# CHE 201 Chemistry Lab Report

Laboratory ExerciseA Heating Curve for Water
The temperature of a substance indicates the kinetic energy (energy of motion) of its molecules. When water
molecules gain heat energy, they move faster and the temperature rises. Eventually, the water molecules gain
sufficient energy to separate from the other liquid molecules and form a gas. The liquid changes to a gas in a
change of state called boiling. When a liquid boils, forming a gas, a horizontal line (plateau) appears on the graph,
as shown in the figure below. This constant temperature is called its boiling point.
Heat of Fusion for Water
Changing state from solid to liquid (melting) requires energy. Ice melts at a constant temperature of 0 °C. At that
melting point, the amount of heat required to melt 1 g of ice is called the heat of fusion. For water, the energy
needed to melt 1 g of ice (at 0 °C) is 334. Joules. That is also the amount of heat released when liquid water
freezes to solid ice at 0 °C.
Melting (0 °C):
H2O(s) + heatfusion (334 J/g) → H2O(l)
H2O(l) → H2O(s) + heatfusion (334 J/g)
Freezing (0 °C):
In this experiment, ice will be added to a sample of water. From the temperature change, the amount of heat lost
by the water sample can be calculated. This is also the amount of heat needed to melt the ice.
Heat (J) lost by water =
=
(gwater) x (ΔT(°C)) x (4.184 J ∙ g-1 ∙ °C-1)
heat (J) gained to melt ice
By measuring the amount of ice that melted, the heat of fusion, in joules/gram, can be calculated.
Heat of fusion (J/g)
1
=
heat (J) needed to melt ice
grams (g) of ice melted
Heat of Vaporization
A similar situation occurs for a substance that changes from a liquid to a gas (vapor). Water boils at 100 °C, its
boiling point. At that temperature, the energy required to convert liquid to gas is called the heat of vaporization.
For water, the energy required to vaporize 1 g of water at 100 °C is 2260 J. This is also the amount of heat
released when 1 gram of steam condenses to liquid at 100 °C.
Boiling (100 °C):
H2O(l) + heatvaporization (2260 J/g) →
Condensation (100 °C):
H2O(g)

H2O(g)
H2O(l) + heatvaporization (2260 J/g)
A Heating Curve for Water
How the data was collected:

Weigh a 250-mL beaker. Record the mass. (4 decimals)
With a 100-mL graduated cylinder, add ~100 mL distilled water to beaker.
Weigh beaker w/water. Record the mass. (4 decimals)
With a digital thermometer, record the temperature of the water (0 min). (1
decimal)
Place on hotplate, add thermometer. Heat on high with constant stirring.
Using stopwatch, record temperature at every minute. (1 decimal)
When water boils, continue to record the temperature each minute for at
least five more minutes.
Turn off the hotplate, unplug and allow the beaker to cool.
Plot the points of temperature vs. time on the graph provided.
Read boiling point temperature from plateau on graph, not from the point
boiling was detected. Average the last five temperature points on the
plateau.
Calculate heat required to change temperature (Q) (watch SFs)
Q = masswater x ∆T x SHliquid water
2
Energy in Changes of State: Heat of Fusion
How the data was collected:

Use two(2) Styrofoam cups as the calorimeter— one inside the other.
Weigh calorimeter and record the mass. (4 decimals)
Use a graduated cylinder to measure ~100 mL distilled water. Pour the water into the calorimeter.
Weigh the calorimeter w/water and record the mass. (4 decimals)
Add the thermometer and record the initial temperature. (1 decimal)
Ice:
• Add a small amount of ice.
• Stir until melted.
• Repeat above two steps until temperature reaches 2-3 ⁰C.
• Quickly remove any unmelted ice.

