Download Key: Sample Test 2

CHEMISTRY 32: Section 1: MTWR 11:00
Name:
____________________________
TA:
____________________________
SS 2001 - Test 2
Student Number:
__ __ __ __ __ __
Lab Section:
A
B
Show your work, include the proper units, and write the answer to a reasonable number of significant
digits. Calculators may be used. No books. No notes (except for the yellow note card). No talking.
Do your own work. Quickly read all of the problems before starting the test; do the easy questions
first!
(10)
1.
Please provide brief and concise responses for each of the following:
(a)
Describe the formation of σ-bonds.
A σ-bond is formed when the orbitals from the two atoms overlap on a line
directly between their nuclei.
(b)
Describe the formation of localized π-bonds.
A π-bond is formed by the two atom’s unhybridized p-orbitals overlap above and
below the line containing the atoms’ σ bond.
(c)
Describe the formation of delocalized π bonds.
A delocalized π bond is formed when unhybridized p-orbitals that are not shared
by two adjacent atoms overlap throughout the molecule thus allowing the “extra”
electrons to be “shared” by multiple atoms.
(d)
Is the π-bond in NO2⎯ localized or delocalized? Briefly defend your answer.
5 + 2(6) + 1 = 18 valence electrons
_
O
N
–
O
O
N
O
Since NO2– exhibits “resonance structures,” it has delocalized a π-bond.
Page 1 of 6
CHEMISTRY 32: Section 1: MTWR 11:00
Name:
(08)
2.
____________________________
↑↓
s
(b)
↑↓
s
3.
Student Number:
__ __ __ __ __ __
Using orbital notation diagrams (boxes and arrows), show how:
(a)
(05)
SS 2001 - Test 2
xenon forms sp3d2 hybrid orbitals
↑↓ ↑↓ ↑↓
p
↑_ ↑_ ↑_ ↑_ ↑_ ↑_
sp3d2
__ __ __ __ __ forms
d
↑_ ↑_ __
d
boron forms sp2 hybrid orbitals
↑_ __ __
p
forms
↑_ ↑_ ↑_
sp2
__
p
Briefly explain the following electron affinity order (least exothermic to more
exothermic):
Na < O < Cl
Going from left to right, the elements have additional valence electrons and are
closer to having an octet. As one get closer to an octet, the addition of an
electron releases more energy.
(08)
4.
A cation with a 3+ charge has the following electron configuration: [Kr] 5s24d10
a.
Identify the element.
Sb3+
b.
Indicate the quantum numbers for this ion’s outermost electron.
4, 2, 2, –1/2
c.
Draw the orbital notation (short form) for a neutral atom for this element.
[Kr] ↑↓
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
↑_ ↑_ ↑_
Page 2 of 6
CHEMISTRY 32: Section 1: MTWR 11:00
Name:
(10)
5.
____________________________
SS 2001 - Test 2
Student Number:
__ __ __ __ __ __
No visible lines in the hydrogen spectrum correspond to an electron moving from the
fifth shell (n=5) to the third shell (n=3). Predict whether the lines would lie in the
ultraviolet or the infrared region and explain your logic. Calculate the wavelength of light
that is emitted. [Note: Visible light ranges from about 400 (violet) to 730 (red) nm.]
c = 3.00 x 108 m/s
RH = 2.18 x 10–18 J
ni = 5
c = λν
h = 6.63 x 10-34 J-s
λ = c/ν
1 nm = 10–9 m
nf = 3
R H #% 1
1&
2.18 x 10"18 J # 1
1&
14
(
! =
=
Hz
2 "
2
"34
2 "
2 ' = " 2.34 x10
$
h $ ni
nf '
6.63 x 10
J•s 5
3
The “–“ indicates that the light was be emitted.
c
3.00 x10 8 m / s
1 nm
! =
=
= 1.28 x 10#6 m x
= 1283 nm
14
"
2.34 x 10 Hz
10#9 m
Since the wavelength, 1283 nm, is greater than 730 nm; the emitted light is in the
infrared region of the electromagnetic spectrum.
(06)
6.
