Quantum Numbers, Molecular Structure and Intermolecular Forces

This assignment delves into various aspects of chemistry, starting with hypothetical quantum numbers in an alternate universe and their implications on electronic configurations and transition metals. It explores wave-particle duality by calculating the wavelength of a macroscopic object. Further, the assignment covers calculations related to the wavelength and energy of photons, followed by determining electron configurations, unpaired electrons, and diamagnetism for various ions and elements. Key concepts like Hund's rule, Pauli Exclusion Principle, ground state, excited state, and quantum numbers are explained. The assignment also addresses trends in electron affinity and ionization energy based on electronic configurations. Lewis structures, VSEPR theory, bond angles, and hybridization are used to predict molecular shapes. Finally, it covers the arrangement of bonds in order of increasing ionic character, lattice energy calculations, and the analysis of intermolecular forces, hydrogen bonding, and dipole moments in various compounds. This resource is available on Desklib, where students can find a wealth of solved assignments and study materials.

Solution
Assignment 2
1. Assume that you travel through space to a strange universe where the allowable values for the
quantum numbers are quite different from those in our own universe. In this new universe, s
orbitals may contain a total of 2 electrons, p orbitals may contain a total of 4 electrons, and d
orbitals may contain a total of 6 electrons (as opposed to our 2, 6, and 10, respectively). The
other rules of electron configuration in this new universe are the same as in our own universe.
Answer the following about elements in this new universe.
a. How many orbitals would there be in each of the s, p, and d levels?
Solution
Each orbital has 2 electrons in it
Therefore,
s level has 1 orbital because it has 2 electrons
p level has 2 orbitals because it has 4 electrons
d level has 3 orbitals because it has 6 electrons
b. Write the rule for the allowable values of ml.
ml = magnetic quantum number
the orbital ml value
s 0
p-1, 0
d-1, 0, +1
c. What is the ground state electronic configuration of element number 10 in this new
universe?
1s22s22p43s2
d. What would be the atomic number of the first transition metal in this new universe?
You can assume the same order of energy levels as in our universe.
First transition metal is Scandium (Sc) whose atomic number is 21
Original electronic configuration of Sc = 1s22s22p63s23p64s23d1
Electronic configuration of Sc in the new universe = 17
Hence, atomic number of the first transition metal in the new universe = 17

2. a. Calculate the wavelength of a 5000 kg elephant travelling at 1.50 m s –1 .
The wavelength formulae =
Planck’s constant, h = 6.63 *10-34 Js
M= 5000 kg
V = 1.50 m/s
By substituting into the formula
= 8.84 *10-38 m
b. Comment on the wave nature of this elephant.
The nature of the wave is that wavelength is smaller
3. a. What is the wavelength (in nanometres) of blue light which has a frequency of 6.62
014 s–1 ?
For light wave we use c = frequency * wavelength
Where c = the speed of light
014 s–1
4.53172 * 10-7 m
b. Calculate the energy, in joules, of a photon with this frequency.
E = hf
= 6.63 *10-34 014
= 4.389 *10-19 J
4. For each of the following species: Cu2+, Mo4+, Ce, Te2–, Sn, V5+ , Sc2+
a. Give the electron configuration.
Cu2+ = 1s22s22p63s23p63d9
Mo4+ = 1s22s22p63s23p64s23d104p64d2
Ce = 1s22s22p63s23p64s23d104p65s24d105p64p15d16s2
Te2- = 1s22s22p63s23p64s23d104p65s24d105p4
Sn = 1s22s22p63s23p64s23d104p65s24d105s25p2
V5+ = 1s22s22p63s23p6
Sc2+ = 1s22s22p63s23p63d1