Record the final temperature. (1 decimal)
Weigh the calorimeter w/water and melted ice. Record the mass. (4 decimals)
Calculate heat lost by water (Qwater) (watch SFs)
Qwater = masswater x ∆T x SHliquid water
The amount of heat lost by the water is equal to the amount of heat absorbed by the ice.
– Qwater

Qice
Calculate the Heat of Fusion ( ∆Hfus )
∆Hfus
3
=
=
Qice / massice
A Heating Curve for Water
1. Mass of Empty 250-mL Beaker
___165.2359______ g
2. Mass of 250-mL Beaker + ~100 mL water
___266.4597______ g
3. Mass of water ( #2 – #1 )
_________________ g
Collect at least 15 data points
4. Time (min)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
4
Temperature (° C)
25.0
28.0
32.0
40.6
45.0
52.3
60.9
68.0
77.5
85.3
92.0
97.5
99.3
100.5
101.2
101.0
101.1
101.1
101.2
101.0
Draw a Heating Curve for Water
(Use a ruler to connect each point to the point before and the point after — NOT a “best-fit” line.)
5. Boiling point of water
(Average the last five temperature points on the plateau.
__________________ °C
6. Initial temperature (at 0 minutes)
__________________ °C
7. Temperature change (∆T) ( #5 – #6 )
__________________ °C
8. Mass of water
__________________ g
( #3 )
9. Joules needed to heat water
5
__________________ J
Energy in Changes of State: Heat of Fusion
1.
Mass of Empty Calorimeter
_____4.4605______ g
2.
Mass of Calorimeter + ~100 mL water
__ __101.1903_____ g
3.
Mass of water ( #2 – #1 )
_________________ g
4.
Initial water temperature
______26.7_______ °C
5.
Final water temperature (after ice added)
_______2.1_______ °C
6.
Mass of calorimeter + water + melted ice
____132.4692_____ g
7.
Temperature change ( #5 – #4 )
(Yes, it will be a negative value.)
_________________ °C
8.
Joules lost by water
(Yes, it will be a negative value.)
_________________ J
9.
Joules needed to melt ice = – ( joules lost by water ) (- #8)
_________________ J
10.
Mass of ice that melted ( #6 – #2 )
_________________ g
11.
Heat of fusion (joules to melt 1 g of ice) ( #9/#10 )
_________________ J/g
6
Practice-Laboratory Questions:
1.
How many joules are required for each of the following changes?
SHliquid water = 4.184 J · g-1 · °C-1
ΔHfus = 334 J · g-1
ΔHvap = 2260 J · g-1
a. heating 56.0 g of water from 21.0°C to 67.0°C
b. melting 18.2 g of ice at 0.0°C
c. boiling 12.1 g of water at 100.0°C
2.
How many kilojoules are released when 2.80 g of water condenses at 100.0°C and cools to 51.0°C?
3.
175.00 g of water was heated from 15.0°C to 88.0°C. How many kilojoules were absorbed by the
water?
4.
How many kilojoules are required at 0.0°C to melt an ice cube with a mass of 25.00 g?
5.
How many kilojoules are required to melt 51.00 g of ice at 0.0°C, and raise the temperature of the
liquid that forms to 58.0°C?
7
CHE 201
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY AND VALENCE BOND THEORY
Laboratory Exercise
SUBMIT PAGES 9-15 ONLY.
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY, AND VALENCE BOND THEORY
I. INTRODUCTION
Because atoms are too small to see with the eye, scientists use models to visualize the physical arrangements of
atoms in molecules and polyatomic ions. These three-dimensional models aid in understanding the shape and
relative position of atoms in a molecule. From this, we can get a better understanding of the molecules polarity,
reactivity and its interaction between other molecules. In this laboratory exercise, you will work with a model kit to
gain a better understanding of the three related theories of molecular bonding and structure.
II. LEARNING OBJECTIVES After completing this experiment, you should feel comfortable with:

