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Strong electrostatic forces between ions Betw. Metal & Nonmetal Crystal lattice is formed
Want to share electrons to achieve full outershell Molecules are formed. Betw. Two non-metals
• Calculate the total # of outer shell electrons • Decide which atom is central (least
electronegative) adding electrons to fill central atoms’ octet.
• Add electrons to give outside atoms full shells, any extra electrons, add to central atom.
• Form multiple bonds so central has octet• Draw lines to represent one shared pair of electrons
Rules for Drawing Lewis Structures
BF F
F
Electron Deficient Atom: Boron does not have an octet
NHH
H
These structures tend to attract other species with a non-bonded pair of electrons!
Using oxidation numbers for assigning charges in molecules may be useful for redox reactions but can lead to an over estimation of actual charges. A more accurate charge assessment can be used in considering Lewis structures.
FC is the difference between the number of valence electrons on the free atom and the number of valence electrons assigned to the atom in the molecule.
Calculating Formal Charge =
(# valence electrons) -( # of Unshared electrons ) - ½(shared electrons)
Which structure is correct?
The one with no formal charges, or the smallest formal charges!
( assign formal charges, now, to figure it out!)
O - S = O O = S - O
O S O
This results in an average of 1.5 bonds between each S and O.
Resonance structuresTwo different structures can be drawn.
Resonance structures
• Benzene, C6H6, is another example of a compound for which resonance structure must be written.
• All of the bonds are the same length.
or
Resonance Structure for NO3-
Arrows do NOT mean that the ion ‘flips’ from one to another, the actual structure is an average of the three!
The distance between two nuclei that includes the shared pair of electrons.
The actual measured distance between the two nuclei is called the bond length. Includes the centers
of + charge with a shared electrons in between.
Double and Triple bonds make this length shorter than a single bond. However, in Nitrate Polyatomic Ion, the bond lengths are the SAME between all three nitrogen-oxygen!
How can the degree of polarity be predicted?
Use Electronegativites to make predictions about the type of bonding in a compound. The greater the difference in electronegativity of the two bonding elements, the greater is the chance for ionic bonding.
The electronegativity difference between two covalently bonded atoms determines the polarity of the bond: The larger the difference, the more polar is the covalent bond. If the difference is zero or very small, the bond is essentially non-polar.
Difference in electronegativity
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1. 1.1 1.2 1.3 1.4 1.5 1.6
0.5 1 2 4 6 9 12 15 19 22 26 30 34 39 43 47
Percent ionic character %
Difference in electronegativity
1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2
51 55 59 63 67 70 74 76 79 82 84 86 88 89 91 92
Percent ionic character %
We can use the difference in electronegativity between two atoms to gauge the polarity of the bonding between them
CompoundF2 HF LiF
Electronegativity Difference 4.0 - 4.0 = 0 4.0 - 2.1 = 1.9 4.0 - 1.0 = 3.0
Type of Bond Nonpolar covalent Polar covalentIonic (non-covalent)
Look at charts available on worksheets
Table of Percent Ionic Character
Topic 8 Bonding(Includes acceptable range)
Dative or Co-ordinate Bonding:
Where one atom does not offer any electrons within the bond. Electrons are only offered by one atom.
Oxygen commonly uses Dative Bonding.
Dipole Moments A method of describing charge concentrations.
If an molecule has a clear center of positive charge away from a center of negative charge, it has a measured dipole moment.
Shapes of Molecules!
Covalent: determined by the number of electron pairs around a central atom
Remember: electron pairs REPEL each other! They will try to get away from each other!
Valence Shell Electron Pair Repulsion Theory lists a set of shapes per # of electron pairs.
VSEPR
*is based on the number of regions of high electron density around a central atom.
*can be used to predict structures of molecules or ions that contain only non-metals by minimizing the electrostatic repulsion between the regions of
high electron density.*can also be used to predict structures of molecules
or ions that contain multiple bonds or unpaired electrons.
*does fail in some cases.
What Is VSEPR?
The Valence Shell Electron Pair Repulsion (VSEPR) model:
VSEPR Rules
1.Draw the Lewis structure for the molecule or ion.
2. Count the total number of regions of high electron density (bonding and unshared electron pairs) around the central atom.
a. Double and triple bonds count as ONE REGION OF HIGH ELECTRON DENSITY.
b. An unpaired electron counts as ONE REGION OF HIGH ELECTRON DENSITY.
c. For molecules or ions that have resonance structures, you may use any one of the resonance structures.
3. Identify the most stable arrangement of the regions of high electron density as ONE of the following:
linear trigonal planar tetrahedral trigonal bipyramidal octahedral
# regions of high electron density
best arrangement description
2 linearcentral atom and regions of high electron
density arranged in straight line
3 trigonal planarregions of high electron density directed at the
corners of an equilateral triangle
4 tetrahedralregions of high electron density directed at the
corners of a tetrahedron
5 trigonal bipyramidal
three regions of high electron density directed at the corners of an equilateral triangle/one
region of high electron density directly above the plane of the triangle and one directly
below the plane
6 octahedralregions of high electron density directed at the
corners of an octahedron
Why isn’t waters’ structure Tetrahedral with bond angles of 109o ?
