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NEET • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Mixed difficulty

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Showing 391-420 of 492 questions

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391 Medium
391 of 492
Application +4 | -1
The mass of one mole of electron is:
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Correct option: A
0.55 mg
Option A (0.55 mg) is correct because the mass of one mole of electrons is the mass of a single electron multiplied by Avogadro's number.
392 Easy
392 of 492
Remembering +4 | -1
According to Bohr's postulates which quantity can take up only discrete values:
View answer and explanation
Correct option: B
Angular momentum
Option B (Angular momentum) is correct because, according to Bohr's postulates, the angular momentum of an electron is quantized and can only take discrete values.
393 Easy
393 of 492
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 (N-Z).
394 Medium
394 of 492
Understanding +4 | -1
Which of the following transitions are not allowed in the normal electronic emission spectrum of an atom?
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Correct option: A
2s→1s2s \to 1s
Option A (2s→1s2s \to 1s) is correct because transitions between s orbitals are forbidden due to the selection rule Δl=±1\Delta l = \pm 1.
395 Easy
395 of 492
Application +4 | -1
In H atom, the electron is de-excited from 5th shell to 1st shell. How many different lines may appear in line spectrum?
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Correct option: C
10
Option C (10) is correct because the number of spectral lines when an electron de-excites from nth shell to the ground state is given by n(n-1)/2. For n=5, it is 5(5-1)/2 = 10.
396 Easy
396 of 492
Remembering +4 | -1
Orbitals processing the same energy are called:
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Correct option: D
Degenerate orbitals
Option D is correct because orbitals possessing the same energy are called degenerate orbitals.
397 Easy
397 of 492
Application +4 | -1
An ion which has 18 electrons in the outermost shell is:
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Correct option: A
Cu+\mathrm{C}{\mathrm{u}^ + }
Option A (Cu+\mathrm{C}{\mathrm{u}^ + }) is correct because Cu⁺ has the electronic configuration [Ar]3d¹⁰, which has 18 electrons in the outermost shell (n=3: 3s² 3p⁶ 3d¹⁰ = 2+6+10 = 18).
398 Medium
398 of 492
Remembering +4 | -1
The frequency of radiation absorbed or emitted when transition occurs between two stationary states with energies E1E_1 (lower) and E2E_2 (higher) is given by
View answer and explanation
Correct option: D
v=E2−E1hv=\frac{E_2-E_1}{h}
Option D (All of the above) is correct because the wave mechanical model of the atom incorporates de Broglie's concept of wave-particle duality, Heisenberg's uncertainty principle, and Schrödinger's wave equation to describe the behavior of electrons in atoms.
399 Medium
399 of 492
Remembering +4 | -1
The magnetic moment of electron in an atom (excluding orbital magnetic moment) is given by:
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Correct option: A
n(n+2)\sqrt {n\left( {n + 2} \right)} Bohr Magneton (or B.M)
Option A (n(n+2)\sqrt {n(n + 2)} Bohr Magneton) is correct because it represents the spin-only magnetic moment formula, where 'n' is the number of unpaired electrons.
400 Hard
400 of 492
Understanding +4 | -1
In case of photoelectric effect, slope of V0\mathrm{V}_0 vs v\mathrm{v} curve is (where V0\mathrm{V}_0-stopping potential, v=\mathrm{v}= subjected frequency)
View answer and explanation
Correct option: B
he\frac{\mathrm{h}}{\mathrm{e}}
Option B is correct. The slope of the V0 vs v curve in the photoelectric effect is h/e, derived from the photoelectric equation.
401 Medium
401 of 492
Remembering +4 | -1
What is the packet of energy called?
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Correct option: B
photon
The correct answer is B (photon) because a photon is defined as a discrete packet of energy of electromagnetic radiation.
402 Hard
402 of 492
Application +4 | -1
The shortest wavelength of H atom in the Lyman series is λ1.{\lambda _1}. The longest wavelength in the Balmer series of He+H{e^ + } is
View answer and explanation
Correct option: C
9λ15\frac{{9{\lambda _1}}}{5}
Option C (9λ15\frac{{9{\lambda _1}}}{5}) is correct because it involves calculating the wavelengths using the Rydberg formula and then finding the ratio.
403 Medium
403 of 492
Remembering +4 | -1
The uncertainty principle and the concept of wave nature of matter was proposed by … and … respectively
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Correct option: A
Heisenberg, de Broglie
Option A (Heisenberg, de Broglie) is correct because Heisenberg proposed the uncertainty principle, and de Broglie proposed the wave nature of matter.
404 Medium
404 of 492
Analyzing +4 | -1
Highest energy will be absorbed to eject out the electron in the
configuration.
View answer and explanation
Correct option: B
1s22s22p31{s^2}2{s^2}2{p^3}
Option B (1s22s22p31{s^2}2{s^2}2{p^3}) is correct because it represents a half-filled p-orbital configuration, which is exceptionally stable and requires more energy to remove an electron.
405 Medium
405 of 492
Understanding +4 | -1
In photoelectric emission the energy of the emitted electrons is:
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Correct option: B
Smaller than that of incident photo
Option B (Smaller than that of incident photo) is correct because some of the energy of the incident photon is used to overcome the work function of the metal, leaving less energy for the kinetic energy of the emitted electron.
406 Easy
406 of 492
Analyzing +4 | -1
Statement (S): Specific charge of α\alpha-particles is twice to that of proton.