Drawing Lewis structures for molecules that obey the octet rule.
Drawing Lewis structures for molecules that may violate the octet rule.
Visualizing the three-dimensional shape of a molecule based on a drawing.
Identifying electron pair geometries (VSEPR).
Applying VSEPR theory to determine the shapes of molecules.
Identifying molecular geometries.
Distinguishing between electronic and molecular geometries.
Applying valence bond theory to the hybridization of atomic orbitals.
Distinguishing between polar and nonpolar molecules.
III. INFORMATION/DISCUSSION
A. LEWIS STRUCTURES
The extraordinary non-reactivity of the Noble gases is attributed to a common electronic configuration of eight
electrons (an “octet”) in an outer or valence shell (highest “n” value). Many chemical reactions and molecular
formulas can be related to this observation as most elements bond (share or transfer elec-trons) to achieve the same
electronic configuration as one the Noble gases. From this observation in atom-ic theory, we can predict the bonding
patterns for atoms based on the distribution of valence electrons in an atom. This process is formalized in a Lewis
structure, providing information in a two-dimensional representation about the bonding patterns in covalent
molecules, which can then be used to predict the three dimensional shapes, bond angle, polarity and ultimately
chemical and physical properties.
The following rules and procedures are a good guide for drawing Lewis Dot Structures:
1.
Write the MOLECULAR FORMULA for the compound.
2.
Determine the total number of VALENCE ELECTRONS available for bonding by:
a.
b.
counting the valence electrons from each element in the compound.
adding one electron for each negative charge or subtracting one electron for each positive charge,
for polyatomic ions.
1
CHE 201
3.
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY AND VALENCE BOND THEORY
ARRANGE ATOMS. For small molecules and polyatomic ions, place the element with the lowest electronegativity in the center and arrange the other atoms around this central atom using the following rules:
a.
b.
Hydrogen (H) is never the central atom.
For acids containing oxoanions, the hydrogen atoms are usually bonded to oxygen (O) atoms that are
bonded to the less electronegative central atom.
4.
CONNECT ATOMS. Attach the atoms together with a ” — ” to signify a two-electron bond between the
atoms.
5.
SATISFY OCTET RULE. Place the remaining electrons, in pairs (lone pairs), around each atom to satisfy the
“octet” requirement (a “duet” for hydrogen). It is best to maximize bonding; if all of the electrons have been
used up without satisfying the octet rule then atoms must share more electrons by forming another bond.
The octet rule must apply for C, N, O and F. For all others, it is a good guideline, but may be violated (see item
#6)
6.
EXCEPTIONS TO OCTET RULE: There are some compounds that contain elements with exceptions to the octet
rule. These may have more than eight electrons, typically when n\$3 (PCl5), or less than eight electrons,
typically Group IIA, (BeCl2), IIIA (BCl3) and, of course, hydrogen. For most of the compounds, the atoms are
bonded by single bonds consisting of one electron from the central atom and one electron from the outer
atom. If there are any extra electrons, they are placed on the central atom as unshared or lone pairs.
7.
FORMAL CHARGE: Sometimes atoms will have extra or not enough electron electrons. This imbalance of
electrons is denoted with a formal charge. A negative formal charge means there are too many electrons on
atom. A positive formal charge means there are not enough electrons on an atom. Different types of
electrons are counted differently.
a.
b.
Non-bonding electrons are counted individually.
Electrons in bonds are counted as being shared and so each pair of electrons in a bond counts as only
one electron.
Assign the formal charges to each atom in the molecule. Take the number of valence electrons of the atom
and subtract the number of bonds and the number of non-bonding electrons.
Formal Charge = (# valence electrons) – (# bonds) – (# non-bonding e’s)
If more than one correct Lewis dot structure exists for a set of atoms in a molecule, assign formal charges
and look for the structure that satisfies the following:
a.
b.
c.
The sum of the formal charges of all atoms in a given molecule or ion must equal the overall charge
on that species.
The structure with formal charges closest to zero are preferred.
If nonequivalent Lewis structures exist, if any negative formal charges are in the struc-ture, the more
negative formal charges must be on the more electronegative atoms.
B. VALENCE SHELL ELECTRON PAIR REPULSION (VSEPR) THEORY
VSEPR states that regions of high electron density around a central atom, such as a bonding or lone pairs of electrons
