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NEET • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Mixed difficulty

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Showing 91-120 of 492 questions

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91 Easy
91 of 492
Remembering +4 | -1
Angular momentum of an electron in an orbital is given by
View answer and explanation
Correct option: B
h2π×l(l+1)\frac{h}{{2\pi }} \times \sqrt {l\left( {l + 1} \right)}
Option B (h2π×l(l+1)\frac{h}{{2\pi }} \times \sqrt {l\left( {l + 1} \right)}) is correct because this is the formula for the magnitude of the orbital angular momentum.
92 Hard
92 of 492
Application +4 | -1
If e=1.60206×10−19C,em=1.75875×1011Ckg−1e=1.60206 \times 10^{-19} {C}, \frac{e}{m}=1.75875 \times 10^{11} {Ckg}^{-1} then the mass of electron is
View answer and explanation
Correct option: B
9.1091×10−31 kg9.1091 \times 10^{-31} {~kg}
Option B (9.1091×10−31 kg9.1091 \times 10^{-31} {~kg}) is correct because it is the mass of an electron calculated using the given values of e and e/m.
93 Hard
93 of 492
Analyzing +4 | -1
Assertion :- When we give electron beam of 10.60 eV energy on hydrogen sample it will
excite to first excited state.

Reason :- The difference between n2n_2 and n1n_1 is 10.2005 eV only.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because the energy difference between the ground state and the first excited state of hydrogen is approximately 10.2 eV, and providing 10.60 eV will excite the electron to that level.
94 Medium
94 of 492
Remembering +4 | -1
The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency ' A\mathrm{A} ' ×1012\times 10^{12} hertz and that has a radiant intensity in that direction of 1′B′\frac{1}{{ }^{\prime} \mathrm{B}^{\prime}} watt per steradian. 'A' and 'B' are respectively
View answer and explanation
Correct option: B
540 and 683
Option B (540 and 683) is correct because the candela is defined based on a monochromatic radiation of frequency 540 × 1012 Hz and a radiant intensity of 1/683 watt per steradian.
95 Easy
95 of 492
Understanding +4 | -1
The specific charge for positive rays is much less than the specific charge for cathode rays. This is because:
View answer and explanation
Correct option: C
Positive rays comprise ionised atoms, whose mass is much higher
Option C (Positive rays comprise ionised atoms, whose mass is much higher) is correct because specific charge is the ratio of charge to mass, and positive rays consist of heavier ionized atoms compared to electrons in cathode rays.
96 Easy
96 of 492
Remembering +4 | -1
The number of nodal planes in a PX{P_X} orbital is
View answer and explanation
Correct option: A
One
Option A (One) is correct because a pxp_x orbital has one nodal plane, which is the yz-plane.
97 Medium
97 of 492
Understanding +4 | -1
Which one of the following is the set of correct quantum numbers of an electron in 3d orbital?
View answer and explanation
Correct option: D
n=3,  l=2,  m=1,  s=+1/2n = 3,\;l = 2,\;m = 1,\;s = + 1/2
Option D (n=3,  l=2,  m=1,  s=+1/2n = 3,\;l = 2,\;m = 1,\;s = + 1/2) is correct because it provides a valid set of quantum numbers for an electron in a 3d orbital: n=3 (3rd energy level), l=2 (d-orbital), m=1 (possible magnetic quantum number for d-orbital), and s=+1/2 (possible spin quantum number).
98 Easy
98 of 492
Analyzing +4 | -1
STATEMENT-1 : Emitted radiation will fall in visible range when an electron jump from n=4n=4 to n=2n=2 in H{H}-atom.

