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Chapter 10
Theories of Covalent BondingMolecular Geometry and Hybridization of Atomic Orbitals
Drawing Lewis Structures1) Place least electronegative element as the central atom. Recognize that C,S,P and N are often central atoms. H and halogens are often bonded to central atoms.
2) Sum the total of valence electrons contributed by each atom in the molecule. Look at the Group number to help you.
3) Place bonds to central atoms using 2-electrons per bond.
4) Place an octet of electrons (octet rule) around bonded atoms remembering that H only has 2 electrons---no octet.
5) Place remaining electrons around central atom which should have an octet if period 2 or less, but could be more than octet if period 3 or higher.
6) Some “rules of thumb” to have at your fingertips.
H forms 1-bond, C forms 4-bonds, N forms 3-bonds, O forms 2-bonds.
Write the Lewis dot and skeletals structure of nitrogen trifluoride (NF3).
Write the Lewis dot and skeletal structures of the carbonate ion (CO3
2-).
Write the Lewis dot and skeletal structures structure of the carbonate ion (BrO3
-).
Write the Lewis dot and skeletal structures structure of the carbonate ion HCN?
Write the Lewis structure of nitrogen trifluoride (NF3).
Step 1 – N is less electronegative than F --> N is central atom!
F N F
F
A = 5 + 21 = 26 valence electrons
Step 3 - Write structure with N central and three bonds and rest non-bonding octet electrons around the central atom.
Step 2 - Count valence electrons = A; Nitrogen = 5, Fluorine = 3 X 7 = 21
octet
octet
octet
octet
Step 4 - Write structure with N central and three bonds and rest non-bonding octet electrons.
Write the Lewis structure of the carbonate ion (CO32-).
Step 1 – C is less electronegative than O, put C in center
Step 2 – Count octet and valence electrons (N and A respectively)
Valence electrons = 4 + 6 + 6 + 6 + 2 = 24 valence electrons
Step 3 - Arrange the atoms draw bonds between C and O atoms and complete octet on C and O atoms. 24 - 8 = 16 non-bonding electrons.
O C O
O ][ 2-
]BrO3[–
Valence e- = 7 + 3(6) + 1 = 26
O Br O
O
Look at the formula sometimes it gives clues to the central atom
H C N
HCNValence e- = 1 + 4 + 5 = 10
Carbon is central atom, watch for hydrogen--1 bond
–
Write the Lewis structure of the carbonate ion (BrO3-).
A concept called “resonance” is used when more than one plausible Lewis structure can be drawn.
Measured bond lengths show they are equal!
O O O O O O••••••
••
••••
••••
••••
••
••2 equally good Lewisstructures
Which structure is correct?
Example: Ozone, O3
O O O O O O••••••
••
••••
••••
••••
••
••
Both are!
O O O••••••••
••a resonance hybrid structure
Carbonate Ion- [CO3] 2-
Benzene, C6H6
C - C single bond: 1.54 Å
C = C double bond: 1.34 Å
C - Bond in C6H6: 1.40 Å
Resonance Structures
Write resonance structures for the nitrate ion, NO3
-.
PLAN:
Write resonance structures for the nitrate ion, NO3
-.
Valence e- = 5 + (3X6) + 1 = 24 e-
These are different two plausible structures...how do we decide?
O C O O C O O C O
A book-keeping method called “formal charge” is used to determine the “best” Lewis structure when multiple structures appear plausible.
1. The best structure is one that minimizes total formal charge. Less or no charge is better--and it must equal net charge of ion or molecule.
2.! The best structure is one that places negative charge on the more electronegative atom.
Assigned Atoms = all from lone pair e! + ! ( bonded e! )
Valence e- 6 4 6 6 4 6 # of Assinged e- 6 4 6 5 4 7 Formal Charge 0 0 0 +1 0 !1
O C O O C O
This structure wins!
To use the concept of formal charge, we draw the plausible Lewis structures and then for each atom determine it’s formal charge.
