1. Unit 3
Bonding and VSEPR

2. October 31, 2016 Happy Halloween!! Today’s Agenda:
Turn in Absorbance Lab
Notes: Bonding and Lewis Structures
Practice: Lewis Structures
HW: Reading Guide

3. Types of Bonding Ionic bonds: electrons are taken
Covalent bonds: electrons are shared
Metallic bonds: electrons are pooled
There are a great deal of attractive and repulsive forces within and between atoms that can interfere or lead to bonding. When the net interactions lead to a reduction of energy, a bond occurs.
In ALL cases, bonding occurs to lower the potential energy (increase coulombic attraction) between the charged particles that compose atoms.

4. Bonding Theories The Lewis Model
Valence bond theory (Valence Shell Electron Pair Repulsion Theory)
Molecular Orbital Theory

5. The Lewis Model Developed by an American (yay!) chemist: G.N. Lewis
Uses valence electrons represented as dots to predict whether atoms will form stable molecules.

6. Drawing Lewis Dot for single atoms/ions
Determine number of valence e-
Place dots (electrons) around the symbol for the atom one at a time
Make sure to place them singly first before pairing them up
For example….
Lets look at
Carbon
Calcium
Selenium
Iodine

7. Lone and Paired electrons
Valence electrons fall into 2 categories:
Paired electrons
Lone electrons
Only lone electrons can form bonds
TYPICALLY, the number of lone electrons=number of bonds formed
TYPICALLY, how many bonds can Oxygen form?
Paired electrons
Lone electrons

8. Illustrating Ionic Bonds using Lewis
Using Lewis dot diagrams, illustrate the bond formed between
Lithium and fluorine
Magnesium and chlorine

9. Illustrating Covalent Bonds using Lewis
Shared electrons are bonding electrons and are indicated using a line.
Let’s look at
Water
Diatomic fluorine

10. Formal Charges
The charge an atom would have if all bonding electrons were shared equally between the bonded atoms.
𝑓𝑜𝑟𝑚𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒=# 𝑜𝑓 𝑣𝑎𝑙𝑒𝑛𝑐𝑒 𝑒−(# 𝑜𝑓 𝑛𝑜𝑛𝑏𝑜𝑛𝑑𝑖𝑛𝑔 𝑒 # 𝑜𝑓 𝑠ℎ𝑎𝑟𝑒𝑑 𝑒)
Example: HCl
Example: Draw the Lewis structure for Chlorine trioxide. Assign formal charges to each atom.

11. Resonance
When one of two or more Lewis structures have the same skeletal formula but different electron arrangements

12. Formal Charges
The sum of all formal charges in a neutral compound must be zero.
The sum of all formal charges in an ion must equal the charge of the ion.
Smaller magnitude of charge is always more stable.
When placing e- in a Lewis structure, always consider that more electronegative atoms prefer a negative formal charge

13. November 1, 2016 Today’s Agenda WB Lewis Structures
Notes: VSEPR Shapes and Polarity
Practice: Lewis Structures, VSEPR Shapes, and Polarity
HW: Reading guide

14. Bond Polarity
When electrons are shared between atoms, the sharing is not always equal.
Unequal sharing of electrons is said to be a polar bond.
A polar covalent bond is an intermediate between a covalent and ionic bond.
Polarity is determined using electronegativity values.

16. Bond Type by ΔEN Electronegativity Difference Bond Type <0.4
Based on the difference in electronegativities of atoms we can predict the type of bond that will form
Electronegativity Difference
Bond Type
<0.4
Nonpolar covalent
Polar covalent
>2.0
Ionic

17. Examples Electronegativity difference= 4.0 – 0.8 = 3.2 IONIC BOND
Potassium Fluoride KF
Electronegativity difference= 4.0 – 0.8 = 3.2
IONIC BOND
Two Oxygen Atoms O2
Electronegativity difference= 3.5 – 3.5 = 0
NON-POLAR COVALENT
Carbon Tetrachloride CCl4
Electronegativity difference= 3.0 – 2.5 = 0.5
POLAR COVALENT

18. What this means for the bond…
With respect to polar covalent bonds, the differences in electronegativity tell us about the sharing of electrons Example: Carbon Tetrachloride (CCl4) Cl has EN = 3.0 C has EN = 2.5 From this, we say that chlorine has stronger attraction for electrons than carbon Thus, electrons will spend more time around the Cl than C