Classify each of the following sets of quantum numbers as valid or invalid. For those
that are invalid, explain.
a.
n = 3, l = 3, ml = +1
Invalid: The 3rd energy level does not have a f-sublevel (l =3)
b.
n = 3, l = 1, ml = 0
Valid
c.
n = 3, l = 0, ml = −1
Invalid: The s-sublevel (l = 0) can only have ml = 0.
(05)
7.
Why do Be, N, and Mg have positive electron affinities?
Be and Mg have filled s-sublevel and N also has a half-filled p-sublevel in their
valence shell. The addition of an extra electron would produce a less stable
anion.
Page 3 of 6
CHEMISTRY 32: Section 1: MTWR 11:00
Name:
(06)
8.
SS 2001 - Test 2
____________________________
Student Number:
Draw the resonance structures for ozone, O3(g).
O
O
O
(10)
9.
__ __ __ __ __ __
Using bond energies, estimate the Δ H of for:
!Hrxn =
# (bond
O
O
O
CH4 (g) + 2 O2 (g) ---> CO2 (g) + 2 H2O (g)
enthalpies of bonds broken) "
# (bond
enthalpies of bonds formed)
On the reactant side, there are four C–H bonds and 2 O=O bonds – these are
broken.
On the product side, there are two C=O bonds and four H–O bonds – these are
formed.
!Hrxn =
[4 mol C – H x (414 kJ / mol) + 2 mol O = O x (498 kJ / mol)] "
[2 mol C = O x (803 kJ / mol) + 4 mol H – O x (464 kJ / mol)] =
= " 810 kJ
(13)
10.
Draw the Lewis structure of the molecule N2F2.
F
N
N
F
Use the Lewis structure to complete the following: The hybridization of nitrogen is
__ sp2 __ (a type). Therefore, the nitrogen has ___ 3 ___ (a number) hybridized
__ sp2 __ (type of) orbital(s) and __ 1 __ (a number) unhybridized __ p __ (type)
orbital(s).
Page 4 of 6
CHEMISTRY 32: Section 1: MTWR 11:00
Name:
(06)
11.
SS 2001 - Test 2
____________________________
Student Number:
__ __ __ __ __ __
List the following isoelectronic species K+, Ca2+, S2⎯, and Cl⎯ in order of increasing
diameter. Briefly, but concisely, explain your reasoning for this order.
Ca2+ < K+ < Cl– < S2–
The above order reflects the fact that as the number of protons in the ion’s
nucleus decreases for the same number of electrons (18), the ionic radii
increases due to the greater electron to proton ratio.
(09)
12.
For CCl4, NCl3, and OF2:
(a)
(b)
Name the molecular geometry associated with each compound.
OF2
bent
NCl3
trigonal pyramidal
CCl4,
tetrahedal
Briefly (but concisely) explain why the bond angles between the central
atom and two adjacent halogen atoms for each compound are
104.5o for OF2
Since unbonded electron pairs occupy more volume, O’s two unbonded
electron pairs compress the F–O–F bond angle.
107o for NCl3
N’s single unbonded electron pair compresses the Cl–N–Cl bond angles.
109.5o for CCl4,
Since C has no unbonded electron pairs, the molecule fits the tetrahedral
geometry with Cl–C–Cl bond angle is 109.5o.
Page 5 of 6
CHEMISTRY 32: Section 1: MTWR 11:00
Name:
(12)
13.
SS 2001 - Test 2
____________________________
Student Number:
__ __ __ __ __ __
Answer the following questions about the iodine in IF4⎯.
Draw the VSEPR (electron-domain) model diagram for IF4⎯.
(a)
_
F
F
I
F
F
(b)
Indicate the number of electron domains surrounding the iodine.
six
(c)
What type of hybridization does iodine exhibit?
sp3d2
(d)
What electron-domain geometry is typically associated with this type of
hybridization?
octahedral
(e)
Identify the molecular geometry for this ion.
square planar
(f)
Explain this apparent violation of the octet rule.
Iodine utilizes some of its d-orbitals to form its six hybrid orbitals
(05)
14.
Why is nitrogen’s first ionization energy greater than oxygen’s?
Nitrogen’s half-filled p-sublevel is more stable than is oxygen’s p-sublevel that
has an unbonded electron pair.
Page 6 of 6