Atom Formal charge = # valence e- - Assigned e- to Atom
What if more than one structure works?Example: Write 3 plausible Lewis structures for the
thiocyanate ion [SCN]–
S C N[ ] –S C N[ ] –
S C N[ ] –
1 2 3
Valence = 6 e- + 4 e- + 5 e- + 1 e- = 16 e-S C N
3-plausible Lewis structures which one is best?
Formal Charge For Multiple Structures
N ] –-2S C N[ ] –
FCS = 6 - 4 -2 = 0FCC = 4 - 0 - 4 = 0FCN = 5 - 6 - 2 = -1
S C[FCS = 6 - 2 -3 = 1FCC = 4 - 0 - 4 = 0FCN = 5 - 6 - 1 = -2
S C N[ ] –
FCS = 6 - 6 -1 = -1FCC = 4 - 0 - 4 = 0FCN = 5 - 2 - 3 = 0
0 0 -1 -1 0 0 0+1
1. Formal charge must sum to charge of ion or molecule.2. N is more electronegative than C or S, it should have a the most negative charge in the “best structure”.3. The most plausible structure has the least amount of formal charge.
Structure on the left is “best” structure!S C N[ ] –0 0 -1
1) Incomplete Octet - rare situation that occurs with Be, B and Al as central atoms.
2) Expanded Octet (the largest class of octet exceptions)-occurs mostly with Period 3 non-metals like P, S and halogens.
3) Odd-number electrons highly reactive species called radicals that have an odd number of electrons (uneven).
There are three major exceptions to the octet rule. Incomplete Octet: Occurs With Group 2A (Be) and 3A (B and Al)
BF3B – 3e-
3F – 3x7e-
24e-
F B F
F
Be – 2e-
2H – 2x1e-
4e-
BeH2 H HBe
Cl Al Cl
Cl
AlCl3Al – 3e-
3Cl – 3x7e-
24e-
Draw Lewis structures for the following
SF6
S – 6e-
6F – 42e-
48e- S
F
F
F
FF
F
••
•• ••
••
P
Cl
ClCl
••••
••••
••
••
Phosphorous trichloride PCl3
P
Cl
Cl
••
••Cl
••••
••
••••
•• •• ••
••Cl
••••
••Cl••
Phosphorous pentachloride
[ICl4]-1P – 5e-
5Cl – 35e-
40e-PCl5
Expanded Octet (the largest class of octet exceptions)-occurs mostly with Period 3 non-metals like P, S and halogens.
N – 5e-
O – 6e-
11e-NO N O
H—C—H
H
•
•O—H ••
••
••
••
••
•N=O
Methyl radical Nitrosyl radical Hydroxide radical
An Odd Number of Valence e- = No octet and radical
Odd-Electron Molecules: Radicals are highly reactive species that have an odd number of electrons (uneven).
Chemists use Valence Shell Electron Pair Repulsion Theory to predict the shapes of molecules using these five electron group geometries.
1. Draw Lewis Structure from chemical formula.
2. Count all electron domains to get AXE code.
4. Match the number of bonding and non-bonding domains to the proper VSEPRT geometry.
3. Group domains into bonding and non-bonding pairs of electrons.
VSEPRT explains the geometry of molecules but NOT how covalent bonds are formed with that geometry.
Molecular formula
Lewis structure
VSEPRTGeometry
Hybrid orbitals
VSEPRT Valence BondTheoryVSEPRTLewis Structure
sp3d2
Octahedral
sp sp2 sp3 sp3d
Linear TrigonalPyramidal
Tetrahedral Trigonal Bipyramidal
Molecular formula
Lewis structure
VSEPRTGeometry
Hybrid orbitals
The goal is to understand geometry (via VSEPRT) and to relate it to a picture of covalent bonding in molecules.
Valence BondTheory
The 3-D geometry of a molecule is one of five basic arrangements of electron groups (domains).
