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NEET • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Mixed difficulty

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Showing 211-240 of 492 questions

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211 Easy
211 of 492
Remembering +4 | -1
Non-directional orbital is
View answer and explanation
Correct option: D
3s
Option D is correct because s-orbitals are spherically symmetrical and thus non-directional, unlike p and d orbitals which have specific directional properties.
212 Medium
212 of 492
Understanding +4 | -1
Which particle may be removed from a stable neutral atom with least energy change?
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Correct option: D
An electron
Option D (An electron) is correct because electrons are the least tightly bound particles in a neutral atom and require the least energy to remove.
213 Easy
213 of 492
Application +4 | -1
The de-Broglie wavelength associated with a particle of mass 10−6 kg10^{-6} {~kg} moving with a velocity of 10 ms−110 {~ms}^{-1} is :
View answer and explanation
Correct option: D
6.63×10−29 m6.63 \times 10^{-29} {~m}
The correct answer is D (6.63×10−29 m6.63 \times 10^{-29} {~m}) because it is calculated by using the de Broglie wavelength formula: λ=hmv\lambda = \frac{h}{mv}.
214 Medium
214 of 492
Analyzing +4 | -1
Which of the following is not permissible arrangement of electrons in an atom
View answer and explanation
Correct option: A
n=3,  l=2,  m=−3,  s=−12n = 3,\,\,l = 2,\,\,m = - 3,\,\,s = - \frac{1}{2}
Option A is the correct answer because for n=3 and l=2, the possible values of m are -2, -1, 0, 1, 2. m=-3 is not permissible.
215 Medium
215 of 492
Remembering +4 | -1
For ionizing an ground state hydrogen atom, the energy needed in electron volts will be :
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Correct option: A
13.613.6
Option A (13.613.6) is correct because the ionization energy of a ground state hydrogen atom is 13.6 eV.
216 Hard
216 of 492
Application +4 | -1
Calculate the energy required for the process Zeff \mathrm{Z}_{\text {eff }} He+(g)→He2+(g)+e−\mathrm{He^{+}(g}) \rightarrow \mathrm{He^{2+}(g)+e}^{-}. The ionisation energy for the H\mathrm{H}-atom in the ground state is 2.18×10−18 J atom−12.18 \times 10^{-18} \mathrm{~J ~atom}^{-1}.
View answer and explanation
Correct option: A
8.72×10−188.72 \times 10^{-18} atoms−1^{-1}
Option A (8.72×10−188.72 \times 10^{-18} atoms−1^{-1}) is correct because the ionization energy for He+ is Z2 times the ionization energy of hydrogen, where Z is the atomic number.
217 Medium
217 of 492
Application +4 | -1
What is the maximum number of electrons in an atom that can have the following quantum numbers n=4,  m1=+1?n = 4,\;{m_1} = + 1?
View answer and explanation
Correct option: D
6
Option D (6) is correct because for n=4 and ml=+1, there are 2 possible values of s (+1/2 and -1/2) for each of the three orbitals (4p, 4d) with ml=+1.
218 Easy
218 of 492
Remembering +4 | -1
All types of electromagnetic radiations possess same:
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Correct option: B
Velocity
Option D (Velocity) is correct because all electromagnetic radiations travel at the same speed (speed of light) in a vacuum.
219 Medium
219 of 492
Analyzing +4 | -1
Which set is not correct?
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Correct option: C
3,1,2,-1/2
Option C (3,1,2,-1/2) is not correct because the value of 'l' (azimuthal quantum number) must be less than 'n' (principal quantum number), and the value of 'm' (magnetic quantum number) must be between -l and +l. Here, l=1, so m can only be -1, 0, or +1, not 2.
220 Easy
220 of 492
Remembering +4 | -1
Wave nature of electrons was demonstrated by
View answer and explanation
Correct option: B
De-Broglie
Option B (De-Broglie) is correct because De-Broglie proposed the wave nature of electrons.
