Molecular Geometry and Bonding Theories The properties of

Molecular Geometry and Bonding Theories

The properties of a molecule depend on its shape 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

bond angles: the angles made by the lines joining the nuclei of a molecule’s atoms carbon dioxide CO 2 180 o methane CH 4 109. 5 o formaldehyde CH 2 O 120 o

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 4 e For ammonia, there are NH 3 – do. . three bonding domains and ma H– N –H ins one nonbonding domain. . . H N H Domains arrange themselves so H H as to minimize their repulsions.

. . The electron-domain geometry is one of five basic arrangements of domains. N H H H -- 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 . .

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

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. . . . – Sn. Cl 3 Cl–. . Sn–Cl. . Cl 26 e–. . . . O 3 O O=O O–. . – 18 e. . Se. Cl 2. . . Cl–Se–Cl –. . . 20 e = O. . . ] . . . –O. . . . [ – EDG: tetrahedral MG: trig. pyramidal EDG: trig. planar MG: bent EDG: tetrahedral MG: bent

28 e– . . . EDG: trig. bipyr. MG: T-shaped . . EDG: octahedral MG: sq. pyramidal . . Cl. F 3 EDG: trig. bipyr. MG: seesaw . . 42 e– . . . IF 5 ] . . . . 34 e– [ EDG: trig. planar MG: trig. planar . . SF 4 (res. ) 24 e– 2– . . CO 32– . . O– C= O. . . F– S –F. . F. . – I –F. . . . F– Cl –F. . F

Cl. . Cl–. . I –Cl. . ICl 4– 36 e– Cl. . – ] . . . . [ . . EDG: octahedral MG: sq. planar 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. MG –CH 3 – –C=O –OH. . tetra. trig. plan. tetra. bent . . EDG . . . H O. . H–C–C–O–H. . H PORTION . . CH 3 COOH

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. , the ideal bond angle for the tetrahedral EDG is 109. 5 o . . H H H . . H H H O 124. 3 o O=C 124. 3 o 107. 0 o 104. 5 o . . Cl. . 111. 4 Cl. . . N COCl 2 H 2 O . . H H C NH 3 . . CH 4 H EDG = trig. plan. ideal = 120 o o

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. . . bond dipoles . . . O=C=O H S H overall dipole = zero overall dipole = (NONPOLAR) (POLAR)

Classify as polar or nonpolar: . . . 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. . . 24 e– . . BCl 3 . . 26 e– . . PCl 3 . . . Cl–. . P–Cl. . . . Cl. . – B C Cl–. . l polar nonpolar

Valence-Bond Theory: merges Lewis structures w/the idea of atomic orbitals (2 s, 3 p, etc. ) Lewis theory says… covalent bonding occurs when atoms share electrons 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 V-B theory is like Velcro: “No overlap, no bond. ”

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

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

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

(LS) In… : F Be F: 2 s 2 p sp 2 p 2 s 2 p sp 2 2 p : : F : : : BF 3 : : Be. F 2 …the central atom …but is ACTUALLY …the result SHOULD be…hybridized to be… then being… B F: : : F : H 2 O H O: H 2 s 2 p sp 3 : F: : F : : P F: : : F: : : F : : PF 5 : : : : Xe. F 2 4 hyb. sp 3 orbs. : CH 4 H H C H H 2 hyb. sp orbs. 2 unhyb. 2 p orbs. 3 hyb. sp 2 orbs. 1 unhyb. 2 p orb. : F Xe F: 3 s 3 p 5 s 5 p 3 d 5 d 4 hyb. sp 3 orbs. sp 3 d 5 hyb. sp 3 d orbs. 3 d 4 unhyb. 3 d orbs. sp 3 d 5 hyb. sp 3 d orbs. 5 d 4 unhyb. 5 d orbs. : : : : Xe. F : F: : : 4 F Xe F : F: : 5 s 5 p 5 d sp 3 d 2 6 hyb. sp 3 d 2 orbs. 5 d 3 unhyb. 5 d orbs.

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

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.

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

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

Single bonds are s bonds. e. g. , C 2 H 6 Double bonds consist of one s and one p. e. g. , C 2 H 4 Triple bonds consist of one s and two p. e. g. , C 2 H 2

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

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

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

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

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. 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: wave functions that describe the locations of electrons in molecules -- these are analogous to atomic orbitals in atoms (e. g. , 2 s, 2 p, 3 s, 3 d, 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.

Hydrogen (H 2) The overlap of two atomic orbitals produces two MOs. (antibonding MO) s*1 s 1 s + 1 s H atomic orbitals s 1 s (bonding MO) H 2 molecular orbitals -- 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 s molecular orbitals.

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

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

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

Second-Row Diatomic Molecules 3 4 Li Be 6. 941 9. 012 B 5 10. 811 C 6 12. 011 7 8 N O 14. 007 15. 999 F 9 18. 998 10 Ne 20. 180 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.

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

s*2 s 2 s 2 s 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. Be Be BO = ½ (4– 4) = 0 Be 2 “NOPE. ”

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

Rule 3 above suggests that, from low energy to high, the 2 p MOs SHOULD follow the order: LOW s 2 p < p*2 p < s*2 p 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. )

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

paramagnetism: describes the attraction of molecules with unpaired e– to a magnetic field diamagnetism: describes substances with no unpaired e– ~ “dielectron”) (“di-” = two; diamagnetic = -- 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

Paramagnetic or diamagnetic? B 2 (10) P C 2 (12) D N 2 (14) D s*2 p p*2 p s 2 p p 2 p s*2 p p 2 p s*2 s s 2 s O 2 (16) P F 2 (18) D s*1 s O 2+ (15) P s 1 s O 22– (18) D C 22– (14) D