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Covalent Bonding And Molecular Geometry 2.How do atoms ... Flipbook PDF
Subclass Geometries: Those in which there are one or more nonbonding pairs of electrons on the central atom. Determine t
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Covalent Bonding And Molecular Geometry
Questions: 1.How can the valence electrons of an atom be represented? 2.How do atoms achieve an octet? 3.How are electrons “shared” in a molecule? 4.How can the geometries of molecules be accounted for? 5.What are some limitations of molecular models?
Lewis (electron) Dot Structures
Attractive and repulsive forces and bond formation
Show the number of valence electrons in an atom.
Gilbert Newton Lewis Note: It doesn’t matter which “side” you put the electrons on as long as the grouping is consistent.
Spectroscopic notation and Lewis structure for the element oxygen.
Question: Draw the Lewis structure for phosphorus. What are its valence electrons?
The valence electrons are 3s23p3. Note that the Lewis structure does not differentiate between s and p sublevel electrons.
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Using Lewis Structures to Identify Covalent Bonds
Example: Show the formation of diatomic chlorine using Lewis Structures.
Showing the formation of diatomic hydrogen using Lewis structures. Note that the “shared pair” of electrons produce a “duet” of electrons for each, giving them the configuration of He.
The formation of compounds using Lewis structures The Lewis structure for the formation of water, H2O Important Note: Lewis structures cannot be used to determine the spatial geometry of a molecule. H2O is NOT linear!
Examples: Show the Lewis structure for the formation of the ionic compounds sodium bromide, NaBr, and sodium sulfide, Na2S.
Important terminology you should know
General Procedure for Writing Lewis Structures for Molecules
(Make sure electrons are always placed in pairs)
Note that the sodium octets are not explicitly shown, even though they would have an octet. Why do you think this is?
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Example:
On Your Own:
Draw the most likely Lewis structure for the formation of the cyanide ion, CN-.
Write the most likely Lewis Structures for the following: Calcium Chloride Hypochlorite ion Carbon monoxide
No octet
octet
Resonance, Formal Charge and Exceptions to the Octet Rule Resonance: Resonance occurs when more than one equivalent, plausible structure is possible for a molecular arrangement. In most cases, a “resonance hybrid” has intermediate bond characteristics indicating they are “shared” between atoms.
Bond order The term bond order refers to how many bonds exist between two atoms. For non-resonant bonds, it is equal to the number of electron pairs shared between the atoms: Examples: Ethyne, C2H2, has a bond order of 3 between the carbons and a bond order of 1 between each carbon and hydrogen H C C H For dimethylketone (2-propanone), C3H6O, the bond order between the carbon and oxygen is 2 and each carboncarbon and carbon-hydrogen bond the order is 1. O H
H C C
C
H
H H
If the molecule exhibits resonance, then the bonds that are resonant are “shared” between the possible resonant bonds
H
Example: Draw the possible resonance structures for selenium dioxide, SeO2 and determine the bond order.
For the carbonate polyatomic ion:
Since the secondary bond can be shared equally among any of the carbon-oxygen bonds, the bond order between each carbon-oxygen bond is 1 1/3
Bond order = 1 ½
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Formal Charge: Used to determine the best possible configuration for a molecular structure.
Example: Is the most likely arrangement for methyl alcohol CH3OH or CH2OH2?
Formal charge is given by: Formal charge = number of normal valence electrons in the atom – the number of nonbonding electrons – ½ the bonding electrons.
Answer:
Ideally:
C = 4 – 0 – 4 = 0; H = 1 – 0 – 1 = 0 and O = 6 – 4 – 2 = 0
Numbers should be as close to zero as possible
In the second case, the formal charge are:
More electronegative atoms should have a more negative formal charge.
C = 4 – 2 – 3 = -1; H = 1 – 0 – 1 = 0 and O = 6 – 2 – 3 = +1
In the first case, the formal charges for the elements are:
Since numbers closer to zero are desirable, the first structure is the most likely. (See page 169 in your text)
Example:
Follow up:
Formaldehyde (which contains a carbon, an oxygen and two hydrogen atoms), a liquid with a disagreeable odor, traditionally has been used to preserve laboratory specimens. Draw the most likely Lewis structure for the compound.