221 Hard
221 of 492
Application +4 | -1
In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is (h=6.6×10−34kg  m2s−1,  mass  of  electron,  em=9.1×10−31kg)\left( {\begin{array}{cccccccccccccccccccc} {h = 6.6 \times {{10}^{ - 34}}kg\;{\mathrm{m}^2}{\mathrm{s}^{ - 1}},\;}\\ {mass\;of\;electron,\;{e_\mathrm{m}} = 9.1 \times {{10}^{ - 31}}kg} \end{array}} \right)
View answer and explanation
Correct option: C
1.92×10−3m1.92 \times {10^{ - 3}}\mathrm{m}
Option C (1.92 × 10-3 m) is correct because it is calculated using the Heisenberg Uncertainty Principle, relating uncertainty in position and momentum.
222 Hard
222 of 492
Understanding +4 | -1
In which orbital electron is most tightly bound to the nucleus?
View answer and explanation
Correct option: B
4p
Option B (4p) is correct because, for a given principal quantum number (n), lower azimuthal quantum number (l) means the electron is more tightly bound to the nucleus due to lower energy.
223 Medium
223 of 492
Analyzing +4 | -1
The value of charge on the oil droplets experimentally observed were −1.6×10−19- 1.6 \times {10^{ - 19}} and −4×10−19- 4 \times {10^{ - 19}} coulomb. The value of the electronic charge, indicated by these results is:
View answer and explanation
Correct option: D
−0.8×10−19- 0.8 \times {10^{ - 19}}
Option D (−0.8×10−19- 0.8 \times {10^{ - 19}}) is correct because it represents the greatest common divisor of the observed charges, suggesting it's the fundamental unit of charge.
224 Easy
224 of 492
Understanding +4 | -1
The difference between ions and atoms is of:
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Correct option: D
All of these
Option D (All of these) is correct because ions differ from atoms in size, electron configuration, and the presence of a net charge.
225 Medium
225 of 492
Analyzing +4 | -1
Assertion: Ratio of anode rays is different for different gases.
Reason: Proton is the fundamental particle present in gases.
View answer and explanation
Correct option: B
Both assertion and reason are true but reason is not the correct explanation of assertion
Option B (Both assertion and reason are true but reason is not the correct explanation of assertion) is correct because anode rays depend on the gas used, and protons are fundamental particles, but the proton presence doesn't explain the varying ratio.
226 Hard
226 of 492
Analyzing +4 | -1
Assertion (A): ss-orbital electron will be more tightly bound to the nucleus than pp-orbital electron.
Reason (R): Zeff Z_{\text {eff }} experienced by the electron decreases with increase of azimuthal quantum number (ll).
View answer and explanation
Correct option: B
Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion
Option B is correct because both the assertion and the reason are correct, but the reason is not the correct explanation of the assertion. s-orbitals are more tightly bound, and Zeff decreases with increasing l, but the binding energy is more directly related to penetration.
227 Medium
227 of 492
Analyzing +4 | -1
The last electron placed in the third (n=3) quantum shell for:
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Correct option: C
Cu
Option C (Cu) is correct because copper has an electronic configuration of [Ar] 3d¹⁰4s¹, meaning the last electron enters the n=3 shell (3d orbital).
228 Easy
228 of 492
Understanding +4 | -1
In order to designate an orbital n in an atom, the number of quantum number required are:
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Correct option: C
3
Option C (3) is correct because to designate an orbital in an atom, we need the principal quantum number (n), the azimuthal quantum number (l), and the magnetic quantum number (ml).
229 Medium
229 of 492
Understanding +4 | -1
When atoms are bombarded with α-particles suffer deflections while others pass through undeflected.
This is because :
View answer and explanation
Correct option: B
The nucleus occupies much smaller volume compared to the volume of the atom
Option B (The nucleus occupies much smaller volume compared to the volume of the atom) is correct because most of the atom is empty space, allowing most alpha particles to pass through undeflected, while the small, dense nucleus causes deflections.
230 Medium
230 of 492
Remembering +4 | -1
The region in the electromagnetic spectrum where the Balmer series lines appear is
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Correct option: D
visible
Option D (visible) is correct because the Balmer series lines are produced by electron transitions that emit photons in the visible region of the electromagnetic spectrum.
231 Hard
231 of 492
Application +4 | -1
Uncertainty in position of a particle of 25 g in space is 10−5m{10^{ - 5}}\mathrm{m}. Hence, uncertainty in velocity (ms−1)\left( {\mathrm{m}{\mathrm{s}^{ - 1}}} \right) is (Planck’s constant h=6.6×10−34Jsh = 6.6 \times {10^{ - 34}}\mathrm{Js})
View answer and explanation
Correct option: A
2.1×10−282.1 \times {10^{ - 28}}
Option C (1.92×10−2m1.92 \times {10^{ - 2}}\mathrm{m}) is correct because it is calculated using the Heisenberg uncertainty principle.
