PHYSICAL CHEMISTRY Total Marks : 25

DPP No. 11 Max. Time : 26 min.

Topic : Atomic Structure

Type of Questions M.M., Min.

Single choice Objective ('–1' negative marking) Q.1 to 3,6,7 (3 marks, 3 min.) [15, 15]

Subjective Questions ('–1' negative marking) Q.8 (4 marks, 5 min.) [4, 5]

Short Subjective Questions ('–1' negative marking) Q.4,5 (3 marks, 3 min.) [6, 6]

1. The mass of a neutron is .................... than the mass of a proton :

(A) slightly less (B) slightly more
(C) exactly equal (D) their masses cannot be compared

2. The ratio of the number of neutrons present in one atom each of C and Si with respect to mass
number of 12 and 28 respectively is :
(A) 3 : 7 (B) 7 : 3 (C) 3 : 4 (D) 6 : 28

3. An atom consist of electrons, protons and neutrons. If the mass attributed to neutron was halved and
that attributed to the electron was doubled, the atomic mass of 6C12 would be approximately :
(A) Same (B) Doubled (C) Halved (D) Reduced by 25%

4. There are 11 protons and 12 neutrons in the nucleus of an atom. Find the atomic number (Z), mass number

(A), number of electrons and the symbol of the element.

15 2 –
5. Calculate the number of protons, electrons and neutrons in 8O
.

6. The total number of electrons in a nitrate ion is :

(A) 31 (B) 62 (C) 32 (D) 63

7. Atomic radius is of the order of 10–8 cm and nuclear radius is of the order of 10 –13 cm. The fraction of
atom that is occupied by nucleus is :

(A) 10–5 (B) 105 (C) 10 –15 (D) None of these

8._ If an atom of an element X contains equal number of protons, neutrons and electrons, and its atomic

number (Z) and mass number (A) are related as : 2A + 3Z = 140, then find the total number of nucleons
present in one atom of element X. Also identify the element.

DPPS FILE # 14
 PHYSICAL CHEMISTRY Total Marks : 44
DPP No. 12 Max. Time : 46 min.

Topic : Atomic Structure

Type of Questions M.M., Min.

Single choice Objective ('–1' negative marking) Q.1,3,5 (3 marks, 3 min.) [9, 9]

Multiple choice objective ('–1' negative marking) Q.2, 4,6 (4 marks, 4 min.) [12, 12]

Short Subjective Questions ('–1' negative marking) Q.10,12 (3 marks, 3 min.) [6, 6]

Comprehension ('–1' negative marking) Q.7 to 9 (3 marks, 3 min.) [9, 9]

Match the Following (no negative marking) (2 × 4) Q.11 (8 marks, 10 min.) [8, 10]

1. The potential energy of the electron present in the ground state of Li2+ ion is represented by :

(r = Radius of ground state)

3e 2 3e 3e 2 3e 2
(A)  4   r (B)  4  r (C)  4   r 2 (D)  4   r
0 0 0 0

2. Which of the following are isotopes :

(i) Atom, whose nucleus contains 20p + 15n (ii) Atom, whose nucleus contains 20p + 17n
(iii) Atom, whose nucleus contains 18p + 22n (iv) Atom, whose nucleus contains 18p + 21n
(A) (i) and (iii) (B) (i) and (ii) (C) (ii) and (iii) (D) (iii) and (iv)

3. Which of the following are isobars :

(i) Atom, whose nucleus contains 20p + 15n (ii) Atom, whose nucleus contains 20p + 20n
(iii) Atom, whose nucleus contains 18p + 17n (iv) Atom, whose nucleus contains 18p + 22n
(A) (i) and (iii) (B) (ii) and (iii) (C) (iii) and (iv) (D) (i) and (iv)

4. Which of the following is/are isotones :

15 16
(A) 2
1 H, 3
1 H (B) 7 N, 8 O (C) 40 40
18 Ar, 20 Ca (D) 3
1 H, 24 He

5. Which of the following are isoelectronic :

(I) CH3+ (II) H3O+ (III) NH3 (IV) CH3–
(A) I and III (B) III and IV (C) I and II (D) II, III and IV

6.* Which of the following are isoelectronic species :

(A) CO32–, NO3– (B) SO42–, PO43– (C) CO2, N2O (D) N3–, Al3+

DPPS FILE # 15
Comprehension # (Q.7 to Q.9)

The approximate size of the nucleus can be calculated by using energy

conservation theorem in Rutherford’s -scattering experiment. If an -
particle is projected from infinity with speed v, towards the nucleus having
z protons, then the -particle which is reflected back or which is deflected
by 1800 must have approached closest to the nucleus. It can be
approximated that -particle collides with the nucleus and gets back. Now,
if we apply the energy conservation equation at initial point and collision
point, then :
(Total Energy)initial = (Total Energy)final
(K.E.)i + (P.E.)i = (K.E.)f + (P.E.)f
(P.E.)i = 0, since P.E. of two charge system separated by infinite distance is zero. Finally the particle stops
and then starts coming back.