Valence Shell Electron Pair Repulsion Theory: the optimum arrangements of a given number of electron domains is the one that minimizes repulsion among them.
Note each shape has a specific “bond angle”
The total number of electron groups (domains) defines one of the five basic geometries.
2 EG
3 EG 4 EG
5 EG 6 EG
The electron geometry is the geometry of all electron domains, whereas the “molecular geometry” describes the geometry of only the atoms bonded to the central atom.
AX3E1 = Tetrahedral electron geometery with 109.5˚ bond angles.
Molecular Geometry is trigonal pyramidal bond angles <109.5˚
An electron group (domain) is either a pair of bonding electrons or a pair of non-bonding electrons surrounding a central atom. Multiple bonds only count as 1-group or domain.
Cl ClBe
F N F
F
P
Cl
Cl
••
••Cl
••••
••
••••
•• •• ••
••Cl
••••
••Cl••
2 electron groupsbonding
4 electron groups3 bonding1 non-bonding
5 electron groups5 bonding 0 non-bonding
O C O
O ][ 2-
3 electron groups3 bonding0 non-bonding
••
•• ••
••
P
Cl
ClCl
••••
••••
••
••
4 electron groups3 bonding1 non-bonding
We count and “code” the bonding/non-bonding information into shorthand called AXE classification.
AX2E0 = AX2A = Central Atom
X = # of BondedDomains
E = # Non-Bonded Domains
It’s implied that E = 0
F N F
F
4 electron groups3 bonding1 non-bonding AX3E1
Cl ClBe 2 electron groupsbonding AX2
How Predict Geometry Using VSEPRT1.Draw a plausible Lewis structure for the molecule.
2.Determine the total number of electron domains and identify them as bonding or lone pairs.
3.Use the total number of electron domains to establish the electron geometry from one of the five possible geometric shapes.
4.Establish the AXnEm designation to establish the molecular geometry (or do both electron and molecular geometry together simultaneously)
5. Remember bond angles in molecules are altered by lone pairs of electrons (repulsion forces reduce angles).
6. Molecules with more than one central atom can be handled individually.
2 Electron Groups = Linear Electron Geometry and 1-Possible Molecular Geometry
Other Examples:CS2, HCN, BeF2
Bond Angle
AX2E0 = AX2
Cl ClBe
S C N
O C O
A = Central AtomX = # of BondedDomains
E = # Non-Bonded Domains
Examples:SO2, O3, PbCl2, SnBr2 A
Examples:SO3, BF3, NO3
-, CO32-
AX3
Bond AngleA
3-Electron Domain
3 Electron Groups = Trigonal Planar Electron Geometry and 2-Possible Molecular Geometries
AX2E1
Examples:
CH4, SiCl4, SO4
2-, ClO4-
NH3
PF3
ClO3
H3O+
H2O
OF2
SCl2
AX4
AX3E1 AX2E2
4 Electron Groups = Tetrahedral Electron Geometry and 3-Possible Molecular Geometries
Bond Angle
SF4
XeO2F2
IF4+
IO2F2-
ClF3
BrF3
XeF2
I3-
IF2-
PF5
AsF5
SOF4
AX5 AX4E1
AX2E3
AX3E2
AxialPosition
EquatorialPosition
5 Electron Groups = Trigonal Bipyramial Electron Geometry and 4-Possible Molecular Geometries
SF6
IOF5
BrF5
TeF5-
XeOF4
XeF4
ICl4-AX4E2
AX6
AX5E1
6 Electron Groups = Octahedral Electron Geometry and 3-Possible Molecular Geometries
bonding-pair vs. bondingpair repulsion
lone-pair vs. lone pairrepulsion
lone-pair vs. bondingpair repulsion< <
Non-bonding electrons repulse bonding electrons and alter the bond angles in molecules.
Electron lone pairs render the normal 109˚ tetrahedral angle less than 109!