232 Medium
232 of 492
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 because beta particles are much lighter than alpha particles and would not be significantly scattered beyond the gold foil due to the relatively small mass of electrons in the atoms.
233 Hard
233 of 492
Application +4 | -1
If the ionisation potential for hydrogen atom is 13.6eV, then the wavelength of light required for the ionisation of hydrogen atom would be:
View answer and explanation
Correct option: D
91.2 nm
Option D (12mhπ\frac{1}{{2m}}\sqrt {\frac{h}{\pi }}) is correct because it is derived from the Heisenberg uncertainty principle with equal uncertainties in position and momentum.
234 Hard
234 of 492
Application +4 | -1
The uncertainty in the momentum of an electron is 1.0×10−5kg  ms−11.0 \times {10^{ - 5}}\mathrm{kg\;m}{\mathrm{s}^{ - 1}} , The uncertainty in its position will be
View answer and explanation
Correct option: C
5.27×10−30m5.27 \times {10^{ - 30}}\mathrm{m}
Option C (5.27×10−30m5.27 \times {10^{ - 30}}\mathrm{m}) is correct because it is calculated using the Heisenberg uncertainty principle.
235 Medium
235 of 492
Application +4 | -1
The ratio of radius of III and IV Bohr's orbits in hydrogen atom is:
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Correct option: C
9∶16
Option C (9∶16) is correct because the radius of the nth Bohr orbit is proportional to n2n^2, so the ratio of the radii is the ratio of the squares of the orbit numbers.
236 Hard
236 of 492
Analyzing +4 | -1
Consider the following statements : 1.Electron density in xy plane in 3dx2−y23{d_{{x^2} - {y^2}}} orbital is zero 2.Electron density in xy plane in 3dz23{d_{{z^2}}} orbital is zero 3.2s orbital has only one spherical node 4.For 2pz2{p_z} orbital yz is the nodal plane The correct statements are
View answer and explanation
Correct option: C
Only 3
Option C is correct because only statement 3 is correct. 2s orbital has one spherical node. The other statements are incorrect regarding electron density and nodal planes.
237 Medium
237 of 492
Application +4 | -1
Which one of the following set of quantum numbers is possible for electron in the ground state of an atom with atomic number 19?
View answer and explanation
Correct option: D
n=4,  l=0,  m=0n = 4,\;l = 0,\;m = 0
Option D (n=4,  l=0,  m=0n = 4,\;l = 0,\;m = 0) is correct because the ground state electronic configuration of an atom with atomic number 19 (Potassium) is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹, and the last electron occupies the 4s orbital, which has n=4, l=0, and m=0.
238 Medium
238 of 492
Remembering +4 | -1
The potential energy of electron revolving in the ground state of H\mathrm{H} atom is
View answer and explanation
Correct option: C
−27.2 eV-27.2\,\mathrm{eV}
Option B (Ultraviolet) is correct because ultraviolet radiation has a higher frequency and therefore higher energy than visible light.
239 Hard
239 of 492
Analyzing +4 | -1
When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA{T_A} (expressed in eV) and de Broglie wavelength λA.{\lambda _A}. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70V is TB=TA−1.50  eV.{T_B} = {T_A} - 1.50\;\mathrm{eV}. If the de Broglie wavelength of these photoelectrons is λB=2λA,{\lambda _B} = 2{\lambda _A}, then which is not correct?
View answer and explanation
Correct option: B
The work function of B is 3.70 eV
Option B (The work function of B is 3.70 eV) is not correct. The work function of B is calculated to be 2.75 eV, not 3.70 eV.
240 Medium
240 of 492
Remembering +4 | -1
The momentum of a photon is p,p, the energy associated with photon is given by:
View answer and explanation
Correct option: B
Ep\frac{E}{p}
Option B (Ep\frac{E}{p}) is correct because the energy of a photon is related to its momentum by the equation E=pcE = pc, so c=Epc = \frac{E}{p}, and since the question is asking for the energy associated with the photon, the correct answer is Ep\frac{E}{p}.
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