Chapter 10 Shapes of Molecules

1. Chapter 10Shapes of Molecules The Three Dimensional Reality of Molecular Shapes

2. The Three Dimensional Reality of Molecular Shapes Each atom, bonding e- pair, and lone e- pair has its own position in 3D space. Position determined by the attractive and repulsive forces that govern all matter.

3. Molecular Shape is Crucial to Life Processes:

4. Molecular Shape is Important in Nanotechnology: Nanotechnology: “Building tiny machines at the molecular level.” Buckyball (fullerenes) – a C60 structure discovered in soot in 1985 Many, many uses: One use may be to “sneak” medicines into cells or through the blood brain barrier, where some substances (medicines) may not normally enter.

5. Li Different elements can have the same number of dots Be Mg Same Group (Column) Section 10.1: Depicting Molecules and Ions with Lewis Dot Structures 2D before 3D Lewis dot structures consist of two parts: (1) Element symbol – nucleus + inner electrons Ex: The element lithium has an element symbol Li (2) Surrounding dots – valence electrons (outer most shell)

6. Electron types: Bonding Pairs & Lone Pairs Bond types: Single, Double, Triple • Steps for writing Lewis Structures for molecules that have only single bonds: • Place the atoms relative to each other, with the atom with the lower group (column) • number in the center. Often, center atom is also atom with lower E.N. • Example: NF3 F N If atoms have same groups number, place atom with higher period (row) number in the center. Example: SO3 F F O S O O Section 10.1: Depicting Molecules and Ions with Lewis Dot Structures

7. Continued F N F F • Steps for writing Lewis Structures for molecules that have only single bonds: • Determine the total number of valence e- available • Example: NF3 N  5 e- F  7 e- (x 3) = 21 e- Total = 26 e- • Draw a single bond from each surrounding atom to the central atom, and subtract • 2 valence e-’s for each bond to find the # of e- remaining • Example: NF3 3 single bonds x 2 e- = 6 e- • e- remaining = total – bonded = 26 – 6 = 20 e- • Distribute the remaining e- in pairs so that each atom has 8 e- (except H, has 2 e-) • 1st: Place lone e- pairs on surrounding (more EN) atoms to give each full valence. • 2nd: If any e- remain, place them on the central atom. • Example: NF3 (5) Check that each has a full valence shell. DONE.

8. F F N F This method works for singly bonded compounds where C, N, and O, and elements in higher periods (rows), are the central atom. F N F F In nearly all compounds: • H atoms form 1 bond • C forms 4 bonds • N forms 3 bonds • O forms 2 bonds • Halogens (F, Cl, Br, I,…) form 1 bond F N F F Section 10.1: Depicting Molecules and Ions with Lewis Dot Structures Lewis Dot Structures are not 3D, so several different depictions are correct. Etc…….

9. Section 10.1: Depicting Molecules and Ions with Lewis Dot Structures In cases of multiple bonds (double, triple), there is an additional step: (6) If a central atom still does not have an octet (after Step 4), make a multiple bond by changing a lone pair from one of the surrounding atoms into a bonding pair to the central atom. Examples: CH4 N2 In cases where there is a polyatomic ion, Lewis Dot Structure is shown in square brackets:

10. Section 10.1: Resonance Structures Have the same relative placement of atoms, but different locations of bonding and lone e- pairs. Example: Nitrite (NO2-) Double bonds change location. Lone pair on O atom changes location. Neither structure depicts nitrite accurately: The two O,N bonds in this compound have bond lengths and bond energies that lie somewhere between the O–N and the O=N bond.

11. Section 10.1: Resonance Hybrids Resonance structures are not real bonding depictions: Structure I Structure II In reality, Structure I and Structure II do not switch back and forth from one instant to the next. The actual nitrite molecule is an average of the two. The e-’s are delocalized over the entire molecule. (Just as e-’s in a metallic bond are delocalized around the entire sea of electrons.)

