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Chapter 2 Structure of Atom Chemistry Worksheet for Class 11
Class 11 Chemistry students should refer to the following printable worksheet in Pdf in Class 11. This test paper with questions and solutions for Class 11 Chemistry will be very useful for tests and exams and help you to score better marks
Class 11 Chemistry Chapter 2 Structure of Atom Worksheet Pdf
Question. The number of protons, neutrons and electrons in 17571Lu, respectively, are
(a) 71, 104 and 71
(b) 104, 71 and 71
(c) 71, 71 and 104
(d) 175, 104 and 71
Answer. A
Question. Be2+is isoelectronic with which of the following ions?
(a) H+
(b) Li+
(c) Na+
(d) Mg2+
Answer. B
Question. Isoelectronic species are
(a) CO, CN–, NO+, C22–
(b) CO–, CN, NO, C2–
(c) CO+, CN+, NO–, C2
(d) CO, CN, NO, C2
Answer. A
Question. The ion that is isoelectronic with CO is
(a) CN–
(b) N2+
(c) O2–
(d) N2–
Answer. A
Question. Which one of the following is not isoelectronic with O2–?
(a) Tl+
(b) Na+
(c) N3–
(d) F–
Answer. A
Question. Which of the following series of transitions in the spectrum of hydrogen atom falls in visible region?
(a) Brackett series
(b) Lyman series
(c) Balmer series
(d) Paschen series
Answer. C
Question. Calculate the energy in joule corresponding to light of wavelength 45 nm.
(Planck’s constant, h = 6.63 × 10–34J s, speed of light,
c = 3 × 108 m s–1)
(a) 6.67 × 1015
(b) 6.67 × 1011
(c) 4.42 × 10–15
(d) 4.42 × 10–18
Answer. D
Question. The value of Planck’s constant is 6.63 × 10–34Js. The speed of light is 3 × 1017nm s–1. Which value is closest to the wavelength in nanometer of a quantum of light with frequency of 6 × 1015 s–1?
(a) 50
(b) 75
(c) 10
(d) 25
Answer. A
Question. The value of Planck’s constant is 6.63 × 10–34J s.The velocity of light is 3.0 × 108 m s–1. Which value is closest to the wavelength in nanometers of a quantum of light with frequency of 8 × 1015s–1 ?
(a) 2 × 10–25
(b) 5 × 10–18
(c) 4 × 101
(d) 3 × 107
Answer. A
Question. For given energy, E = 3.03 × 10–19joules corresponding wavelength is
(h = 6.626 × 10–34J sec, c = 3 × 108 m/sec)
(a) 65.6 nm
(b) 6.56 nm
(c) 3.4 nm
(d) 656 nm
Answer. B
Question. What will be the longest wavelength line in Balmer series of spectrum?
(a) 546 nm
(b) 656 nm
(c) 566 nm
(d) 556 nm
Answer. C
Question. According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?
(a) n = 6 to n = 1
(b) n = 5 to n = 4
(c) n = 6 to n = 5
(d) n = 5 to n = 3
Answer. D
Question. The energy of second Bohr orbit of the hydrogen atom is –328 kJ mol–1; hence the energy of fourth Bohr orbit would be
(a) – 41 kJ mol–1
(b) –82 kJ mol–1
(c) –164 kJ mol–1
(d) –1312 kJ mol–1
Answer. B
Question. 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 × 10–18J atom–1 and h = 6.626 × 10–34J s)
(a) 1.54 × 1015 s–1
(b) 1.03 × 1015 s–1
(c) 3.08 × 1015 s–1
(d) 2.00 × 1015 s–1
Answer. B
Question. In hydrogen atom, energy of first excited state is–3.4 eV. Then find out K.E. of same orbit of hydrogen atom.
(a) +3.4 eV
(b) +6.8 eV
(c) –13.6 eV
(d) +13.6 eV
Answer. C
Question. Who modified Bohr’s theory by introducing elliptical orbits for electron path?
(a) Rutherford
(b) Thomson
(c) Hund
(d) Sommerfeld
Answer. B
Question. The Bohr orbit radius for the hydrogen atom (n = 1) is approximately 0.530 Å. The radius for the first excited state (n = 2) orbit is (in Å)
(a) 4.77
(b) 1.06
(c) 0.13
(d) 2.12
Answer. C
Question. In a Bohr’s model of an atom, when an electron jumps from n = 1 to n = 3, how much energy will be emitted or absorbed?
(a) 2.389 × 10–12ergs
(b) 0.239 × 10–10 ergs
(c) 2.15 × 10–11 ergs
(d) 0.1936 × 10–10 ergs
Answer. A
Question. The radius of hydrogen atom in the ground state is 0.53 Å. The radius of Li2+ion (atomic number = 3) in a similar state is
(a) 0.53 Å
(b) 1.06 Å
(c) 0.17 Å
(d) 0.265 Å
Answer. D
Question. The spectrum of He is expected to be similar to that
(a) H
(b) Li+
(c) Na
(d) He+
Answer. D
Question. If r is the radius of the first orbit, the radius of nth orbit of H-atom is given by
(a) rn2
(b) rn
(c) r/n
(d) r2n2
Answer. D
Question. In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is
(a) 211.6 pm
(b) 211.6 p pm
(c) 52.9 p pm
(d) 105.8 pm
Answer. B
Question. A 0.66 kg ball is moving with a speed of 100 m/s.