(electron region), arrange as far apart as possible in 3-D space from other electron regions. Single, double or triple
bonds and lone pairs are each counted as a single electron region.
2
CHE 201
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY AND VALENCE BOND THEORY
The VSEPR Rules:
1.
Write the LEWIS STRUCTURE.
2.
Count the number of ELECTRON REGIONS around the central atom (or any atom).
(Single, double, triple bonds = 1 VSEPR region)
(Non-bonding (lone) pairs of electrons = 1 VSEPR region)
3.
Pick the VSEPR/ELECTRONIC GEOMETRY. Note that this electronic geometry is based only on the number
of electron regions, regardless of what they are (triple bond, lone pair, etc.). Consult Table A to assign these
geometries and the corresponding bond angles.
(Arrange the VSEPR pairs to minimize repulsion)
Table A: VSEPR Electronic Geometries
VSEPR/Electronic
Number of Electron Regions
4.
Ideal Bond Angle(s)
Geometry
2
Linear
180E
3
Trigonal Planar
120E
4
Tetrahedral
109.5E
5
Trigonal Bipyramidal
180E/120E/90E
6
Octahedral
180E/90E
Determine the IDEAL MOLECULAR GEOMETRY. The molecular geometry is determined by what we can
actually “see” — the atoms bonded to the central atoms, but not the lone pairs. The molecular geometry is
set by the electronic geometry. Consult Table B to assign these geometries.
(Account for any unfilled positions in the electron pair geometry, i.e. non-bonding pairs/lone pairs)
Table B: VSEPR Molecular Geometries
Number of non-bonding or lone pairs on an atom
Total # of
e- regions
0
2
Linear
3
1
2
Trigonal Planar
Bent
Linear
4
Tetrahedral
Trigonal Pyramidal
Bent
5
Trigonal
Bipyramidal
See-Saw
T-Shaped
Linear
6
Octahedral
Square Pyramidal
Square Planar
T-shaped
3
3
CHE 201
5.
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY AND VALENCE BOND THEORY
Adjust angles to recognize STERIC (size) EFFECTS. This gives rise to slight deviations to the ideal bond angles.
Multiple Bonds —
double, triple bonds take up more space than single bonds, therefore
angles involving them will be somewhat larger.
Non-Bonding/Lone Pairs — Lone Pair electrons take up much more space than bonding
pairs, compressing the angles between other, bonding pairs.
The molecular geometries (the actual geometry of the atoms) might best be explained with the diagrams in your text.
C. VALENCE BOND THEORY
VSPER bonding theory does not explain the whole picture when it comes to bonding and associated properties of
a molecule. Valence bond theory attempts to fill this gap, and works together with VSEPR theory to explain how
carbon bonds. Carbon is able to form four bonds with other atoms, based on its valence, but uses two different types
of electrons (s and p) of different energies to do this. Based on VSEPR geometries, a tetrahedral carbon atom in CH4
should have the same bond angle between hydrogens, and all bonds should be of the same length and energy. To
accomplish this, carbon must mix or hybridize its four atomic orbitals of different energies and redistribute them into
four hybrid atomic orbitals of equal energy (Figure 1).
These new hybrid orbitals are of intermediate energy, falling in between the energies of the original orbitals. The
electrons are then redistributed in the new hybrid orbitals according to Hund’s Rule, where they can now accept
bonding electrons from other atoms to form bonds.
The number of hybrid atomic orbitals must equal the number of atomic orbitals used to make the hybrid orbitals:
One 2s and three 2p orbitals form four sp3 hybrid orbitals. In CH4, these hybrid orbitals interact with the electrons
in the 1s orbitals of Hydrogen to make four equivalent C-H bonds.
It is assumed that O and N atoms follow the same hybridization as carbon. Table C can be used to determine the
hybridization of a central atom using VSPER electronic regions.
4
CHE 201
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY AND VALENCE BOND THEORY
Table C: Hybridization
# of VSEPR
regions
2
3
4
5
6
Electronic
Geometry
linear
trigonal
planar
tetrahedral
trigonal
bipyramidal
octahedral
Atomic Orbitals
s, p
s, p, p
s, p, p, p
s, p, p, p, d
s, p, p, p, d, d
Hybridization
sp
sp2
sp3
sp3d
sp3d2
IDEAL Bond
Angles
180E
120E
109.5E
90E,120E,
180E
90E, 180E
Example: From the molecular formula for nitrite, NO2-1 determine the Lewis structure, VSEPR (electronic) geometry,
hybridization, molecular shape and ideal bond angles.
A. Drawing the Lewis Structure
Step 1:
Determine the number of valence electrons total in the structure
N: 1 x 5 = 5 and O: 2 x 6 = 12
Total = 17 But the ion is a negatively charged ion, so add 1 more electron for a total of 18 valence
electrons.
Step 2:
Put the least electronegative atom in the center, and “err” on the side of symmetry (nature likes
symmetry so make the atoms in the structure symmetrical). Connect the central atom to the others
with a ” — ” indicating a bond for two electrons.
O—N—O
