1. Bonding

2. Ionic Bonds: Formed between ions. Transfer of electrons occurring. Covalent Bonds: Molecules form between atoms that share electrons. Metallic Bonds: Form multiple bonds using the free flowing electrons of a metal.

3. Atoms tend to gain and lose electrons until they have 8 valence electrons similar to noble gases. This is a stable electron configuration as shown experimentally by the high ionization energy and low electron affinity.

4.  An atom with a low ionization energy reacts with an atom with high electron affinity.  The electron moves.  Opposite charges hold the atoms together.

6.  Lattice energy - the energy associated with making a solid ionic compound from its gaseous ions.  M + (g) + X - (g)  MX(s)  This is the energy that “pays” for making ionic compounds.  Energy is a state function so we can get from reactants to products in a round about way.

7.  First sublime NaNa(s)  Na(g)  H = 109 kJ/mol  Ionize Na(g) Na(g)  Na + (g) + e -  H = 495 kJ/mol  Break F-F Bond½F 2 (g)  F(g)  H = 77 kJ/mol  Add electron to F F(g) + e -  F - (g)  H = -328 kJ/mol  Formation of solid Na+ + F-  NaF (s)  H = -575 kJ/mol  Lattice energy Na(s) + ½F 2 (g)  NaF(s)  H = -928 kJ/mol

9.  Electrons are shared by atoms.  Ionic and Covalent are extremes.  In between are polar covalent bonds.  The electrons are not shared evenly.  One end is partially positive, the other negative. Indicated using small delta 

10. H - F ++ -- ++ -- ++ -- ++ -- ++ -- ++ -- ++ -- ++ -- + -

11. ++ -- ++ -- ++ -- ++ -- ++ -- ++ -- ++ -- ++ -- - +

12. Electronegativity difference Bond Type Zero Intermediate Large Covalent Polar Covalent Ionic Covalent Character decreases Ionic Character increases

13.  The electrons in each atom are attracted to the nucleus of the other.  The electrons repel each other,  The nuclei repel each other.  They reach a distance with the lowest possible energy.  The distance between is the bond length.

14. 0 Energy Internuclear Distance Bond Length

15. 0 Energy Internuclear Distance Bond Energy

16.  The ability of an electron to attract shared electrons to itself.  Tends to increase left to right.  Decreases as you go down a group.  Noble gases aren’t discussed.  Difference in electronegativity between atoms tells us how polar.

17.  A molecule with a center of negative charge and a center of positive charge is dipolar (two poles),  We say that is has a dipole moment.  Center of charge doesn’t have to be on an atom.  Will line up in the presence of an electric field. H - F ++ --

18. % Ionic Character Electronegativity difference 25% 50% 75% LiBr LiF HCl

19.  The forces that cause a group of atoms to behave as a unit.  Why?  Due to the tendency of atoms to achieve the lowest energy state.  It takes 1652 kJ to dissociate a mole of CH 4 into its ions  Since each hydrogen is hooked to the carbon, we get the average energy = 413 kJ/mol

20.  We made some simplifications in describing the bond energy of CH 4  Each C-H bond has a different energy. CH 4  CH 3 + H  H = 435 kJ/mol CH 3  CH 2 + H  H = 453 kJ/mol CH 2  CH + H  H = 425 kJ/mol CH  C + H  H = 339 kJ/mol  Each bond is sensitive to its environment.

21.  single bond one pair of electrons is shared.  double bond two pair of electrons are shared.  triple bond three pair of electrons are shared.  More bonds, shorter bond length.

22.  We can find  H for a reaction.  It takes energy to break bonds, and end up with atoms (+).  We get energy when we use atoms to form bonds (-).  If we add up the energy it took to break the bonds, and subtract the energy we get from forming the bonds we get the  H.  Energy and Enthalpy are state functions.

23. 2 CH 2 = CHCH 3 +2NH 3 O2O2 +  2 CH 2 = CHC  N +6 H 2 O C-H 413 kJ/mol C=C 614kJ/mol N-H 391 kJ/mol O-H 467 kJ/mol O=O 495 kJ/mol C  N 891 kJ/mol C-C 347 kJ/mol

24.  Simple model, easily applied.  A molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms. Three Parts 1) Valence electrons using Lewis structures 2) Prediction of geometry using VSEPR 3) Description of the types of orbitals

25.  Shows how the valence electrons are arranged.  One dot for each valence electron.  A stable compound has all its atoms with a noble gas configuration.  Hydrogen follows the duet rule.  The rest follow the octet rule.  Bonding pair is the one between the symbols.