1 Kq1q2 1 2e  ze 4Kze 2
mv 2 + 0 = 0 +  mv 2 = K  R= 2
2 R 2 R m v 

Thus the radius of nucleus can be calculated using above equation. The nucleus is so small a particle that
we can’t define a sharp boundary for it. Experiments show that the average radius R of a nucleus may be
written as:
R = R0(A)1/3
where R0 = 1.2 × 10 m –15

A – mass number of atom

R – radius of nucleus

7. If the diameter of two different nuclei are in the ratio 1:2, then their mass number are in the ratio :
(A) 1:2 (B) 8:1 (C) 1:8 (D) 1:4
8. An -particle with speed v 0 is projected from infinity and it approaches up to r0 distance from a nuclei.
Then, the speed of -particle which approaches upto 2r0 distance from the nucleus is :
v0 v0
(A) 2 v0 (B) (C) 2v 0 (D)
2 2
9. Radius of a particular nucleus is calculated by the projection of -particle from infinity at a particular
speed. Let this radius is the true radius. If the radius calculation for the same nucleus is made by another
-particle with half of the earlier speed, then the percentage error involved in the radius calculation is :
(A) 75% (B) 100% (C) 300% (D) 400%
10. With what velocity should an -particle travel towards the nucleus of a Copper atom, so as to arrive at a
distance of 10–13 m from the nucleus of Copper atom. (At. No. of Cu = 29). (Take 40 = 6.32)

11. Column-I Column-II

(A) Frequency (p) Linear distance travelled by a wave per unit time.
(B) Wavelength (q) Number of waves passing through a point in one second.
(C) Time period (r) Linear distance between starting and end point of one complete
wave.
(D) Speed (s) Time taken for one complete wave to pass through a point.

12. For a wave, frequency is 10 Hz and wavelength is 2.5 m. How much linear distance will it travel in 40
seconds ?

DPPS FILE # 16
 PHYSICAL CHEMISTRY Total Marks : 31
DPP No. 13 Max. Time : 35 min.

Topic : Atomic Structure

Type of Questions M.M., Min.

Single choice Objective ('–1' negative marking) Q.4,9 (3 marks, 3 min.) [6, 6]
Subjective Questions ('–1' negative marking) Q.1,6,7,8 (4 marks, 5 min.) [16, 20]
Short Subjective Questions ('–1' negative marking) Q.2,3,5 (3 marks, 3 min.) [9, 9]

1. Visible spectrum contains light of following colours "Violet - lndigo - Blue - Green - Yellow - Orange - Red"
(VIBGYOR).
Its frequency ranges from Violet (7.5 × 1014 Hz) to Red (4 × 1014 Hz). Find out the maximum wavelength (in
Å) in this range.

2. For a broadcasted electromagnetic wave having frequency of 1200 KHz, calculate number of waves that
will be formed in 1 km distance (wave number per km).

3. (a) If volume of nucleus of an atom V is related to its mass number A as V  An , find the value of n.

(b) If the frequency of violet radiation is 7.5 × 1014 Hz, find the value of wavenumber (  ) (in m–1) for it.

4. The ratio of the energy of a photon of wavelength 3000 Å to that of a photon of wavelength 6000Å respectively
is:
(A) 1 : 2 (B) 2 : 1 (C) 3 : 1 (D) 1 : 3

5. Assume that 10–17 J of light energy is needed by the interior of the human eye to see an object. How many
photons of green light ( = 310 nm) are needed to generate this minimum energy ?

6. A photon of 300 nm is absorbed by a gas and then, it re-emits two photons and attains the same initial
energy level. One re-emitted photon has wavelength 500 nm. Calculate the wavelength of other photon re-
emitted out.