This is an ion — an easy way to indicate this is to place the
structure in brackets and put the charge on the outside of the bracket.
Step 3:
Give all atoms an octet by placing lone pair electrons around
all species.
Step 4:
Count up the electrons on your structure — to be sure you
have the same number of electrons as you previously
counted in Step 1 and you have a valid Lewis structure that
follows the octet rule. In this example, if we count up the
number of electrons, we see that we have put 20 electrons
on the structure. Too many electrons on a structure means that you most likely need to include
double or triple bonds. To reduce the number of electrons in the structure and maintain octets
around the atoms you can add a second bond between two atoms and remove a lone pair of
electrons from each atom on the double bond. You cannot simply erase electrons — because then
your species do not have an octet.
5
CHE 201
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY AND VALENCE BOND THEORY
Step 5:
Do another count of electrons, optimizing bonding and
maintaining octets until the correct number of valence
electrons is represented in the structure. This structure now
has 18 electrons (e-) — which is what we need. In addition,
each atom still “feels” as if it has 8 e- surrounding it.
Step 6:
Assign the formal charges to each atom in the molecule. Take the number of valence electrons of the
atom and subtract the number of bonds and the number of non-bonding electrons.
Formal Charge = (# valence electrons) – (# bonds) – (# non-bonding electrons)
For the nitrite ion, there is only one nonequivalent Lewis dot
structure. The formal charges are distributed, from left to
right, with -1, 0 and 0. The sum of the formal charges equals
the -1 charge on the ion. The formal charge distribution
obeys the rules, in that the more negative formal charge
is on the O, which is more electronegative than N.
B. Determining the VSEPR/Electronic Geometry and Ideal Bond Angle
Count the number of electron regions around the central atom (N); all single, double, triple bonds and any lone pairs
each count as ONE region.
On the N atom:
1.
one single bond,
2.
one double bond, and
3.
one lone pair of electrons.
This adds up to three electron regions around the N atom. From Table A for three electron regions, the VSPER shape
is trigonal planar, with ideal bond angles of 120E.
Table A: VSEPR Electronic Geometries
Number of
Electron Regions
VSEPR/Electronic
Geometry
Ideal Bond Angle(s)
2
Linear
180E
3
Trigonal planar
120E
4
Tetrahedral
109.5E
C. Determining the Molecular Shape/Geometry
This molecular geometry/shape is based on the electron regions/VSEPR shape. The nitrite structure has two bonding
regions (a single bond and a double bond) and one lone-pair (non-bonding) region. Using Table B, the molecular
shape for three total regions (3rd row) and with one non-bonding region (3rd column) is bent. Note the ideal bond
angles remain as 120E.
6
CHE 201
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY AND VALENCE BOND THEORY
Table B: VSEPR Molecular Geometries
Number of non-bonding or lone pairs on an atom
Total # of
e- regions
0
2
Linear
3
1
2
3
Trigonal Planar
Bent
Linear
4
Tetrahedral
Trigonal Pyramidal
Bent
5
Trigonal
Bipyramidal
See-Saw
T-Shaped
Linear
6
Octahedral
Square Pyramidal
Square Planar
T-shaped
D. Determining the Hybridization
Again, based on the number of electron regions, using Table C, three total electron regions (3rd column) show an
sp2 hybridization for the N in the nitrite ion.
Table C: Hybridization
# of VSEPR
regions
2
3
4
5
6
Electronic
Geometry
linear
trigonal
planar
tetrahedral
trigonal
bipyramidal
octahedral
Hybridization
sp
sp2
sp3
sp3d
sp3d2
E. Determining Polarity vs. Nonpolarity
Refer to Table D. Of the two atoms involved in the bond,
subtract the smaller electronegativity value from the
larger. When the difference is 0.5 to 1.8 between the
central atom and an attached atom, the bond is polar. If
the difference is less than 0.5, the bond is nonpolar.
In the molecular geometry of the molecule, if the polar
bonds in a molecule are symmetrical and the dipoles
cancel, the net polarity overall the molecule is nonpolar.
When the dipoles do not cancel, molecule or ion is polar.
In Table E are examples of how molecules can have
either a net polarity or no net polarity. “X” and “Y” are
different elements and “A” is the central atom in the
structure.
Table D: Relative Electronegativities of Elements
7
CHE 201
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY AND VALENCE BOND THEORY
Table E: Evaluating Polarity vs. Nonpolarity of Molecules
Use the molecular models to build the molecules to further visualize the three dimensional structures of the
molecules in the assignment.
PROCEDURE:
On the following pages, for each molecular formula, draw the Lewis dot structure and indicate:

