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Chemistry(I)
Midterm Test
November 10, 2006
Total Score: 100 points
Time: 2 hours (18:00-20:00)
TAs will NOT answer questions regarding the content of the test.
No calculators allowed.
1. From the following list of observations,
i) emission spectrum of hydrogen
ii) the photoelectric effect
iii) scattering of alpha particles by metal foil
iv) diffraction
v) cathode “rays”
choose the one that most clearly supports the following conclusion:
(a) Electrons have wave properties. (2%) iv
(b) Electrons in atoms have quantized energies. (2%) i
(c) The mass of the atom is located mainly in the nucleus. (2%) iii
(d) Atoms contain electrons. (2%) v
2. As the electron in a hydrogen atom is moved from an excited 4d orbital to a 1s
orbital, the energy of the photon emitted is 2.04 x 10–18 J. Calculated the energy of the
photon released if the electron is moved from an excited 3d orbital to each of the
following orbitals: (a) 2s, (b) 2p, and (c) 3s. (8%).
En = –hR / n2
ΔE = Efinal – Einitial
–18
–2.04 x 10 J = –hR (1/1 – 1/16) hR = 2.18 x 10–18 J
(a) –hR (1/4 – 1/9) = –2.18 x 10–18 J x 0.139 = –3.02 x 10–19 J
The energy of the photon emitted is 3.02 x 10–19 J. (4%)
(b) The energy of the photon emitted is 3.02 x 10–19 J, because orbitals in the
hydrogen atom with the same principle quantum number n are degenerate (i.e. they
have the same energy). (2%)
(c) Zero. In a hydrogen atom, no photon would be emitted on moving between
orbitals possessing the same n. (2%)
3. For the electrons on a carbon atom in the ground state, decide which o the
following statements are true. If false, explain why. (8%)
(a) Zeff for an electron in a 1s-orbital is the same as Zeff for an electron in a 2s-orbital.
(b) Zeff for an electron in a 2s-orbital is the same as Zeff for an electron in a 2p-orbital.
(c) The electrons in the 2s-orbital have the same value for the quantum number ms.
(d) The electrons in the 2p-orbitals have spin quantum numbers ms of opposite sign.
(a) False. The 2s-electron will be shielded by the electrons in the 1s-orbital and will
thus experience a lower Zeff. (2%)
(b) False. The 2p-electrons experience a lower Zeff, because they do not penetrate to
the nucleus as the 2s-electrons do. (2%)
(c) Because the electrons are in the same orbital, they must have opposite spin
quantum number ms (The Pauli exclusion principle). (2%)
(d) The electron configuration for C is 1s22s22p2. The two 2p-electrons occupy
different p-orbitals with parallel spins (i.e. spin quantum numbers have the same sign,
Hund’s rule). (2%)
4. Discuss why a function of the type A cos (kx) is not an appropriate solution for the
particle in a one-dimensional box. (5%)
If we set x = 0 and use cos (0) = 1 in ψ = A cos (kx), we find ψ(0) = A ≠ 0. This
violates the boundary condition that ψ(x) must be zero at the edges of the box.
5. Which element has the higher electron affinity, carbon or nitrogen? Justify your
answer. (5%)
The electron affinity of carbon is greater than that of nitrogen; indeed, the latter is
negative. (2%)
We may expect more energy to be released when an electron enters the N atom,
because the effective nuclear charge for the outmost electrons of the neutral N atom is
greater. However, when C– forms from C, the additional electron occupies an empty
2p-orbital. On the other hand, when N– forms from N, the additional electron must
pair with an electron in a p-orbital; it experiences more repulsion and less effective
nuclear charge. Therefore, electron affinity of nitrogen is lower than that of carbon.
(3%)
6. Which of the compounds below has bonds with the most covalent character? (2%)
Please explain it. (3%)
(a) KCl
(b) NaBr
(c) Al2O3
(d) MgS
Answer:
Al2O3 has bonds with the most covalent character, because Al is a smallest and
highly charged cation.