7. Find out the number of photons emitted by a 60 watt bulb in one minute, if wavelength of an emitted photon
is 620 nm.

8. If a photon having wavelength 620 nm is used to break the bond of A2 molecule having bond energy
144 KJ mol–1, then find the % of energy of photon that is converted into kinetic energy of A atoms.
[hc = 12400 eVÅ ,1 eV/atom = 96 KJ/mol]

9. A certain dye absorbs light of certain wavelength and then fluorescence light of wavelength 5000 Å.
Assuming that under given conditions, 50% of the absorbed energy is re-emitted out as fluorescence and
the ratio of number of quanta emitted out to the number of quanta absorbed is 5 : 8, find the wavelength of
absorbed light (in Å) : [hc = 12400 eVÅ ]
(A) 4000 Å (B) 3000 Å (C) 2000 Å (D) 1000 Å

DPPS FILE # 17
 PHYSICAL CHEMISTRY Total Marks : 37
DPP No. 14 Max. Time : 41 min.

Topic : Atomic Structure

Type of Questions M.M., Min.

Single choice Objective ('–1' negative marking) Q.3 to Q.9 (3 marks, 3 min.) [21, 21]
Subjective Questions ('–1' negative marking) Q.1 to Q.2 (4 marks, 5 min.) [8, 10]
Match the Following (no negative marking) (2 × 4) Q.10 (8 marks, 10 min.) [8, 10]
1. In I experiment, electromagnetic radiations of a certain frequency are irradiated on a metal surface ejecting
photoelectrons having a certain value of maximum kinetic energy. However, in II experiment,on doubling
the frequency of incident electromagnetic radiations, the maximum kinetic energy of ejected photoelectrons
becomes three times. What percentage of incident energy is converted into maximum kinetic energy of
photoelectrons in II experiment ?
2. The potential difference applied on the metal surface to reduce the velocity of photoelectron to zero is
known as Stopping Potential. When a beam of photons of wavelength 40 nm was incident on a surface
of a particular pure metal, some emitted photoelectrons had stopping potential equal to 18.6 V, some
had 12 V and rest had lower values. Calculate the threshold wavelength (0) of the metal (in Å) assum-
ing that at least one photoelectron is ejected with maximum possible kinetic energy. (hc = 12400 eVÅ)
3. For which of the following species, Bohr model is not valid :
(A) He+ (B) H (C) Li2+ (D) H+
4. Wavelength of radiations emitted when an electron in a H-like atom jumps from a state A to C is 2000 Å
and it is 6000 Å, when the electron jumps from state B to state C. Wavelength of the radiations emitted
when an electron jumps from state A to B will be :
(A) 2000 Å (B) 3000 Å (C) 4000 Å (D) 6000 Å

5. If the radius of the first Bohr orbit of the H atom is r, then for Li2+ ion, it will be :
(A) 3r (B) 9r (C) r/3 (D) r/9
6. In a certain electronic transition in the Hydrogen atom from an initial state i to a final state f, the difference
in the orbit radius (ri  rf) is seven times the first Bohr radius. Identify the transition :
(A) 4  1 (B) 4  2 (C) 4  3 (D) 3  1
7. The velocity of electron in the ground state of H atom is 2.184 × 108 cm/sec. The velocity of electron in
the second orbit of Li2+ ion in cm/sec would be :
(A) 3.276 × 108 (B) 2.185 × 108 (C) 4.91 × 108 (D) 1.638 × 108

8. The potential energy of the electron present in the ground state of Li2+ ion is represented by :
3e 2 3e 3e 2 3e 2
(A)  (B)  (C)  (D) 
4  0 r 4 0 r 4  0 r 2 4  0 r
2h
9. If the angular momentum of an electron in a Bohr orbit is , then the value of potential energy of this

electron present in He+ ion is :
(A) – 13.6 eV (B) – 3.4 eV (C) – 6.8 eV (D) – 27.2 eV.
10. Match the following :
En = total energy, n = angular momentum, Kn = K.E. , Vn = P.E., Tn = time period, rn = radius of nth orbit
Column () Column ()
–y
(A) En  rn /Z, then y is (p) 1/2
(B) n  nx ,then x is (q) – 2
En
(C) Value of V is (r) – 3
n
Zt
(D) Tn  , t & m are respectively (s) 1
nm

DPPS FILE # 18
 PHYSICAL CHEMISTRY Total Marks : 25
DPP No. 15 Max. Time : 25 min.

Topic : Atomic Structure

Type of Questions M.M., Min.