Number of electronic regions around the central atom;

Hybridization around central atom;

Electronic geometry around central atom;

Number of lone pairs of electrons (nonbonding pairs) around central atom;

Molecular geometry around central atom;

IDEAL existing bond angles around central atom (Note: Do not report anything other than 90E, 109.5E, 120E
and/or 180E and do not report bond angles that do not exist in the structure draw n); and

Whether the molecule is internally polar or nonpolar.
You may need to assign formal charges to determine the correct arrangement of the electrons. The octet rule must
apply for C, N, O and F. Molecules marked with an asterisk (*) do not obey the octet rule.
8
CHE 201
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY AND VALENCE BOND THEORY
Name _____________________________________________
Molecular
Formula
NH3
5+3(1)= 8e(# valence e-)
Date _________________________
Lewis Structure
Number of
Electron Regions
Hybridization
Electronic
Geometry
Number of
Lone Pairs
Molecular
Geometry
IDEAL
Existing
Bond Angles
Polar or
Nonpolar
..
H—N—H
|
H
4
sp3
tetrahedral
1
trigonal
pyramidal
109.5E
Polar
H2O
CH2F2
9
CHE 201
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY AND VALENCE BOND THEORY
Name _____________________________________________
ra luce lM
o
Formula
Lewis Structure
Number of
Electron Regions
*OPI3
(violates
octet rule)
SO32-
*AlBr63(violates
octet rule)
10
Date _________________________
Hybridization
Electronic
Geometry
Number of
Lone Pairs
Molecular
Geometry
IDEAL
Existing
Bond Angles
Polar or
Nonpolar
CHE 201
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY AND VALENCE BOND THEORY
Name _____________________________________________
ra luce lM
o
Formula
Lewis Structure
Number of
Electron Regions
*S3
(violates
octet rule)
*TeO42(violates
octet rule)
*ClBr4(violates
octet rule)
11
Date _________________________
Hybridization
Electronic
Geometry
Number of
Lone Pairs
Molecular
Geometry
IDEAL
Existing
Bond Angles
Polar or
Nonpolar
CHE 201
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY AND VALENCE BOND THEORY
Name _____________________________________________
ra luce lM
o
Formula
Lewis Structure
Number of
Electron Regions
*IF4
(violates
octet rule)
*SF4
(violates
octet rule)
*BrF
5
(violates
octet rule)
12
Date _________________________
Hybridization
Electronic
Geometry
Number of
Lone Pairs
Molecular
Geometry
IDEAL
Existing
Bond Angles
Polar or
Nonpolar
CHE 201
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY AND VALENCE BOND THEORY
Name _____________________________________________
ra luce lM
o
Formula
Lewis Structure
Number of
Electron Regions
Date _________________________
Hybridization
Electronic
Geometry
*PF5
(violates
octet rule)
CH2O
C
C
|
C2H6O
O
(CH3OCH3)
|
C
13
Number of
Lone Pairs
Molecular
Geometry
IDEAL
Existing
Bond Angles
Polar or
Nonpolar
CHE 201
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY AND VALENCE BOND THEORY
Name _____________________________________________
ra luce lM
o
Formula
Lewis Structure
Number of
Electron Regions
Date _________________________
Hybridization
Electronic
Geometry
C
|
C
|
O
C2H5OH
(CH3CH2OH)
C6H6
C
(C’s form
a ring)
CO2
14
Number of
Lone Pairs
Molecular
Geometry
IDEAL
Existing
Bond Angles
Polar or
Nonpolar
CHE 201
MOLECULAR GEOMETRY: LEWIS STRUCTURES, VSEPR THEORY AND VALENCE BOND THEORY
Name _____________________________________________
ra luce lM
o
Formula
Lewis Structure
Number of
Electron Regions
*XeBr2
(violates
octet rule)
*SeF6
(violates
octet rule)
15
Date _________________________
Hybridization
Electronic
Geometry
Number of
Lone Pairs
Molecular
Geometry
IDEAL
Existing
Bond Angles
Polar or
Nonpolar

