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NEET Chemistry MCQ
Structure Of Atom Mcq Neet Chemistry
Quiz 6
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Q.1
The set of quantum numbers not applicable to an electron
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a) 1, 1, 1, 1/2
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b) 1, 0, 0, 1/2
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c) 1, 0, 0 , -1/2
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d) 2, 0, 0, 1/2
Explanation
The value of l can never be equal to n If n = 1, l = 1 which is not possible Answer:(a)
Q.2
The frequency of a wave is 6 × 1015 s–Its wave number would be
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a) 105 cm–1
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b) 2 × 107 m–1
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c) 2 × 107 cm–1
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d) 2 × 105 cm–1
Explanation
Wave number = 1/wave length λ = c/ν (ν is frequency) Wave number = ν /c Wave number = 6 × 1015 / 3×108 Wave number =2 × 107 m–1 Answer:(b)
Q.3
Atomic number and mass number of an element M are 25 and 52 respectively. The number of electrons, protons and neutrons in M2+ ion are respectively
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a) 25, 25 and 27
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b) 25, 27 and 25
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c) 27, 25 and 27
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d) 23, 25 and 27
Explanation
Atomic number = 25 Mass number = 52 Number of proton = Atomic number = 25 Number of neutron = (Mass number – Atomic number) 52 – 25 = 27 Number of e– = Number of protons, but M2+ ion means 2e– are removed Number of e– = Number of protons – 2 = 25 – 2 = 23 Answer:(d)
Q.4
The maximum number of electrons in an atom which can have n = 4 is
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a) 4
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b) 8
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c) 16
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d) 32
Explanation
Number of electrons = 2n2 (n = shell number) For 4th shell = 2 × (4)2 = 32 electrons Answer:(d)
Q.5
Maximum number of electrons in a subshell with l = 3 and n = 4 is
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a) 10
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b) 12
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c) 14
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d) 16
Explanation
n =4, l = 3 means 4f for l = 3, m = –3, –2, –1, 0, 1, 2, 3 = 7 orbital Therefore, maximum 14 electrons are present. Answer:(c)
Q.6
Assertion : Zn(II) salts are diamagnetic. Reason : Zn2+ ion has one unpaired electron.
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a) Both Assertion and Reasoning are true and Reasoning is correct explanation of Assertion
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b) Both are correct but Reason is not explanation of Assertion
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c) Assertion is correct but Reason is wrong
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d) Both Assertion and Reason are incorrect.
Explanation
Answer:(c)
Q.7
Which of the following has maximum number of unpaired electrons?
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a) Mg2+
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b) Ti3+
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c) Fe2+
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d) Mn2+
Explanation
Mg2+ = 10 = 1s2, 2s2, 2p6 number of unpaired electrons = 0 Ti3+ = 19 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d1 number of unpaired electrons =1 Fe2+ = 24 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d6 number of unpaired electrons =4 Mn2+ = 23 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d5 number of unpaired electrons =5 maximum number Answer:(d)
Q.8
A certain metal when irradiated with light (ν = 3.2 × 1016 Hz) emits photo electrons with twice kinetic energy as did photo electrons when the same metal is irradiated by light (ν = 2.0 × 1016 Hz). Calculate ν0 ofelectron?
