Chemistry I
Midterm Exam
18:00-21:00, 18 April, 2014 (Total Score: 100 points)
1.
A simple harmonic oscillator comprises a particle (mass = m) that experiences a restoration force
F = –k∙x (x: the position of the particle, k: the force constant). According to classical physics,
the potential energy of the particle is expressed V(x) = (1/2)k∙x2 . With the above information,
please show
(a) the equation of motion according to Newton’s 2nd law;
(4%)
(b) the Schrödinger’s wave equation according to quantum theory.
(5%)
2.
“O” (true) or “X” (false) You gain 2 points for a correct answer, but lose 1 additional
point for a wrong answer. 每小題兩分，答錯倒扣 1 分，未作答不給分不倒扣 (16% total)
(a) If a blue and a red laser pointers emit light of same power, the blue one ejects more photons
(b)
(c)
(d)
(e)
than the red one does in a unit time.
The wavelength of an electron is smaller than that of a proton if they travel at the same speed.
The spectrum of a blackbody radiation shifts toward the left (in the scale of frequency) as
the temperature increases.
If one electron is sent at a time to a double-slit, there will be no diffraction pattern no matter
how many electrons are sent to the double slit.
In an experiment that illustrates the photoelectric effect, the kinetic energy of emitted
electrons increases with the intensity of the incident light.
1
(f) If a hydrogen atom at an excited state emits a photon at the frequency of , the energy of that
excited state is E = h
The following two questions refer to a particle (mass = m) in a one-dimension box (width = L).
(g) If the particle is in the 2nd excited state, its kinetic energy is h2/(2mL2) .
(h) A wavefunction of a higher energy has more nodes than that of a lower energy.
3.
“O” (true) or “X” (false)
(a)
(b)
(c)
(d)
(1% each, 20% total)
答錯不倒扣
2
An atom with the electronic configuration [Kr]5s 4d10 has 58 electrons.
An ion Fe3+ has 23 electrons.
Gallium (Ga) is a transition metal atom.
Fe2+ and Ni3+ are isoelectronic.
(e) Curium (Cm) and thorium (Th) belong to lanthanides.
(f) Scandium (Sc) has 1 valence electrons.
(g) Effective nuclear charge Zeff can be larger than the actual nuclear charge Z.
(h)
(i)
(j)
(k)
(l)
(m)
(n)
Atomic radius of sodium (Na) is larger than that of potassium (K).
Atomic radius of Li is larger than the ionic radius of Li+.
The first ionization of the Cl atom is the largest among elements.
Electron affinity of oxygen (O) is larger than that of fluorine (F).
Cobalt (Co) atom and nickel (Ni) atom are both paramagnetic.
S+ has more unpaired electrons than S−.
The element with 80 electrons is diamagnetic.
(o) The total number of p electrons in tellurium is 20.
(p) Three elements are liquid on hot Taiwanese day: mercury (Hg), bromine (Br), and indium
(In).
(q) Noble gases do not form any chemical compounds.
(r) Se2− is larger than Te2−.
(s) Zn is not considered to be a transition metal.
(t) The 3s, 3p, and 3d orbitals in helium (He) have the same energy.
4.
Two atoms have the electron configurations 1s22s22p6 and 1s22s22p63s1. The first ionization
energy of one of them is 2080 kJ/mol, and that of the other is 496 kJ/mol. Give the name or
symbol of each element and match the ionization energies with the given electron configurations.
(5%)
5.
For carbon monoxide molecule:
(a) Give the Lewis structure and show formal charges.
(3%)
(b) Sketch the orbital overlap in the VB description. In this description, where do the lone pairs
reside?
(3%)
2
(c) Draw the molecular energy diagram and predict its electron configuration and bond order.
(4%)
6.
With the help of the molecular orbital energy level diagram, compare the relative stability of the
following species and indicate their magnetic properties: O2, O2+, O2–, and O22–.
(10%)
7.
MO description for the H2 molecule:
1
[𝜙 (1) + 𝜙𝐵,1𝑠 (1)][𝜙𝐴,1𝑠 (2) + 𝜙𝐵,1𝑠 (2)]
2 𝐴,1𝑠
where 1 and 2 denote the two electrons; A and B denote the two hydrogen atoms. 𝜙 is the wave
function of an atomic orbital.
