Atomic Structure Chemistry
CHEMISTRY
Assignment – 1
Atomic Structure
Questions Dalton’s, Rutherford’s atomic model, 9 The spectrum of He-atom may be considered
based on Fundamental Particles similar to the spectrum of -
(A) H (B) Li +
1. The study of cathode rays (i.e. electronic (C) Na (D) He+
discharge through gases) shows that - 10 Supposing the energy of fourth shell for
(A) Alpha particles are heavier than protons hydrogen atom is - 50 a.u. (arbitrary unit). What
(B) All forms of matter contain electrons would be its ionization potential -
(C) All nuclei contain protons (A) 50 (B) 800
(D) e/m is constant (C) 15.4 (D) 20.8
2 Proton is - 11 Supposing the ionization energy of hydrogen
(A) Nucleus of deuterium atom is 640 eV. Point out the main shell having
(B) Ionised hydrogen molecule energy equal to – 40 eV -
(C) Ionised hydrogen atom (A) n = 2 (B) n = 3
(D) An -particle (C) n = 4 (D) n = 5
3 Which is not deflected by magnetic field - 12 A 1-kW radio transmitter operates at a
(A) Neutron frequency of 880 Hz. How many photons per
(B) Electron and Neutron second does it emit -
(C) Proton (A) 1.71 x 1021 (B) 1.71 x 1033
(D) Electron 23
(C) 6.02 x 10 (D) 2.85 x 1026
4 According to Dalton’s atomic theory, an atom
13 On Bohr’s stationary orbits -
can –
(A) Electrons do not move
(A) Be created
(B) Electrons move emitting radiations
(B) Be destroyed
(C) Energy of the electron remains constant
(C) Neither be created nor destroyed
(D) None (D) Angular momentum of the electron is h/2
5 Rutherford’s experiment on scattering of alpha 14 The value of Bohr radius of hydrogen atom is -
particles showed for the first time that atom has- (A) 0.529x 10–7cm (B) 0.529x 10-8cm
(A) Electrons (B) Protons (C) 0.529x 10-9cm (D) 0.529x 10–10 cm
(C) Nucleus (D) Neutrons 15 On the basis of Bohr’s model, the radius of the
6 - particles are represented by – 3rd orbit is -
(A) Lithium atoms (B) Helium nuclei (A) Equal to the radius of first orbit
(C) Hydrogen nucleus (D) None of these (B) Three times the radius of first orbit
Questions (C) Five times the radius of first orbit
based on Bohr’s Atomic Model (D) Nine time the radius of first orbit
7 The energy of electron in first Bohr’s orbit of H- 16 The correct expression derived for the energy of
atom is –13.6 eV. What will be its potential an electron in the nth energy level is for H-atom-
energy in n = 4th orbit - 2 2 me 4 2 2 me 4
(A) En = 2 2
(B) En = –
(A) – 13.6 eV (B) –3.4 eV n h nh 2
(C) –0.85 eV (D) –1.70 eV 2 2 me 2 2 2 me 4
8 The frequency of line spectrum of sodium is (C) En = – 2 2 (D) En= –
n h n 2h 2
5.09 x 1014 sec–1. Its wave length (in nm) will 17 Ionization energy for hydrogen atom in ergs,
be - [c = 3 × 108 m/sec]- Joules and eV respectively is –
