Download Chemistry 110 Lab

MOLECULAR GEOMETRY
EXPERIMENT 4
INTRODUCTION
Molecules are held together by forces called chemical bonds. One type of bond, the covalent bond, is
the result of a sharing of electrons between two atoms. Mathematically, bonds arise from the overlap
of the wave functions representing the atomic orbitals of each of the atoms involved in bonding.
The atomic orbitals (s, p, d, f,...) are not sufficient to explain the variety of geometries found in
molecules. The idea of hybridized orbitals has been developed to fill that gap. Hybrid orbitals arise
from the mathematical mixing of two or more atomic orbitals.
During this laboratory you will examine models of hybrid orbitals and investigate the molecular
geometries associated with each of them.
It is important to remember that:
!
The actual geometry of a molecule is determined by experiment.
!
Hybridization is a way of explaining experimental observations.
The models you are using are subject to many limitations. Note these as you go along. You will be
asked to list three in your conclusion to the experiment (PART 2).
IMPORTANT NOTES
!
You do not have to write a summary of the procedure for this experiment.
!
You don’t have to wear a lab coat or safety glasses.
!
Part 2 is given like an exam, demonstrators will not answer any question. Part 1 is not graded,
demonstrators will answer all questions you may have.
everything in part 1, you are ready for part 2 !
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When you are comfortable with
THEORY
Only a brief overview is presented here since most of the theory has already been covered in the
lectures. Refer to your textbook, chapter 10 (p.396-407), and chapter 11 (p.424-434) if needed.
In this experiment, you will deal with molecules which can be thought to consist of 3 parts when
considering their geometry:
! the central atom (A),
! the ligands which are bound to the central atom (X),
! the lone pairs of electrons in the valence shell of the central atom (E).
They can be put together to create a molecule AXmEn where the subscripts indicate the number of
ligands (m), and the number of lone pairs (n).
The geometry of a molecule AXmEn is determined by two factors. The first one is the geometry of the
orbitals of the central atom. This geometry can be found by looking for the appropriate m+n in the
table below.
m+n
Orbital geometry
Molecular geometry
n=0
n=1
n=2
2
Linear
Linear
3
Trigonal planar
Trigonal
planar
Bent
4
Tetrahedral
Tetrahedral
Trigonal
pyramidal
Bent
5
Trigonal bipyramidal
Trigonal
bipyramidal
Sawhorse
T-shaped
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octahedral
octahedral
Square
pyramidal
Square planar
n=3
Linear
The second is the number of ligands bound to the central atom. If n=0 (no lone pairs), then the
molecular geometry corresponds to the orbital geometry. However, when the value of n is not zero,
the molecular and orbital geometries differ. The obvious difference is the absence of a ligand at a
position which is occupied by a lone pair. The molecular geometry of AX2E1 is then bent, because one
position is missing a ligand in the trigonal planar orbital geometry. To understand why the bond angle
in this molecule is not 120°, one needs to study a more subtle difference which arises from the
differing volume requirements of bonding electrons and lone pair electrons.
The bond between a ligand and the central atom contains two electrons, which are under the attraction
of the two nuclei. Lone pairs, on the other hand, are attracted to only one nucleus. As such, the orbitals
they occupy do not extend, out from the atom, as far as a bonding orbital. But, their orbitals are wider
than bonding orbitals. To visualize this, picture a balloon as a bonding orbital with the central atom
located at the knot of the balloon. Now make the balloon an orbital occupied by a lone pair by pushing
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on the end of the balloon. The balloon does not extend out quite so far and at the same time it gets
fatter near the knot. If you had started with three balloons in a trigonal planar arrangement and had
pushed on the end of one of the balloons you would see that the other two balloons were pushed closer
together (decreasing the angle between the two balloons, to a value lower than 120°). In terms of
orbitals and electrons, this contraction arises because the repulsion between a lone pair of electrons
and a pair of bonding electrons is greater than the repulsion between two pairs of bonding electrons.
This contraction effect of lone pairs is cumulative. This is apparent in the bond angles found in AX3E1
and AX2E2 molecules as you will be discussing in this experiment. When two or more lone pair of
electrons are present in a molecule it has been found that they are located are located as far away as
possible from any other pair.
APPARATUS
1 set of orbital models
Models
s orbital: represented by a WHITE sphere.
p and d orbitals: represented by RED and BLUE tear-drop shaped lobes.
hybrid orbitals: represented by YELLOW or ORANGE tear-drop shaped lobes.
Model representation:
Styrofoam lobe or sphere
Rod
= bonding electrons
= single electrons
= non bonding electrons = lone pair of electrons
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PROCEDURE
PART 1 - Identify the models set up on the wooden stand and answer the questions pertaining to each
model. Then, consider the thirteen molecules identified in the first five tables. Those are the reference
molecules. Answer questions 1 to 14 (below) for each molecule, and fill in tables 1-5. The
demonstrators will assist you as much as needed to understand the reference molecules.
