CH107: Physical Chemistry
12 November, 2014
End Semester Exam 2 Hours, 30 Minutes
Answer all questions. Total 30 Marks. Answer each question in a separate page. Calculators may be used. Use PEN to
write all answers, including sketches. Read questions carefully and keep answers to-the-point. Provide arguments to
earn full credit.
h=6.626x10-34Js; c=3x108 ms-1;me=9.1x10-31kg;mp=1.672x10-27 kg; e=6x10-19C; 1eV=1.6x10-19 J
1A.
Draw the polar plot for the angular part of the wavefunction Y (θ ) = (3cos
the calculated data points on the polar plot provided.
2
θ −1) , using the table provided. Show
3
1B.
Why is α(1)α(2) an acceptable 2-electron spin function, while α(1)β(2) is not?
1C.
Explain in the light of MO theory whether the carbon or the oxygen end of CO is used for coordinate bond formation.
1
3 2
2A.
The probability density for the electron in a 2s orbital where
. Assume that the nuclear diameter for H is
ψ2s =
1
 
32π  ao 
1

r  − r 2ao
 2 −  e
, is maximum at
 a0 
2×10−15 m . Using this assumption, calculate the total probability of
finding the electron in the nucleus if it occupies the 2s orbital. {Hint: (nuclear diameter <<
2B.
1
)}
2
Qualitatively sketch the bonding and anti-bonding MOs of a homonuclear diatomic molecule formed by a linear
combination of two dX2-Y2orbitals, where the internuclear axis is along the x-axis. Define axes and put proper signs on
the lobes of individual atomic orbitals. Qualitatively plot the overlap integral (S(R)) as a function of the internuclear
distance R in the case of the bonding MO formed by a linear combination of two dX2-Y2orbitals (for homonuclear
diatomic) .
3
3A.
Explain the reasons why the most intense transition in vibrational spectroscopy of a heteronuclear diatomic molecules is
from v=0 to v=1state, whereas in rotational spectrum the transition from J=0 to J=1 is not the most intense?
2
3B.
The vibration frequency of 1H35Cl is 2990.6 cm-1; without calculating the bond force constant, estimate the vibrational
frequencies for 1H37Cl, 2D35Cl and 2D37Cl and explain the results obtained.
2
3C.
There are two kinds of molecular motion with energies in the range of 100kCal and 1kCal. Which of these motions is
faster?
1
4A.
What are the intermolecular forces that can be modeled using the Lennard Jones Potential?
1
4B.
For the Lennard Jones potential calculate for what value of ‘r’ (intermolecular distance) the force between the molecules
will be maximum? Does it coincide with the position of the minima observed for this potential? Explain graphically.
3
4C.
Which of the following molecules will show a microwave rotational spectrum? CH4, CH3Br, CH2Br2and SBr6
1
5.
For CH2Cl2 molecule (provided bond angle H-C-H 112.0° (see Figure) and Cl-C-Cl 111.78°)
obtain the normalized wavefunctions of all the hybrid orbitals (in terms of 2s and 2 p atomic
orbitals on the central carbon atom) for the given geometry.Hence evaluate the bond angle HC-Cl for the given molecule.
6.
H!
+x!
H!
+y!
5
!(Cl!atoms!in!!!!the!xz!plane)
Construct the labeled MO energy level diagram for BeH2 molecule considering s-p mixing in Be. Write the expressions
for the two bonding orbitals in terms of linear combination of the atomic orbitals. Sketch the two corresponding bonding
MO’s with showing proper signs and label them according to their symmetries.
5
Roll No:
θ
00
100
200
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
Y (θ )
Division:
Polar Graph
Tutorial batch:
Answer 1A. Polar Plot
Table correct
1 mark
2 or more values wrong in the Table
-0.5 mark
4 or more values wrong in the Table
0 mark
If left blank
0 marks
Plotting Graph Correctly
If both sides plotted (360 degree)
2 marks
-1 mark
If the negative part of the function plotted on opposite side (Mode value not considered) -1 mark
If no or less dots
-0.5 mark
1B) Ans(a)α(1)α(2)
is acceptable 2 electron spin function because the electrons are
indistinguishable and that aspect is maintained. However in the case of α(1)(2) we are assigning
specific spins to each electron and the indistinguishabilty criteria is not met.