The associated wavelength will be (h = 6.6 × 10–34 J s)
(a) 6.6 × 10–32 m
(b) 6.6 × 10–34 m
(c) 1.0 × 10–35 m
(d) 1.0 × 10–32 m
Answer. C
Question. The measurement of the electron position is associated with an uncertainty in momentum,which is equal to 1 × 10–18 g cm s–1. The uncertainty in electron velocity is (mass of an electron is 9 × 10–28 g
(a) 1 × 105 cm s–1
(b) 1 × 1011 cm s–1
(c) 1 × 109 cm s–1
(d) 1 × 106 cm s–1
Answer. C
Question. Given : The mass of electron is 9.11 × 10–31 kg, Planck constant is 6.626 × 10–34 J s, the uncertainty involved in the measurement of velocity within a distance of 0.1 Å is
(a) 5.79 × 105 m s–1
(b) 5.79 × 106 m s–1
(c) 5.79 × 107 m s–1
(d) 5.79 × 108 m s–1
Answer. B
Question. 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)
(a) 5.27 × 10–30 m
(b) 1.05 × 10–26 m
(c) 1.05 × 10–28 m
(d) 5.25 × 10–28 m
Answer. A
Question. The de Broglie wavelength of a particle with mass 1 g and velocity 100 m/s is
(a) 6.63 × 10–35 m
(b) 6.63 × 10–34 m
(c) 6.63 × 10–33 m
(d) 6.65 × 10–35 m
Answer. C
Question. 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 m s–1.
The minimum uncertainty in the measurement of the momentum of the helium atom is
(a) 8.0 × 10–26 kg m s–1
(b) 80 kg m s–1
(c) 50 kg m s–1
(d) 5.0 × 10–26 kg m s–1
Answer. D
Question. Uncertainty in position of an electron (Mass = 9.1 × 10–28 g) moving with a velocity of 3 × 104 cm/s accurate upto 0.001% will be (Use h/(4p) in uncertainty expression where h = 6.626 × 10–27 erg second)
(a) 5.76 cm
(b) 7.68 cm
(c) 1.93 cm
(d) 3.84 cm
Answer. C
Question. Which of the following statements do not form a part of Bohr’s model of hydrogen atom?
(a) Energy of the electrons in the orbits are quantized.
(b) The electron in the orbit nearest the nucleus has the lowest energy.
(c) Electrons revolve in different orbits around the nucleus.
(d) The position and velocity of the electrons in the orbit cannot be determined simultaneously.
Answer. D
Question. 4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is
(a) 5f > 6p > 4d > 5p
(b) 5f > 6p > 5p > 4d
(c) 6p > 5f > 5p > 4d
(d) 6p > 5f > 4d > 5p
Answer. B
Question. Orbital having 3 angular nodes and 3 total nodes is
(a) 5p
(b) 3d
(c) 4f
(d) 6d
Answer. C
Question. Which one is the wrong statement?
(a) The uncertainty principle is ΔE × Δt ≥ h/4π
(b) Half filled and fully filled orbitals have greater stability due to greater exchange energy, greater symmetry and more balanced arrangement.
(c) The energy of 2s-orbital is less than the energy of 2p-orbital in case of hydrogen like atoms.
(d) de-Broglie’s wavelength is given by λ = h/mv,where m = mass of the particle, v = group velocity of the particle.
Answer. C
Question. How many electrons can fit in the orbital for which n = 3 and l = 1?
(a) 2
(b) 6
(c) 10
(d) 14
Answer. A
Question. Which of the following pairs of d-orbitals will have electron density along the axes?
(a) dz2, dxz
(b) dxz, dyz
(c) dz2, dx2 – y2
(d) dxy, dx2 – y2
Answer. C
Question. Two electrons occupying the same orbital are distinguished by
(a) azimuthal quantum number
(b) spin quantum number
(c) principal quantum number
(d) magnetic quantum number.
Answer. B
Question. Which is the correct order of increasing energy of the listed orbitals in the atom of titanium?
(At. no. Z = 22)
(a) 4s 3s 3p 3d
(b) 3s 3p 3d 4s
(c) 3s 3p 4s 3d
(d) 3s 4s 3p 3d
Answer. C
Question. The number of d-electrons in Fe2+ (Z = 26) is not equal to the number of electrons in which one of the following?