𝜓MO ≈ 𝜎1𝑠 (1)𝜎1𝑠 (2) =
VB theory treatment:
𝜓VB = N[𝜙𝐴,1𝑠 (1)𝜙𝐵,1𝑠 (2) + 𝜙𝐵,1𝑠 (1)𝜙𝐴,1𝑠 (2)]
Under what condition do they become identical?
(5%)
8.
Write the Lewis structures, VSEPR formula (ABnEm), determine the molecular geometries (in
English) and identify the hybrid orbitals used by the phosphorus atom in each of the following
species: (a) PCl4+ ; (b) PCl6– ; (c) PCl5 ; (d) PCl3 .
(8%)
9.
There are three different dichlorobenzenes, C6H4Cl2, which differ in the relative positions of the
chlorine atoms.
(a) Write down the Lewis structures of the isomers;
(b) Which of the three forms are polar?
(c) Which has the largest dipole moment?
(3%)
(1%)
(1%)
10. The conventional unit of electric dipole moment is Debye (D), which is defined so that a single
negative charge separated from a single positive charge by 100 pm has a dipole moment of 4.8 D.
If the dipole moment of HCl is 1.1 D, please estimate the percentage of the partial charge in HCl
providing the equilibrium distance of the H-Cl is 127.46 pm.
(4%)
11. Butadiene (CH2CHCHCH2) is the simplest conjugated organic molecule. There are four
delocalized molecular orbitals (MOs) which are contributed from the four atomic p orbitals (AOs)
of each carbon atom forming the four  orbitals using the LCAO-MO model. Draw these four
MOs (1 –4) in the appropriate energy order and indicate the corresponding bonding character
(bonding, antibonding or nonbonding) and the number of the nodal planes for each MO. Using
building-up principle to fill the four electrons in the four  orbitals and indicate the electron
configuration for the ground state of butadiene.
(8%)
3
102B Chemistry (II) midterm exam
Answer
1.
(a)
−𝑘 ∙ 𝑥(𝑡) = 𝑚 ∙
𝑑 2 𝑥(𝑡)
𝑑𝑡 2
(b)
ℏ2 𝑑2 𝜓(𝑥) 1 2
−
+ 𝑘𝑥 ∙ 𝜓(𝑥) = 𝐸 ∙ 𝜓(𝑥)
2𝑚 𝑑𝑥 2
2
2.
3.
每題答對得 2 分，未作答不給分，答錯倒扣 1 分，倒扣至本大題得分 0 分為止
a
b
c
d
e
f
g
h
X
X
X
X
X
X
X
O
每題 1 分，答錯不倒扣
a
b
c
d
e
f
g
h
i
j
X
O
X
X
X
X
X
X
O
X
k
l
m
n
o
p
q
r
s
t
X
O
O
O
X
X
X
X
O
X
解題，毋須回答
(a)
(FALSE), Cd has 48 electrons.
(b)
(TRUE), Fe has 26 electrons so Fe3+ has 23 electrons.
(c)
(FALSE) Ga is a p-block atom.
(d)
(FALSE) Fe2+ has 24 electrons and Ni3+ has 25 electrons.
(e)
(FALSE) They are actinides.
(f)
(FALSE) Sc has 3 valence electrons.
(g)
(FALSE) We have Zeff = Z – , where 0 ≤  ≤ Z.
(h)
(FALSE) Both elements are in the same group (1A) but K is lower.
(i)
(TRUE) Li+ is much smaller than Li since it has no electrons in the 2s orbitals.
(j)
(FALSE) The first ionization of many elements (Ne, Ar, He, F) is larger.
(k)
(FALSE) Cl, F, and Br have the largest electron affinity of all elements.
(l)
(TRUE) Co has 7 d electrons and Ni has 8 d electrons which only partially fill the d shell.
(m) (TRUE) S+ has configuration [Ne]3s2p3 with 3 unpaired electrons. S− has configuration [Ne]3s2p5 with only 1
unpaired electron.
(n)
(TRUE) Hg has configuration [Xe]s2d10 with no unpaired electrons.
(o)
(FALSE) Te has the subshells 2p, 3p and 4p fully filled with 3 x 6 electrons plus 4 electrons in the 5p subshell.