Why isn’t a structure with the hydrogens bonded to the oxygen desirable for formaldehyde? 4-4-1/2 (4) = -2
Answer: The most likely structure has the H’s bonded to the carbon and the oxygen double bonded on the carbon. (Formal charges for all atoms are zero)
Atoms with Less than an Octet
Answer:
H | .. C .. = O | H
6-0-1/2(8) = +2
The formal charge on the carbon would be -2 and the oxygen would be +2. Numbers of +/-2 are fairly large and having the more electronegative atom (oxygen) as +2 is not desirable
Example: Show, using formal charge, which of the two structures for BeCl2 is the most likely.
Typical for atoms of: Beryllium (2-pair of electrons) Boron (3-pair of electrons) Hydrogen (1-pair of electrons)
Cl = Be = Cl Or Cl – Be – Cl In the first case, the formal charges would be: Be = 2 – 0 -4 = -2 and Cl = 7 – 4 – 2 = +1 In the second case Be = 2 – 0 – 2 = 0 and Cl = 7 – 6 – 1 = 0 The second form is the most likely. In the first case the FC’s are not as close to zero and the more electronegative atom, Cl, is more positive.
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Molecules with more than an Octet: Expanded Valences
There are three possible resonance structures for the cyanate ion NCO-. Using atom formal charges, decide which is the most important of these structures.
Most common with nonmetal atoms that contain valence electrons in an energy level that has a d-sublevel. Examples: Draw the most likely structure for: sulfur hexafluoride phosphorus pentachloride Explain whether or not it is likely that NF5 would exist.
.. .. .. .. - [:N=C=O:]- [:N≡C-O:][:N-C≡O:] .. ..
NF5 is not likely to exist because N’s valence electrons are in the second energy level which does not have a d-sublevel
Determine what is wrong with each of these structures
H H
F
C C
N F
N C
F
H
O
Sn
O
H
O
F
B
F
F
H
C
F
VSEPR Theory Stands for Valence Shell Electron Pair Repulsion Theory Accounts for the geometry of most molecules Premise: Premise Electron pairs around an atom (usually a central atom is the one of interest) arrange themselves to be as geometrically far apart as possible to minimize electron repulsions.
O
F
VSEPR Geometries (Major Classes) Examples: Predict the geometries of the following molecules Perspective Diagram
a. NH4+ b. SF6 c. CH4 d. AsCl5 e. BI3 Answers: a. tetrahedral b. octahedral c. tetrahedral d. trigonal bipyramidal e. trigonal planar (plane triangular)
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When determining the geometry of a molecule, treat
multiple bonds as though they were single bonds.
Electronic geometries refer to the arrangement of electron
Examples: Predict the geometry of the following molecules
The molecular geometries refer to the actual physical appearance of a molecule and may be a major class (no nonbonding pairs on the central atom) or a subclass (one or more nonbonding pairs on the central atom)
a. CO32-
b. SO42-
c. CO2
pairs around the central atom and will always be a major class.
Answers: a. Trigonal planar (plane triangular) b. Tetrahedral (No multiple bonds) c. Linear
Subclass Geometries: Those in which there are one or more nonbonding pairs of electrons on the central atom.
