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NEET Chemistry MCQ
Structure Of Atom Mcq Neet Chemistry
Quiz 4
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Q.1
The nitrogen atom has 7 protons and 7 electrons, the nitride ion (N3-) will have
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a) 7 protons and 10 electrons
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b) 4 protons and 7 electrons
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c) 4 protons and 10 electrons
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d) 10 protons and 7 electrons
Explanation
Answer: (a)
Q.2
Which of the following has maximum number of unpaired electrons
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a)Cu2+
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b) Cr3+
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c)Ni2+
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d)Fe3+
Explanation
For Cu, Z=29 Thus Cu2+ number of electrons=29-2=27Electron configuration [Ar]3d9 thus one unpairedFor Cr, Z=24 , Thus Cr3+ , number of electrons=24-3=21Electron configuration [Ar]3d3 thus Three unpaired electronsFor Ni, Z=28 , Thus Ni2+ , number of electrons=28-3=26Electron configuration [Ar]3d8 thus two unpaired electronsFor Fe, Z=26, Thus Fe3+ , number of electrons=26-3=23Electron configuration [Ar]3d5 thus Five unpaired electronsAnswer: (d)
Q.3
Assertion : In a multielectron atom, the electrons in different sub-shells have different energies. Reason : Energy of an orbital depends upon n + l value.
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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.4
Isoelectronic species are
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a) CO, CN–, NO+, C22–
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b) CO–, CN, NO, C2–
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c) CO+, CN+, NO–, C2
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d) CO, CN, NO, C2
Explanation
Isoelectronic species have same number of electrons CO, CN–, NO+, C22– all have 14 electrons Answer:(a)
Q.5
The ratio of the energy required to remove an electron from the first three Bohr’s orbits of hydrogen is
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a) 3 : 2 : 1
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b) 9 : 4 : 1
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c) 36 : 9 : 4
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d) 1 : 4 : 9
Explanation
For hydrogen atom Energy E ∝ 1/n2 Answer:(c)
Q.6
Be2+ is isoelectronic with which of the following ions ? [AIPMT-2014]
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a) H+
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b) Li+
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c) Na+
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d) Mg2+
Explanation
Be2+ have 2 electron and Li+have 2 electron Answer:(b)
Q.7
Assertion : The charge to mass ratio of the particles in anode rays depends on nature of gas taken in the discharge tube. Reason : The particles of anode rays carry positive charge.
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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:(b)
Q.8
If the uncertainty in the position of electron is zero, the uncertainty in its momentum would be
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a) Zero
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b) Greater than h/4π
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c) Less than h/4π
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d) Infinite
Explanation
Δx × Δp =h/4π If Δx =0 implies Δp = Infinite Answer:(d)
Q.9
Which of the following is not permissible arrangement of electrons in an atom? [AIPMT (Prelims)-2009]
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a) n = 5, l = 3, m = 0, s =1/2
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b) n = 3, l = 2, m = –3, s = -1/2
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c) n = 3, l = 2, m = –2, s = -1/2
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d) n = 4, l = 0, m = 0, s = +1/2
Explanation
Value of m (orbital) depends upon l i.e., it cannot be more than 'l '. Therefore is wrong. Answer:(b)
Q.10
Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states? [AIPMT (Prelims)-2009]
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a) 3d5, 4s1
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b) 3d5, 4s2
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c) 3d2, 4s2
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d) 3d3, 4s2
Explanation
Answer:(b)
Q.11
Which one of the following ions has electronic configuration [Ar]3d6? (At. nos. Mn = 25, Fe = 26, Co = 27, Ni = 28)
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a) Co3+
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b) Ni3+
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c) Mn3+
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d) Fe3+
Explanation
Electronic configuration [Ar]3d63+ have 24 electrons Ni3+ = 28 – 3 have 24 electrons Mn3+ = 25 – 3 have 24 electrons Fe3+ = 26 – 3 have 24 electrons Answer:(a)
Q.12
If n = 6, the correct sequence of filling of electrons will be [AIPMT (Prelims)-2011]
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a) ns → np →(n - 1)d →(n - 2)f
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b) ns →(n - 2)f →(n -1)d→ np
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c) ns →(n - 1)d →(n - 2)f → np
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d) ns → (n - 2)f → np → (n - 1)d
Explanation
Put thevalues of n and l in abouve option to calculate (n+l) If n+l is more thenmore energy If two (n+l)havesame values then more the value of n more the energy Answer:(b)
Q.13
What will be the ratio of the wavelength of the first line to that of the second line of Paschen series of H atom?