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a) 1.2 × 1014 Hz
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b) 8 × 1015 Hz
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c) 1.2 × 1016 Hz
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d) 4 × 1012 Hz
Explanation
K. E. = h(ν – ν0 ) K.E. of photoelectrons when ν = 3.2 × 1016 Hz K.E1= h (3.2 × 1016 – ν0) K E. of photoelectron when ν = 2.0 × 1016 Hz KE2 = h(2.0 × 1016 – ν0) According to question KE1 = 2KE2 ∴ h(3.2 × 1016 – ν0) = 2h(2.0 × 1016 – ν0) 3.2 × 1016 – ν0 = 4.0 × 1016 – 2ν0 ν0= 4.0 ×1016 – 3.2 × 1016 = 0.8 × 1016 Hz = 8 × 1015 Hz Answer:(b)
Q.9
Uncertainty in the momentum of an electron is 1.0×10-5 kg m-the uncertainty in its position will be ( h=6.62 ×10-34 kg m2 s-1) ... [ AFMC 1998]
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a) 1.05 ×10-28 m
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b) 1.05 ×10-26 m
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c) 5.27 ×10-30 m
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d)5.25 ×10-28 m
Explanation
use formula Answer:(c)
Q.10
The principal and azimuthal quantum number of electrons in 4f orbitals are
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a) 4, 2
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b) 4, 4
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c) 4, 3
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d) 3, 4
Explanation
for s l= 0 for p, l = 1 for d. l = 2 for f, l = 3 Thus for 4f electron n =4 and l=3 Answer:(c)
Q.11
ratio of kinetic energy and potential energy of an electron in a Bohr orbit like hydrogen - like species is
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a) 1/2
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b) -1/2
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c)1
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d)-1
Explanation
Answer:(b)
Q.12
If the energy difference between the ground state and excited state of an atom is 4.4 × 10–19 J. The wavelength of photon required to produce this transition is
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a) 4.5 × 10–7 m
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b) 4.5 × 10–7 nm
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c) 4.5 × 10–7 Å
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d) 4.5 × 10–7 cm
Explanation
ΔE = [Excited state - ground state] = 4.4 × 10-19 J According to Plank's quantum theory energy od single photon Answer:(a)
Q.13
Two particles A and B are in motion. If the wavelength associated with particle A is 5 × 10–8 m ; calculate the wavelength associated with particle B if its momentum is half of A.
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a) 5 × 10–8 m
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b) 10–5 cm
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c) 10–7 cm
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d) 5 × 10–8 cm
Explanation
λ ∝ 1/λ λA = 5 × 10–8m ∴ λB = 10–7m ∴ λB = 10–5cm Answer:(b)
Q.14
Assertion : Angular momentum of an electron in an atom is quantized. Reason : In an atom only those orbits are permitted in which angular momentum of the electron is a natural number multiple of h/2π
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a) Both Assertion and Reasoning are true and Reasoning is correct explanation of Assertion
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b) Both are correct but Reason is not explanation of Assertion
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c) Assertion is correct but Reason is wrong
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d) Both Assertion and Reason are incorrect.
Explanation
Answer:(a)
Q.15
hat is the maximum number of orbitals that can be identified with the following quantum numbers? n = 3, l = l, m l= 0 [AIPMT-2014]
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a) 1
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b) 2
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c) 3
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d) 4
Explanation
It represents 3p orbital Answer:(a)
Q.16
If r1is the radius of the first orbit of hydrogen atom, then the radii of second, third and fourth orbits in term of r1 are
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a) r12, r13, r14
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b) 4r1, 9r1, 16r1
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c) 8r1, 27r1, 64r1
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d) 2r1, 6r1, 8r1
Explanation
By substitutingvalues Z = 1 and n = 2, 3 and 4 We get answer 4r1, 9r1, 16r1 Answer:(b)
Q.17
The wavelength of radiation emitted, when in He+ electron falls from infinity to stationary state would be (R = 1.097 × 107 m–1)
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a) 2.2 × 10–8 m
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b) 2.2 × 10–9 m
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c) 120 m
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d) 22 × 107 m
Explanation
Z =2 , n1= 1 and n2 =∞ Onsubstituting above values in equation λ = 2.2 × 10–8 m Answer:(a)
Q.18
The time taken by the electron in one complete revolution in the nth Bohr’s orbit of the hydrogen atom is
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a) Inversely proportional to n2
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b) Directly proportional to n3
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c) Directly proportional to h/2π
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d) nversely proportional to n/h
Explanation
Period T = Circumference /velocity Answer:(b)
Q.19
The bond order for a species with the configurationσ1s2, σ*1s2, σ2s2, σ*2s2, σ2px1 will be ... [ Haryana CEET 1991]
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a) 1
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b) ½
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c) zero
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d) 1
Explanation
bonding electrons Nb 2+2+1=5 Antibonding electrons Na=2+ 2=4 ∴ B.O.=(5-4)/2=1/2 Answer: (b)
Q.20
A p-orbital can accommodate upto
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a) Four electrons
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b) Six electrons
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c) Two electrons
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d) Eight electrons
Explanation
In any orbital maximum two electrons can accommodate A p-orbital can accommodate upto two electrons Answer:(c)
Q.21
The four quantum numbers of valence electron of potassium are
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a) 4, 0, 1, 1/2
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b) 4, 1, 0, 1/2
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c) 4, 0, 0, 1/2
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d) 4, 1, 1, 1/2
Explanation
K = 19 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s1 last e–Last electron 4s1 ∴ n = 4, l = 0, m = 0, s =1/2 Answer:(c)
Q.22
Quantum numbers for some electrons are given below A : n = 4, l = 1 B : n = 4, l = 0 C : n = 3, l = 2 D : n = 3, l = 1 The correct increasing order of energy of electrons
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a) A < B < C < D
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b) D < C < B < A
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c) D < B < C < A
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d) C < B < A < D
Explanation
According to Pauli exclusion principle (1) Larger the (n + l); Larger will be energy (2) Same value of (n + l) ; bigger n ; more will be energy Answer:(c)
Q.23
Which of the following electronic level would allow the hydrogen to absorb a photon but not emit a photon?