STATEMENT-2 : Balmer series radiations belong to visible range for hydrogen atom only.
View answer and explanation
Correct option: A
If both the statement are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-3
Option A (If both the statement are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-3) is correct because the Balmer series corresponds to transitions to n=2n=2, which results in emitted radiation falling in the visible range.
99 Easy
99 of 492
Remembering +4 | -1
Splitting of spectral lines when atoms are subjected to strong electric field is called:
View answer and explanation
Correct option: B
Stark effect
Option B (Stark effect) is correct because the splitting of spectral lines in a strong electric field is known as the Stark effect.
100 Easy
100 of 492
Application +4 | -1
The Bohr's energy equation for H atom reveals that the energy level of a shell is given by E=−13.58/n2  eV.E = - 13.58/{n^2}\;\mathrm{eV}. The smallest amount that an H-atom will absorb, if in ground state is:
View answer and explanation
Correct option: D
10.19 eV
Option D (10.19 eV) is correct because it represents the energy difference between the ground state (n=1) and the first excited state (n=2) of the hydrogen atom.
101 Hard
101 of 492
Application +4 | -1
What is the mass of a photon of sodium light with a wavelength of 5890 A0?(h=6.63×10−27erg−s){A^0}?(h = 6.63\mathrm{ \times }{10^{ - 27}}erg - s)
View answer and explanation
Correct option: D
3.752×10−33g3.752\mathrm{ \times }{10^{ - 33}}g
Option D (3.752×10−33g3.752\mathrm{ \times }{10^{ - 33}}g) is correct because it represents the mass of a photon calculated using the energy-mass equivalence and the energy of a photon.
102 Medium
102 of 492
Understanding +4 | -1
Which electronic level will allow the hydrogen atom to absorb photon but not to emit?
View answer and explanation
Correct option: A
1s
Option A (1s) is correct because an electron in the 1s orbital is in the ground state and can only absorb a photon to move to a higher energy level.
103 Medium
103 of 492
Analyzing +4 | -1
Assertion: The boundary surface for ∣ψ∣2|\psi|^2 and ∣ψ∣|\psi| are identical.

Reason : From the value of ∣ψ∣2|\psi|^2 at different points within an atom, it is not possible to predict the region around the nucleus where electron will most probably be found.
View answer and explanation
Correct option: C
If A is true but R is false
Option C is correct because the assertion is true, but the reason is false. The boundary surfaces are indeed identical, but the reason provided about predicting electron location is incorrect.
104 Hard
104 of 492
Understanding +4 | -1
Consider the following statements.
I. ∣ψ∣2|\psi|^2 is a measure of electron density at a point in an atom.

II. Radial probability function (=4πr2R2ψ2)\left(=4 \pi r^2 R^2 \psi^2\right) gives the probability of finding the electron at a distance rr (atomic radius) from the nucleus regardless of direction.

III. The shape of an orbital is defined as a surface of constant probability density that encloses some large fractions of the probability of finding the electron.

Select the correct statements.
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because all three statements accurately describe the concepts of electron density, radial probability function, and orbital shape in quantum mechanics.
105 Easy
105 of 492
Analyzing +4 | -1
The number of molecules or ions from the following, which do not have odd number of electrons are _____
(A) NO2\mathrm{NO}_{2} \quad (B) ICl4−\mathrm{ICl}_{4}^{-} \quad(C) BrF3\mathrm{BrF}_{3} \quad (D) ClO2\mathrm{ClO}_{2} \quad (E) NO2+\mathrm{NO}_{2}^{+} \quad (F) NO\mathrm{NO}
View answer and explanation
Correct option: A
3
The correct answer is 3. ICl4−ICl_4^-, BrF3BrF_3, and NO2+NO_2^+ have an even number of electrons. The others have an odd number of electrons.
106 Hard
106 of 492
Analyzing +4 | -1
The frequency v of certain line of the Lyman series of the atomic spectrum of hydrogen satisfies the following conditions:
(i) It is the sum of the frequencies of another Lyman line and a Balmer line.
(ii) It is the sum of the frequencies of a certain line, a Lyman line, and a Paschen line.
(iii) It is the sum of the frequencies of a Lyman and a Paschen line but no Bracket line.
To what transition does v correspond?
View answer and explanation
Correct option: D
n2=4  to  n1=1{n_2} = 4\;\mathrm{to\;}{n_1} = 1
Option D (n2=4  to  n1=1{n_2} = 4\;\mathrm{to\;}{n_1} = 1) is correct because it satisfies all three given conditions for the frequency v.
107 Medium
107 of 492
Understanding +4 | -1
Which of the following statement is not correct about the characteristics of cathode rays.
View answer and explanation
Correct option: D
Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube
Option D (Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube) is not correct because the characteristics of cathode rays depend only on the accelerating voltage and not on the gas.
108 Easy
108 of 492
Remembering +4 | -1
The best metal to be used for photoemission is:
View answer and explanation
Correct option: C
Cesium
Option C (Cesium) is correct because Cesium has the lowest work function among the given options, making it the most efficient for photoemission.
109 Easy
109 of 492
Analyzing +4 | -1
S-I: Line spectrum of Li+2\mathrm{Li}^{+2} and He+\mathrm{He}^{+}are identical.