19. Partial Charges
A slight separation of positive and negative charges are called “partial charges” and represent them as δ+ or δ-
Example: CCl4
Chlorine with greater EN will have greater attraction of e- and thus will have partial negative charge δ-
Carbon with lower EN will have less attraction of e- and thus will have partial positive charge δ+
Shown as
δ+C-Cl δ-

20. Dipoles
When the bond is separated into partial positive and negative charges we call this bond a dipole
We represent dipole with a vector arrow that points to the more electronegative atom
Example CCl4
δ+C-Cl δ-

21. Dipole Moments The magnitude of the dipole is the dipole moment (μ)
μ=qr
Where q is the charge of the proton or electron and r is the distance
Measured in debyes (D)
How far the electrons being unequally shared are being pulled away from the nucleus

22. Valence Shell Electron Pair Repulsion
Lone pairs of electrons bonding pairs repel one another because of Coulombic forces.
The repulsion determines the geometry of the molecule because the electrons want to be as separated as possible.
Electron pairs will repel more than bonding electrons.

23. We use our information on polar bonds to predict whether molecules will be polar or non- polar
We also must know our VSEPR shapes in order to do this!!

24. VSEPR Shapes See handout and Table 10.1 on page 434
Electron groups: any bonding on lone groups of electrons (dots and lines)
Bonding groups: any bonding group on the central atom (lines; double and triple bonds count as one bonding group)

25. Polar Molecules occur when…
There are polar bonds in the molecule AND
The molecule has an asymmetrical shape OR
The molecule’s peripheral atoms in a symmetrical shape have different electronegativities.

26. Carbon Tetrachloride CCl4
Determine bond type
EN difference= 3.0 – 2.5 = 0.5
Thus is POLAR COVALENT
Determine partial charges
Cl has greater EN than C
Our partial charges are:
Tetrahedral shape according to VSEPR

27. Partial negative charges are symmetrically distributed around the molecule.
This results in a molecule that is NON-POLAR

28. Water H20 Electronegativity difference= 3.5 – 2.1 = 1.4
Determine bond type
Electronegativity difference= 3.5 – 2.1 = 1.4
Thus is POLAR COVALENT
Determine partial charges
O has higher EN and H has lower EN
Our partial charges are:
If we include the dipoles
Bent shape according to VSEPR

29. The distribution of electrons is asymmetrical.
This is where VSEPR is important!
Water has two partially positive ends and one partially negative end
The distribution of electrons is asymmetrical.
We say that the molecule is POLAR

30. Carbon Dioxide CO2 Determine partial charges If we include the dipoles
Determine bond type
Electronegativity difference= 3.5 – 2.5 = 1.0
Thus is POLAR COVALENT
Determine partial charges
O has greater EN than C
Our partial charges are:
If we include the dipoles
Linear shape according to VSEPR

31. The dipoles created in this molecule are symmetrical (both ends are negative) and will cancel each other out.
This molecule is said to be NON-POLAR

32. Polarity of Molecules Checklist
Are there polar bonds in the molecule?
Is the molecule symmetrical?
Are all the peripheral (terminal) atoms the same electronegativity?

33. Examples to Try
Determine whether the following molecules will be polar or non-polar
SI2
CH3F

35. Bond energy
The energy required to break 1 mole of a bond in the gas phase.
Average bond energies can be used to estimate the enthalpy change of a reaction.
Enthalpy: the energy associated with the breaking and forming bonds in a chemical reaction
Breaking a bond is endothermic (positive bond energy; needs energy)
Forming a bond is exothermic (negative bond energy; releases energy)

36. Bond Lengths
The average length of a bond between two particular atoms in a large number of compounds.
In general bond lengths vary between single, double and triple bonds.
Single bond> double bond> triple bond

37. Ionic structures
Ionic compounds form a crystal lattice Evenly spaced, alternating positive and negative ions

38. Lattice Energy
The energy released with the formation of an ionic crystalline lattice
Formation is highly exothermic.