Double-bonds and/or triple bonds in molecules also decrease bond angles in molecules (think repulsion by electron rich regions).
CC
H
H
H
H
CC
H
H
H
H
Predicted Bond Angles Actual Bond Angles
122°
120° 116°
Double-bond vs. Single-bond repulsion >
Single-bond vs, Single-bond repulsion
The electron geometry is the geometry of all electron domains, whereas the “molecular geometry” describes the geometry of only the atoms bonded to the central atom.
AX3E1 = Tetrahedral electron geometery with 109.5˚ bond angles.
Molecular Geometry is trigonal pyramidal bond angles <109.5˚
Predicting Molecular ShapesDraw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) PF3 and (b) COCl2.
<109.50
Predicting Molecular ShapesDraw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) PF3 and (b) COCl2.
2. Count the electron domains and find electron geometry and molecular from core 5 electron domain shapes (using AXE designation and sub-shapes)
5. The F-P-F bond angles should be <109.50 due to the repulsion of the nonbonding electron pair.
3. There are 4 electron domains so the electron geometry is tetrahedral
4. The designation is AX3E1 so the molecular geometry is trigonal pyramidal.
1. Count the valence electrons and draw Lewis structure for PF3: VE = 5 + 3(7) = 26 e-
Predicting Molecular Shapes with Two, Three, or Four Electron Groups
(b) For COCl2, C has the lowest EN and will be the center atom.
Predicting Molecular Shapes with Two, Three, or Four Electron Groups
(b) For COCl2, C has the lowest EN and will be the center atom.There are 24 valence e-, 3 atoms attached to the center atom.
124.50
1110
Type AX3
5. The Cl-C-Cl bond angle will be less than 1200 due to the electron density of the C=O.
2. Count the electron domains and establish electron geometry from 5 shapes3. There are 3 electron domains so the electron geometry is trigonal planar4. The molecular geometry designation is AX3E0 so the molecular geometry is also trigonal planar (no lone pairs).
1. Draw the Lewis structure
Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) SbF5 and (b) BrF5.
Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) SbF5 and (b) BrF5.
(a) SbF5 - 40 valence e-; all electrons around central atom will be in bonding pairs; shape is AX5 - trigonal bipyramidal.
(b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX5E, square pyramidal.
More Than One Central Atom
• In acetic acid, CH3COOH, there are three central atoms.• We assign the geometry about each central atom
separately.
What is the geometryaround these atoms?
Take one atom at a time and apply the rules of electron domains.
ethaneCH3CH3 ethanol
CH3CH2OH
More Than One Central Atom
tetrahedral electron domain and molecular geometry
Determine the shape around each of the central atoms in acetone, (CH3)2C=O.
Find the shape of one atom at a time after writing the Lewis structure.
tetrahedral tetrahedral
trigonal planar
>1200
<1200
Predicting the Molecular Shape With Multiple Central Atoms
Electronegativity is an element’s inherent property to draw electrons to itself when chemically bonded to another atom in a molecule. The units are dimensionless (all relative measurements to Li).
RankFONClBr
Differences in elements electronegativity between bonding atoms result in the formation of polar-covalent bonds and net dipole moments in molecules.
Net Dipole MomentNo Net Dipole Moment
Polar BondPolar Bond
Polar Bond Polar Bond
Think of the dipole moment as a molecule with separated charges + and -.
For a poly-atomic molecule we must consider the vector sum of polar bonds in the molecule to see if there is a net dipole moment.
No NetDipoleMoment
DipoleMoment
DipoleMoment
DipoleMoment
No NetDipoleMoment
Valence Bond Theory explains covalent bonding by the spatial overlap of atomic orbitals on bonding atoms and the sharing of electron pairs.
Electrons that must have opposite spins.