S-II: Isoelectronic species produce identical spectrum.
View answer and explanation
Correct option: A
Both S-I and S-II are correct
Option B (Both S-I and S-II are incorrect) is correct because Line spectrum of Li+2Li^{+2} and He+He^{+} are not identical and Isoelectronic species do not produce identical spectrum.
110 Medium
110 of 492
Analyzing +4 | -1
If Ee,Eα {E_e},{E_\alpha}\, and Ep{E_p} represent the kinetic energies of an electron, alpha particle and a proton respectively, each moving with same de-Broglie wavelength then
View answer and explanation
Correct option: D
Ee>Ep>Eα{E_e} > {E_p} > {E_\alpha }
Option D (Ee>Ep>Eα{E_e} > {E_p} > {E_\alpha }) is correct because for the same de Broglie wavelength, kinetic energy is inversely proportional to mass, and the electron has the smallest mass, followed by the proton, and then the alpha particle.
111 Medium
111 of 492
Understanding +4 | -1
The mass of electron moving with velocity of light is:
View answer and explanation
Correct option: C
Infinite
Option C (Infinite) is correct because, according to the theory of relativity, as an object approaches the speed of light, its mass increases towards infinity.
112 Easy
112 of 492
Application +4 | -1
For a metal ion, the calculated magnetic moment is 4.90 BM. This metal ion has____number of unpaired electrons.
View answer and explanation
Correct option: A
4
Option A (4) is correct because the magnetic moment is related to the number of unpaired electrons by the formula μ=n(n+2)\mu = \sqrt{{n(n + 2)}} BM, where n is the number of unpaired electrons.
113 Easy
113 of 492
Remembering +4 | -1
Which of the subshell has double dumb-bell shape?
View answer and explanation
Correct option: C
d
Option C (d) is correct because d-orbitals, except for dz2d_{z^2}, have a double dumbbell shape.
114 Medium
114 of 492
Remembering +4 | -1
Which of the following observations was not correct during Rutherford’s scattering experiment?
View answer and explanation
Correct option: C
A large number of the α\alpha-particles were bounced back
Option A (Inversely proportional to momentum) is correct because the de Broglie wavelength is given by λ=hpλ = \frac{{h}}{{p}}, where h is Planck's constant and p is the momentum of the particle.
115 Medium
115 of 492
Application +4 | -1
If the wavelength of an electromagnetic radiation is 2000Å, what is its energy in ergs?
View answer and explanation
Correct option: A
9.94×10−129.94 \times {10^{ - 12}}
Option A is correct. The energy is calculated using E = hc/λ, converting the result to ergs, which gives approximately 9.94 × 10-12 ergs.
116 Easy
116 of 492
Remembering +4 | -1
Which d-orbital does not have four lobes?
View answer and explanation
Correct option: C
dz2{d_{{z^2}}}
Option C is correct because the dz² orbital has two lobes along the z-axis and a donut-shaped ring in the xy-plane, not four lobes.
117 Medium
117 of 492
Analyzing +4 | -1
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason
R.
Assertion A: Energy of 2s orbital of hydrogen atom is greater than that of 2s orbital of lithium.
Reason R : Energies of the orbitals in the same subshell decrease with increase in the atomic number.
In the light of the above statements, choose the correct answer from the options given below.
View answer and explanation
Correct option: A
Both A and R are true and R is the correct explanation of A.
Option A (Both A and R are true and R is the correct explanation of A) is correct because the energy of an orbital decreases with increasing atomic number due to the increased nuclear attraction.
118 Easy
118 of 492
Understanding +4 | -1
Which one is not true for the cathode rays?
View answer and explanation
Correct option: D
They are electromagnetic waves
Option D (They are electromagnetic waves) is correct because cathode rays are streams of electrons, which are particles, not electromagnetic waves.
119 Hard
119 of 492
Application +4 | -1
The energy of electron in first orbit of He+\mathrm{H}{\mathrm{e}^ + } is (RH=−871.6×10−20  J).\left( {{R_\mathrm{H}} = - 871.6 \times {{10}^{ - 20}}\mathrm{\;J}} \right). The energy of electron in the first orbit of H is:
View answer and explanation
Correct option: C
−217.9×10−20  J- 217.9 \times {10^{ - 20}}\;\mathrm{J}
Option C (−217.9×10−20  J- 217.9 \times {10^{ - 20}}\;\mathrm{J}) is correct because the energy of an electron is proportional to Z2Z^2, where Z is the atomic number.
120 Medium
120 of 492
Remembering +4 | -1
According to Bohr’s theory, angular momentum of electron in N – shell is
View answer and explanation
Correct option: C
2hπ\frac{{2h}}{\pi }
Option C (2hπ\frac{{2h}}{\pi }) is correct because, according to Bohr's theory, the angular momentum of an electron is quantized and given by L=nh2πL = n\frac{h}{{2\pi}}, where n is the principal quantum number.
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