1a-topic 5 - shape 201640 topic&5:&structure!andshape! lewis structures!...
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Topic 5: Structure and Shape
Lewis structures Lewis structures are a means of determining stable electron arrangements in molecules. It considers the valence electrons of an atom only. A stable arrangement is one in which each atom has achieved a noble gas electron configuration by distribution of the electrons as bond pairs or lone pairs (non-‐bonded pairs). A noble gas electron configuration is 2 for hydrogen and 8 for C, N, O and F. This is sometimes called The Octet Rule. To draw a Lewis Structure you need to know which atoms are bonded to which. Then: 1. Add up the total number of valence electrons present, add or subtract electrons to
account for any charge. 2. Join the appropriate atoms using an electron pair for each bond. 3. Distribute the remaining electrons to result in an octet of electrons on each atom
(except hydrogen that always has two electrons associated with it). 4. If there are too few electrons to give every atom an octet, move non-‐bonded pairs
between atoms to give multiple bonds. 5. If there are electrons left over after forming octets, place them on the central atom. 6. Indicate the overall charge. Example: Draw the Lewis structure of NH3.
• Total number of valence electrons: 5 (N) + 3×1 (H) = 8 • N is at the centre, three N-‐H bonds require 3×2 electrons:
• Electrons remaining = 8 (valence) – 6 (bonding) = 2; place lone
pair on nitrogen: • Nitrogen has ‘octet’: 3×2 electrons (in bonds) + 2 electrons (lone
pair). Example: Draw the Lewis structure of OH–
• Total number of valence electrons: 6 (O) + 1 (H) + 1 (charge) = 8 • One O-‐H bond requires 2 electrons:
• 6 electrons remaining; three lone pairs:
• Oxygen has ‘octet’: 2 electrons (in bond) + 6 electrons (lone
pairs). • Indicate charge outside brackets.
Example: Draw the Lewis structure of O2
• Total number of valence electrons: 2x6 (O) = 12 • One O-‐O bond requires 2 electrons: • 10 electrons remaining; 5 lone pairs: • One O atoms has only 6 electrons so share electrons from the
other atom to form a double bond. Now both O atoms are associated with an ‘octet’ of electrons:
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1. Draw the Lewis structures of: CH4, NH4+, H2O and N2
The ‘octet rule’ predicts all stable compounds of C, N, O & F (C, N, O & H are the principle components of organic molecules which represent the vast majority of known compounds). In contrast compounds of Be and B may have less than 8 electrons and are referred to as ‘electron deficient’. Example: Draw the Lewis structures of BeCl2
• Total number of valence electrons: 2 (Be) + 2x7 (Cl) = 16 • Two Be-‐Cl bonds requires 4 electrons:
• 12 electrons remaining; three lone pairs on each Cl: • While Be does not have an ‘octet’ further sharing of electrons
with Cl to form double bonds is not found as Be is recognised as ‘electron deficient’.
2. Draw the Lewis structure of BF3
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Elements in the third period and beyond have available d orbitals as well as s & p orbitals and so may form stable compounds with greater than 8 electrons surrounding these atoms. Typically these are observed in compounds of Cl, Br, I, P and S where stable bonding arrangements containing 8, 10 or 12 electrons around these atoms are observed. Example: Draw the Lewis structure of ClF3
• Total number of valence electrons: 7 (Cl) + 3x7 (F) = 28 • Three Cl-‐F bonds requires 6 electrons:
• 22 electrons remaining; three lone pairs on each F and one on Cl
gives each atom an octet: • There are still 2 electrons remaining. These form an additional lone
pair on the Cl. While F can only have a maximum of 8 electrons (valence shell is the n = 2 shell which can accommodate only 8 electrons), Cl may have more than 8 electrons (valence shell is the n = 3 shell which can accommodate 18 electrons though in practice we observe bonding with 8, 10 or 12 electron arrangements).
3. Draw the Lewis structures of PCl3, PCl5, SF6 and ICl4-‐
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Where more than one structure may be drawn, resonance occurs where the actual structure is a weighted combination of all possible structures and the electrons are delocalised in the molecule/ion. The Lewis structures are drawn connected with a double headed arrow. Example: Draw the Lewis structure of CO3
2–
• Total number of valence electrons: 4 (C) + 3x6 (O) + 2 (charge) = 24 • Three C-‐O bonds requires 6 electrons:
• 18 electrons remaining; try three lone pairs on each O:
• C does not have an octet so share a pair from one of the O atoms:
• Equally likely to share a lone pair from any of the three O atoms so
there are three resonance structures:
When resonance occurs…
• The actual structure is an average of all possible Lewis structures. • The actual structure is a single structure. • The actual structure is more stable than predicted from the Lewis structure. • There is no easy way of drawing the actual structure!