(a) d-electrons in Fe (Z = 26)
(b) p-electrons in Ne (Z = 10)
(c) s-electrons in Mg (Z = 12)
(d) p-electrons in Cl (Z = 17) (2015, Cancelled)
Answer. D
Question. The angular momentum of electron in ‘d’ orbital is equal to
(a) 2 √3h
(b) 0 h
(c) √6 h
(d) √2 h
Answer. C
Question. What is the maximum number of orbitals that can be identified with the following quantum numbers?
n = 3, l = 1, ml = 0
(a) 1
(b) 2
(c) 3
(d) 4
Answer. A
Question. What is the maximum numbers of electrons that can be associated with the following set of quantum numbers?
n = 3, l = 1 and m = –1
(a) 4
(b) 2
(c) 10
(d) 6
Answer. B
Question. The outer electronic configuration of Gd (At. No. 64) is
(a) 4f 55d46s1
(b) 4f 75d16s2
(c) 4f 35d56s2
(d) 4f 45d56s1
Answer. B
Question. Maximum number of electrons in a subshell with l = 3 and n = 4 is
(a) 14
(b) 16
(c) 10
(d) 12
Answer. A
Question. The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is
(a) 5, 1, 1, +1/2
(b) 6, 0, 0, +1/2
(c) 5, 0, 0, +1/2
(d) 5, 1, 0, +1/2
Answer. C
Question. The total number of atomic orbitals in fourth energy level of an atom is
(a) 8
(b) 16
(c) 32
(d) 4
Answer. B
Question. If n = 6, the correct sequence for filling of electrons will be
(a) ns → (n – 2)f → (n – 1)d → np
(b) ns → (n – 1)d → (n – 2)f → np
(c) ns → (n – 2)f → np → (n – 1)d
(d) ns → np → (n – 1)d → (n – 2)f
Answer. A
Question. Maximum number of electrons in a subshell of an atom is determined by the following
(a) 2l + 1
(b) 4l – 2
(c) 2n2
(d) 4l + 2
Answer. D
Question. Which of the following is not permissible arrangement of electrons in an atom?
(a) n = 5, l = 3, m = 0, s = +1/2
(b) n = 3, l = 2, m = –3, s = –1/2
(c) n = 3, l = 2, m = –2, s = –1/2
(d) n = 4, l = 0, m = 0, s = –1/2
Answer. B
Question. Consider the following sets of quantum numbers:
n l m s
(i) 3 0 0 +1/2
(ii) 2 2 1 +1/2
(iii) 4 3 –2 –1/2
(iv) 1 0 –1 –1/2
(v) 3 2 3 +1/2
Which of the following sets of quantum number is not possible?
(a) (i), (ii), (iii) and (iv)
(b) (ii), (iv) and (v)
(c) (i) and (iii)
(d) (ii), (iii) and (iv)
Answer. B
Question. The orientation of an atomic orbital is governed by
(a) principal quantum number
(b) azimuthal quantum number
(c) spin quantum number
(d) magnetic quantum number.
Answer. D
Question. The following quantum numbers are possible for how many orbitals?
n = 3, l = 2, m = +2
(a) 1
(b) 2
(c) 3
(d) 4
Answer. A
Question. For which of the following sets of four quantum numbers, an electron will have the highest energy?
n l m s
(a) 3 2 1 +1/2
(b) 4 2 – 1 +1/2
(c) 4 1 0 –1/2
(d) 5 0 0 –1/2
Answer. B
Question. Electronic configuration of calcium atom can be written as
(a) [Ne]4p2
(b) [Ar]4s2
(c) [Ne]4s2
(d) [Kr]4p2
Answer. B
Question. In a given atom no two electrons can have the same values for all the four quantum numbers. This is called
(a) Hund’s Rule
(b) Aufbau principle
(c) Uncertainty principle
(d) Pauli’s Exclusion principle.
Answer. D
Question. For azimuthal quantum number l = 3, the maximum number of electrons will be
(a) 2
(b) 6
(c) 0
(d) 14
Answer. D
Question. The order of filling of electrons in the orbitals of an atom will be
(a) 3d, 4s, 4p, 4d, 5s
(b) 4s, 3d, 4p, 5s, 4d
(c) 5s, 4p, 3d, 4d, 5s
(d) 3d, 4p, 4s, 4d, 5s
Answer. B
Question. The electronic configuration of Cu (atomic number 29) is
(a) 1s2 2s22p6 3s23p6 4s23d9
(b) 1s2 2s22p6 3s23p63d10 4s1
(c) 1s2 2s22p6 3s23p6 4s24p6 5s25p1
(d) 1s2 2s22p6 3s23p6 4s24p63d3
Answer. B
Question. The total number of electrons that can be accommodated in all the orbitals having principal quantum number 2 and azimuthal quantum number 1 are
(a) 2
(b) 4
(c) 6
(d) 8
Answer. C
Question. An ion has 18 electrons in the outermost shell, it is
(a) Cu+
(b) Th4+
(c) Cs+
(d) K+
Answer. A
Question. Number of unpaired electrons in N2+ is/are
(a) 2
(b) 0
(c) 1
(d) 3
Answer. C
Question. The maximum number of electrons in a subshell is given by the expression
(a) 4l – 2
(b) 4l + 2
(c) 2l + 2
(d) 2n2
Answer. B
Question. The number of spherical nodes in 3p orbitals are/is
(a) one
(b) three
(c) none
(d) two
Answer. A
Please click on below link to download CBSE Class 11 Chemistry Structure of Atom Worksheet Set A
Chapter 2 Structure of Atom CBSE Class 11 Chemistry Worksheet
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