Together 22 electrons, not 20.
4
(p)
(FALSE) Indium melting temperature is 156°C. The third element that is liquid in Taiwan summer is gallium
with the melting temperature of 30°C.
(q)
(FALSE) Many examples of chemical compound of Xe and Kr have been observed experimentally. The light
noble gas elements indeed may not form stable chemical compounds.
(r)
(FALSE) Se2− has fully filled the 4p subshell while the larger Te2− has also the 5p subshell filled.
(s)
(TRUE) Zn has the d subshell completely filled behaving more like Ca rather than a transition metal.
(t)
(FALSE) Orbitals with the same quantum number n but different quantum number l in many-electron atoms
have in general different orbital energy.
4.
configuration 1s22s22p6 → neon
or Ne
configuration 1s22s22p63s1 → sodium
(1%)
or Na
(1%)
ionization energy of 2080 kJ/mol → neon and ionization energy of 496 kJ/mol → sodium
(3%)
5.
(a) :C:::O:
or :C≡O:
or
formal charges C [4-(2 +6/2) =-1]; O [6-(2+6/2)=+1]
(3%)
(b) The bonding orbitals, depicted below, are due to the overlap of the p-orbitals (or sp hybrid
orbitals).
The lone pairs of electrons remain in the 2s orbitals (or sp hybrid orbitals) on
each atoms (not shown in the illustration below).
(圖 2%、文字 1%)
(c) 課本上出現多種座標方向選擇與標示法，故本題答案只要 ,  orbital、bondingantibonding 和電子數目寫對即可
(2s)2 (2s*)2 (2px)2 (2py)2 (2pz)2
(2s)2 (2s*)2 (2p)4
(2%)
bond order = (8-2)/2 = 3
(2%)
or (2s)2 (2s*)2 (x)2 (y)2 (2p)2
or
5
, 順序畫反扣 1 分
(2%)
6.
O2 has a bond order of 2 and is paramagnetic (two unpaired electrons).
(2%)
O2+ has a bond order of 2.5 and is paramagnetic (one unpaired electrons).
(2%)
O2– has a bond order of 1.5 and is paramagnetic (one unpaired electrons).
(2%)
O22– has a bond order of 1 and is diamagnetic.
(2%)
Based on molecular orbital theory, the stability of these molecules increases as follows:
O22– < O2– < O2 < O2+
(2%)
7.
 MO   1s (1)   1s (2)

1
2

A,1s
(1)  B ,1s (1)
1
2

A,1s
(2)  B ,1s (2)
1
 A,1s (1) B,1s (2)  B,1s (1) A,1s (2)   A,1s (1) A,1s (2)  B,1s (1) B,1s (2)
2
  VB  H  H   H  H 



  VB   IONIC
The wavefunction determined from molecular orbital (MO) theory contains the same terms as the
wavefunction determined from valence bond (VB) theory. The wavefunction determined from
MO theory also contains terms corresponding to both electrons being on the same atom. If the
likelihood of the two electrons being on the same atom is negligible (such as for a homonuclear
diatomic molecule), then the wavefunctions from the two theories are identical.
6
8.
各 2 分，錯一項扣一分 扣完為止
(a) VSEPR formula: AB4
;
molecular shape: tetrahedral
;
hybridization: sp3
(b) VSEPR formula: AB6
;
molecular shape: octahedral
;
hybridization: sp3d2
(c) VSEPR formula: AB5
;
molecular shape: trigonal bipyramidal
;
hybridization: sp3d
molecular shape: trigonal pyramidal
;
hybridization: sp3
(d) VSEPR formula: AB3E ;
9.
Cl
Cl
Cl
Cl
Cl
(a)
Cl
結構中苯環畫
亦可
(b) The first two isomers are polar.
(c) The first one has largest dipole moment.
10. μ = e ∙ R = δq × 127.46 = 1.1 D ; 100q × 100 = 4.8D
⇒ δ = 1.1⁄1.2746⁄4.8 = 0.18 ; partial charge ~18%
7
11. 共 8 分，錯一項扣一分 扣完為止
Electron configuration of butadiene in the ground state: (1)2(2)2
8