b. Give the number of unpaired electrons.
Cu2+ = 1 unpaired electron
Mo4+ = 2 unpaired electrons
Ce = 2 unpaired electrons
Te2- = 2 unpaired electrons
Sn = 2 unpaired electrons
V5+ = 0 unpaired electron
Sc2+ = 1 unpaired electron
c. Indicate which would be diamagnetic.
Ce
V5+
5. In your own words, give two- or three-line explanations of the differences between the
following pairs. Do not just quote the textbook definition of each. Explain the
difference. Give appropriate examples for each.
a. Orbital and subshell
Orbital contains up to two electrons and has a particular shape, for example 2pz is the z- aligned
orbital in the p subshell in the second shell, while subshell is a group of orbitals with
particular properties like shape and angular momentum, labelled s, p, d, f, g etc
b. Hund’s rule and the Pauli Exclusion Principle
Pauli Exclusion Principle: no two electrons in the same atom can have exactly the same energy,
therefore no two electrons in the same atom can have identical set of quantum number,
while
Hund’s rule: electrons occupy the maximum number of orbital possible; on pairing cannot be
done until all the orbital are single filled
c. Ground state and excited state
ground state is an atom in which the total energy of the electrons cannot be lowered by
transferring one or more electrons to different orbital, while excited state is an atom in
which the total energy of the electrons can be lowered by transferring one or more
electrons to different orbitals
d. ml and ms
m l distinguishes the orbital available within a subshell
m s denotes the size of the orbital

6. Each electron in an atom may be described by a set of four quantum numbers. For each of
the following, write down each different set of quantum numbers possible such that each
set contains the given value. (The answer for part (a) has been done for you below.)
For part (a), all the possible sets of quantum numbers are:
n = 3, l = 1, ml = +1, ms = +
n = 3, l = 1, ml = +1, ms = –
n = 3, l = 1, ml = 0, ms = +
n = 3, l = 1, ml = 0, ms = –
n = 3, l = 1, ml = –1, ms = +
n = 3, l = 1, ml = –1, ms = –
I value sublevel
0 s
1 p
2 d
ml value
ml value for s – sublevel is 0. (s sublevel has 1 orbital)
ml value for d- sublevel are -2, -1, 0, +1, +2(d sublevel has 5 orbitals)
a. n = 3, l = 1
n = 3, I = 1, ml = -1, 0, +1, s = +1/2, -1/2
Therefore, possible sets are
n = 3, I = 1, ml = -1, s= +1/2 or -1/2
n = 3, I = 1, ml = 0, s= +1/2 or -1/2
n = 3, I = 1, ml = +1, s= +1/2 or -1/2
b. n = 3, l = 2
n = 3, I = 2, ml = -2, -1, 0, +1, +2 s = +1/2, -1/2
Therefore, possible sets are
n = 3, I = 2, ml = -2, s= +1/2 or -1/2
n = 3, I = 2, ml = -1, s= +1/2 or -1/2
n = 3, I = 2, ml = 0, s= +1/2 or -1/2
n = 3, I = 2, ml = +1, s= +1/2 or -1/2
n = 3, I = 2, ml = +2, s= +1/2 or -1/2
1
2
1
2
1
2
1
2
1
2
1
2

c. n = 4, l = 2
n = 4, I = 2, ml = -2, -1, 0, +1, +2 s = +1/2, -1/2
Therefore, possible sets are
n = 4, I = 2, ml = -2, s= +1/2 or -1/2
n = 4, I = 2, ml = -1, s= +1/2 or -1/2
n = 4, I = 2, ml = 0, s= +1/2 or -1/2
n = 4, I = 2, ml = +1, s= +1/2 or -1/2
n = 4, I = 2, ml = +2, s= +1/2 or -1/2
d. n = 3, l = 2, ml = –2
n = 4, I = 2, ml = -2, s = +1/2, -1/2
e. n = 3, l = 0, ml = 0
n = 3, I = 0, ml = 0, s = +1/2, -1/2