Explanation (E): Specific charge is given by e/m.e / m.
View answer and explanation
Correct option: B
S is wrong but E is correct
Option B (S is wrong but E is correct) is the correct answer because the specific charge of an alpha particle is half that of a proton, and the explanation of specific charge being e/m is correct.
407 Easy
407 of 492
Understanding +4 | -1
Dimensions of Planck’s constant are:
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Correct option: D
energy × time
Option D (energy × time) is correct because Planck's constant (h) relates energy and frequency (E = hf), so h = E/f, and since frequency is inverse of time, h has dimensions of energy × time.
408 Easy
408 of 492
Understanding +4 | -1
When beryllium is bambarded with α\alpha - particles extremely penetrating 396s which cannot be deflected by electrical or magnetic field are given out. These are
View answer and explanation
Correct option: C
A bean of neutrons
Option C (A bean of neutrons) is correct because when beryllium is bombarded with alpha particles, neutrons are emitted, which are neutral and highly penetrating.
409 Easy
409 of 492
Understanding +4 | -1
The electric configuration of element with atomic number 24 is
View answer and explanation
Correct option: D
1s2,2s22p6,3s23p63d5,4s11{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5},4{s^1}
Option D (1s2,2s22p6,3s23p63d5,4s11{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5},4{s^1}) is correct because Chromium (Cr) with atomic number 24 has an exception to Hund's rule and Aufbau principle, where it achieves a more stable half-filled d-orbital configuration.
410 Medium
410 of 492
Application +4 | -1
The energy per mole of photon of electromagnetic radiation of wavelength 4000 Å is:
View answer and explanation
Correct option: A
3.0×1012erg3.0 \times {10^{ 12}}\mathrm{erg}
Option A (3.0×1012erg3.0 \times {10^{ 12}}\mathrm{erg}) is correct because it represents the energy per mole of photons calculated using Planck's equation and Avogadro's number.
411 Easy
411 of 492
Remembering +4 | -1
In the atomic spectrum of hydrogen the series of lines observed in the visible region is:
View answer and explanation
Correct option: A
Balmer series
Option A (Balmer series) is correct because the Balmer series corresponds to transitions to the n=2n=2 energy level, which results in emission lines in the visible region of the electromagnetic spectrum.
412 Easy
412 of 492
Application +4 | -1
The volume of a proton is approximately;
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Correct option: B
1.5×10−38cm31.5 \times {10^{ - 38}}\mathrm{c}{\mathrm{m}^3}
Option B (1.5×10−38cm31.5 \times {10^{ - 38}}\mathrm{c}{\mathrm{m}^3}) is correct because it is the approximate volume of a proton, calculated using its radius.
413 Medium
413 of 492
Remembering +4 | -1
Amongst the following statements, select the set having statements which was proposed by Dalton.
(1) All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.
(2) When gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same T & P\mathrm{T} \,\&\, \mathrm{P}
(3) Chemical reactions involve reorganization of atoms. These are neither created nor destroyed in a chemical reaction.
(4) Matter consists of indivisible atoms
View answer and explanation
Correct option: B
(1), (3), (4)
Option B ((1), (3), (4)) is correct because Dalton's atomic theory proposed that all atoms of an element are identical, chemical reactions involve reorganization of atoms, and matter consists of indivisible atoms.
414 Medium
414 of 492
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?
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Correct option: C
S1>S2{S_1} > {S_2}
Option C (S1>S2{S_1} > {S_2}) is correct because electrons (cathode rays) have a much larger charge-to-mass ratio than positive ions (positive rays).
415 Medium
415 of 492
Remembering +4 | -1
Which principle/rule limits the maximum number of electrons in an orbital to two?
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Correct option: B
Pauli’s exclusion principle
Option B (Pauli’s exclusion principle) is correct because it states that no two electrons in an atom can have the same set of four quantum numbers, limiting each orbital to two electrons with opposite spins.
416 Easy
416 of 492
Understanding +4 | -1
The ratio of e/m,i.e., specific charge for a cathode ray:
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Correct option: B
Is constant
Option B (Is constant) is correct because the specific charge (e/m) of a cathode ray (electron) is a fundamental constant and does not depend on the gas in the discharge tube or the cathode material.
417 Hard
417 of 492
Analyzing +4 | -1
Statement (S): All s-orbital in H\mathrm{H}-atom corresponds to a non-zero probability density at nucleus

Explanation (E): The probability density is given by: ψ2\psi^2 and ψ∝e−Zr2a0\psi \propto e^{-\frac{Z r}{2 a_0}}
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 s-orbitals have a non-zero probability density at the nucleus, and the given equation explains this behavior.
418 Easy
418 of 492
Application +4 | -1
A certain radio station broadcasts on a frequency of 980 KHz, what is the wavelength of electromagnetic radiation broadcast by the radio station
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Correct option: A
306 m
Option A (306 m) is correct because wavelength is calculated using the formula wavelength = speed of light / frequency.
419 Hard
419 of 492
Application +4 | -1
Uncertainty in the position of an electron (mass=9.1×10−31kg)\left( {\mathrm{mass} = 9.1 \times {{10}^{ - 31}}\mathrm{kg}} \right) moving with a velocity 300 ms−1\mathrm{m}{\mathrm{s}^{ - 1}} accurate upon 0.001% will be
(h=6.63×10−34  Js)\left( {h = 6.63 \times {{10}^{ - 34}}\mathrm{\;Js}} \right)
View answer and explanation
Correct option: C
1.92×10−2m1.92 \times {10^{ - 2}}\mathrm{m}
Option D (91.2 nm) is correct because it is calculated using the formula λ = hc/E, where E is the ionization potential.
420 Easy
420 of 492
Analyzing +4 | -1
Assertion : In case of isoelectronic ions the ionic size increases with the increase in atomic number.

Reason : The greater the attraction of nucleus, greater is the ionic radius
View answer and explanation
Correct option: D
If the assertion and reason both are false.
Option D is correct because both the assertion and the reason are false. For isoelectronic ions, ionic size decreases with increasing atomic number, and greater nuclear attraction leads to a smaller ionic radius.
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