1. Molecular Geometry and Bonding Theories

2. The properties of a molecule depend on its shape and and the nature of its bonds. In this unit, we will discuss three models. (1) a model for the geometry of molecules -- valence-shell electron-pair repulsion (VSEPR) theory (2) a model about WHY molecules form bonds and WHY they have the shape they do -- valence-bond theory (3) a model of chemical bonding that deals with the electronic structure of molecules -- molecular orbital (MO) theory

3. CO 2 bond angles: the angles made by the lines joining the nuclei of a molecule’s atoms CH 4 CH 2 O carbon dioxidemethaneformaldehyde 180 o 120 o 109.5 o

4. VSEPR electron domain: a region in which at least two electrons are found -- they repel each other because… they are all (–) bonding domain: 2-to-6 e – that are shared by two atoms; they form a… covalent bond nonbonding domain: 2 e – that are located on a single atom; also called a… lone pair For ammonia, there are three bonding domains and one nonbonding domain. N–HH– H.. N H H H 4 e – domains NH 3 Domains arrange themselves so as to minimize their repulsions.

5. The electron-domain geometry is one of five basic arrangements of domains. -- it depends only on the total # of e – domains, NOT the kind of each domain The molecular geometry describes the orientation of the atoms in space. -- it depends on how many of each kind of e – domain N H H H..

6. Total # of Domains Electron-Domain Geometry Possible Molecular Geometries 2 3 4 5 6 linear linear (CO 2 ) trigonal planar trigonal planar (BF 3 ), bent (NO 2 ) tetrahedral tetrahedral (CH 4 ), trigonal pyramidal (NH 3 ), bent (H 2 O) trigonal bipyramidal trig. bipyramidal (PCl 5 ), linear (XeF 2 ) seesaw (SF 4 ), T-shaped (ClF 3 ) octahedral octahedral (SF 6 ), sq. pyr. (BrF 5 ), square planar (XeF 4 ) “atoms – axial”

7. To find the electron-domain geometry (EDG) and/or molecular geometry (MG), draw the Lewis structure. Multiple bonds count as a single domain. Predict the EDG and MG of each of the following. SnCl 3 – 26 e – [ ] Sn–ClCl–.. Cl –.. EDG: tetrahedral MG: trig. pyramidal O3O3 18 e – EDG: trig. planar MG: bent.. O –O O=.. SeCl 2 20 e – EDG: tetrahedral MG: bent Cl–Se–Cl.. O O– O=..

8. CO 3 2– 24 e – [ ] C= OO–.. O 2– EDG: trig. planar MG: trig. planar SF 4 34 e – EDG: trig. bipyr. MG: seesaw.. S –F F–.. F F IF 5 42 e – I.. F –F.. –F.. F–.. F EDG: octahedral MG: sq. pyramidal ClF 3 28 e – Cl –F F–.. F EDG: trig. bipyr. MG: T-shaped (res.)

9. ICl 4 – 36 e – EDG: octahedral MG: sq. planar [ ].. I –Cl Cl–.. Cl.. – For molecules with more than one central atom, simply apply the VSEPR model to each part. Predict the EDG and MG around the three interior atoms of ethanoic (acetic) acid. H–C–C–O–H.. H HO CH 3 COOH PORTIONEDGMG tetra.tetra. –CH 3 trig. plan.trig. plan. tetra.bent C=O – –.. –OH..

10. Nonbonding domains are attracted to only one nucleus; therefore, they are more spread out than are bonding domains. The effect is that nonbonding domains (i.e., “lone pairs”) compress bond angles. Domains for multiple bonds have a similar effect. e.g., CH 4 NH 3 H 2 O N H H H.. C H H H H O H H Cl O=C Cl.. the ideal bond angle for the tetrahedral EDG is 109.5 o 111.4 o EDG = trig. plan. ideal = 120 o COCl 2 124.3 o 124.3 o 109.5 o 107.0 o 104.5 o

11. Polarity of Molecules A molecule’s polarity is determined by its overall dipole, which is the vector sum of the dipoles of each of the molecule’s bonds. Consider CO 2 v. H 2 S... O=C=O.. S H H bond dipoles overall dipole = bond dipoles (NONPOLAR) (POLAR) zero

12. Boron can be extracted by the electrolysis of molten boron trichloride. Boron is an essential nutrient for plants, and is also a primary component of control rods in nuclear reactors. Classify as polar or nonpolar: PCl 3 BCl 3 26 e – P –Cl Cl–.. Cl.. polar.. B –Cl Cl–.. Cl.. nonpolar 24 e –

13. Valence-Bond Theory: merges Lewis structures w /the idea of atomic orbitals (2s, 3p, etc.) V-B theory says… covalent bonding occurs when valence orbitals of adjacent atoms overlap; then, two e – s of opposite spin (one from each atom) combine to form a bond Lewis theory says… covalent bonding occurs when atoms share electrons V-B theory is like Velcro: “No overlap, no bond.”