Single choice Objective ('–1' negative marking) Q.1 to Q.6,8 (3 marks, 3 min.) [21, 21]
Multiple choice objective ('–1' negative marking) Q.7 (4 marks, 4 min.) [4, 4]

1. An electron in a H–like atom jumps from a higher energy level ‘n’ to ground state by emitting two successive
photons of wave numbers 5.25 × 108 m–1 and 7.25 × 108 m–1. If the same electron undergoes the same
transition by emitting a single photon, then the wavelength of this photon is :
(A) 32.84 Å (B) 8 Å (C) 0.125 Å (D) 0.03 Å

2. The ratio of the difference in energy between the first and second Bohr orbit to that between the second
and third Bohr orbit in a H-like species is :

1 1 4 27
(A) (B) (C) (D)
2 3 9 5

3. The radii of two of the first four Bohr orbits of the Hydrogen atom are in the ratio 1 : 4. The energy difference
between them may be :
(A) Either 12.09 eV or 3.4 eV (B) Either 2.55 eV or 10.2 eV
(C) Either 13.6 eV or 3.4 eV (D) Either 3.4 eV or 0.85 eV

4. The ratio of radius of two different orbits in a H-atom is 4 : 9. Then, the ratio of the frequency of revolution
of electron in these orbits is :
(A) 2 : 3 (B) 27 : 8 (C) 3 : 2 (D) 8 : 27

5. According to Bohr’s theory, the ratio of electrostatic force of attraction acting on electron in 3rd orbit of He+
x
3
ion and 2 orbit of Li ion is   . Then, the value of x is :
nd 2+
2
(A) 7 (B) –6 (C) 6 (D) –7
6. Suppose a hypothetical H-like atom produces a blue, yellow, red and violet line in emission spectrum.
Match the above lines with their corresponding possible electronic transition :
Colour of spectral lines Possible corresponding transitions
(A) Blue (p) 6  3
(B) Yellow (q) 2  1
(C) Red (r) 5  2
(D) Violet (s) 4  3
(A) (A)  r , (B)  p , (C)  s , (D)  q (B) (A)  r , (B)  s , (C)  q, (D)  p
(C) (A)  p , (B)  r , (C)  s , (D)  q (D) (A)  p , (B)  r , (C)  q, (D)  s

7. If the binding energy of 2nd excited state of a hypothetical H-like atom is 12 eV, then :
(A) I excitation potential = 81 V (B) II Excitation energy = 96 eV
(C) Ionisation potential = 192 V (D) Binding energy of 2nd state = 27 eV

8. Wave number of a spectral line for a given transition is x cm –1 for He+ ion. Then, its value for Be3+ ion
(isoelectronic of He+) for same transition is :
x
(A) x cm–1 (B) 4x cm–1 (C) cm–1 (D) 2x cm–1
4

DPPS FILE # 19
 PHYSICAL CHEMISTRY Total Marks : 37
DPP No. 16 Max. Time : 43 min.

Topic : Atomic Structure

Type of Questions M.M., Min.

Single choice Objective ('–1' negative marking) Q.1 to Q.3 (3 marks, 3 min.) [9, 9]
Multiple choice objective ('–1' negative marking) Q.4 (4 marks, 4 min.) [4, 4]
Subjective Questions ('–1' negative marking) Q.5,6,8,9 (4 marks, 5 min.) [16, 20]
Match the Following (no negative marking) (2 × 4) Q.7 (8 marks, 10 min.) [8, 10]

1. If numerical value of mass and velocity are equal for a particle, then its de-Broglie wavelength in terms of
K.E. is :
mh h
(A) (B) (C) both are correct (D) none is correct.
2K.E. 2mK.E.

2. A wavelength of 400 nm of electromagnetic radiation corresponds to :

(A) frequency () = 7.5 × 1014 Hz (B) wave number(  ) = 2.5 × 106 m–1.
(C) momentum of photon = 1.66 × 10–27 kg ms–1 (D) all are correct values.
3. In one experiment, a proton having initial kinetic energy of 1 eV is accelerated through a potential differ-
ence of 3 V. In another experiment, an -particle having initial kinetic energy 20 eV is retarded by a poten-
tial difference of 2 V. The ratio of de-Broglie wavelengths of proton and -particle is :
(A) 2 6 : 1 (B) 8 : 1 (C) 4 : 1 (D) 2 2 : 1

4.* When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maxi-
mum kinetic energy (K.E)A and de-Broglie wavelength is A. The maximum kinetic energy of photoelec-
trons liberated from another metal B by photons of energy 4.7 eV is (KE)B, where (KE)B = (KE)A –1.5 eV.
If the de-Broglie wavelength of these photoelectrons is B (= 2A), then :
(A) The work function of metal A is 2.25 eV (B) The work function of metal B is 4.20 eV
(C) (KE)A = 2 eV (D) (KE)B = 2.75 eV
5. Average life time of an electron in hydrogen atom excited to n = 2 state is 10–8 s. Find the number of
revolutions made by the electron on the average, before it jumps to the ground state.
6. The ionisation energy of He+ ion is 19.6 × 10–18 J per ion. Calculate the energy of the first stationary state of
Li2+ ion.
7. Match the following :
Column () Column ()
(A) Binding energy of 5th excited state of Li2+ sample (p) 10.2 V
(B) st excitation potential of H-atom (q) 3.4 eV
(C) 2nd excitation potential of He+ ion (r) 13.6 eV
(D) I.E. of H-atom (s) 48.4 V