(A) 510 nm (B) 420 nm (A) 21.8 x 10–12, 218 x 10–20 , 13.6
(C) 589 nm (D) 622 nm
(B) 13.6 x 218 x 10–20, 21.8 x 10–13
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Atomic Structure Chemistry
(C) 21.8 x 10–20, 13.6 , 21.8 x 10–13 (C)
21R H
(D)
21R H
(D) 21.8 x 10–13, 13.6, 21.8 x 10–20 100 100
18 The velocity of an electron in the third orbit of 27 Wave number of a spectral line for a given
hydrogen atom – transition is x cm–1 for He+ , then its value for
(A) 7.28 x107 cm sec–1 Be3+ for the same transition is -
(B) 7.08 x 107 cm sec–1 (A) 4x cm–1 (B) x cm–1
(C) 7.38 x 107cm sec–1 (C) x/4 cm –1 (D) 2x cm–1
(D) 7.48 x107cm sec–1 28 A photon was absorbed by a hydrogen atom in
19 The ionization energy of a hydrogen atom is its ground state and the electron was promoted
13.6eV. The energy of the third-lowest to the fifth orbit. When the excited atom
electronic level in doubly ionized lithium ion returned to its ground state, visible and other
(Z = 3) is - quanta were emitted. Other quanta are -
(A) –28.7 eV (B) –54.4 eV (A) 2 1 (B) 5 2
(C) –122.4 eV (D) –13.6 eV (C) 3 1 (D) 4 1
20 Difference between Cl atom and Cl – ion is of : 29 Wave-length of the first line of Paschen Series
(A) Proton (B) Neutron 1
hydrogen spectrum is 912Å -
(C) Electron (D) Proton and electron R
21 For ionising an excited hydrogen atom, the
(A) 18761(Å) (B) 2854 (Å)
energy required in eV will be -
(C) 3452 (Å) (D) 6243 (Å)
(A) `3.4 or less
30 Electronic transition in He+ ion takes from n2 to
(B) More than 13.6
n1 shell such that :
(C) Little less than 13.6
2n2 + 3n1 = 18 ….(i)
(D) 13.6
2n2 – 3n1 = 6
22 A gas absorbs a photon of 300 nm and then re-
Then what will be the total number of photons
emits two photons. One photon has a
emitted when electrons transit to n1 shell ?
wavelength 600 nm. The wavelength of second
(A) 21 (B) 15
photon is -
(C) 20 (D) 10
(A) 300 nm (B) 400 nm
31 If the shortest wavelength in Lyman series of H
(C) 500 nm (D) 600 nm
atom is x, then longest wavelength in Balmer
23 The energy of a photon of radiation having
wavelength 300 nm is - series of He+ is -
(A) 6.63 × 10–29 J (B) 6.63 × 10–19 J 9x 36 x
(A) (B)
(C) 6.63 × 10–28 J (D) 6.63 × 10–17 J 5 5
24 For H– atom, the energy required for the x 5x
removal of electron from various sub-shells is (C) (D)
given as under– 4 9
3s 32 Which of the following expressions represents
3p 3d n= the spectrum of Balmer series(If n is the
E1 0 principal quantum number of higher energy
E2 level) in Hydrogen atom -
0
E3 R (n – 1)(n 1)
0 (A) v cm 1
The order of the energies would be – n2
(A) E1 > E2 > E3 (B) E3 > E2 > E1 R (n – 2)(n 2)
(B) v cm 1
(C) E1 = E2 = E3 (D) None of these 4n 2
R (n – 2)(n 2)
Questions
Hydrogen Sperctrum (C) v cm 1
based on n2
R (n – 1)(n 1)