39. Trends in Lattice Energy
Remember Coulomb’s Law?!
𝐸= 1 4π ϵ 0 𝑞 1 𝑞 2 𝑟
Lattice energies become more exothermic
with increasing magnitude of ionic charge
With decreasing atomic radius

40. Properties of Ionic Compounds
High melting points
High boiling points
Conductive when dissolved in water
Brittle and hard

41. Properties of Covalent Compounds
Low melting points
Low boiling points
Not conductive when dissolved in water
Generally more soft and malleable
(covalent compounds are made up of a specific group of atoms, not a “network” of atoms like ionic bonds)

42. Metallic Bonding
Electron sea model: when metal atoms bond together to form a solid, each metal atom donates one or more electrons to an electron sea.
These free moving electrons are called delocalized electrons.
The electrons move freely among positive ions, thus making metals good conductors of electricity and heat
Since there are no dedicated “bonds” in the electron sea, metals are malleable and ductile

43. Practice
Which ionic compound would have the most exothermic lattice energy: CaCl2 or MgCl2? Explain.

44. Practice
Which compound would you predict to have a higher melting point: sodium carbonate or sucrose? Explain.

45. Calculating enthalpy of rxn from bond energy
Enthalpy: the energy associated with the breaking and forming bonds in a chemical reaction
∆ 𝐻 𝑟𝑥𝑛 =𝛴 ∆ 𝐻 ′ 𝑠 𝑏𝑜𝑛𝑑𝑠 𝑏𝑟𝑜𝑘𝑒𝑛 + 𝛴(∆ 𝐻 ′ 𝑠 𝑏𝑜𝑛𝑑𝑠 𝑓𝑜𝑟𝑚𝑒𝑑)

46. Valence Bond Theory
Electrons reside in quantum-mechanical orbitals, localized on individual atoms.
Either in s, p, d, and f orbitals
Or hybridized atomic orbitals
A bond is a result of the overlap of two half-filled orbitals (or, sometimes, the overlap of a completely filled and empty orbital

48. Orbital Hybridization
Standard atomic orbitals are combined to form new atomic orbitals called hybrid orbitals.
Orbitals in a molecule are different from orbitals in an atom (shapes and energies change)
In hybrid orbitals, there is greater overlap and minimized energy.
Focus: central atoms and their hybridization

49. Predicting Hybridization
The number of standard atomic orbitals always equals the number of hybrid orbitals formed
Use VSEPR shapes to predict the hybridization.

50. sp3 hybridization Example: CH4
Carbon only has 2 half-filled orbitals to bond with hydrogen, yet we see 4 bonds
Carbon’s 2s and 2p orbitals hybridize to form 4 sp3 hybrid orbitals (which all have the same energy)
The four hybrid orbitals are all half-filled, allowing carbon to make four bonds.

51. Practice
If hydrogen is used as fuel, it could be burned according to the equation:
H2 (g) + ½ O2 (g)  H20 (g)
Use average bond energies (p. 410) to calculate the enthalpy of this reaction. If the enthalpy of combustion of methane is -804 kJ/mol , which fuel yields more energy per mole?
-243 kj/mol
p. 422 # 103

52. sp3 hybridization Example: CH4
Carbon only has 2 half-filled orbitals to bond with hydrogen, yet we see 4 bonds
Carbon’s 2s and 2p orbitals hybridize to form 4 sp3 hybrid orbitals (which all have the same energy)
The four hybrid orbitals are all half-filled, allowing carbon to make four bonds.

53. sp2 hybridization *double bonds
one s orbital and three p orbitals hybridize to form three sp2 orbitals and one unhybridized p orbital
This will form two single bonds and one double bond
Example: carbon in H2CO

54. Sigma and pi bonds
Sigma (σ) bonds are formed from the direct, end-to-end overlap of orbitals
Pi (π) bonds are formed from the side by side overlap of p orbitals

55. sp hybridization *triple bonds
One s orbital and three p orbitals hybridize to form two sp orbitals and two unhybridized p orbitals
Forms one single bond and one triple bond
Example: carbon in C2H2

56. sp3d and sp3d2
One s, three p, and one d  five sp3d and four unhybridized d
Shape: trigonal bipyramidal
One s, three p, and two d  six sp3d2 and three unhybridized
Shape: octahedral

58. Practice
Write a hybridization and bonding scheme for BrF5. Sketch the structure, including overlapping orbitals, and label all bonds (sigma or pi). Look at examples 10.6 and 10.7 to help you.
p. 476 #63

59. Practice
Write a hybridization and bonding scheme for N2H2. Sketch the structure, including overlapping orbitals, and label all bonds (sigma or pi). (Hint: both nitrogens are hybridized)