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a) 3s
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b) 2p
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c) 2s
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d) 1s
Explanation
1s-orbital is the ground state Further emission is not possible i.e. de excitation not possible Answer:(d)
Q.24
The energy absorbed by each molecule (A2) of a substance is 4.4 × 10–19J and bond energy per molecule is 4.0 × 10–19 J. The kinetic energy of the molecule per atom will be [AIPMT (Prelims)-2009]
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a) 2.2 × 10–19 J
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b) 2.0 × 10–19 J
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c) 4.0 × 10–20J
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d) 2.0 × 10–20J
Explanation
KE = Energy observed by molecule – Energy required to break one bond Now A2 have two atoms Thus Energy per atom= KE/2 Answer:(d)
Q.25
The uncertainty in momentum of an electron is 1 × 10–5 kg-m/s. The uncertainty in its position will be (h = 6.62 × 10–34 kg-m2/s)
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a) 5.27 × 10–30 m
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b) 1.05 × 10–26 m
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c) 1.05 × 10–28 m
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d) 5.25 × 10–28 m
Explanation
Use equation Δx × Δp = h/4π Answer:(a)
Q.26
For which of the following options m = 0 for all orbitals?
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a) 2s, 2px, 3dxy
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b) 3s ,2pz ,3dz2
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c) 3s ,2pz ,3dx2 -y2
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d) 3s, 3px, 3dxy
Explanation
Answer:(b)
Q.27
The number of lobes in most of the d-orbitals are
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a) 6
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b) 8
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c) 10
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d) 4
Explanation
Answer:(d)
Q.28
The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H = 2.18 × 10–18 J atom–1 and h = 6.625 × 10–34 Js)
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a) 1.54 × 1015 s–1
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b) 1.03 × 1015 s–1
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c) 3.08 × 1015 s–1
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d) 2.00 × 1015 s–1
Explanation
Here E1 is ionization energy of H = 2.18 × 10–18 J atom–1 Now energy difference emitted as radiation givenby E=hν Answer:(c)
Q.29
The total number of atomic orbitals in fourth energy level of an atom is [AIPMT -2011]
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a) 4
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b) 8
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c) 16
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d) 32
Explanation
Number of orbitals = n2 Answer:(c)
Q.30
Calculate the energy in joule corresponding to light of wavelength 45 nm: (Planck's constant h = 6.63 × 10–34 Js; speed of light c = 3 × 108 ms–1) [AIPMT-2014]
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a) 6.67 × 1015
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b) 6.67 × 1011
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c) 4.42 × 10–15
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d) 4.42 × 10–18
Explanation
Use formula E = hc/λ Answer:(d)
Q.31
Number of neutron in heavy hydrogen atom is [ MP PMT]
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a)0
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b) 1
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c)2
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d)3
Explanation
Answer: (b)
Q.32
Which of the following statement is incorrect [ BHU 1985]
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a) The charge on an electron electron and on protons are equal and opposite
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b) Neutron has no charge
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c)Electrons and protons have the same weight
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d)The mass of a proton and neutron are nearly identical
Explanation
Answer: (c)
Q.33
The maximum number of electrons in f -subshell is .... [ IIT Ranchi 1988]
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a) 2
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b) 8
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c)14
0%
d)32
Explanation
for f ; l=3 thus number of electrons=2(l+1)=2(3+1)=14 Answer:(c)
Q.34
A photon in X region is more energetic than in the visible region; X is [ BHU 1981]
0%
a) IR
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b) UV
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c) Microwave
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d) Radiowave
Explanation
Answer: (b)
Q.35
Which of the following are possible values of n , l and m for an atom having maximum value of m=+2? [ Pb. PMT 1985]
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a)n=4, l=3, m=+2
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b) n=3, l=2, m=-2
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c)n=3, l=3, m=+2
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d)n=4, l=3, m=-2
Explanation
For Max. m=+2 means l=2, which can be so when n=3 NOw when l=2, m=-2, -1, 0, +1, +2Answer: (b)
Q.36
The increasing value of e/m ( charge /mass) is .... [ IIT 1984]