25 The wave number of the first line of Balmer (D) v cm 1
4n 2
series of hydrogen is 15200 cm–1. The wave
Questions
number of the first Balmer line of Li2+ ion is- based on Quantum Numbers
(A) 15200cm–1 (B) 60800 cm–1
33 The maximum number of electrons in a principal
(C) 76000 cm–1 (D) 136800 cm–1 shell is -
26 The wavelength of the third line of the Balmer
(A) 2n (B) 2n2
series for a hydrogen atom is -
21 100 (C) 2 (D) 2 n
(A) (B) 34 Which of the following statements concerning
100R H 21 R H
the four quantum numbers is false -
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Atomic Structure Chemistry
(A) n gives idea of the size of an orbital 1 1
(B) gives the shape of an orbital (A) 3, 2, –2, (B) 3, 3, 1, –
2 2
(C) m gives the energy of the electron in the
1 1
orbital (C) 3, 2, 1, (D) 3, 1, 1,
(D) s gives the direction of spin of the 2 2
electron in an orbital
35 How many electrons can fit into the orbitals that Questions Shapes of orbitals, Nodal Plane &
comprise the 3rd quantum shell n = 3 - based on surface
(A) 2 (B) 8
(C) 18 (D) 32 44 Which of the following sets of quantum numbers
36 The shape of the orbital is given by - is correct for an electron in 4 f-orbital?
(A) Spin quantum number 1
(B) Magnetic quantum number (A) n = 4, l = 3, m = +4, s = +
2
(C) Azimuthal quantum number
(D) Principal quantum number 1
(B) n = 4, l = 4, m = –4, s = –
37 The set of quantum numbers not applicable for 2
an electron in an atom is – 1
(A) n = 1, = 1, m = 1, s = + ½ (C) n = 4, l = 3, m = +1, s = +
2
(B) n = 1, = 0, m = 0, s = + 1/2
1
(C) n = 1, = 0, m = 0, s = – ½ (D) n = 3, l = 2, m = –2, s = +
(D) n = 2, = 0, m = 0, s = + 1/2 2
38 Maximum number of electrons in a subshell is 45 In which of the following pairs is the probability
given by - of finding the electron in xy-plane zero for both
(A) (2 + 1) (B) 2(2 +1) orbitals ?
(C) (2 +1)2 (D) 2(2 + 1)2 (A) 3d yz ,4d x 2 – y 2 (B) 2p z , dz 2
39 The magnetic quantum number for valence (C) 4dzx, 3pz (D) All of these
electron of sodium atom is -
46 For 4py orbital : There are -
(A) 3 (B) 2
(C) 1 (D) Zero nodal plane = ........ and azimuthal quantum
40 Which one of the following represents an number =
impossible arrangement – (A) 1, 0 (B) 0, 1
n m s (C) 1, 1 (D) 2, 1
(A) 3 2 –2 1/2 47 The maximum probability of finding electron in
(B) 4 0 0 1/2 the dxy orbital is -
(C) 3 2 –3 1/2 (A) Along the x axis
(D) 5 3 0 1/2 (B) Along the y axis
41 The set of quantum number for the 19th (C) At an angle of 45º from the x and y axis
electron in chromium is - (D) At an angle of 90º from the x and y axis
(A) n=4, =0, m=0, s =+1/2 or -1/2 Questions
basedon Electronic Configuration
(B) n=3, =2, m=1, s=+1/2 or -1/2
(C) n=3, =2, m= -1, s=+1/2 or -1/2 48 An electron has a spin quantum number
(D) n=4, =1, m=0, s=+1/2 or -1/2 + 1/2 and a magnetic quantum number –1. It
.42 The electronic configuration together with the cannot be present in -
quantum number of last electron for lithium is - (A) d-Orbital (B) f-Orbital
1 (C) s-Orbital (D) p-Orbital
(A) 1s22s1 2, 0, 0 +
2 49 If the electronic structure of oxygen atom is
1 1 2p
(B) 1s22s1 2, 0, 0 + or – 2 2
written as 1s , 2s it would violate-
2 2
1 (A) Hund’s rule
(C) 1s22s02p1 2, 1, 0 ± (B) Paulis exclusion principle
2
(C) Both Hund’s and Pauli’s principles
2 1 1
(D) 2s 2s 2,1, 0 ± (D) None of these
2 50 The energy of an electron of 2py orbital is -
43 Four sets of values of quantum numbers
(n, , m and s) are given below. Which set does (A) Greater than 2px orbital
not provide a permissible solution of the wave (B) Less than 2pz orbital
equation - (C) Equal to 2s orbital
(D) Same as that of 2px and 2pz orbitals
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Atomic Structure Chemistry
ANSWER KEY
Assignment - 1
Atomic Structure
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
D C A C C B D C B B C B C B D D A A D C
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
A D B C D B A A A D A B B C C C A B D C
41 42 43 44 45 46 47 48 49 50
A B B C C C C C A D
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46597906, 8595703399, E.mail :-info@mbition.in
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