8. The IP of H-atom is 13.6 V. It is exposed to electromagnetic waves of wavelength 1026 Å and then, it gives
out induced radiations. Find the wavelength of all possible induced radiations.
9. The ionization energy of a Hydrogen like species is 4 Rydberg. What is the radius of the first orbit of
this atom ? (Given : Bohr radius of hydrogen = 5.3 × 10 11 m; 1 Rydberg = 2.2 × 1018 J)

DPPS FILE # 20
 PHYSICAL CHEMISTRY Total Marks : 36
DPP No. 17 Max. Time : 38 min.

Topic : Atomic Structure

Type of Questions M.M., Min.

Single choice Objective ('–1' negative marking) Q.1 to Q.4,10 (3 marks, 3 min.) [15, 15]
Multiple choice objective ('–1' negative marking) Q.5 (4 marks, 4 min.) [4, 4]
Comprehension ('–1' negative marking) Q.6 to Q.8 (3 marks, 3 min.) [9, 9]
Match the Following (no negative marking)(2 × 4) Q.9 (8 marks, 10 min.) [8, 10]

1. The wavenumber of the spectral line of shortest wavelength of Balmer series of He+ ion is :
(R = Rydberg's constant)
(A) R (B) 3R (C) 4R (D) 4R/9

2. Last line of the Lyman series of H-atom has frequency 1 , last line of Lyman series of He + ion has
frequency  2 and 1st line of Lyman series of He+ ion has frequency  3 . Then :
(A) 4 1 =  2 +  3 (B) 1 = 4  2 +  3 (C)  2 =  3 – 1 (D)  2 = 1 +  3
3. If 1 and 2 are respectively the wavelengths of the series limit of Lyman and Balmer series of Hydrogen
atom, then the wavelength of the first line of the Lyman series of the H-atom is :
 2 – 1 1 2
(A) 1 – 2 (B)  1 2 (C) 1 2
(D)  – 
2 1
4. STATEMENT -1: We can use two photons successively of 1240 Å and 2000 Å wavelength in order to
ionise a H atom from ground state.
STATEMENT -2: Sum of the energies of both the photons is greater than IE of H atom.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
5. Which of the following statements is/are INCORRECT :
(A) All spectral lines belonging to Balmer series in Hydrogen spectrum lie in visible region.
(B) If a light of frequency  falls on a metal surface having work function h , photoelectric effect will take
place only if   0.
(C) The number of photoelectrons ejected from a metal surface in photoelectric effect depends upon the
intensity of incident radiations.
4
(D) The series limit wavelength of Balmer series for H-atom is , where R is Rydberg's constant.
R

Comprehension # (Q.6 to Q.8)]

The only electron in the hydrogen atom resides under ordinary conditions
in the first orbit. When energy is supplied, the electron moves to higher
energy orbit depending on the amount of energy absorbed. When this
electron returns to any of the lower orbits, it emits energy. Lyman series is
formed when the electron returns to the lowest orbit, while Balmer series
is formed when the electron returns to second orbit. Similarly, Paschen,
Brackett and Pfund series are formed when electron returns to the third,
fourth and fifth orbits from higher energy orbits respectively (as shown in
figure)

DPPS FILE # 21
 Maximum number of lines produced when an electron jumps from nth level to ground level is equal to
n(n  1)
. For example, in the case of n = 4, number of lines produced is 6.
2
(4  3, 4  2, 4  1, 3  2, 3  1, 2  1). When an electron returns from n2 to n1 state, the number of
lines in the spectrum will be equal to :
(n2  n1 )(n 2  n1  1)
2
If the electron comes back from energy level having energy E2 to energy level having energy E1, then the
difference may be expressed in terms of energy of photon as :
hc
E2 – E1 = E ,  , E = h ( - frequency)
E
Since h and c are constants, E corresponds to definite energy; thus each transition from one energy level
to another will produce a light of definite wavelength. This is actually observed as a line in the spectrum of
hydrogen atom.