26. 1. Sum the valence electrons. 2. Use a pair to form a bond between each pair of atoms. 3. Arrange the rest to fulfill the octet rule (except for H and the duet). 4. Place left over electrons on the central atom. 5. If there are not enough electrons to give the central atom an octet, try multiple bonds. H2OH2O  CO 2  CN -

27.  The difference between the number of valence electrons on the free atom and that assigned in the molecule.  We count half the electrons in each bond as “belonging” to the atom.  Molecules try to achieve as low a formal charge as possible.  Negative formal charges should be on electronegative elements.  SO 4 2-

28.  Sometimes there is more than one valid structure for an molecule or ion.  Use double arrows to indicate it is the “average” of the structures.  It doesn’t switch between them.  Localized electron model is based on pairs of electrons, doesn’t deal with odd numbers.  NO 3 -  NO 2 -

29.  XeO 3  NO 4 -3  SO 2 Cl 2

30. 1. Odd numbers of electrons  NO 2. Atoms with fewer than octet  Be and B often do not achieve octet  BF 3 3. Atoms with more than octet  Third row and larger elements can exceed the octet  SF 6  Use 3d orbitals?

31.  When we must exceed the octet, extra electrons go on central atom.  ClF 3  XeO 3  ICl 4 -  BeCl 2

32.  Lewis structures tell us how the atoms are connected to each other.  They don’t tell us anything about shape.  The shape of a molecule can greatly affect its properties.  Valence Shell Electron Pair Repulsion Theory allows us to predict geometry  Molecules take a shape that puts electron pairs as far away from each other as possible.

33.  Have to draw the Lewis structure to determine electron domains.  bonding  nonbonding lone pair  Lone pair takes more space.  Multiple bonds count as one domain.  The number of domains determines  bond angles  underlying structure  The number of atoms determines  actual shape

34. Electron Domains Bond Angles Underlying Shape 2180° Linear 3120° Trigonal Planar 4109.5° Tetrahedral 5 90° & 120° Trigonal Bipyramidal 6 90°Octagonal

36. Electron Domain  Just count the number of electron domains and the geometry corresponds to that number. Molecular Shape  Electron domain geometry is many times not the shape of the molecule. Here we are worried about just the atoms.

37.  Multiple molecular geometries can result from the same electron geometry.

38.  Non-bonding pairs take up more space than bonding pairs and therefore change the angles within the atom.

39.  Multiple bonds create a high electron density on one side of the molecule which also changes angles.

40.  Linear  Trigonal Planar  Bent  Tetrahedral  Trigonal pyramidal  Trigonal bipyramidal  Seesaw  T-shaped  Octahedral  Square pyramidal  Square planar

41.  Individual bonds can have bond dipoles but the overall molecules polarity may be different.

42.  Individual bond dipoles (vectors) can be added to give the entire molecular dipole.

44. Beryllium: Has no single occupied orbitals therefore should not bond. If it absorbs energy to excite an electron to 2p it has 2 electrons that can form a bond.

45. The s and p orbitals can combine to make a hybrid orbital called the sp hybridized orbital. -sp orbitals have a linear position and is consistent with molecular geometries. -it has 2 degenerate orbitals.

46.  Other hybrid orbitals that can be made include sp 2 and sp 3. This occurs when there is more than one p orbital overlapping with an s orbital.

47.  Sigma (σ) bonds overlap head to head and electron density is in line with the internuclear axis.  Pi (π) bonds overlap side to side and electron density is above and below internuclear axis.

48.  Single bonds are all sigma bonds.  Multiple bonds have one sigma bond and the rest are pi.

50.  Pi electrons are not localized over one N-O bond but delocalized over all 3 bonds.  This is related to resonance.

51.  In MO theory, we invoke the wave nature of electrons.  If waves interact constructively, the resulting orbital is lower in energy: a bonding molecular orbital.

52.  In hydrogen the electrons go into the bonding orbital because it is lower in energy.  The bond order is one half the difference between the number of bonding and antibonding electrons.  For hydrogen, with two electrons in the bonding MO and none in the antibonding MO, the bond order is 2-0/2 = 1

53.  In the case of He 2, the bond order would be 1212 (2 − 2) = 0 Therefore, He 2 does not exist.

54.  For atoms with both s and p orbitals, there are two types of interactions:  The s and the p orbitals that face each other overlap in  fashion.  The other two sets of p orbitals overlap in  fashion.

55.  The resulting MO diagram looks like this (Fig. 9.41).  There are both  and  bonding molecular orbitals and  * and  * antibonding molecular orbitals.