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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 391-420 of 592 questions

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391 Easy
391 of 592
Remembering +4 | -1
Energy of atomic orbitals in a particular shell is in the order
View answer and explanation
Correct option: A
s<p<d<fs < p < d < f
Option A (s<p<d<fs < p < d < f) is correct because, for a given shell, the energy of atomic orbitals increases with the azimuthal quantum number (l), where s (l=0), p (l=1), d (l=2), and f (l=3).
392 Medium
392 of 592
Application +4 | -1
The atomic number of an element is 35 and its mass is 81. The number of electrons in its outermost shell is
View answer and explanation
Correct option: C
7
Option C (7) is correct because the electronic configuration of the element with atomic number 35 has 7 electrons in its outermost shell.
393 Easy
393 of 592
Understanding +4 | -1
One require energy En{E_n} to remove a nucleon from a nucleus and an energy Ee{E_e} to remove an electron from the orbit of an atom, then:
View answer and explanation
Correct option: C
En>Ee{E_n} > {E_e}
Option C is correct because the energy required to remove a nucleon from a nucleus (En{E_n}) is much greater than the energy required to remove an electron from the orbit of an atom (Ee{E_e}) due to the strong nuclear force.
394 Medium
394 of 592
Understanding +4 | -1
The radius of first orbit of H\mathrm{H}-atoms is 0.529A∘0.529\mathop {\text{A}}\limits^ \circ. The radius of first orbit of D\mathrm{D}-atoms should be
View answer and explanation
Correct option: B
slightly less than 0.529A∘0.529\mathop {\text{A}}\limits^ \circ
Option B (slightly less than 0.529A∘0.529\mathop {\text{A}}\limits^ \circ) is correct because the radius depends on the reduced mass, which is slightly larger for deuterium than for hydrogen, leading to a slightly smaller radius.
395 Medium
395 of 592
Understanding +4 | -1
In unielectron system, the wave number of any spectral line is directly proportional to
View answer and explanation
Correct option: C
1n12−1n22\frac{1}{n_1^2}-\frac{1}{n_2^2}
Option C (1n12−1n22\frac{1}{n_1^2}-\frac{1}{n_2^2}) is correct because the wave number is directly proportional to the difference in the inverse squares of the principal quantum numbers of the initial and final energy levels.
396 Easy
396 of 592
Understanding +4 | -1
If n=3,n = 3, then which value of ′l′'l' is correct
View answer and explanation
Correct option: D
All of them
The correct answer is D (All of them) because for a given principal quantum number n=3, the azimuthal quantum number l can have values from 0 to n-1, which includes 0, 1, and 2.
397 Easy
397 of 592
Understanding +4 | -1
An atom having even number of electrons may be:
View answer and explanation
Correct option: C
Diamagnetic or paramagnetic
Option C (Diamagnetic or paramagnetic) is correct because atoms with an even number of electrons can have all electrons paired (diamagnetic) or some unpaired (paramagnetic).
398 Medium
398 of 592
Understanding +4 | -1
If S1{S_1} be the specific charge (e/m) of cathode ray and S2{S_2} be that of positive rays, then which is true?
View answer and explanation
Correct option: C
S1>S2{S_1} > {S_2}
Option C (S1>S2{S_1} > {S_2}) is correct because the specific charge (e/m) is inversely proportional to mass, and electrons (cathode rays) have a much smaller mass than positive ions.
399 Medium
399 of 592
Remembering +4 | -1
Cathode rays carry/are
View answer and explanation
Correct option: B
−-ve charge
Option B (-ve charge) is correct because cathode rays are streams of electrons, which are negatively charged particles.
400 Hard
400 of 592
Application +4 | -1
Be3+\mathrm{Be}^{3+} and a proton are accelerated by the same potential, their de-Broglie wavelengths have the ratio (assume mass of proton = mass of neutron) :
View answer and explanation
Correct option: D
1:331: 3 \sqrt{3}
Option D (1:331: 3 \sqrt{3}) is correct because the de Broglie wavelength is inversely proportional to the square root of the mass and charge, leading to the given ratio after calculation.
401 Medium
401 of 592
Application +4 | -1
A 100 watt bulb emits monochromatic light of wavelength 400 nm, the number of photons emitted per second by this bulb is
View answer and explanation
Correct option: A
2×1020 s−12 \times {10^{20}}\,{s^{ - 1}}
Option A (2×1020 s−12 \times {10^{20}}\,{s^{ - 1}}) is correct because it is the number of photons emitted per second calculated using the power and wavelength of the light.
402 Easy
402 of 592
Remembering +4 | -1
No two electrons can have the same values of &hellip;. quantum numbers.