12. Section 10.1: Resonance Hybrids Sometimes implied (no indication). Sometimes indicated by dotted lines. Example: Ozone (O3) Example: Benzene (C6H6)

13. Draw a Lewis Dot Structure for: CCl4 CCl2F2 CH4O NH3O C2H6O (no O–H bonds) 10.6 10.8 Resonance Structures: 10.10 10.12

14. Section 10.1: Which is the more important resonance structure? All resonance forms contribute equally when central atom has surrounding atoms that are all the same. When atoms surrounding the central atom are not the same: One resonance form may “weight” the average (In other words, it “counts more” than the other forms.) Determining the most important resonance form: Determine each atom’s formal charge – the charge it would have if the bonding electrons were equally shared,

15. Section 10.1: Which is the more important resonance structure? Formal charge – Determined for each atom in a compound # valence e- - (# unshared valance e- + ½ # shared valence e-) Example: Ozone (O3) Formal charge for OA = 6 – (4 – ½*4) = 0 Formal charge for OB = 6 – (2 – ½*6) = +1 Formal charge for OC = 6 – (6 – ½*2) = -1 In the case of ozone, both resonance structures (I and II) have the same formal charges (but on different O atoms) so they contribute equally to the resonance hybrid. Formal charges must sum to the total charge on the chemical species.

16. Section 10.1: Which is the more important resonance structure? Formal charge – Determined for each atom in a compound Recall – In nearly all compounds: • H atoms form 1 bond • C forms 4 bonds • N forms 3 bonds • O forms 2 bonds • Halogens (F, Cl, Br, I,…) form 1 bond When formal charge is 0, an atom has its usual # of bonds. Ozone Example: Formal charge for OA = 6 – (4 – ½*4) = 0 Formal charge for OB = 6 – (2 – ½*6) = +1 Formal charge for OC = 6 – (6 – ½*2) = -1

17. Section 10.1: Which is the more important resonance structure? What about a case where the resonance structures do not contribute equally? Occurs when there are different atoms around the central atom. Example: Cyanate ion, NCO- Criteria for choosing the more important resonance structure: (1) Smaller formal charges (+ or -) are preferable to larger ones. Resonance form I is out. (2) The same nonzero formal charges on adjacent atoms are not preferred. Not applicable to this example. (3) A more negative formal charge should reside on a more electronegative atom. O is more EN than N, so Resonance form III is the winner.

18. Section 10.1: Formal charge is not the same as the ON What is the difference? Formal charge - Bonding e-’s are assigned equally to the atoms, so that each atom has ½ Formal charge = # valence e- - (# lone pair e- + ½ # bonding e-) Oxidation number - Bonding e-’s are assigned completely to the more EN atom O.N. = # valence e- - (# lone pair e- + # bonding e-)

19. Section 10.1: Exceptions (Limitations) to the Octet and Formal Charge Rules (1) e- deficient molecules – gaseous molecules containing Be or B as central atom Halogens much more EN than Be or B = Formal charge rules make sharing of extra lone pairs by halogens unlikely Be: 4 e- B: 6 e- (2) Odd e- molecules– central atom has odd # of valence e- free radicals – very reactive (b/c very unstable): often react with each other to pair up their lone e- (make you age) (3) Expanded valence shells – more than 8 valence e-; occurs only where d orbitals are available  Row 3 or higher (Review of orbital types: p289 – 295; s – 2 e-, p – 6 e-, d – 10 e-, f – 14 e-)

20. Resonance Structures: 10.14a 10.16b Other suggested problems: 10.15 10.17 10.19 10.24

21. Formal charge = # valence e- - (# unshared valance e- + ½ # shared valence e-) Criteria for choosing the more important resonance structure: (1) Smaller formal charges (+ or -) are preferable to larger ones. Resonance form I is out. (2) The same nonzero formal charges on adjacent atoms are not preferred. Not applicable to this example. (3) A more negative formal charge should reside on a more electronegative atom. O is more EN than N, so Resonance form III is the winner.

22. Section 10.2: Valence-shell electron-pair repulsion (VSEPR) theory Molecular shape is important in many, many scientific disciplines. Medicine: Receptors Nanotechnology: Membrane Transport

23. Section 10.2: Valence-shell electron-pair repulsion (VSEPR) theory Molecular shape is important in many, many scientific disciplines. Ecology: Talking trees Jack Schultz, Chemical Ecologist

24. Section 10.2: Valence-shell electron-pair repulsion (VSEPR) theory Lewis Dot Structures, 2D (Blueprint) VSEPR, 3D (House)

25. Section 10.2: Valence-shell electron-pair repulsion (VSEPR) theory Each group of valence electrons around a central atom is located as far away as possible from the others in order to minimize repulsions. e- group can be: a single bond, double bond, triple bond, lone pair, lone e-

27. Section 10.3: Molecular Shape and Molecular Polarity In molecules with more than 2 atoms: Shape and bond polarity determine molecular polarity. Molecules with only 2 atoms. Relative electronegativities of the two atoms determine polarity.