7. Explain the following on the basis of electronic configuration.
a). The electron affinity of S is -200 kJ.mol–1 and of Cl is -348 kJ.mol–1.
S : Atomic number : 16 :
Electronic configuration : [ Ne] 3s2 3p4
Cl : Atomic number : 17
Electronic configuration : [ Ne] 3s2 3p5
There is only one electron necessary for Cl to complete its octet and form nearest noble gas
configuration which is stable. But in case of S there are 2 electrons required to complete octet so
Cl gives more enegy than S.
b). The first ionization energy of Li is 520 kJ.mol–1 and of Cs is 376 kJ.mol–1.
Li : Atomic number = 3
Electronic configuration : 1s2 2s1
Cs : Atomic number of Cs = 55
Electronic configuration = [Xe] 6s1
Both have outer electron in s orbital but 6s orbital is far away from nucleus than 2s. So there is
relative difference in electrostatic force of attraction between positively charged nucleus and S
electron. So Cs needs low energy to remove its electron from its outer orbital than Li.
So Ionization energy of Li is 520 kJ /mol and that of Cs is 376 kJ /mol
c. The electron affinity of Si is -120 kJ.mol–1 and of P is -74 kJ.mol–1.
Atomic number of Si = 14 , Electronic configuration : [Ne ] 3s2 3p2
Atomic number of P = 15 : Electron configuration : [Ne ] 3s2 3p3
If we look at the p orbital of P which is half filled and we know half filled orbital gets extra
stability than unsymmetrical filling. Carbon has only two electrons in its 2p orbital so it give
lower energy when we add electrons than P
So electron affinity of carbon is – 120 kJ /mol and that of P is 74 kJ /mo

.
8. a. Arrange the following elements in order of increasing atomic radius:
Cl, Tl, Ga, Br
Cl > Ti > Ga > Br
b. Arrange the following in order of decreasing radius:
Te2–, Xe, Cs +, Ba2+, I–
Decreasing radius: Te2- > I- > Xe > Cs+ > Ba2+
c. Arrange the following in order of increasing first ionization energy:
Br, Ge, As, Kr, Se
Increasing 1st ionization energy Kr <Br < Se < As < Ge
d. Arrange the oxides of the following elements in increasing order of acidity:
Al, P, S, Mg, Cl
Increasing order of acidity Cl < S < P < Al < Mg

Assignment 3
1. Draw Lewis structures for each of the following species.
a. Ca3(PO4)2
b. N2O4 (all oxygen atoms are terminal)
c. C2H4O

⏟ ⏞ ⏞
d. ClO2– (Cl is the central atom)
e. SCl6
2. Use the VSEPR method to predict the molecular shape of the following species. Show all
your reasoning.
a. CO32 –

b. CHCl3
c. IO2F2–
d. IF4–

e. OCS
f. PCl3
g. BrF3

3. a. Arrange the following bonds in order of increasing percent ionic character:
Li—F, K—O, N—N, S—O, and Cl—F.
N—N < S—O < Cl—F < K—O < Li—F
b. Calculate the lattice energy for solid CaBr2 from the following information:
• the heat of sublimation of Ca is 121 kJ/mol
• the first ionization energy for Ca is 590 kJ/mol
• the second ionization energy for Ca is 1145 kJ/mol
• the heat of vaporization for liquid Br2 is 315 kJ/mol (the standard state of
bromine is as a liquid)
• the bond dissociation energy for gaseous Br2 is 193 kJ/mol
• the electron affinity for gaseous bromine atoms is 2(–324 kJ/mol) = -648
kJ/mol
• the overall heat of formation of solid CaBr2 is –675 kJ/mol.
Summing up all the energies gives the overall energy change for the formation of
CaBr2
= (121 + 590 + 1145 + 315 + 193 + (-68) + (-LE) = -675 kJ/mol
1716 – LE = -675 kJ/mol
LE = 2391 kJ/mol
4. Use the VSEPR or the hybridization method to predict the bond angles in the following
molecules. Show your reasoning.
a. SO2