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a) e, p, n, α
0%
b) n. p, e, α
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c)
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d)n, α, p, e
Explanation
Charge on n is zero magnitude of charge on e=p but mass pf p > e Charge on α=2P but mass of α=4(p) Answer: (d)
Q.37
The radius of atom nucleus is of the order of...... [ IIT 1985]
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a) 10-12 m
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b) 10-8 m
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c) 10-15 m
0%
d) 10-10 m
Explanation
Answer:(c)
Q.38
In the given atom no two electrons can have the same value for all the four quantum numbers. This is called [ CPMT 92]
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a) Hund's rule
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b) Aufbau principle
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c) Uncertainty principle
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d) Pauli's exclusion principle
Explanation
Answer: (d)
Q.39
Cathode rays are [ JIPMER 1987]
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a)Electromagnetic waves
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b) Radiation
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ac)stream of α particles
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d)stream of electrons
Explanation
Answer: (d)
Q.40
Which statement is correct about proton [ MP PMT 1985]
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a) Proton is nucleus of Deuterium
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b) proton is α particle
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c)Proton is ionised hydrogen atom
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d)proton is ionised hydrogen
Explanation
Answer: (d)
Q.41
When the azimuthal quantum number is l=1, the shape of the orbital will be ..[ EAMCET 1982]
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a)Spherical
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b) dumbell
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c)double dumbell
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d)more complicated
Explanation
Answer:(b)
Q.42
There is no difference between 2p and a 3p orbital regarding [ BHU 1981]
0%
a) shape
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b) size
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c) energy
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d) value of n
Explanation
Answer: (a)
Q.43
Bohr's model can explain ..... [ IIT 1985]
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a)Spectrum of Hydrogen atom only
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b) Spectrum of atom or ion containing one electron only
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c)the spectrum of hydrogen molecule
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d)The solar spectrum
Explanation
Answer: (b)
Q.44
The electron level which allows the hydrogen to absorb photon but not to emit is ... [ IIT1984]
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a) 1s
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b) 2s
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c)2p
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d)3d
Explanation
1s is ground state of Hydrogen thus electron in 1s absorbs energy and electron goes to higher energy levelAnswer: (a)
Q.45
Neutral atom ( atomic number > 1) consists of [ CPMT 1982]
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a) Only protons
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b) Neutrons + Protons
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c)Neutrons + electrons
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d)Neutrons + electrons + protons
Explanation
Answer:(d)
Q.46
The quantum number which is designed by letters s, p, d and f instead of number is [ BHU 1980]
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a) n
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b) l
0%
c) ml
0%
d) ms
Explanation
Answer: (b)
Q.47
The number of unpaired electrons in carbon atom is .. [ MLNR 1987]
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a)1
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b) 2
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c)3
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d)4
Explanation
C=1s, 2s2, 3p2Answer: (b)
Q.48
Rutherford's alpha particle scattering experiment eventually led to the conclusion that ... [ IIT 1986]
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a) mass and energy are related
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b) electron occupy space around the nucleus
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c)Neutron are buried deep in the nucleus
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d)the point of impact with matter can be precisely determined
Explanation
Answer: (b)
Q.49
The maximum number of electrons in a subshell for which l=3 i9s .... [ EAMCET 1982]
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a) 14
0%
b) 10
0%
c)8
0%
d)4
Explanation
Number of orbital=(2l+1)=(2×3 + 1)=7Each orbital can have 2 electrons thus 7 ×2=14 Answer:(a)
Q.50
" The exact path of electron in 2p-orbital cant be determined" this statement is based on [DPMT 1981]
0%
a) Hund's rule
0%
b) Bohr's rule
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c) Uncertainty principle
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d) Aufbau principle
Explanation
Answer: (c)
0 h : 0 m : 1 s
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