Ans(b) A 2-electron spin function is acceptable if under exchange of electronic coordinate the
wavefunction remains either symmetric or antisymmetric. In that respect α(1)α(2) is acceptable.
Exchange operator E
E [α(1)α(2) ]  α(2)α(1)  α(1)α(2) same function back, symmetric
E [α(1) (2)]  α(2) (1)   α(1) (2) same function is not back, not acceptable, neither symmetric
or antisymmetric
1 mark for either answer. No part marking if concept is not correct
1C) The carbon is used for coordinate bond formation. Writing this without valid explanation ( All
valid explanations should involve MO theory) 0 marks
Caron end is used for coordinate bond formation because the electrons in the HOMO participate in
coordinate bond formation for example with Metals. This is because when a MO picture of CO is
drawn, the HOMO is centred on carbon atom that is the MO wavefunction has coefficients which
have higher contribution from the carbon atom relative to the oxygen atom. The CO HOMO is like a
non-bonding MO.
1 mark if logic is correct otherwise 0, if oxygen end mentioned. No coordinate bond is between
carbon and oxygen atom!! All those answers are also 0 marks
MODEL ANSWERS FOR QUESTION 2
2a. The probability density for the electron in a 2s orbital where
is maximum at
. Assume that the nuclear diameter for H is
. Using this
assumption, calculate the total probability of finding the electron in the nucleus if it occupies the
2s orbital. {Hint: (nuclear diameter <<
)}
Solution:
2a. Total probability of finding an electron within the nucleus when it is in 2s state (orbital) is
3
2π
rN
π
2
2
r  − r ao
d
Sin
d
r
2
−
φ
θ
θ

∫ 0∫
∫  a0  e dr
0
1  1 
P =Ψ
 
∫ dτ =
32π  ao  0
2
1 MARK
OR
 1
P = ∫ Ψ × 4π r dr ≈ Ψ × 4π r × ( rN ) = 
 32π
0
rN
2
2
2
2
N
3
2
 1  
r  − r ao 
 × 4π rN3
  2 −  e
a
a

0 
 o 
1 MARK
OR
2
 1  1 3 
r  − r ao  4 3
 × Vol Nuc 
 × π rN
P ≈  Ψ
=
  2 −  e
a0 
 32π  ao  
 3
2
1 MARK
Since rN << a0, r/a0 0 for r close to rN. So, approximate:
3
2
−r
1  1  
r 
1  1 
1
2
ao
=
Ψ
≈
  × 2 −  ×e
 3  × (2) × 1 ~
32π  ao  
a0 
32π  a0 
8π a03
2
r
r
N
N
1
1
1  rN3  1  rN3   rN 
2
2
×
×
×
×
×  ~  3 ~ 
2
2
=
=
π
P≈
r
dr
r
dr
3
3
8π a03
2
2
a
a
0
0
 3  6  a0   a0 
0
0
∫
∫
0.5 MARKS
3
0.5 MARKS
Answer : P ~ 10−14 − 10−16
Partial Marking Scheme for Q2A:
Incorrect limits (such as ∞ instead of rN or –rN to +rN, etc.) or constants incorrect (such as 4π
factor missing or Ψ2 expression partially correct at starting equation):
0.5 Marks
Ψ instead of Ψ2 in starting equation, or r2 omitted (in integral) of starting equation 0 Marks
Only one approximation  ( 2 − r a0 )2 → 4 OR e−r ao → 1  used instead of both:
0 Marks
Answer within 10-13 to 10-17 due to simple calculation error (method correct):
1.5 Marks


2b. Qualitatively sketch the bonding and anti-bonding MOs of a homonuclear diatomic molecule
formed by a linear combination of two dX2-Y2 orbitals, where the internuclear axis is along the xaxis. Define axes and put proper signs on the lobes of individual atomic orbitals. Qualitatively
plot the overlap integral (S(R)) as a function of the internuclear distance R in case of bonding MO
formed by a linear combination of two dX2-Y2 orbitals (for homonuclear diatomic) .