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a) 256 : 175
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b) 175 : 256
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c) 15 : 16
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d) 24 : 27
Explanation
First line of paschen series n1= 3, n2= 4 Second line of paschen series n1= 3, n3= 5 Answer:(a)
Q.14
Which of the following statement concerning the four quantum number is incorrect?
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a) n gives the size of an orbital
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b) l gives the shape of an orbital
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c) m gives the energy of the electron in orbital
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d) s gives the direction of spin of electron in the orbital
Explanation
m = represents the orientation of orbital in magnetic field. m = orbitals Answer:(c)
Q.15
The energy of second Bohr orbit of the hydrogen atom is –328 kJ mol–1; hence the energy of fourth Bohr orbit would be [AIPMT (Prelims)-2005]
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a) –41 kJ mol–1
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b) –1312 kJ mol–1
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c) –164 kJ mol–1
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d) –82 kJ mol–1
Explanation
E = 1312/n2 kJ mol–1 for hydrogen Answer:(d)
Q.16
he geometry of sp3d2 hybrid orbitals is
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a) linear
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b) pentagonal bipyramidal
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c)trigonal bipyramidal
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d)octahedral
Explanation
Answer:(d)
Q.17
The d-orbitals involved in dsp2 hybridization is
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a) dxy
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b) dz2
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c) dx2 - y2
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d) dzx
Explanation
Answer: (c)
Q.18
Assertion : Electronic energy for hydrogen atom of different orbitals follow the sequence : 1s < 2s = 2p < 3s = 3p = 3d. Reason : Electronic energy for hydrogen atom depends only on n and is independent of ‘l’ & ‘m’ values.
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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.19
A transition element X has a configuration (Ar)3d4 in its +3 oxidation state. Its atomic number is
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a) 22
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b) 19
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c) 25
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d) 26
Explanation
(Ar)3d4 Numberelectrons = 18+4=22 Now this atom have +3 oxudation state i.e. it has lost three electron ∴ total electroninatom = 22+3 = 25 Answer:(c)
Q.20
What will be the longest wavelength line in Balmer series of spectrum of H-atom?
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a) 546 nm
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b) 656 nm
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c) 566 nm
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d) 556 nm
Explanation
All the wavelength are in visible region i.e. between 400 nm to 760 nm. Therefore maximum wavelength line will be 656 nm. Answer:(b)
Q.21
Zeeman effect refers to the
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a) Splitting of the spectral lines in a magnetic field
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b) Splitting up of the spectral lines in an electrostatic field
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c) Emission of electrons from metals when light falls on it
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d) Random scattering of α-particles by gold foil
Explanation
Splitting of line in magnetic field is known as Zeeman effect Answer:(a)
Q.22
Assertion : The number of angular nodes in 3zd2 is zero. Reason : Number of angular nodes of atomic orbitals is equal to value of l.
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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:(b)
Q.23
Which of the following set of quantum number is possible?
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a) n = 4, l = 2, m = –2, s = –2
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b) n = 4, l = 4, m = 0, s =1/2
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c) n = 4, l = 3, m = –3, s =1/2
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d) n = 4, l = 0, m = 0, s = 0
Explanation
Option 1 not possible because s can never have –2 value Option 2 not possible because n and l cannot have same value Option 4 not possible because s cannot have zero value Correct answer = c Answer:(c)
Q.24
The position of both, an electron and a helium atom is known within 1.0 nm. Further the momentum of the electron is known within 5.0 × 10–26 kg ms–The minimum uncertainty in the measurement of the momentum of the helium atom is
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a) 8.0 × 10–26 kg ms–1
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b) 80 kg ms–1
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c) 50 kg ms–1
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d) 5.0 × 10–26 kg ms–1
Explanation
Δx × Δp = h/4π ( constant irrespective of any particle) Now it is given "position of both, an electron and a helium atom is known within 1.0 nm" or Δx is same for electron and a helium atom ∴ Δp must be same for electron and helium atom Answer:(d)
Q.25
If kinetic energy of a proton is increased nine times, the wavelength of the de-Broglie wave associated with it would become
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a) 3 times (2) (3)
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b) 9 times
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c) (1/3) times
