Chem 130 – Second Exam
Name_______________________________
On the following pages you will find seven questions covering varies topics ranging from the structure of molecules, ions, and solids to different models for explaining bonding. Read each question
carefully and consider how you will approach it before you put pen or pencil to paper. If you are
unsure how to answer a question, move to another; working on a new question may suggest an approach to the one that is more troublesome. If a question requires a written response, be sure that
you answer in complete sentences and that you directly and clearly address the question.
Partial credit is willingly given on all problems so be sure to answer all questions!
Question 1 _____/32
Question 5 _____/10
Question 2 _____/10
Question 6 _____/10
Question 3 _____/10
Question 7 _____/18
Question 4 _____/10
Total _____/100
Potentially useful equations and constants:
c = λν
E = hν = hc/λ
⎛ 1
1
1
= 1.09737 × 10 − 2 nm⎜⎜ 2 − 2
λ
⎝ n1 n2
KE = hν – W
FC a = Va − N a −
Ba
2
𝑉∝
⎞
⎟⎟
⎠
𝑄! 𝑄–
𝑑
OX a = Va − N a − Ba × (0 if least EN; 1 if most EN)
⎛
EN a
δ a = Va − N a − B a ⎜⎜
⎝ EN a + EN b
h = 6.626×10–34 Js
⎞
⎟⎟
⎠
c = 2.998×108 m/s
NA = 6.022×1023 mol–1
Also available on separate handouts are a table of electronegativity values based on average valance
electron energies, a table of packing possibilities based on relative sizes of cations and anions, and a
periodic table.
Problem 1. For each of the following molecules or ions, draw any one valid Lewis structure of your
choosing (it need not be the “best” structure). Annotate your structure by indicating the formal
charge on each atom. Finally, give the name for the bonding geometry around the underlined central
atom, predict whether the molecule or ion is polar (P) or non-polar (NP), and provide the idealized
bond angle for the stated bonds.
Molecule or
Ion
CCl2F2
Lewis Structure
There are 4 + 2(7) + 2(7) = 32
electrons in the molecule and four
substituents around the central
atom, requiring four electron domains. The only possible LS has
single bonds from C to Cl and
from C to F. All formal charges
are zero.
IO2F2–
There are 7 + 2(6) + 2(7) + 1 = 34
electrons and four substituents
around the central atom, requiring
five electron domains with the lone
pair and two oxygens (larger than the
fluorines) in axial positions. Using
single bonds to O and F gives formal
charges of zero for F and –1 for O
and +1 for I.
ICl4–
There are 7 + 4(7) + 1 = 36 electrons
and four substituents around the central atom, requiring six electron domains. The only possible LS has single bonds from I to Cl. All formal
charges on Cl are zero, leaving a
formal charge of –1 on I.
XeOF4
There are 8 + 6 + 4(7) = 42 electrons
and five substituents around the central atom, requiring six electron domains. With a double bond between
Xe and O, and single bonds between
Xe and F, all formal charges are zero.
Bonding
Geometry
Polar or
Non-Polar?
Ideal Bond
Angle for …
…an F–C–Cl
bond is….
tetrahedral
polar
109.5°
…an O–I–O
bond is….
see-saw
polar
120°
…a Cl–I–Cl
bond is….
square
planar
non-polar
90°
…an F-Xe-F
bond is…
square
pyramidal
polar
90° or 180°
Problem 2. The element Z forms the molecular compound ZBr3 with a trigonal pyramidal electron
domain geometry. Identify one example of an element that could be Z. In no more than three sentences, clearly explain your reason for picking this element.
A trigonal pyramidal geometry means that there are three bonds between Z and Br, and a lone-pair
of electrons on Z. The structure, therefore, has a total of 26 electrons. Each bromine provides seven
electrons, for a total of 21, leaving five electrons for Z. Possibilities for Z are nitrogen (N), phosphorous (P), arsenic (As), antimony (Sb), and bismuth (Bi).