Here, according to p.79 of our textbook, I list the difference of electronegativity
between the anion and cation for the selection of the above question as your
reference:
KCl
:
3.2-0.82 = 2.38
NaBr :
3.0-0.93= 2.07
Al2O3 :
3.4-1.6= 1.8
MgS
:
2.6-1.3=1.3
Here, according to p.41, I list the ionic radii of the anion and cation for the
selection of the above question as your reference:
K+
138pm,
Cl- 181pm
Na+
102pm,
Br- 196pm
Al3+
54pm,
O2- 140pm
Mg2+ 72pm,
S2- 184pm
7. If the following crystallize in the same type of structure, which has the highest
lattice energy? (2%) Please explain it. (3%)
(a) NaBr
(b) CaO
(c) Al2O3
(d) MgBr2
Answer:
Al2O3 has the highest lattice energy. In addition to Al2O3 has the highest charge
for both the cation and anion, it has the smallest ionic radius for both the cation
and the anion among these four ionic compounds.
8. The following Lewis structure was draw for a Period 3 element. Identify the
element. (2%) Please explain it by using the formula of the formal charge. (3%)
O
Cl
E O
Cl
Answer:
E = S (sulfur). Because this is a neutral molecule, the summation of the formal
charges of all atoms should be equal to zero. The formal charges of O and Cl are
equal to zero, therefore the formal charge of E is equal to zero (F=0). Since there
are no lone pair electrons ( L=0 ) but only 6 covalent bonds ( B = 12 bonding
electrons) linked to E, E should have 6 valence electrons (V=6) according to the
formula of the formal charge ( F = V – (L+ 0.5*B). Sulfur is the only element of
the Period 3, which has 6 valence electrons.
9. Answer the following questions.
甲、There are 8 possible resonance structures for SO3. Among these 8 possible
resonance structures for SO3, please pick up 3 structures and draw their
Lewis structures: the first SO3 structure must contain only one double bonds,
the second one must contain two double bonds, and the third one must
contain three double bonds. (9%)
乙、Please indicate whether they obey the octet rule or not. (3%)
丙、Please write the formal charge on each atom and identify the structure of the
lowest energy. (6%)
Answer:
O S O
O
O S O
O
or O S O
O
or O S O
O
O S O
O
or or O S O
O S O
O
O
It obeys the octet
rule.
-1
+2
It does not obey
the octet rule.
-1
0
+1
-1
It does not obey
the octet rule.
0
0
0
O S O
O S O
O S O
O
O
O
0
0
0
the lowest energy 10. (14 points total) Allene (C3H4) is a nonpolar molecule.
(a)
Please draw allene’s Lewis structure. (2 points)
H
H
|
|
C=C=C
|
|
H
H
(2 points)
(b)
Discuss hybridization of the carbon atoms of allene. (6 points)
H
H
|
|
Ca=Cb=Cc
|
|
H
H
Ca and Cc are sp2 hybridized. (4 points)
Cb is sp hybridized. (2 points)
(c)
Is allene a planar molecule? Explain your reason. (6 points)
No. (2 points)
The central carbon atom Cb has to use different p orbitals to bond to the outer
carbon atoms Ca and Cc. The p orbitals of Cb are perpendicular. (2 points)
The plane containing Ca and the H atoms bonded to Ca has to be perpendicular to
the plane containing Cc and the rest of the H atoms. (2 points)
See below for allene’s structure.
11.(12 points total) Determine the structure and polarity of the molecules in the
following table.
linear angular trigonal trigonal T‐ planar pyramidal shaped tetrahedral seesaw square trigonal square octa‐ Polar planar bipyramidal pyramidal
hedral
(Yes or No) PCl4F X Y BF3 X N H2O X Y SF4 X Y IF5 X Y ClF3 X Y (1 point for each correct answer, 12 points total)
12. (12 points total) H4 is a hypothetical square planar molecule. Its LCAO-MOs are
drawn below.
Determine the number of nodal planes for each of φ1 - φ4. (4 points)
(d)
φ1
φ2
φ3
φ4
0
1
1
2
(1 point each)
(e)
Which orbital are degenerate? (2 points)
φ2 and φ3 (2 points)
Draw a molecular orbital energy level diagram for H4. (4 points)
(2 points for energy levels, 2 points for putting electrons correctly)
(f)
_____ φ4
φ2 __↑__ __↑__ φ3
_↑↓_ φ1
(g)
Is H4 likely to be a diamagnetic or paramagnetic molecule? (2 points)
paramagnetic (2 points)
+
+
+
+
+
+
-
-
φ1
φ2
+
-
-
+
+
-
+
-
φ3
φ4