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d) (1/6) times
Explanation
de-Brogli wavelength interms of kinetic energ E is given by From equation when Kinetic energy E. of the electron increased 9 times. The de-Broglie wavelength decreased by 1/3 times. Answer:(c)
Q.26
Two electrons in K shell will not have
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a) Same principal quantum number
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b) Same azimuthal quantum number
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c) Same magnetic quantum number
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d) Same spin quantum number
Explanation
As K shell is the 1st shell and have maximum two electron. Therefore, to Pauli's exclusion principal two electrons can't have the same value of all the four quantum number. Therefore, can't have same spin quantum number Answer:(d)
Q.27
The momentum of a particle which has a de-Broglie wavelength of 0.1 nm is
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a) 3.2 × 10–24 kg ms–1
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b) 4.3 × 10–22 kg ms–1
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c) 5.3 × 10–22 kg ms–1
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d) 6.62 × 10–24 kg ms–1
Explanation
Use forula λ = h/p, p=mv Answer:(d)
Q.28
An electron jumps from lower orbit to higher orbit, when
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a) Energy is released
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b) Energy is absorbed
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c) No change in energy
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d) It radiates energy
Explanation
When e– jumps from lower to higher energy level absorbed. When e– comes from higher to lower energy level released Answer:(b)
Q.29
For a ‘p’ electron, the orbital angular momentum is
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a) √6 ħ
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b) √2 ħ
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c) ħ
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d) 2 ħ
Explanation
Orbital angular momentum For p-electron value of l = 1 Answer:(b)
Q.30
The total number of subshells in fourth energy level of an atom 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
Answer:(a)
Q.31
Total number of spectral lines in UV region, during transition from 5th excited state to 1st excited state
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a) 10
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b) 3
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c) 4
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d) zero
Explanation
As 1st excited state means n1= 2 For 5th excited state means n2= 6 electron will transit between 6th level to 2nd level, whichisa Balmar series, is in visible range No transition will be upto 1st level. Because no line will appear in Lyman series i.e. UV region. Answer:(d)
Q.32
Which of the following electronic in a transition hydrogen atom will require the largest amount of energy?
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a) n = 1 to n = 2
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b) n = 2 to n = 3
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c) n = 1 to n = ∞
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d) n = 3 to n = 5
Explanation
Largest amount of energy is required for the transition between n = 1 to n = ∞ Answer:(c )
Q.33
Which combinations of quantum numbers n, l, m and s for the electron in an atom does not provide a permissible solutions of the wave equation?
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a) 3, 2, 2, 1/2
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b) 3, 3, 1, -1/2
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c) 3, 2, 1, 1/2
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d) 3, 1, 1,-1/2
Explanation
n = 3, l = 3 [Not possible because value of l can never be equals to n] Answer:(b)
Q.34
The measurement of the electron position is associated with an uncertainty in momentum, which is equal to 1 × 10–18 g cm s–The uncertainty in electron velocity is, (Mass of an electron is 9 × 10–28g) [AIPMT (Prelims)-2008]
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a) 1 × 1011 cm s–1
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b) 1 × 109 cm s–1
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c) 1 × 106 cm s–1
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d) 1 × 105 cm s–1
Explanation
Answer:(b)
Q.35
The number of radial nodes in 4s and 3p orbitals are respectively
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a) 2, 0
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b) 3, 1
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c) 2, 2
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d) 3, 2
Explanation
Number of radial nodes = n - l -1 For 4s,n = 4, l =0 ∴ number of radialnodes = 4-0-1=3 For 3p,n=3, l =1 ∴ number of radialnodes = 3-1-1=1 Answer:(b)
Q.36
Which shell would be the first to have ‘g’ sub-shell?
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a) L
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b) M
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c) N
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d) O
Explanation
For g-subshell l = 4; the value of n will be = (l + 1) = 5 for K, n =1 for L, n =2 for M, n =3 for N, n =4 for O, n =1 ∴ For n = 5 corresponding is 'o' shell and it contain 'g' subshell. Answer:(d)
Q.37
Maximum number of electrons in a subshell with l = 3 and n = 4 is [AIPMT (Prelims)-2012]
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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 represents 4f subshell having 7 orbitals ∴ Total number of electrons = 14 Answer:(c)
Q.38
Assertion : The orbitals having equal energy are known as degenerate orbitals. Reason : The three 2p orbitals are degenerate in the presence of external magnetic field.