14. Consider H 2, Cl 2, and HCl... H Cl[Ne] H2H2 Cl 2 HCl (unpaired e – is in 1s orb.) (unpaired e – is in 3p orb.) = orbital overlap region (responsible for bond)

15. There is always an optimum distance between two bonded nuclei. At this optimum distance, attractive (+/–) and repulsive (+/+ and –/–) forces balance. H–H distance Energy 0 optimum dist. (min. PE) H 2 molecule

16. Hybridized Orbitals V-B theory can’t explain some observations about molecular compounds without the concept of hybridized orbitals. i.e., covalent (NM/NM) compounds

17. In… BeF 2 (LS) …the central atom SHOULD be… …but is ACTUALLY hybridized to be… …the result then being… 2 hyb. sp orbs. 2 unhyb. 2p orbs. 2s2p BF 3 3 hyb. sp 2 orbs. 1 unhyb. 2p orb. 2s2p sp sp 2 2p CH 4 4 hyb. sp 3 orbs. 2s2p sp 3 H2OH2O 4 hyb. sp 3 orbs. 2s2p sp 3 PF 5 3s3p sp 3 d 3d 5 hyb. sp 3 d orbs. 4 unhyb. 3d orbs. XeF 2 sp 3 d 5s5p5d 5 hyb. sp 3 d orbs. 4 unhyb. 5d orbs. XeF 4 sp 3 d 2 5s5p5d 6 hyb. sp 3 d 2 orbs. 3 unhyb. 5d orbs. : BeF : : : F : : B : : : F : F : : : F : : CHH H H OH H : : P : : : F : F : : : F : : : : : F : : : F : XeF : : : F : : : : : : F : : : F : : : : F :: F :: : :

18. KEY: EDG  hybridization of central atom linear  sp trig. planar  sp 2 tetrahedral  sp 3 trig. bipyr.  sp 3 d octahedral  sp 3 d 2

19. sp hybridization sp hybridization occurs when one s and one of the three p orbitals hybridize. It results in the creation of two sp hybrid orbitals (orange). The other two p orbitals (purple) are unhybridized; they retain the “figure-8” or “dumbbell” shape.

20. Multiple Bonds  (sigma) bonds are bonds in which the e – density is along the internuclear axis. -- These are the single bonds we have considered up to this point. -- e.g., s-s, s-p, or p-p overlap, and also p-sp hybrid overlap Multiple bonds result from the overlap of two p orbitals (one from each atom) that are oriented perpendicularly to the internuclear axis. These are  (pi) bonds. sigma (  ) = single pi, multiple, unhybridized N H H H.. The bonds in ammonia are  bonds.

21.  bonds are generally weaker than  bonds because  bonds have less overlap. CC H H H H For ethene (C 2 H 4 )… C=C – – – – H H H H For each carbon atom, there are 3 sp 2 hybridized orbitals; these form  bonds (- - - -) with C or H. Where unhybridized orbitals overlap, a  bond ( ) is formed.

22. Single bonds are  bonds. Double bonds consist of one  and one . Triple bonds consist of one  and two . e.g., C 2 H 2 e.g., C 2 H 4 e.g., C 2 H 6

23. Experiments indicate that all of C 2 H 4 ’s atoms lie in the same plane. This suggests that  bonds introduce rigidity (i.e., a reluctance to rotate) into molecules. --  bonding does NOT occur with sp 3 hybridization, only sp and sp 2 --  bonding is more prevalent with small molecules (e.g., C, N, O) (Big atoms don’t allow enough p-orb. overlap for  bonds to form.)

24. resonance structures. Delocalized  Bonding Localized  bonds are between… -- e.g., C 2 H 2, C 2 H 4, N 2 Delocalized  bonds are “smeared out” and shared among… -- These are common for molecules with… -- The electrons involved in these bonds are delocalized electrons. The nitrate ion (NO 3 – ) has delocalized electrons. [ ] N=OO–.. O – > two atoms. two atoms only.

25. Consider benzene, C 6 H 6. -- Each carbon atom is ___ hybridized. -- This leaves... a 2p orbital on each C that is oriented to plane of m’cule. C C C C CC H H H H H H -- 6 e – shared equally by 6 C atoms sp 2

26. Which of the following exhibit delocalized bonding? H 3 C–C–CH 3 O = SO 4 2– SO 2 sulfate ion sulfur dioxide 18 e – 32 e – O–S=O.. “YEP.” H H–C–C–C–H O =.. – – H – H H – “NOPE.” “NOPE.” [ ] S –O O–.. O O 2– 24 e – propanone (res.)

27. Molecules respond to the many wavelengths of light. The wavelengths that are absorbed and then re-emitted determine an object’s color, while the wavelengths that are NOT re-emitted raise the temperature of the object. Molecular Orbitals -- The VSEPR and valence-bond theories don’t explain the excited states of molecules, which come into play when molecules absorb and emit light. -- This is one thing that the molecular orbital (MO) theory attempts to explain.