Determine the electronic and molecular geometries of the following molecules: a. CH3b. SF2 c. I3d. NO3e. XeF4 f. BrF5 g. H2O h. H3O+ i. OPF (Phosphenous fluoride)
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Summary of VSEPR Geometries
Molecular Polarity
Electron pairs
Caused by unequal sharing of electrons between atoms Determined through the difference in electronegativity between two bonding atoms (∆EN). The greater ∆EN is, the more polar the bond is Electrons are pulled towards the more EN atom. Bond differences in ∆EN greater than about 1.5 to 1.7 are considered ionic in nature Melting point, boiling point, electrostatic properties and more are affected by bond polarity
Electronegativity and Bond Polarity
Dipoles A polar bond is known as a dipole
Electron Cloud (shared electrons)
Mathematically, the strength of the dipole is calculated by multiplying the amount of charge separation times the distance between them. This is known as a dipole moment. A dipole moment is represented on a Lewis dot structure by an arrow with a cross on the tail pointing towards the more electronegative atom in the bond. Also, small delta + and – symbols (δ+, δ-) are used to designate the polar ends of the bond
+
+
δ+
δDipole
Green with greater Green and atom red atoms with equal electronegativity= =nonpolar polar electonegativities
Dipoles A molecule can have polar bonds but be nonpolar overall depending on the symmetry of the molecule Example: Carbon tetrachloride is nonpolar even thought there are individual dipoles between the central carbon and the chlorines. The net effect of the geometry is to cancel out the dipole nature of the molecule
Cl
δ-
δ+ δ+ δ-
C
δ+ δ-
Cl δ+
Cl
Cl δ-
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Clues that a Molecule Might be Polar One or more nonbonding pairs of electrons on the central atom Dissimilar atoms attached to the central atom These clues must be applied cautiously as they depend on the ∆EN values and the location of the nonbonding electron pairs
Examples: Draw the Lewis structures for the following molecules and indicate the direction of the net dipole if applicable:
a. BrF5 b. ICl4c. ClF3 d. AsF5
δ+
e. NH3 δ-
δ+
Net Dipole (Vector sum of the dipoles)
δ-
Answers
Valence Bond Theory Valence electrons take part in chemical bonding through the hybridization (blending) of an atom’s orbitals. Accounts for the geometries of molecules Does not occur in all atoms (e.g. the hydrides of some group VI nonmetals and the larger group V nonmetals do not (e.g. H2S)). The actual geometry must always be considered (determined through bond angles) A bond forms between two atoms when a pair of electrons with their spins paired is shared by two overlapping atomic orbitals
Hybridization The blending of existing orbitals in an atom to produce a new set of “hybridized” orbitals for bonding
The number of orbitals taking part in the hybridization determines the number of hybridized orbitals that are created.
Hybridization of carbon
Molecular structure of ammonia with nitrogen sp3 hybridized.
Introduction
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Disregard +/designation
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The first bond between any two atoms is known as a sigma bond (σ).
For instance, consider the molecule ethene, C2H4
Multiple bonds consist of an end-on hybridized orbital bond (called a sigma, σ, bonds) and one or two unhybridized p orbitals (called pi, π, bonds) that overlap sideways
Each central carbon has sp2 hybridization (3 sigma bonds) and one unhybridized p orbital (pi bond).
The sigma bond, being an end-on bond is stronger than the pi bond
The geometry of each central atom (carbon) is trigonal planar The hydrogens are not hybridized
Nonbonding electron pairs usually exist in hybridized orbitals as well (i.e. they are part of the hybridization).
H σ
σ
H
An ethene, C2H4, molecule
σ
Cπ C
H σ σ
H
An ethyne, C2H2, molecule
Note that both the upper and lower overlap of the unhybridized p-orbital constitutes only one π bond.
Example: Describe the hybridization of the following molecules with respect to the central atom and identify the number of sigma (σ) and pi (π) bonds: BF3
Answers: BF3
Boron is sp2 hybridized and has 3σ bonds
PCl5 Phosphorus is sp3d hybridized and has 5σ bonds. NH4+ Nitrogen is sp3 hybridized and has 4σ bonds.
PCl5
NH3 Nitrogen is sp3 hybridized and has 3σ bonds (and one lone pair in a hybridized orbital)
NH4+
SF6
NH3
CH2O Carbon is sp2 hybridized and has 3σ bonds and 1π bond (between it and the oxygen atom)
SF6
CO2 Carbon is sp hybridized and has 2σ bonds and 2π bonds
CH2O
Sulfur is sp3d2 hybridized and has 6σ bonds
CO2
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