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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.39
Among the following which one is not paramagnetic? [Atomic numbers; Be = 4, Ne = 10, As = 33, Cl = 17]
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a) Ne2+
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b) Be+
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c) Cl–
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d) As+
Explanation
Ne+2 = 8 = 1s2, 2s2, 2p4 = 1 unpaired electron Be+1 = 3 = 1s2, 2s = 1 unpaired electron Cl– = 18 = 1s2, 2s2, 2p6, 3s2, 3p6 = 0 unpaired electron As+= 32 = 1s2, 2s2, 2p6, 3s2, 3p6 4s2, 3d10, 4p3 = 3 unpaired electron If electrons are unpaired then it is not paramagnetic Answer:(c)
Q.40
The de Broglie wavelength of a particle with mass 1 g and velocity 100 m/s is
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a) 6.63 × 10–35 m
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b) 6.63 × 10–34 m
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c) 6.63 × 10–33 m
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d) 6.65 × 10–35 m
Explanation
Use formula λ= h/p Answer:(c)
Q.41
Number of spectral lines falling in Balmer series when electrons are de-excited from nth shell will be given as
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a) (n – 2) in UV
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b) (n – 2) in visible region
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c) (n – 3) in near IR
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d) (n – 3) in far IR
Explanation
Answer:(b)
Q.42
Electronic energy is negative because
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a) Electron has negative charge
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b) Energy is zero near the nucleus and decreases as the distance from nucleus increases
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c) Energy is zero at infinite distance from the nucleus and decreases as the electron comes towards nucleus
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d) These are interelectronic repulsions
Explanation
At infinite distance energy of electron will be zero and it will decreases as the electron approaches towards nucleus Answer:(c)
Q.43
The electronic configuration of gadolinium (Atomic No. 64) is
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a) [Xe]4f3 5d5 6s2
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b) [Xe]4f 6 5d2 6d2
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c) [Xe]4f8 5d9 6s2
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d) [Xe] 4f 7 5d1 6s2
Explanation
Gd have exceptional configuration e– will enter in 5d because 4f have 7 electrons and have half filled stability Answer:(d)
Q.44
The energies E1and E2 of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths i.e. λ1 and λ2 will be
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a) λ1 = λ2/2
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b) λ1 = λ2
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c) λ1 = 2λ2
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d) λ1 = 4λ2
Explanation
From formula E = hC/λ E ∝ 1/λ Answer:(c)
Q.45
The correct order of energy difference between adjacent energy levels in H atom
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a) E2– E1 > E3– E2 > E4– E3
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b) E2– E1 > E4 – E3 > E3 – E2
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c) E4 – E3 > E3 – E2 > E2– E1
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d) E3 – E2 > E4– E3 > E2 – E1
Explanation
∴ E2– E1 > E3– E2 > E4– E3 Alternatively as the distance from the nucleus increases the value of Δ E (energy difference between two shell) decreases Answer:(a)
Q.46
In the presence of magnetic field, the possible number of orientations for an orbital of azimuthal quantum number 3, is
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a) Three
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b) One
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c) Five
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d) Seven
Explanation
When magnetic field is applied subshell will give orbital i.e., l = 3 m = –3, –2, –1, 0, +1, +2, +3, therefore 7 orbitals are possible Answer:(d)
Q.47
Maximum number of electrons in a subshell of an atom is determined by the following [AIPMT (Prelims)-2009]
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a) 2l+1
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b) 4l- 2
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c) 2n2
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d) 4l + 2
Explanation
otal number of subshells =(2l+1) ∴ Maximum number electron in a subshell =2(2l+1) =4l+2 Answer:(d)
Q.48
The orbital angular momentum of electron in 4s orbital is
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a) h/4π
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b) zero
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c) h/2π
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d) 2.5h/2π
Explanation
For 4s electron the value of l = 0 ∴ [orbital angular momentum = zero] Answer:(b)
Q.49
Given : The mass of electron is 9.11×10–31 kg. Planck’s constant is 6.626×10–34 Js, the uncertainty involved in the measurement of velocity within a distance of 0.1 Å is [AIPMT (Prelims)-2006]
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a) 5.79 × 106 ms–1
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b) 5.79 × 107 ms–1
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c) 5.79 × 108 ms–1
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d) 5.79 × 105 ms–1
Explanation
Δx × Δp = h/4π Answer:(a)
Q.50
Which of the following electronic configuration is not possible?
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a) 2p3
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b) 2d5
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c) 4s1
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d) 5f8
Explanation
Value of l cannot be greater or equal to n. For 2d n = 2, l = 2 not possible Answer:(b)
0 h : 0 m : 1 s
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