28. molecular orbitals: wave functions that describe the locations of electrons in molecules -- these are analogous to atomic orbitals in atoms (e.g., 2s, 2p, 3s, 3d, etc.), but MOs are possible locations of electrons in molecules (not atoms) -- MOs, like atomic orbitals, can hold a maximum of two e – with opposite spins -- but MOs are for entire molecules MO theory is more powerful than valence-bond theory; its main drawback is that it isn’t easy to visualize.

29. Hydrogen (H 2 ) The overlap of two atomic orbitals produces two MOs. 1s H atomic orbitals + H 2 molecular orbitals  1s (bonding MO) (antibonding MO)  * 1s -- The lower-energy bonding molecular orbital concentrates e – density between nuclei. -- For the higher-energy antibonding molecular orbital, the e – density is concentrated outside the nuclei. -- Both of these are  molecular orbitals.

30. Energy-level diagram (molecular orbital diagram) 1s  1s  * 1s H atom H 2 m’cule H atom 1s H atomic orbitals + H 2 molecular orbitals  1s (bonding MO) (antibonding MO)  * 1s

31. Consider the energy-level diagram for the hypothetical He 2 molecule… 1s  1s  * 1s He atom He 2 m’cule He atom 2 bonding e –, 2 antibonding e – No energy benefit to bonding. He 2 molecule won’t form.

32. no bond bond order = ½ (# of bonding e – – # of antibonding e – ) -- the higher the bond order, the greater the bond stability -- a bond order of... 0 = single bond 1 = double bond 2 = triple bond 3 = -- MO theory allows for fractional bond orders as well. What is the bond order of H 2 + ? 1 e – total 1 bonding e –, zero antibonding e – BO = ½ (1–0) = ½ James Bond 7 =007 =

33. Second-Row Diatomic Molecules 1. # of MOs = # of combined atomic orbitals 2. Atomic orbitals combine most effectively with other atomic orbitals of similar energy. 3. As atomic orbital overlap increases, bonding MO is lowered in energy, and the antibonding MO is raised in energy. 4. Both the Pauli exclusion principle and Hund’s rule apply to MOs. Li 3 6.941 Be 4 9.012 B 5 10.811 C 6 12.011 N 7 14.007 O 8 15.999 F 9 18.998 Ne 10 20.180

34. Li 2 Li Li  1s  * 1s 1s 2s  * 2s  2s Use MO theory to predict whether Li 2 and/or Be 2 could possibly form. BO = ½ (4–2) = 1“YEP.”

35. Be 2 Be Be 2s  * 2s  2s BO = ½ (4–4) = 0 “NOPE.” Bonding and antibonding e – cancel each other out in core energy levels, so any bonding is due to e – in bonding orbitals of outermost shell.

36. Molecular Orbitals from 2p Atomic Orbitals The 2p z orbitals overlap in head-to-head fashion, so these bonds are...  bonds. -- the corresponding MOs are:  2p and  * 2p The other 2p orbitals (i.e., 2p x and 2p y ) overlap in sideways fashion, so the bonds are...  bonds. -- the corresponding MOs are:  2p (two of these) and  * 2p (two of these) z x y

37. Rule 3 above suggests that, from low energy to high, the 2p MOs SHOULD follow the order:  2p <  2p <  * 2p <  * 2p LOW ENERGY HIGH ENERGY General energy-level diagrams for MOs of second-row homonuclear diatomic molecules... don’t fit on this slide. (And so, they’re on the next one… …if that’s all right with you.) (And if not, you can… …quit and go make pancakes.)

38. Here, the interaction between the 2s of one atom and the 2p of the other is strong. The orbital energy distribution is altered. Here, the interaction is weak. The energy distribution is as expected. (1s MOs are down here) For B 2, C 2, and N 2... For O 2, F 2, and “Ne 2 ”...  2s  * 2s  2p  * 2p  2p  * 2p  2s  * 2s  2p  * 2p  * 2p  2p same for both “Mr. B” (or Mr. C) (or Mr. N)

39. paramagnetism: describes the attraction of molecules with unpaired e – to a magnetic field diamagnetism: describes substances with no unpaired e – -- such substances are VERY weakly (almost unnoticeably) repelled by a magnetic field Use the energy diagrams above to tell if diatomic species are paramagnetic or diamagnetic. paramagnetism of liquid oxygen (“di-” = two; diamagnetic = “dielectron”) ~

40. Paramagnetic or diamagnetic? B 2 C 2 N 2 O 2 F 2 O 2 + O 2 2– C 2 2– (10) (12) (14) (16) (18) (15) (18) (14)  2s  * 2s  2p  * 2p  2p  * 2p  2s  * 2s  2p  * 2p  * 2p  2p  1s  * 1s  1s  * 1s P D D P D P D DP D D P D P D D