 1 1
Wave number of line is given by the formula   RZ 2  2  2  .
 n1 n 2 
where R is Rydberg constant (R = 1.1 × 107 m–1)
(i) First line of a series : It is called ‘line of longest wavelength’ or ‘line of lowest energy’.
(ii) Series limit or last line of a series : It is the line of shortest wavelength or line of highest energy.
6. In a hydrogen like sample, electrons are in a particular excited state. If electrons make transition upto
1st excited state, then it produces maximum 15 different types of spectral lines. Then, electrons were
initially in :
(A) 5th state (B) 6th state (C) 7th state (D) 8th state
7. The difference between the wave number of 1st line of Balmer series and last line of Paschen series
for Li 2+ ion is :

R 5R R
(A) (B) (C) 4R (D)
36 36 4

8. In a single isolated atom of hydrogen, electrons make transition from 4 th excited state to ground state
producing maximum possible number of wavelengths. If the 2 nd lowest energy photon is used to
further excite an already excited sample of Li 2+ ion, then transition will be :
(A) 12  15 (B) 9  12 (C) 6  9 (D) 3  6

9. Match the following :

List-I List-II
(A) From n = 6 upto n = 3 (In H-atom sample) (p) 10 lines in the spectrum
(B) From n = 7 upto n = 3 (In H-atom sample) (q) Spectral lines in visible region
(C) From n = 5 upto n = 2 (In H-atom sample) (r) 6 lines in the spectrum
(D) From n = 6 upto n = 2 (In H-atom sample) (s) Spectral lines in infrared region

3Rc
10. A photon of frequency cannot be emitted from which of the following transitions :
4
(Given : R = Rydberg's constant, c = speed of light)
(A) From 5 upto 1 transition in a sample of H– atom.
(B) From 6 upto 1 transition in a sample of He+ ion.
(C) From 7 upto 3 transition in a sample of Li2+ ion.
(D) From 8 upto 3 transition in a sample of He+ ion.

DPPS FILE # 22
 PHYSICAL CHEMISTRY Total Marks : 30
DPP No. 18 Max. Time : 33 min.

Topic : Atomic Structure

Type of Questions M.M., Min.

Single choice Objective ('–1'negative marking) Q.1 to Q.4,7,8 (3 marks, 3 min.) [18, 18]
Subjective Questions ('–1' negative marking) Q.5 to Q.6,9 (4 marks, 5 min.) [12, 15]
1. In a mixture of sample of H-atoms and He+ ions, electrons in all the H-atoms and He+ ions are present in
n = 4th state. Then, find total number of spectral lines obtained when all the electrons make transition from
n = 4 upto ground state :
(A) 12 (B) 6 (C) 11 (D) 16
2. What would be the maximum number of emission lines for atomic hydrogen that you would expect to see
with the naked eye, if the only electronic energy levels involved are those shown in the Figure :

(A) 4 (B) 6 (C) 5 (D) 15

3. A sample of H–like ion is in a particular excited state n2. The electron in it makes back transition upto a lower
excited state n1 producing a maximum of 10 different spectral lines. The change in angular momentum of
h
electron corresponding to maximum frequency line is expressed as y J-s. Then, the value of y is :
4
(A) 2 (B) 4 (C) 8 (D) 6
4. In a sample of H–atom, electron makes transition from lower state n1 to higher state n2 by absorbing
photons emitted by another sample of Li2+ ions from 12  3 transition. The electron in H–atom then makes
back transition from state n2 to ground state by emitting all posible photons. Then, the number of lines in
infrared region in emission spectrum of H–atom sample is :
(A) 3 (B) 2 (C) 1 (D) 0
5. In a sample containing a finite number of H- like atoms, electrons make transition from n = 6 upto
ground state producing a total of 10 different spectral lines. Find the minimum number of atoms that
must have been present in the sample.
6. An ion (atomic number Z), isoelectronic with Hydrogen, is in nth excited state. This ion emits two photons of
energies 10.2 eV and 17eV successively to return to first exited state. It can also emit two photons of
energies 4.25 and 5.95 eV successively to return to second excited state. What is value of n and Z ?
7. Determine the de-Broglie wavelength associated with an electron in the 3rd Bohr's orbit of He+ ion :
(A) 10 Å (B) 2 Å (C) 5 Å (D) 1 Å
8. If the radius of first Bohr’s orbit of H-atom is x, which of the following is the INCORRECT conclusion :
(A) The de-Broglie wavelength of electron in the third Bohr orbit of H-atom = 6 x.
(B) The fourth Bohr’s radius of He+ ion = 8x.
(C) The de-Broglie wavelength of electron in third Bohr’s orbit of Li2+ = 2x.
(D) The second Bohr’s radius of Be2+ = x
9. An electron is present in nth orbit of Be3+ ion such that its de-Broglie wavelength is same as associated with
the electron in 3rd orbit of C5+ ion. Find the distance between two adjacent crests of wave associated with
motion of an electron in nth orbit of Be3+ ion.