4. Draw the Lewis structure of O3, HCO2
-‐ and NO3-‐
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Molecular Geometry The valence shell electron pair repulsion model (VSEPR model) assumes that electron pairs repel one another. This produces a set of geometries which depend only on the number of valence shell electron pairs and not on the atoms present. To determine the molecular geometry: 1. Draw the Lewis structure. 2. Count the number of electron areas (that is bond pairs and lone pairs but count each
multiple bond as one electron area). 3. Arrange electron areas to minimise repulsion. This is the electron pair distribution. 4. Position the atoms to minimise the lone pair -‐ lone pair repulsion if > 1 lone pair. 5. Name the geometry from the atom positions. This is the molecular geometry and may
be different from the electron pair distribution. 6. For large molecules repeat the process for each atom that is attached to at least two
other atoms to build up a picture of the overall shape.
Blackman Figure 5.13, 5.16, 5.19
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In summary:
No of Electron Pairs (Lewis Structure)
Arrangement of Electron Pairs
No of σ Bond Pairs
No of Lone Pairs
Molecular geometry Examples
2
2 0 Linear BeCl2, CO2, N3-‐
3
3 2
0 1
Trigonal planer
Angular
BCl3, SO3, CO32-‐
SO2, O3, NO2
-‐
4
4 3 2
0 1 2
Tetrahedral
Trigonal pyramid
Angular
CH4, NH4+, PO4
3-‐
H3O+, NH3, XeO3
H2O, NH2
-‐, ClO2-‐
5
5 4 3 2
0 1 2 3
Trigonal bipyramidal
"See-‐saw"
T-‐shaped
Linear
PCl5
SF4, PBr4-‐
ClF3, XeF3+
ICl2-‐, XeF2
6
6 5 4
0 1 2
Octahedral
Square pyramidal
Square planar
SF6, SiF62-‐, AsF6-‐
IF5, SF5-‐, SbF52-‐
ICl4-‐, XeF4
Example: What is the molecular geometry of CO2?
• Lewis structure is: • Two areas of electrons about the central atom, no lone pairs, therefore linear.
Example: What is the molecular geometry of CO3
2–?
• Lewis structure shows resonance but examining any one of the resonance structures will give the correct answer:
• Three areas of electrons about the central atom, no lone pairs, therefore triganol planar.
Example: What is the molecular geometry of ClF3?
• Lewis structure is: • Five areas of electrons about the central atom, so electron pair
distribution is trigonal bipyramidal. There are two lone pairs which results in a molecular geometry of T-‐shaped.
X
X
X
X
X
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5. Draw the Lewis structures and determine the molecular geometry of the following species:
CH4
CH3
–
NH4
+
NH3
NH2
–
H3O+
H2O
6. Draw the Lewis structures and determine the molecular geometry for the following molecules: a. CF4 b. AsCl5 c. SeH6
7. For the following molecules, draw the Lewis structure(s) and predict the molecular shape.
a. PBr4–
b. I3−
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Dipole Moments Any bond between two different atoms will be polar as a result of the electronegativity difference between the atoms. A molecule has a permanent dipole moment if it contains polar bonds and is not a symmetrical shape. Examples of polar molecules:
Examples of non-‐polar molecules:
8. Complete the following table. Molecule Lewis structure Molecular geometry Is there a dipole moment?
HCl
SO2
SO3
SF4
IF5
SF6
NH H
HOH HH Cl
H
CCl Cl
ClC
H H
O
O C OCl
CCl Cl
ClH H
F
S
FF F
FF
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Intra- & Inter-molecular forces We have examined intramolecular forces – those that act within a molecule. There are the forces responsible for covalent bonds and, more loosely, the ionic bonds and metallic bonds that hold compounds together. They are STRONG (typically 80-‐4000 kJ mol-‐1). When you break and form these “bonds” you are also performing a chemical reaction. There are also non-‐bonding forces that generally occur between molecules and so are termed inter-‐molecular forces. These are largely WEAK (typically 0-‐50 kJ mol-‐1) and when you make and break these you are mainly performing physical changes eg boiling a liquid. There are a range of intermolecular forces: Ion-‐Dipole
- dipoles align (partially or fully) in the electric field of an ion - typical energy 40-‐600 kJ mol-‐1
Dipole-‐dipole
- dipoles align (partially or fully) in the electric field of the neighboring dipole. - typical energy 5-‐25 kJ mol-‐1
Ion-‐induced dipole - a dipole is induced in a polarizable molecule by the presence of an ion. - typical energy: 3-‐15 kJ mol-‐1
Dipole – induced dipole
- a dipole is induced in a polarizable molecule by the presence of another dipole. - typical energy: 2-‐10 kJ mol-‐1
Dispersion or London forces
- two polarizable molecules induce transient dipole moments in each other. - typical energy: 0.05 – upwards kJ mol-‐1.