PART 2 - You will be given a set of three test molecules. Demonstrators will not answer any
questions about those molecules. You will follow the same procedure as for the reference molecules,
and write your answers in table 6. During PART 2, you will be allowed to use the models and the
periodic chart we provide, nothing else. It is highly recommended that you make sure your answers
for the reference molecules (PART 1) are correct before doing PART 2…
The following 14 questions are to be answered for all molecules.
1. Identify the central atom.
2. How many ligands are bonded to the central atom?
3. Use box diagrams to write the ground state electronic configuration of the central atom
valence shell electrons.
4. How many electrons are single in the ground state configuration?
5. Use the box diagrams to represent the excited state for which the number of single electrons
matches the number of ligands. (Only if the number of ligands is higher than the number of
single electrons.)
6. On the excited state diagram, circle the orbitals that will become hybrid orbitals. The circle
should include all the occupied orbitals. If no excited state is required for that molecule, draw
the circle on the ground state diagram.
7. Use box diagrams of the hybrid orbitals to rewrite the electronic configuration.
8. Indicate the number and type of the hybrid orbitals formed?
9. How many bonding electron pairs are found around the central atom in the molecule?
10. How many lone electron pairs are found around the central atom in the molecule?
11. How many electron pairs total are found around the central atom in the molecule?
12. What is the molecular geometry of the molecule?
13. What are the bond angles in this molecule?
14. Is the molecule polar or nonpolar?
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sp ORBITALS
The answers for sp orbitals are provided as an example to help you get started.
Examine the model of sp hybrid orbitals. Answer the 14 questions and compare your answers with the
ones in Table 1.
BeCl2 is one molecule in which the central atom has sp hybridized orbitals. It was determined
experimentally that BeCl2 is a linear molecule in which both Be"Cl bonds are identical. The model
can be thought to represent Be in BeCl2.
sp2 ORBITALS
Examine the model of the sp2 hybrid orbitals and complete Table 2.
BCl3 is one example of a molecule whose central atom has sp2 hybridized orbitals. Experiments have
determined that BCl3 is a trigonal planar molecule in which all B-Cl bonds are identical, and all Cl-BCl bond angles are equal to 120°.
SnCl2 is another example of a molecule whose central atom is sp2 hybridized. However, in this
molecule one of the hybrid orbitals is occupied by a lone pair of electrons. Experiments have
determined that in the gas phase, this is an angular molecule, with the Cl-Sn-Cl bond angle slightly
less than 120°, and with both bonds equal in length and strength. Look at the model again. Remove
one of the orbitals from the sp2 model and let the rod represent a lone pair of electrons. The lone pair
of electrons tend to repel the bonding pairs more than the bonding pairs repel each other.
sp3 ORBITALS
Examine the model of the sp3 hybrid orbitals and complete Table 3.
CH4 is one example of a molecule whose central atom has sp3 hybridized orbitals. I was determined
experimentally, that all C-H bonds are identical and that all H-C-H bond angles are 109.5°.
NH3 is another example of a molecule whose central atom has sp3 hybridized orbitals. However, in
this molecule one of the hybrid orbitals is occupied by a lone pair of electrons. Experiments have
determined that all bonds are equivalent, and that the H-N-H bond angles are 107.3°. Let the model
represent N by removing one orbital and allowing the rod to represent the lone pair.
H2O is yet another example of a molecule whose central atom has sp3 hybridized orbitals. Two of the
hybrid orbitals are occupied by lone pairs of electrons. Experiments have it was determined that both
H-O bonds are equivalent, and that the H-O-H bond angle is 104.5°. Let the model represent O by
removing a second orbital and allowing the two exposed rods to represent lone pairs.
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Look at the model again. Use the following information to explain to yourself the trend in the bond
angles of these three molecules.
!
!
!
The repulsions between a bonding pair and a lone pair (BP#LP) are greater than those
between two bonding pairs (BP#BP).
Where more than one lone pair is present, the repulsions between two lone pairs (LP#LP) are
greater again than the BP#LP repulsions.
The magnitude of the repulsion therefore decreases in the order LP#LP $ BP#LP $
BP#BP, and these repulsions determine the geometry and bond angles of the molecule.
sp3d ORBITALS
Examine the model of the sp3d hybrid orbitals and complete Table 4. Note that Table 4 has one extra
row identified as Lone pair(s) position(s). The geometry of the molecule is dependent upon which
orbital(s) contain(s) the lone electron pair(s). Therefore, complete the diagram by adding the lone
pair(s) at the appropriate relative position, and only then, determine the molecular geometry. To help
you in this, consider the following rules:
!
two electron pairs forming a 90% angle with the central atom are said to be close neighbors,
!
two electron pairs forming a 120% angle with the central atom are said to be far neighbors,
!
since a lone pair is known to occupy more space than a bonding pair, it will preferably be in the
position where the number of close neighbor is minimum.
Take the model in your hands and remove one of the two identical axial orbitals from the model and let
the rod represent the lone pair of electrons. Observe that this lone pair has three close neighbors.
Replace that orbital and remove one of the three identical planar orbitals and let that rod represent the
lone pair of electrons. Observe that this lone pair has two close neighbors.