View answer and explanation
Correct option: D
Four
Option D (Four) is correct because, according to Pauli's exclusion principle, no two electrons in an atom can have the same set of four quantum numbers.
403 Easy
403 of 592
Remembering +4 | -1
Which of the following is not a fundamental particle?
View answer and explanation
Correct option: D
X-rays
Option D (X-rays) is correct because X-rays are electromagnetic radiation, not a fundamental particle like electrons, protons, and neutrons.
404 Medium
404 of 592
Application +4 | -1
Magnetic moments of V(Z=23),Cr⁡(Z=24)V(Z=23), \operatorname{Cr}(Z=24) and Mn⁡(Z=25)\operatorname{Mn}(Z=25) are x,y,zx, y, z. Hence
View answer and explanation
Correct option: C
x<z<yx < z < y
Option C (x<z<yx < z < y) is correct because the magnetic moment depends on the number of unpaired electrons, and the electronic configurations lead to this order.
405 Hard
405 of 592
Application +4 | -1
Calculate the wavelength of light required to break the bond between two chlorine atoms in a chlorine molecule. The Cl−Cl\mathrm{Cl}-\mathrm{Cl} bond energy is 243KJ/mol.(h=6.6×10−34Js)243 \mathrm{KJ} / \mathrm{mol} .\left(h=6.6 \times 10^{-34} \mathrm{Js}\right)
View answer and explanation
Correct option: A
4.91×10−7m4.91 \times 10^{-7} \mathrm{m}
Option A (4.91×10−7m4.91 \times 10^{-7} \mathrm{m}) is correct because the wavelength is calculated using the energy required to break the Cl-Cl bond and the Planck's equation.
406 Easy
406 of 592
Application +4 | -1
Which of the following pairs of nucleides are isodiaphers ?
View answer and explanation
Correct option: D
2555Mn{ }_{25}^{55} \mathrm{Mn} and 3065Zn{ }_{30}^{65} \mathrm{Zn}
Option D (2555Mn{ }_{25}^{55} \mathrm{Mn} and 3065Zn{ }_{30}^{65} \mathrm{Zn}) is correct because isodiaphers have the same difference between the number of neutrons and protons (A-2Z).
407 Medium
407 of 592
Understanding +4 | -1
Which of the following sets of quantum numbers is correct for an electron in 4f-orbital.
View answer and explanation
Correct option: A
n=4,  l=3,  m=+1,   s=+12n = 4,\,\,l = 3,\,\,m = + 1,\,\,\,s = + \frac{1}{2}
Option A (n=4,  l=3,  m=+1,  s=+12n = 4,\,\,l = 3,\,\,m = + 1,\,\,s = + \frac{1}{2}) is correct because for a 4f orbital, n=4, l=3, m can be any integer from -3 to +3, and s can be +1/2 or -1/2.
408 Hard
408 of 592
Application +4 | -1
The ionisation enthalpy of hydrogen atom is 1.312×106Jmol−11.312 \times {10^6}\mathrm{Jmo}{\mathrm{l}^{ - 1}}. The energy required to excite the electron in the atom from n1=1  to  n2=2{n_1} = 1\;\mathrm{to }\;{n_2} = 2 is
View answer and explanation
Correct option: D
9.84×105  J  mol−19.84 \times {10^5}\mathrm{\;J\;mo}{\mathrm{l}^{ - 1}}
The correct answer is 9.84×105  J  mol−19.84 \times {10^5}\mathrm{\;J\;mo}{\mathrm{l}^{ - 1}} because it is calculated by finding the difference in energy levels between n=1 and n=2 using the ionization enthalpy.
409 Hard
409 of 592
Application +4 | -1
Oxygen consists of O16,O17  and  O18{\mathrm{O}^{16}},{\mathrm{O}^{17}}\mathrm{\;and\;}{\mathrm{O}^{18}} isotopes and carbon consists of isotopes of C12{\mathrm{C}^{12}} and C13{\mathrm{C}^{13}}. Total number of CO2C{O_2} molecules possible are:
View answer and explanation
Correct option: B
12
Option C (18) is incorrect. There are 3 isotopes of Oxygen and 2 isotopes of Carbon. Since each CO2 molecule has one carbon and two oxygen atoms, we need to consider all combinations. The correct answer is 6.
410 Easy
410 of 592
Understanding +4 | -1
The electronic configuration of element with atomic number 24 is
View answer and explanation
Correct option: D
1s2,2s22p6,3s23p63d54s11 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^5 4 s^1
Option D (1s2,2s22p6,3s23p63d54s11 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^5 4 s^1) is correct because Chromium (atomic number 24) has an exception to Hund's rule, leading to a half-filled d-orbital.