1s1 + 1s1
Bonding in F2
1s1 + 2p5
Bonding in H2
Bonding in HF
2p5 + 2p5
Major Points and Themes of Valence Bond Theory
1. Pauli Exclusion Principle Holds: 2-electrons per overlapped bond with opposing spins.
2. Greater orbital overlap gives stronger bonds. Depends on orbital shapes and how they overlap.
3. Bonding is accounted by mixing or blending or “hybridization of pure valance atomic orbitals”.
4. The number of hybrid orbitals formed equals the number of atomic orbitals combined.
5. The types of hybrid orbitals combined varies with the types of orbitals mixed or blended together. USE VSEPRT to help!
sp3d2
Octahedral
sp sp2 sp3 sp3d
Linear Trigonalplanar
Tetrahedral Trigonal Bipyramidal
Molecular formula
Lewis structure
VSEPRTGeometry
Hybrid orbitals
Connect the dots and it becomes easy to see and understand.
Valence Bond Theoryexplains how bonds are made
We use “pure atomic orbitals” (think ground state electronic structure and those orbitals) to describe bonding in some molecules.
Bonding in F2
1s1 + 1s22s22p5
Bonding in HF
2p5 + 2p5
1s22s22p5 + 1s22s22p5
1s1 + 2p5
sp3
hybridized orbitals
hybridization
Bonding in carbon presents a problem as combining atomics orbitals fails. Valance Bond Theory solves this by allowing the blending or mixing of pure atomic orbitals in a process called hybridization.
Pure atomic orbitals (valence orbitals)only two bond are possible in the ground
state but we don’t observe CH2
By hybridizing 4 bonds are possible.
By combing or mixing different numbers of pure atomic orbitals we make “hybrids” that match one of the VSEPRT geometries. For example 1 pure s orbital + 1 p-orbital combine to give and two “sp hybrids” that when superimposed form a linear geometry for bonding.
s-orbital + p-orbital --> 2 sp hybrid orbitals -->
s-orbital + Two p-orbital --> 3 sp2 hybrids = Trig Planar
2-superimposed sphybrid orbitals
s-orbital + Three p-orbitals -> Four sp3 hybrids = Tetrahedral
sp3 hybrid orbitals
The process of combining pure atomic orbitals to form “hybrid orbitals” on central bonding atoms in a molecule is called hybridization.
1. The number of hybrid orbitals obtained equals the number of atomic orbitals mixed.
2. The name of and shape of a “hybrid orbital” varies with the types of atomic orbitals mixed. (s + p vs s + two p)3. Each hybrid orbital has a specific geometry that matches one of five VSEPRT shapes (show below).
sp3d2
Octahedral
sp sp2 sp3 sp3d
Linear TrigonalPlanar
Tetrahedral Trigonal Bipyramidal
Some generalized rules and comments on VBT and the formation of hybridized orbitals.
sp3d2
Octahedral
sp sp2 sp3 sp3d
Linear TrigonalPyramidal
Tetrahedral Trigonal Bipyramidal
Molecular formula
Lewis structure
VSEPRTGeometry
Hybrid orbitals
The logic is connected all the way to Lewis and VSEPRT
ElectronGeometry
Molecular Geometry AXnEm Hybridization
Linear Linear AX2 sp
Trigonal planar
Trigonal planar V-shaped bent
AX3
AX2E1sp2
TetrahedralTetrahedral
Trigonal pyramidal V-shaped bent
AX4
AX3E1 AX2E2
sp3
Trigonal bipyramidal
Trigonal bipyramidalSeesaw
T-shaped Linear
AX5
AX4E1
AX3E2
AX2E3
sp3d
OctahedralOctahedral
Square pyramidal Square planar
AX6
AX5E1
AX4E2
sp3d2
ElectronGeometry
Molecular Geometry AXnEm Hybridization
Linear Linear AX2 sp
Trigonal planar
Trigonal planar V-shaped bent
AX3
AX2E1sp2
Tetrahedral
Tetrahedral Trigonal pyramidal
V-shaped bent
AX4
AX3E1 AX2E2
sp3
Trigonal bipyramidal
Trigonal bipyramidalSeesaw
T-shaped Linear
AX5
AX4E1
AX3E2
AX2E3
sp3d
OctahedralOctahedral
Square pyramidal Square planar
AX6
AX5E1
AX4E2
sp3d2
Determine the VSEPRT geometry, the bond angles and the hybridization of each indicated atom in the following molecule? How many sigma and pi bonds are in the molecule?