DPPS FILE # 23
 PHYSICAL CHEMISTRY Total Marks : 27
DPP No. 19 Max. Time : 27 min.

Topic : Atomic Structure

Type of Questions M.M., Min.

Single choice Objective ('–1' negative marking) Q.1 to Q.9 (3 marks, 3 min.) [27, 27]

1. Which of the following quantum numbers has not been derived from Schrodinger wave equation :
(A) Principal quantum number (n) (B) Subsidiary quantum number ( )
(C) Magnetic quantum number (m) (D) Spin quantum number (s)

2. Which d -orbital does not has four lobes :

(A) d x 2 – y 2 (B) dxy (C) dyz (D) d z2

3. The total number of subshells in nth main energy level are :

(A) n2 (B) 2n2 (C) (n–1) (D) n.

4. Which of the following orbital does not make sense :

(A) 3d (B) 3f (C) 5p (D) 7s.

5. The maximum number of electrons that can be accomodated in s, p and d-subshells respectively are :
(A) 2 in each (B) 2, 6 and 6 (C) 2, 6 and 10 (D) 2, 6 and 12.

6. Any p-orbital can accommodate upto :

(A) four electrons (B) two electrons with parallel spin
(C) six electrons (D) two electrons with opposite spin.

7. In which transition, the change in de-Broglie wavelength of electron is maximum :

(A) n = 8  n = 6 (B) n = 5  n = 4 (C) n = 3  n = 2 (D) n = 2  n = 1

8. S1 : Photoelectric effect can be explained on the basis of wave nature of electromagnetic radiations.

S2 : An orbital represented by n = 2,  = 1 is dumb-bell shaped.

S3 : dxy orbital has zero probability of finding electrons along X-axis and Y-axis.

(A) FTF (B) FTT (C) TFT (D) TFF

9. S1 : According to Bohr model, the angular momentum of revolving electron is directly proportional to the
atomic
number of H-like species bearing the electron.

S2 : An orbital cannot accomodate more than 2 electrons.

S3 : All orbitals have directional character.

(A) FTF (B) TFF (C) FFT (D) TTF

DPPS FILE # 24
 PHYSICAL CHEMISTRY Total Marks : 35
DPP No. 20 Max. Time : 37 min.

Topic : Atomic Structure

Type of Questions M.M., Min.

Single choice Objective ('–1' negative marking) Q.1 to Q.4 (3 marks, 3 min.) [12, 12]
Comprehension ('–1' negative marking) Q.5 to Q.9 (3 marks, 3 min.) [15, 15]
Subjective Questions ('–1' negative marking) Q.10 to Q.11 (4 marks, 5 min.) [8, 10]

1. The orbital angular momentum corresponding to n = 4 and m = –3 is :

h 6h 3h
(A) 0 (B) (C) (D)
2 2 

2. Spin magnetic moment of Xn+ (Z = 26) is 24 B.M. Hence number of unpaired electrons and value of n
respectively are :
(A) 4, 2 (B) 2, 4 (C) 3, 1 (D) 0, 2

3. Spin magnetic moments of V (Z = 23), Cr (Z = 24), Mn (Z = 25) are x, y, z respectively. Hence :

(A) x = y = z (B) x < y < z (C) x < z < y (D) z < y < x

4. Which of the following sets of quantum numbers can be correct for an electron in 4f-orbital :
1 1
(A) n = 3,  = 2, m = –2, s = + (B) n = 4,  = 4, m = –4, s = –
2 2
1 1
(C) n = 4,  = 3, m = +1, s = + (D) n = 4,  = 3, m = +4, s = +
2 2
Comprehension # (Q.5 to Q.9)

Azimuthal quantum number () : It describes the shape of electron cloud and the number of subshells in
a shell.
* It can have values from 0 to (n – 1)
* value of  subshell
0 s
1 p
2 d
3 f
* Number of orbitals in a subshell = 2 + 1

h  h 
* Orbital angular momentum L = ( 1) =   (   1)    2 
2  
Magnetic quantum number (m) : It describes the orientations of the subshells. It can have values from –l
to + l including zero, i.e., total (2l + 1) values. Each value corresponds to an orbital. s-subshell has one orbital,

DPPS FILE # 25
 p-subshell three orbitals (px, py and pz), d-subshell five orbitals (d xy , d yz , d zx , d x 2  y 2 , d z 2 ) and f-subshell has

seven orbitals.
Spin quantum number (s) : It describes the spin of the electron. It has values +1/2 and –1/2 signifying
clockwise spinning and anticlockwise rotation of electron about its own axis.

h 1
Spin of the electron produces angular momentum equal to S = s(s  1) where s = + .
2 2

n n
Total spin of an atom =  or 
2 2
where n is the number of unpaired electron.