411 Medium
411 of 592
Application +4 | -1
A quantity of 1.0 g1.0 \mathrm{~g} hydrogen atoms are excited to radiations. The study of spectra indicates that 45%45 \% of the atoms are in third energy level and 40%40 \% of atoms in second energy level and the rest in ground state. The total energy (in kJ\mathrm{kJ}) evolved, when all the atoms return to ground state is (Ionization energy of hydrogen atom is 1310 kJ/mol1310 \mathrm{~kJ} / \mathrm{mol})
View answer and explanation
Correct option: A
917
Option A (917 kJ) is correct because it is calculated by summing the energy released when atoms transition from the third and second energy levels to the ground state.
412 Medium
412 of 592
Application +4 | -1
The uncertainty in position and velocity of a particle are 10−11 m10^{-11} \mathrm{~m} and 5.27×10−24 ms−15.27 \times 10^{-24} \mathrm{~ms}^{-1}, respectively. The minimum mass of the particle (in kg\mathrm{kg} ) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the minimum mass is calculated using the Heisenberg uncertainty principle.
413 Easy
413 of 592
Remembering +4 | -1
The electronic configuration of silver atom in ground state is:
View answer and explanation
Correct option: C
[Kr]4d10,5s1\left[ {\mathrm{Kr}} \right]4{d^{10}},5{s^1}
Option C ([Kr][\mathrm{Kr}]4{d10},5{s1}) is correct because silver's electronic configuration has a completely filled d-orbital and a single electron in the s-orbital for stability.
414 Medium
414 of 592
Analyzing +4 | -1
If Rutherford would have used β\beta-particles instead of α\alpha-particles in his Gold-leaf experiment, which of the following observations would definitely not have been made:
View answer and explanation
Correct option: A
β\beta-particles scattering beyond the gold foil.
Option A is correct. Beta particles, being negatively charged, would be attracted towards the positively charged nucleus, so they would not scatter *beyond* the gold foil.
415 Medium
415 of 592
Analyzing +4 | -1
Which set of quantum number are not possible
View answer and explanation
Correct option: B
nlms221+1/2\begin{array}{cccccccccccccccccccc} n&l&m&s\\ 2&2&1&{ + 1/2} \end{array}
Option B (n=2, l=2, m=1, s=+1/2) is not possible because the value of l cannot be equal to n; l must be less than n.
416 Medium
416 of 592
Application +4 | -1
A 25 Watt bulb emits monochromatic yellow light of wavelength of 0.57μm0.57 \mu \mathrm{m}. What will be the rate of emission of quanta per second?
View answer and explanation
Correct option: A
7.17×10197.17 \times 10^{19}
Option A (7.17×10197.17 \times 10^{19}) is correct because it is the rate of emission of quanta per second calculated using the energy of a photon and the power of the bulb.
417 Medium
417 of 592
Application +4 | -1
Total number of spectral lines in UV region, during transition from 5th 5^{\text {th }} excited state to 1st 1^{\text {st }} excited state
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Correct option: D
Zero
Option D (1) is correct because the percentage of ionization is calculated using the formula α=KaC×100\alpha = \sqrt{\frac{K_a}{C}} \times 100.
418 Easy
418 of 592
Analyzing +4 | -1
Assertion : Total number of orbitals associated with principal quantum number n=3n=3 is 6 .

Reason : Number of orbitals in a shell equals to 2n2 n.
View answer and explanation
Correct option: D
If the assertion and reason both are false.
Option D is correct. The assertion is false because the total number of orbitals with n=3 is 9 (1s + 3p + 5d), and the reason is false because the number of orbitals in a shell is n².
419 Medium
419 of 592
Application +4 | -1
An electron is accelerated from a very low velocity (∼\sim zero speed) by the application of a potential difference of V\mathrm{V} volts. If the de Broglie wavelength should change (i.e., decrease) by 1.0%1.0 \% what percent increase in V\mathrm{V} causes it-
View answer and explanation
Correct option: D
2%2 \%
Option D (2%) is correct because the de Broglie wavelength is inversely proportional to the square root of the potential difference.
420 Easy
420 of 592
Remembering +4 | -1
Pauli's exclusion principle states that
View answer and explanation
Correct option: D
All the four quantum numbers of twoelectrons in an atom cannot be equal
Option D (All the four quantum numbers of twoelectrons in an atom cannot be equal) is correct because Pauli's exclusion principle states that no two electrons in an atom can have the same set of all four quantum numbers.
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