PF5 is one example of a molecule whose central atom has sp3d hybridized orbitals. Large atoms (third
row and beyond) can accommodate more than eight electrons around the nucleus; that is, more than
four orbitals. It was determined experimentally, that the three P—F bonds lie in the same plane, at
120o to each other. The other two bonds are slightly longer and perpendicular to the three bonds in the
plane.
SF4 is an example of a molecule whose central atom has sp3d hybridized orbitals with a lone pair of
electrons occupying one of the hybrid orbitals. Experiments have determined that the geometry of that
molecule was sawhorse.
ClF3 is an example of a molecule whose central atom has sp3d hybridized orbitals with lone pairs of
electrons occupying two of the hybrid orbitals. It was experimentally determined to be a T-shaped
molecule with bond angles of 87.5%.
XeF2 is an example of a molecule whose central atom has sp3d hybridized orbitals with lone pairs
occupying three of the hybrid orbitals. Experiments have determined that the geometry of that
molecule was linear.
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sp3d2 ORBITALS
Examine the model of the sp3d2 hybrid orbitals and complete Table 5.
SF6 is one example of a molecule whose central atom is sp3d2 hybridized. Experiments have
determined that the geometry of that molecule was octahedral, with all bond angles are 90%.
ClF5 is an example of a molecule whose central atom is sp3d2 hybridized, with a lone pair of electrons
occupying one of the hybrid orbitals. Experiments have determined that the geometry of that molecule
was: square-base pyramid, with bond angles of 90%.
XeF4 is an example of a molecule whose central atom is sp3d2 hybridized, with lone pairs of electrons
occupying two of the hybrid orbitals. Experiments have determined that the geometry of that molecule
was square planar, with bond angles of 90%.
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PART 1
Table 1 " sp orbitals
BeCl2
Central atom
Be
Number of ligands
Ground state
configuration
2
2s
Single electrons in
ground state
Excited state
configuration
2p
0
2s
Hybridized orbitals
configuration
Hybrid orbitals
2p
sp
2p
2 sp orbitals
Electron pairs
bonding
2
lone
0
total
2
Molecular geometry
linear
Bond angles
180%
Polar or nonpolar
nonpolar
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PART 1
Table 2 " sp2 orbitals
BCl3
SnCl2
Central atom
Number of ligands
Ground state
configuration
2s
2p
5s
5p
2s
2p
5s
5p
Single electrons in
ground state
Excited state
configuration
Hybridized orbitals
configuration
sp2
2p
sp2
Hybrid orbitals
Electron pairs
bonding
lone
total
Molecular geometry
Bond angles
Polar or nonpolar
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5p
PART 1
Table 3 " sp3 orbitals
CH4
NH3
H2O
Central atom
Number of ligands
Ground state
configuration
2s
2p
2s
2p
2s
2p
2s
2p
2s
2p
2s
2p
Single electrons in
ground state
Excited state
configuration
Hybridized orbitals
configuration
sp3
sp3
Hybrid orbitals
Electron pairs
bonding
lone
total
Molecular geometry
Bond angles
Polar or nonpolar
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sp3
PART 1
Table 4 " sp3d orbitals
PF5
SF4
XeF2
ClF3
Central atom
Ligands
Ground state
configuration
3s
3p
3d
3s
3p
3d
3s
3p
3d
5s
5p
5d
3s
3p
3d
3s
3p
3d
3s
3p
3d
5s
5p
5d
Single
electrons
Excited state
configuration
Hybrid orbitals
configuration
sp3d
3d
sp3d
3d
Hybrid orbitals
Electron pairs
bonding
lone
total
Lone pair(s)
position(s)
Molecular
geometry
Bond angles
Polar or
nonpolar
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sp3d
3d
sp3d
5d
PART 1
Table 5 " sp3d2 orbitals
SF6
ClF5
XeF4
Central atom
Ligands
Ground state
configuration
3s
3p
3d
3s
3p
3d
5s
5p
5d
3s
3p
3d
3s
3p
3d
5s
5p
5d
Single electrons
Excited state
configuration
Hybrid orbitals
configuration
sp3d2
3d
sp3d2
3d
Hybrid orbitals
Electron pairs
bonding
lone
total
Lone pair(s)
position(s)
Molecular
geometry
Bond angles
Polar or
nonpolar
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sp3d2
5d
PART 2 - hand in before leaving the lab
Molecular Geometry
Experiment 4
Name
First
Last
McGill ID number
Demonstrator
Lab Section (day + AM or PM)
Date
QUESTIONS
1. List three ways in which models used in this experiment are inaccurate representations of orbitals.
2. Explain why the H–O"H bond angle in water (104.5%) is less than the H"N"H bond angle in ammonia
(107.3%).
3. Explain why the ClF3 molecule is not a trigonal planar molecule.
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PART 2 - hand in before leaving the lab
Table 6 " test molecules
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Central atom
Ligands
Ground state
configuration
Single electrons
Excited state
configuration
Hybrid orbitals
configuration
Hybrid orbitals
Electron pairs
bonding
lone
total
Lone pair(s)
position(s)
Molecular
geometry
Bond angles
Polar or
nonpolar
14
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