Because of the two lone pairs of electrons
Reduces the bond angles
A step beyond VSEPR The localized Electron Model
Electron Cloud Hybridization
* describes how the role of orbitals can mold the shape of molecules
sp3, sp2, sp, sp3d, sp3d2
sigma and pi bonds
and
Carbon in methane: CH4
Consider Carbon: 1s2 2s2 2p2
(1s2 electrons don’t enter into the bonding)
Bonding electrons: 2s2 2px1 2py
1 2pz0
Since the first two electrons are in s orbitals, the other two electrons are in p orbitals, the shape of the clouds
and the bonding angles should be different.
But all four bond angles are actually the same!
2s2 2px1 2py
1 2pz0
Carbon combines the s and three p orbitals into four new orbitals, so electron can be alone:
If this arrangement is true, then the bond angles would all the same: 109o, perfect tetrahedral.
1s2 (sp3)1 (sp3)1 (sp3)1 (sp3)1
Carbon’s new electron configuration is:
sp3 sp3 sp3 sp3
Ammonia HAS bond angles of 107o NH3
Nitrogen’s electron configuration: 1s2 2s2 2px
1 2py1 2pz
1
If hydrogen bonded with each of the p’s, normally, then an angle of 90o would be expected.
So, there’s hybridization goin’ on…
N’s NEW electron configuration: 1s2 (sp3)2 (sp3)1 (sp3)1 (sp3)1
unshared pair ! (takes up more room!)
H2S bond angles are close to 90o, what orbitals are involved?
Sulfur: 1s2 2s2 2p6 3s2 3p4
3p4 = 3px2 3py
1 3pz1
Hydrogen’s 1s1’s must bond in the two p’s of sulfur and since the angles are close to 90o, safe to say NO hybridization is goin’ on.
Double and Triple Bonds provide some problems
The REAL structure of a double bond is more complex and changes hybridization!
sp2 hybridzation
Ethylene C2H4 H2C = CH2
Carbon: 1s2 2s2 2px1 2py
1 2pz0
Double bond counts as only one bond. The pi bond is NOT counted in hybridization! so there are only 3 bonds to consider! only the 2s and two p orbitals are used!
2s and 2px and 2py = sp2
sp hybridization
Carbon Dioxide: CO2 O = C = O
Only two orbitals are affected and will hybridize
Carbon: 1s2 2s2 2px1 2py
1 2pz0
Turns into: 1s2 (sp)1 (sp)1 2py1 2pz
1
sigma bonds pi bonds
sp3d hybridization
PCl5 Phosphorous provides 5 electrons to different atoms of Chlorine.
P: 1s2 2s2 2p6 3s2 3px1 3py
1 3pz1 3dz2
0
(sp3d)1 (sp3d)1 (sp3d)1 (sp3d)1 (sp3d)1
Uses one s, three p and one d orbitals = 5
WAS: s p p p d
sp3d2 hybridization
SF6 Sulfur provides 6 electrons for bonding.
S: 1s2 2s2 2p6 3s2 3px2 3py
13pz13dz2
0 3dx2-y20
(sp3d2)1 (sp3d2)1 (sp3d2)1 (sp3d2)1 (sp3d2)1 (sp3d2)1
Uses one s, three p and two d orbitals = 6
Nuclear areas are left exposed in a Hydrogen molecule
These exposed areas allow attractions to the + charged nucleus! These attractions are hydrogen bonds!
These attractions are greatest between Hydrogen and nitrogen, fluorine or oxygen.
Hydrogen bonds are responsible for: A. High boiling points for substances that have them (water!)
B. Substances with them are more viscous
The increased attraction make it more difficult to separate the particles of these substances.
When molecules that have dipole moments come together, they will line themselves up using opposite ends.
Very small electrostatic forces that are formed formed when two molecules come close to each other, the electrons of the second molecule are repelled away from that end of the molecule to the end. This induces a pseudo + and pseudo – end in the molecule. These ends can act as temporary dipoles.
Increase in Boiling points of moving down the Halogens, making them harder to separate.
A.Giant Molecular Structures Diamond and Graphite
Carbon atoms bonded covalently in a continuous network!
Diamond based on Tetrahedral unit, graphite has a layered structure.
All covalent bonds are strong and lend these molecules to have high melting and boiling points.
Cool properties of graphite:
Conducts electricity in only one plane! Electrons move in one layer but not between layers.
The layers are held together very weakly so the layers can be separated easily, making it a great lubricant.
B. Molecular StructuresIodine
Some substances, like Iodine, are held in lattice structures by very weak dispersion forces. This provides Iodine with a low melting point. Iodine has no charged particles so it doesn’t carry an electrically current.
C. Ionic Structures
Lattice structures created with charged particles. These substances are held together by strong bonds and have high melting and boiling points because of that.
Hydration of these structures……….