28. Section 10.3: Molecular Shape and Molecular Polarity In molecules with more than 2 atoms: Dipole moments Dipole moments – a measure of molecular polarity magnitude of partial charges on ends of a molecule (in coulombs) x distance between them Behavior of Molecules With and Without Dipole Moments Electric field: Polar molecules (which have a dipole moment) orient with partial charges towards oppositely charged electric plates. Molecules with out a dipole moment will not orient themselves in any particular direction. *Molecules with no dipole moment can be polar.

29. Section 10.3: Molecular Shape and Molecular Polarity Dipole moments: When molecular shape influences polarity Large ∆EN between C (EN = 2.5) and O (EN = 3.5)  C = O bonds are polar CO2 molecule is linear  Two identical bond polarities are counterbalanced (in other words, they cancel each other out) As a result, CO2 has no net dipole moment.

30. Section 10.3: Molecular Shape and Molecular Polarity Dipole moments: When molecular shape influences polarity H2O (like CO2) also has two identical molecules bonded to the central atom. However, H2O (unlike CO2) has a dipole moment. Bond polarities are not canceled out because of the shape of the water molecule: V-shaped rather than linear. The O end of the molecule is more negative than the H ends

31. Section 10.3: Molecular Shape and Molecular Polarity When different molecules have the same shape, the nature of the atoms Surrounding the central atom can have a major effect on polarity. CCl4 – does not have a dipole CHCl3 – has a dipole

32. Effect of Molecular Polarity on Behavior Example: Boiling point of NH3 versus PH3 Why is NH3 boiling point higher? Also determines reaction behavior: NH3 + H+ NH4+ (p392)

33. A closer look at molecular shapes: double bonds and lone pairs Bond Angles: Idealized vs. Actual

34. A closer look at molecular shapes: double bonds Effect of double bondson bond angle when surrounding atoms are different. Rule: The double bond, with its greater e- density, repels the two single bonds more strongly than they repel each other.

35. A closer look at molecular shapes: lone pairs Effect of lone pairson bond angle. Rule: Lone pairs repel bonding pairs more strongly than bonding pairs repel each other.

36. A closer look at molecular shapes: a few more details Bond angles for: Equatorial groups = 90º Axial groups = 120º General Rule: The greater the bond angle, the weaker the repulsion. In this case: Equatorial-equatorial repulsions are weaker than axial- rquatorial repulsions. Implication: Lone pairs, which exert stronger repulsions, will tend to occupy equatorial positions.

37. A closer look at molecular shapes: a few more details General Rule: The greater the bond angle, the weaker the repulsion. Implication: If two lone pairs are present, they will always occupy opposite vertices (furthest apart)

38. More VSEPR Practice (1) Lewis dot structure (dominant resonance form – calculate formal charge if needed) (2) 3-D geometry (Table 10.9) (3) Molecular polarity Bond dipoles cancel? Lone pairs present? Different surrounding atoms? GeH2 PCl5 SF4 ClF3 XeF2 SF6 BrF5 XeF4 NH4+ SO4- Additional Optional Homework Problems: 10.30, 10.35, 10.38, 10.39, 10.51, 10.53, 10.54, 10.97

39. Chemical Reaction vs. Physical Interactions (Chapter 12) N has greater EN than P: Why? Why is EN is inversely proportion to atomic radius? e- shielding Boiling point of NH3 versus PH3: Why is NH3 boiling point higher? The N – H bonds between NH3 molecules matter, not the N – H bonds within the NH3 molecule. EN is inversely proportion to atomic radius.

40. Summary So Far: Overall Polarity of a Molecule If a molecule is polar, it will: (1) have a net dipole moment (2) orient itself in an electric field If a molecule is nonpolar, it will: (1) not have a net dipole moment (2) will move about randomly in an electric field • Steps used to determine molecular polarity: • Draw the 2-D Lewis dot structure to determine the • number and types (single, double, triple) of bonds • present, and any lone pairs present. • *When dealing with several options (resonance structures), • determine the dominant resonant structure by: • 1. calculating formal charge for the atoms of each • molecular possibility that you are evaluating • 2. using the three criteria for selecting the dominant • resonance structure based on formal charge • (2) Determine the 3-D shape of the molecule • (3) Determine the overall polarity of the molecule

41. Is HCl symmetrical? Is CO2 symmetrical? In addition to the wavelength of energy interaction with the molecule The symmetry of the molecule (Lewis dot structure + VSEPR!!!) will also determine whether a photon can be absorbed. *Symmetry is NOT a net dipole moment: • Symmetrical = mirror image • Asymmetrical = not mirror image