Determine the electron domain, molecular geometry, the bond angles and the hybridization of each indicated atom in the following molecule? How many sigma and pi bonds are in the molecule?
tetrahedral, 180, sp3
sp3
sp2
sp
sp2
bent, <109.5, sp3
trig planar 120˚, sp2
linear 180˚, sp
Atomic OrbitalsMixed
# Hybrid OrbitalsFormed
HybridShape
Linear AX2
Trig Planar AX3
Tetrahedral AX4
Trig Bypyr AX5
Octahedral AX6
s + p s + 2 p s + 3 p s + 3 p + d s + 3 p + 2d
Two sp Three sp2 Four sp3 Five sp3d Six sp3d2
Orbitals Leftover for Pi bonds
Two p one p none Four d Three d
Linking VSEPRT To Valence Bond Theory Hybrids
2s
--The number of hybrid orbitals formed is equal to the number of “pure orbitals” combined!
--When superimposed the “sp-hybrid” give us bonding orbitals for a linear molecules.
An sp hybrid is formed from the combination of a one pure 1s orbital and a one 2p orbital from a central bonding atom producing two new orbitals called sp orbitals.
s-orbital p-orbitalTwo sp hybrid orbitals
sp hybrid orbitals superimposed
Hybridization
s + p Hybridization = 2 sp
Show the bonding scheme and hybridized orbitals used in BeCl2
2 unhybridized unoccupied p-orbitals
After hybridization we have on the central atom, 2 pure p-orbitals and two sp hybrids.
2 “left-over” p-orbitalshybridization
Isolated Be Atom
Hybridized Be Atom
Show the bonding scheme and hybridized orbitals in BeCl2
two sp hybrids on Be
two lone p-orbitals
sp2 = Triginal planar geometry, 120˚ bond angle
3-atomic orbitals, s and two p’s combine to form 3-sp2 hybrid orbitals
An sp2 hybrid is formed from the combination of a one pure 1s orbital and a two 2p orbitals from a central bonding atom producing two new orbitals called sp2 orbitals.
Superimposed Hybrid orbitals form a triginal planar geometry
sp hybrid:Ethylyne: HC!CH:Linear
sp hybrid orbitals
Lone p orbitals that are not hybridized
Sigma bonds (! bonds) and Pi bonds (" bonds)are two different types of covalent chemical bonds that form as a result of end to end spatial overlap of atomic orbitals or hybridized orbitals (! bonds) or side to side overlap on bonding atoms (" bonds)
Lone p orbitals that were not hybridized on each carbon atom are able to form Pi bonds in a “side to side” overlap. A pair of electrons is shared in this region of space.
sp2 hybrid orbitals on each carbon atom use end to end overlap to form a sigma bond.
" bonds overlap side to side
sp hybrid:Ethylyne: HC!CH:Linear Example 2: sp2 hybridizaton scheme BF3.
Boron Orbital Box Diagram
Boron Hybrid Box Diagram
Bonding of pure p-orbital in F with sp2 hybridized orbitals in BF3
Tetrahedral geometry = sp3 hybrid orbitals
sp3 = Tetrahedral geometry = 109.5˚ bond angle
Note the number of hybrids formed is the number of atomic orbitals combined!
combine to generatefour sp3 orbitals
which are representedcollectively as: sp3
Example: sp3 orbital hybridization: CH4.
the four sp3 hybrid orbitals form a tetrahedral shape
sp3 hybridization mixes one 2s orbital with three 2p orbitals to produce four sp3 orbitals on each carbon atom. End to end overlap with a 1s orbital from H gives four sigma bond in CH4.
