delta E = 2.18 \times {10^{ - 18}}\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right] J, where, n1=1,2,3,  …{n_1} = 1,2,3,\; \ldots and n2=2,3,4,…{n_2} = 2,3,4, \ldots The spectral lines correspond to Paschen series are
View answer and explanation
Correct option: C
n1=3{n_1} = 3 and n2=4,5,6{n_2} = 4,5,6
The correct answer is C (n1=3{n_1} = 3 and n2=4,5,6{n_2} = 4,5,6) because the Paschen series corresponds to transitions where the electron falls to the n=3 energy level.
168 Easy
168 of 592
Analyzing +4 | -1
Statement (S): Dipositive zinc ion exhibits paramagnetism due to the loss of two electrons from 3d3 d-orbitals of neutral atom.

Explanation (E): Paramagnetism is due to the presence of unpaired electron.
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 dipositive zinc ion is diamagnetic due to the complete filling of d-orbitals, and paramagnetism is indeed due to unpaired electrons.
169 Medium
169 of 592
Understanding +4 | -1
Which shell would be the first to have ' gg ' sub-shell?
View answer and explanation
Correct option: D
O\mathrm{O}
Option D (O) is correct because the 'g' subshell corresponds to l=4, and the first shell to have l=4 is the n=5 (O) shell.
170 Easy
170 of 592
Understanding +4 | -1
Statement (S): All the dd-orbitals are identical in shape.

Explanation (E): All the pp-orbitals are identical in shape.
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 not all d-orbitals have the same shape, while all p-orbitals do have the same dumbbell shape, differing only in orientation.
171 Medium
171 of 592
Analyzing +4 | -1
Among the following transitions in hydrogen and
hydrogen-like ion spectrum, which one emits light
of longest wavelength ?
View answer and explanation
Correct option: C
n=4\mathrm{n}=4 to n=3\mathrm{n}=3 for He+\mathrm{He}^{+}
The transition with the smallest energy difference will emit light of the longest wavelength. Among the given options, n=4 to n=3 for He+ has the smallest energy difference considering the Z2 factor, making option C the correct answer.
172 Easy
172 of 592
Remembering +4 | -1
Heisenberg’s uncertainty principle rules out the exact simultaneous measurement of
View answer and explanation
Correct option: D
Position and momentum
Option D (Position and momentum) is correct because Heisenberg's uncertainty principle states that it is impossible to simultaneously know the exact position and momentum of a particle.
173 Medium
173 of 592
Analyzing +4 | -1
Of the following transitions in hydrogen atom, the one which gives an absorption line of lowest frequency is :
View answer and explanation
Correct option: B
n=3  to  n=8n = 3\;\mathrm{to\;}n = 8
The correct answer is B (n=3  to  n=8n = 3\;\mathrm{to\;}n = 8) because the energy difference between these levels is the smallest among the absorption transitions, resulting in the lowest frequency absorption line.
174 Medium
174 of 592
Remembering +4 | -1
The radiation having maximum wavelength is
View answer and explanation
Correct option: B
Radio waves
Option B (Radio waves) is correct because radio waves have the longest wavelength compared to UV rays, X-rays, and IR rays.
175 Medium
175 of 592
Application +4 | -1
The uncertainty in the position of an electron (mass == 9.1×10−28 g9.1 \times 10^{-28} {~g} ) moving with a velocity of 3.0×104 cm s−13.0 \times 10^4 {~cm} {~s}^{-1} accurate upto 0.011%0.011 \% will be
View answer and explanation
Correct option: C
0.175 cm0.175 {~cm}
Option C (0.175 cm0.175 {~cm}) is correct because it is calculated using the Heisenberg uncertainty principle, relating uncertainty in position and momentum.
176 Medium
176 of 592
Application +4 | -1
The electronic velocity in the fourth Bohr's orbit of hydrogen is v. The velocity of the electron in the first orbit would be:
View answer and explanation
Correct option: A
4v
The correct answer is 4v because the electronic velocity in a Bohr orbit is inversely proportional to the principal quantum number n.
177 Medium
177 of 592
Application +4 | -1
Ratio of radii of second and first Bohr orbits of H – atom
View answer and explanation
Correct option: B
4
Option B (4) is correct because the radius of the nth Bohr orbit is proportional to n², so the ratio of the radii of the second and first Bohr orbits is 2²/1² = 4.
178 Easy
178 of 592
Understanding +4 | -1
In hydrogen spectrum least energetic transition of electrons are found in:
View answer and explanation
Correct option: D
Pfund series
Option D (Pfund series) is correct because the Pfund series corresponds to transitions to the n=5 energy level, which results in the lowest energy transitions in the hydrogen spectrum compared to Lyman, Balmer, and Bracket series.
179 Easy
179 of 592
Understanding +4 | -1
If the value of azimuthal quantum number is 3 , the possible values of magnetic quantum number would be:
View answer and explanation
Correct option: C
0,±1,±2,±30, \pm 1, \pm 2, \pm 3
Option C (0,±1,±2,±30, \pm 1, \pm 2, \pm 3) is correct because for an azimuthal quantum number of 3, the magnetic quantum number can range from -3 to +3, including 0.
180 Medium
180 of 592
Application +4 | -1
Arrange α−\alpha - particle (α)(\alpha ), ,electron (e−),({e^ - }), proton (p) and neutron (n) in increasing order of their e/m value
View answer and explanation
Correct option: B
n<α<p<e−n < \alpha < p < {e^ - }
The correct answer is B (n<α<p<e−n < \alpha < p < {e^ - }). The e/m ratio depends on the charge and mass of the particle.
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