The magnetic moment of an atom, s = n (n  2) B.M.

n – number of unpaired electrons, B.M. (Bohr magneton)

5. A d-block element has total spin value of +3 or –3. Then, the spin only magnetic moment of the
element is approximately :
(A) 2.83 B.M. (B) 3.87 B.M. (C) 5.9 B.M. (D) 6.93 B.M.

x
6. Spin only magnetic moment of 25 Mn ion is 15 B.M. Then, the value of x is :
(A) 1 (B) 2 (C) 3 (D) 4

7. Spin only magnetic moment of 2+

26 Fe ion is same as :
(A) 2+ 4+
26 Fe (B) 24 Cr (C) 28 Ni (D) All of these

h
8. Orbital angular momentum of an electron is 3 . Then, the number of orientations of this orbital in


space are :
(A) 3 (B) 5 (C) 7 (D) 9

9. The correct order of the magnetic moment of [ 25Mn4+, 24 Cr

3+ ,
26 Fe
3+ ] is :
(A) Fe3+
> =Cr3+ Mn4+ (B) Fe3+
> > Cr3+ Mn4+
(C) Cr = Mn > Fe3+
3+ 4+ (D) Fe > Mn > Cr3+
3+ 4+

10. What is the maximum possible number of electrons in an atom with (n +  = 7) ?

11. Predict total spin for each configuration :

(a) 1s2 (b) 1s2 2s2 2p6 (c)1s2 2s2 2p5

(d) 1s2 2s2 2p3 (e) 1s2 2s2 2p6 3s2 3p6 3d5 4s2.

DPPS FILE # 26
 PHYSICAL CHEMISTRY Total Marks : 28
DPP No. 21 Max. Time : 32 min.

Topic : Atomic Structure

Type of Questions M.M., Min.

Single choice Objective ('–1' negative marking) Q.1 to Q.4 (3 marks, 3 min.) [12, 12]
Subjective Questions ('–1' negative marking) Q.5 to Q.8 (4 marks, 5 min.) [16, 20]

1. In the following electronic configuration, some rules have been violated :

I : Hund II : Pauli's exclusion III : Aufbau

(A) I and II (B) I and III (C) II and III (D) I, II and III

2. What is the potential difference through which an electron, with a de Broglie wavelength of 1.5 Å should be
accelerated, if its de Broglie wavelength has to be reduced to 1 Å :
(A) 110 volts (B) 70 volts (C) 83 volts (D) 55 volts

3. X2+ is isoelectronic with sulphur and has (Z + 2) neutrons (Z is atomic no. of element X). Hence, mass
number of X2+ is :
(A) 34 (B) 36 (C) 38 (D) 40

4. Which of the following compounds is isoelectronic with [NH3  BH3] :

(A) B2H6 (B) C2H6 (C) C2H4 (D) C3H6

5. A neutral atom of an element has 2K, 8L, 9M and 2N electrons. Find out the following :
(a) Atomic number of element (b) Total number of s electrons
(c) Total number of p electrons (d) Total number of d electrons
(e) Number of unpaired electrons in element

6. Calculate :
(a) the value of spin only magnetic moment of Co3+ ion (in BM).
(b) the number of radial nodes in a 3p-orbital.
(c) the number of electrons with (m = 0) in Mn2+ ion.
(d) the orbital angular momentum for the unpaired electron in V4+.

7. An element undergoes a reaction as shown :

X + e–  X – Energy released = 30.876 eV
The energy released, is used to dissociate 8 g of H2 molecules equally into H+ and H*, where H* is in an
excited state, in which the electron travels a path length equal to four times its debroglie wavelength.
(a) Determine the least amount (moles) of ‘X’ that would be required.
Given: I.E. of H = 13.6 eV/atom
Bond energy of H2 = 4.526 eV/molecule.
(b) Why is the amount of X calculated in the above question ‘least’?

8. A compound of Vanadium has a spin magentic moment 1.73 BM. Work out the electronic configuration of
the Vanadium ion in the compound.

DPPS FILE # 27