CH4
This is the ground stateconfiguration of valence atomic orbitals
Example 3: sp3 hybrid orbitals in H2O.
What is the electronic geometry?What is the molecular geometry?What orbitals contribute to bonding?
Note the lone pairs occupy 2-of the sp3 orbitals
sp3 hybridization mixes one 2s orbital with three 2p orbitals to produce four sp3 orbitals. The e- are distributed throughout the hybrids ready for bonding. End to end overlap with a 1s orbital from H gives four sigma bond in CH4.
sp3 is tetrahedral shape. In water we have AX2E2
What is the electron geometry, the molecular geometry at each carbon atom? Use that information to determine the hybridization around each carbon atom in nicotinic acid? How many sigma and pi bonds are in nicotinic acid?
Example 2: sp3 hybridization in NH3.
Tetrahedral Electron Geometry AX3E1Trigonal Pyramidal Molecular Geometry
sp3d hybridization in PCl5.
Isolated P atom
Trigonal Bipyramidal Electron Geometry AX5E0Trigonal BiPyramidal Molecular Geometry
The sp3d2 hybrid orbitals in SF6Octahedral Electron Geometry AX6E0Octahedral Molecular Geometry
Describe the types of bonds and orbitals in acetone, (CH3)2CO and in CO2 and in HCN?
Molecular formula
Lewis structure
VSEPRTGeometry
Hybrid orbitals
Step 1 Step 2 Step 3
Describe the types of bonds and orbitals in acetone, (CH3)2CO.PLAN:
Draw the Lewis structures to ascertain the arrangement of groups and shape at each central atom. Postulate the hybrid orbitals taking note of geometries predicted from VSEPRT. Draw the orbitals and show overlap.
SOLUTION:
sp3 hybridized
sp3 hybridized
sp2 hybridized
" bonds# bond
Postulating Hybrid Orbitals in a Molecule
PROBLEM: Use partial orbital diagrams to describe mixing of the atomic orbitals of the central atom leads to hybrid orbitals in each of the following:
PLAN: Use the Lewis structures to ascertain the arrangement of groups and shape of each molecule. Postulate the hybrid orbitals. Use partial orbital box diagrams to indicate the hybrid for the central atoms.
(a) Methanol, CH3OH (b) Sulfur tetrafluoride, SF4
SOLUTION: (a) CH3OH The groups around C are arranged as a tetrahedron.O also has a tetrahedral arrangement with 2 nonbonding e- pairs.
Postulating Hybrid Orbitals in a Molecule
(a) Methanol, CH3OH
SOLUTION: (a) CH3OH The groups around C are arranged as a tetrahedron.O also has a tetrahedral arrangement with 2 nonbonding e- pairs.
single C atom hybridized C atom single O atom hybridized
O atom
11-
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Postulating Hybrid Orbitals in a Molecule
(b) SF4 has a seesaw shape with 4 bonding and 1 nonbonding e- pairs.
S atomhybridized
S atom
Bond order is the number of bonds between two bonded atoms.
S C NS
F
F
F
FF
F
N N
Bond order = 3C-N: Bond order = 2
S-C: Bond order = 2
S-FBond order = 1
– Single bond between 2 atoms = order = 1– Double bond between 2 atoms = order = 2– Triple bond between 2 atoms = order = 3
Higher bond orders give shorter bond lengths and require more energy to break a bond.
Bond LengthsTriple bond < Double Bond < Single Bond
Note how bond energies (energy required to break a bond) goes up as bond order increases.
Ethane (CH3CH3) sp3 hybrid
both C are sp3 hybrids
H and C are s-sp3 overlaps to " bonds
sp3-sp3 C overlap forms a " bond
sp hybridization showing " and # bonds in acetylene (C2H2).
overlap in one position - "
p overlap - #