• Dispersion forces exist between ALL molecules. The force increases in strength with
molecular mass. • Forces associated with permanent dipoles are found only in substances with overall
dipole moments (polar molecules). Their existence adds to the dispersion forces. • When comparing substances of widely different masses, dispersion forces are usually
more significant than dipolar forces. • When comparing substances of similar molecular mass, dipole forces can produce
significant differences in molecular properties (e.g. boiling point). 9. Match the five forces above to the following situations:
1. Liquid chloroform (CHCl3). 2. Petrol (a mixture of non-‐polar hydrocarbons). 3. Oxygen dissolved in water. 4. A solution of sodium chloride. 5. Oxygen interacting with the iron of haemoglobin.
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Hydrogen bonds Hydrogen bonding arises from an unusually strong and directed dipole-‐dipole force. When H is bonded to a very electronegative element (F, O, N) the bond is polar covalent. H is unusual because with only one electron, it leaves a partially exposed nucleus (H has no other core electrons to shield the nucleus). The bond can be thought of as forming between the hydrogen atom and the lone pairs of the F, N, or O in HF, NH3 and H2O, respectively. Water is perhaps the most unusual liquid. Each water molecule is H-‐bonded to FOUR other water molecules (donating 2 H-‐atoms and accepting two H-‐atoms to the lone pairs), forming a tetrahedral network (ice) and a loose tetrahedral network (liquid).
Measuring Intermolecular Forces The melting transition from solid to liquid is not a good way of measuring intermolecular forces, as the molecules are held together by attractive forces in both phases. In order to get an idea of the strength of the interaction, we need to separate the molecules. This requires a transition to a gas, i.e. vaporisation or sublimation. A measure of this is the molar enthalpy of vaporisation. An indirect measure is the boiling point. 10. Explain the trend in the table in terms of the type and size of IM Forces present.
Substance ΔHvap (kJ mol-‐1)
Molar mass Intermolecular forces present
Butane, CH3(CH2) 2CH3 22 58
Diethyl ether, (C2H5)-‐O-‐(C2H5) 27 74
Methanol, CH3OH 38 32
Ethanol, CH3CH2OH 43 46
Water, H2O 44 18
Blackman Figure 6.31
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11. Which of the following molecules can be expected to show hydrogen-‐bonding with another molecule of the same substance? Draw the H-‐bonding interactions for those that will.
C2H6
CH3OH
CH3F
(glycine – an amino acid)
12. Circle the molecule in the following pairs that is likely to have the higher boiling point
and identify the intermolecular forces present in that molecule.
CO2 and SO2
CCl4 and CHCl3
CH3OCH3 and CH3CH2OH
Ar and Xe
H2 and F2
O
CH3C CH3 H2N
CH2
COH
O
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Phase changes The normal sequence of phases we expect to see on warming is Solid → Liquid → Gas The phase change (eg solid to liquid) occurs at a sharply defined temperature for most materials. However, particularly for covalent materials that possess a rod like structure, an intermediate phase displaying some characteristics of both solids and liquids may occur. These are liquid crystals.
13. In the pictures below, each ellipse represents a molecule. Label the pictures that correspond to (a) the solid phase and (b) the liquid phase.
14. In smectic and nematic liquid crystals, the molecules are orientated to be roughly parallel. In smectic liquid crystals, the molecules are also roughly arranged in layers. There are two types of smectic liquid crystal. Labels the pictures that correspond to (a) smectic and (b) nematic liquid crystals.
15. In the smectic A phase, the average orientation of the molecule is normal to the layer. In the smectic C phase, the average orientation of the molecule is at an angle to the layer. Labels the pictures that correspond to (a) smectic A and (b) smectic C liquid crystals.
16. Summarise the phase transitions by redrawing the pictures in the table below.
Examples of liquid crystal materials are:
Solid Smectic C Smectic A Nematic Liquid
Regular crystal lattice in three dimensions
Molecules still in layers but orientate at an angle to the layer
Molecules still in layers but orientated on average normal to the layer
Preferential orientation along the long axis of the molecule but no layers discernible
Random orientation of molecules within the liquid
SC OO
CN