Solution 2b:
MO and AO shapes
correct:
0.5x2 = 1 MARK
AO and MO signs
Correct:
0.5x2 = 1 MARK
0 Marks for showing π-bond formation along with σ-bond (bonding MO) or δ-bond formation
0 Marks for π-bond formation due to wrong choice of AOs such as dxy/dyz/dxz
0 Marks for choosing incorrect internuclear axis or incorrect orientation of dx2-y2 AOs
Plot of S(R) against R for bonding situation
Qualitative shape correct, clearly showing minima (not flat line) followed by maxima: 1 MARK
If value of S is not maximum at R = 0, or S is negative at any R (cross x-axis):
0 MARKS
3A.
Explain the reasons why the most intense transition in vibrational spectroscopy of a heteronuclear diatomic molecules is
from v=0 to v=1state, whereas in rotational spectrum the transition from J=0 to J=1 is not the most intense?
2
The energy gap for vibrational levels is in hundreds to thousands of cm-1. According to Boltzmann distribution, only the
v=0 is populated in such a case and any upward transition must originate from this level. The selection rule is v = + 1.
In case of rotational levels, however, the higher levels are also populated, as the energy gap is 10 cm-1 at most. Besides,
the population of each level is weighted by a degeneracy factor of (2J +1). So, the most intense transition originates at J
> 0.
Marking Scheme: If all the three factors: Boltzmann distribution, selection rule and degeneracy are mentioned in an
appropriate context, then 2
Any two of the above: 1.5
Any one of the above: 1
Selection rule: 1
3B.
The vibration frequency of 1H35Cl is 2990.6 cm-1; without calculating the bond force constant (assume that it is same for
all the moleucles), estimate the vibrational frequencies for 1H37Cl, 2D35Cl and 2D37Cl and explain the results obtained.
if k is constant:
2
0.5
Values of m in atomic units:
1
H35Cl : 35/36; 1H37Cl: 37/38; 2D35Cl: 70/37 2D37Cl: 74/39
0.5
Vibrational frequencies:
1
H37Cl:
2
D35Cl:
2
D37Cl:
0.5. Award the mark even if a wrong
relationship has been used but the calculation
has been done correctly, according to the wrong
relationship.
Additional 0.5 if
is calculated correctly
Explanation:
For a diatomic molecule, the isotope effect is much more prominent for the lighter atom, as the fractional change in
mass is significantly more.
0.5
3C.
There are two kinds of molecular motion with energies in the range of 100 kCal and 1 kCal. Which of these motions is
faster?
The one with higher energy. Frequency is proportional to energy gap. Time involved is inversely proportional to
frequency.
(Correct identification without justification: full marks)
1
1
Marking scheme
4.A. Dipole-dipole interaction, dipole-induced dipole interaction and induced dipole-induced
dipole interaction (dispersion) can be modelled by Lenard Jones potential.
For writing any 2 correct examples
⟶1
For writing 1 correct example
⟶ 0.5
For writing 1 wrong example
⟶ -0.5
So for writing 1 correct, 1wrong example ⟶ 0
4.C. To show microwave rotational spectrum a molecule must have a net non-zero dipole
moment. Thus CH3Br and CH2Br2 will show rotational spectrum while CH4 and SBr6 will
not.
For writing 2 correct
⟶1
For writing 1 correct
⟶ 0.5
For writing 1 wrong
⟶ -0.5
So for writing 1 correct, 1wrong ⟶ 0
4.B.
Force
For not writing this expression ⟶ -0.5
For correct expression of F ⟶ 0.5
For wrong sign ⟶ -0.5
To find the ‘r’ for which force is greatest we need to calculate
⟶ 0.5
For only writing condition
For getting correct expression of
⟶1
For correct value of r ⟶ 0.5
This does not coincide with the minima for the potential which comes from
and
For writing this in statement or showing in graph ⟶ 0.5
For showing the graph for V(r) vs r ⟶ 0.5
No marks awarded or deducted for plotting F(r) vs r
Graph drawn by pencil is not accepted.
Even if someone does not get correct answer, for following correct steps, partial marks have
been awarded.
5. For CH2X2 molecule (provided bond angle H-C-H 112.0° (see
Figure) ) obtain the normalized wavefunctions of all the hybrid
orbitals (in terms of
and
atomic orbitals on the central
carbon atom) for the given geometry. Hence evaluate the bond
angle H-C-X for the given molecule.