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Instead, we offer bonuses, discounts, and free services to make your experience outstanding.
How it works
Receive a 100% original paper that will pass Turnitin from a top essay writing service
step 1
Fill out the order form and provide paper details. You can even attach screenshots or add additional instructions later. If something is not clear or missing, the writer will contact you for clarification.
Pro service tips
How to get the most out of your experience with Writall
One writer throughout the entire course
If you like the writer, you can hire them again. Just copy & paste their ID on the order form ("Preferred Writer's ID" field). This way, your vocabulary will be uniform, and the writer will be aware of your needs.
The same paper from different writers
You can order essay or any other work from two different writers to choose the best one or give another version to a friend. This can be done through the add-on "Same paper from another writer."
Copy of sources used by the writer
Our college essay writers work with ScienceDirect and other databases. They can send you articles or materials used in PDF or through screenshots. Just tick the "Copy of sources" field on the order form.
Testimonials
See why 20k+ students have chosen us as their sole writing assistance provider
Check out the latest reviews and opinions submitted by real customers worldwide and make an informed decision.
Criminal Justice
Absolutely LOVE the essay I received. I really appreciate it so much.
Customer 454561, February 16th, 2021
Psychology
Thanks
Customer 453933, January 13th, 2021
Psychology
I received a 100 on this work.. I am very pleased and know that I have someone that knows what they are doing and follows instructions!!!! Great work !!! Thank you, Thank you !!
Customer 454497, September 3rd, 2020
unbelievable!!
Customer 454811, January 29th, 2022
Nursing
Outstanding
Customer 455459, October 8th, 2022
Other
Thank you
Customer 454677, March 18th, 2021
Psychology
This writer was suburb compared to the first one.
Customer 454893, September 18th, 2021
Technology
Thank you
Customer 454677, April 27th, 2021
Other
Amazing writer. Truly the best.
Customer 454983, February 19th, 2022
Health Care
GREAT. PERFECT TIMELINE.
Customer 455425, April 22nd, 2023
Exactly what I wanted to see in the paper. Thanks.
Customer 455001, October 20th, 2021
Economics
Excellent!! Great Work Exactly what I needed!!!!
Customer 455199, December 15th, 2021
11,595
Customer reviews in total
96%
Current satisfaction rate
3 pages
Average paper length
37%
Customers referred by a friend