Problem 3. Consider the molecules and ions listed below. The underlined central atoms in three of
these species use the same type of hybrid orbitals to form bonds with the remaining atoms; the
fourth species uses a different set of hybrid orbitals. Circle the species that is different and indicate
the type of hybrid orbitals it uses. In no more than three sentences, clearly explain your reason for
picking this species.
NO2–
CO32–
BF3
NH3
Lewis structures for the ions NO2– and CO32–, and the molecule BF3 have identical geometries
consisting of three electron domains. Each has a trigonal planar electron domain geometry and,
therefore, each has the same hybridization.
Ammonia, NH3, has four electron domains (three of which are bonding), which requires a tetrahedral arrangement of electron domains and sp3 hybridization.
Problem 4. Shown to the right is the molecular orbital
diagram for the molecule BO. Complete the molecular
orbital diagram by filling in the electrons for each atom
and for the molecule. Based on your molecular orbital
diagram, what is BO’s bond order? Is BO a paramagnetic or a diamagnetic compound (circle your answer
below)? In no more than three sentences, clearly explain the reasons for your answers.
The bond order for BO is 2.5.
BO is: paramagnetic
diamagnetic
The bond order is the difference between the number
of bonding and antibonding electrons, divided by two;
thus (7 – 2)/2 or 2.5. There is a single unpaired electron in the σ2p orbital, so BO is paramagnetic.
Problem 5 The following compounds are generally considered covalent: HCl, ICl, and SCl2. As we
have seen, a pure covalent bond is rare. Rank these compounds from the one showing the least ionic
character to the one showing the greatest ionic character. Place your answers in the table and show
any relevant work and/or explanation in the space below the table.
least ionic character
←→
greatest ionic character
SCl2
ICl
HCl
Ionicity is given by the difference in electronegativities (ΔEN) between the two elements. The greatest ΔEN is for HCl (0.57), so it has the most ionic character, and the smallest ΔEN is for SCl2 (0.28),
so it has the least ionic character.
Problem 6. The following compounds have similar ionic characters, but differ in the degree of metallic and covalent character: MgH2, BaSi2 and ZnS. Rank these compounds from the one showing
the most metallic character to the one showing the most covalent character. Place your answers in
the table and show any relevant work and/or explanation in the space below the table.
most metallic character
←→
most covalent character
BaSi2
MgH2
ZnS
Covalency is given by the average electronegativity of the elements. The largest average electronegativity is ZnS (2.095), so it has the most covalent character, and the smallest average electronegativity
is for BaSi2 (1.40), which it has the most metallic character.
Problem 7. Shown below are three cross-sections through the unit cell of calcium titanate, also
known as the mineral perovskite.
What is the empirical formula for calcium titanate based on this unit cell? Be sure to clearly explain
how you arrived at this formula.
Each of the eight calcium ions is on a corner and contributes 1/8th each to the unit cell for a total of
one calcium ion. The six oxygen ions are on faces and contribute ½ each to the unit cell, for a total
of three oxygen ions. The single titanium ion is in the middle of the unit cell and contributes itself
wholly to the unit cell. The empirical formula for calcium titanate, therefore, is CaTiO3.
The titanium ion can be considered to sit in two types of holes – a hole in a lattice defined by calcium ions and a hole in a lattice defined by oxygen ions. For each, state the type of hole in which the
titanium ion sits and what percentage of these holes are filled.
For a lattice of calcium ions, titanium sits in a cubic hole and occupies 100% of these
holes.
For a lattice of oxygen ions, titanium sits in an octahedral hole and occupies 100% of
these holes.
To how many oxygen ions is each calcium ion coordinated? Briefly explain how you arrived at your
answer in no more than three sentences and/or with an appropriate sketch.
Each calcium is coordinated to 12 oxygen ions. To see this, consider the figure
on the left, which shows a cross-section through the xy-plane of eight unit cells.
The black circle is a calcium ion on the corner of these eight unit cells. Each
open circle is an oxygen ion sitting on the face shared by two unit cells. We can
see here that the coordination number within the xy-plane is four. We can also
draw cross-sections through the xz-plane and the yz-plane, each of which also
will contain four oxygen ions, giving a total of 12 oxygen ions coordinated to
this single calcium ion.