Ans:
The two equivalent hybrids will be given as,
[
= N [Cosθ .ψ
φ Hsp13 = N Cosθ .ψ 2 p x + Sinθ .ψ 2 p y − α .ψ 2s
φ Hsp23
2 px
− Sinθ .ψ 2 p y − α .ψ 2s
]
] (1 Marks) (Grading scheme: Both hybrid correct = 1 Marks, for one incorrect = 0.5 Marks, for signs wrong €
= 0.5 Marks, no marks deducted for wrong sign in front of ψ2s , for both incorrect = 0 Marks)** These two hybrids must be orthogonal to each other, i.e., φ Hsp13 φ Hsp23 = 0 hence 2
Coming to Cos2θ = −α or α = 0.612 (by appropriate method = 0.5 Marks) ((Even if ** this is wrong totally but appropriate method followed to obtain the α value will still get 0.5 marks, Any other correct method will also get 0.5 €
marks but someone writing that in all the hybrids s will contribute equally will get 0 Marks for this part)) €
[
= N [0.559.ψ
φ Hsp13 = N 0.559.ψ 2 p x + 0.829.ψ 2 p y − 0.612.ψ 2s
or,
φ Hsp23
2 px
− 0.829.ψ 2 p y − 0.612.ψ 2s
]
] To get N, we have to use the normalization condition, €
i.e., φ Hsp13 φ Hsp13 = 1 and N=0.853 [
= [0.477.ψ
φ Hsp13 = 0.477.ψ 2 p x + 0.707.ψ 2 p y − 0.522.ψ 2s
]
] €
or, €
(Grading scheme: Both hybrid correct = 1 Marks, for one incorrect = 0.5 Marks, for only signs wrong = 0.5 Marks, for only coefficients wrong = 0.5 Marks, no marks deducted for wrong sign in front of ψ2s , for both sign and coefficients incorrect = 0 Marks, even if ** is wrong, correct method followed will be 0.5 marks, no appropriate method shown = 0 Marks) φ Hsp23
2 px
− 0.707.ψ 2 p y − 0.522.ψ 2s
(1 Marks) The other two hybrids will be given as, [
= [ −0.522.ψ
φ Xsp13 = −0.522.ψ 2 p x + 0.707.ψ 2 p z − 0.477.ψ 2s
φ Xsp23
€
2 px
− 0.707.ψ 2 p z − 0.477.ψ 2s
]
] (1 Marks) (Grading scheme: Both hybrid correct = 1 Marks, for one incorrect = 0.5 Marks, for only signs wrong = 0.5 Marks, for only coefficients wrong = 0.5 Marks, no marks deducted for wrong sign in front of ψ2s , for both sign and coefficients incorrect = 0 Marks, even if ** is wrong, correct method followed will be 0.5 marks, no appropriate method shown = 0 Marks) To find out the H-­C-­X bond angle, we have to consider the angle between φ Hsp13 and φ Xsp13 or any such combination of the un-­‐equivalent hybrids. (0.5 Marks) i.e., €
[
≈ [ −0.522.ψ
φ Hsp13 ≈ 0.477.ψ 2 p x + 0.707.ψ 2 p y
φ Xsp23
2 px
+ 0.707.ψ 2 p z
sp
€
]
3
If ω is the angle between φ H1 and
Cosω =
] With no s contribution (0.5 Marks) φ Xsp13 then, 0.477 × (−0.522)
2
(0.477)
+ (0.707) 2 . (−0.522) 2 + (0.707) 2
€
or, ω = 109.4 0
€
= −0.332
(0.5 Marks) (Grading scheme: unequal hybrids taken = 0.5 Marks, no s contribution considered = 0.5 Marks, then for angle value found within 910 to 1190 = 0.5 Marks. If equal hybrids taken = 0 Marks but no s contribution considered = 0.5 Marks, then for angle value found within 910 to 1190 = 0.5 Marks. If unequal hybrids taken = 0.5 Marks but with s contribution considered = 0 Marks, then for angle value found within 910 to 1190 = 0.5 Marks. If equal hybrids taken = 0 Marks, with s contribution considered = 0 Marks, then for angle value found within 910 to 1190 = 0 Marks. no appropriate method shown = 0 Marks)