Suggested HW: Ch 9: 25, 29, 39, 43, 72 (For 25 and 43, you are illustrating the hybridization of the atomic orbitals into hybrid orbitals and the overlapping of these hybrid orbitals as described in the examples provided) Lecture 11 Covalent Bonding Pt 3: Hybridization (Ch )

Introduction We now know that atoms can bond covalently through the sharing of electrons VSEPR theory helps us predict molecular shapes. But, it does not explain what bonds are, how they form, or why they exist. In ch 9, chemical bonding will be explained in terms of orbitals

Covalent Bonding Is Due to Orbital Overlap In a covalent bond, electron density is concentrated between the nuclei. Thus, we can imagine the valence orbitals of the atoms overlapping The region of orbital overlap represents the covalent bond

Overlapping Valence Orbitals Recall s and p orbitals (ch 5) S orbitals are spherical. L = 0, m L = 0 Max of 2 electrons P orbitals consist of two lobes of electron density. L= 1, m L = -1, 0, 1 (3 suborbitals) Max of 6 electrons pxpx pypy S pzpz

Forming Sigma ( σ ) Bonds σ Covalent bond 1s 1 H + H + H + H + σ Two overlapping atomic orbitals form a molecular bonding orbital. Plus sign indicates phase of electron wave, NOT CHARGE A sigma (σ) bonding orbital forms when s-orbitals overlap. stabilization (energy drop) Energy

Introduction to Hybridization Imagine the molecule CH 4. We know that carbon has 4 valence electrons (2s 2 2p 2 ). However, when we fill our orbitals in order as according to Hund’s rule, we notice there are only enough unpaired electrons to make two bonds. Stay mindful of the fact that a covalent bond involves the sharing of unpaired electrons 2s 2 2p 2 C 1s 1 4 H ENERGY 1s 1 X

Four sp 3 hybrid orbitals sp 3 Hybridization So how does CH 4 form? How can carbon make 4 bonds? To make four bonds, carbon hybridizes four of its atomic orbitals. This creates four equivalent sp 3 hybrid orbitals, each containing one unpaired electron. 2s 2 2p 2 The name “sp 3 ” originates from the fact that the hybrid orbitals form as a result of the mixture of 1 s-orbital and 3 p-orbitals. Thus, each sp 3 orbital is 25% s character and 75% p character ENERGY

Formation of Sigma Bonding Orbitals sp 3 hybrid orbitals ENERGY 1s 1 C 4H σ bonding orbitals atomic s-orbitals

The addition of an s-orbital to a p z orbital is shown above. The s orbital adds constructively to the (+) lobe of the p z orbital and adds destructively to the lobe that is in the opposite phase (-). The symbols indicate phase, not charge. Whenever we mix a certain number of s and p atomic orbitals, we get the same number of molecular orbitals. This is called the principle of conservation of orbitals. s pzpz + = z z Illustration of Orbital Hybridization

Illustration of sp 3 Hybrid Orbitals and Orbital Overlap 4 σ-bonds The four hybrid orbitals arrange themselves tetrahedrally.

sp 2 Hybridization The BH 3 molecule gives us an example of sp 2 hybrid orbitals. Once again, we have a situation where we don’t have enough bonding sites to accommodate all of the hydrogens. (Remember, B is electron deficient!) 2s 2 2p 1 B 1s 1 3 H ENERGY X

sp 2 Hybridization So, to make 3 bonding sites, 3 hybrid molecular orbitals are formed by mixing the 2s-orbital with two 2p-suborbitals. This forms an sp 2 orbital. Each of these three hybrid orbitals are one- third s-character, and two-thirds p-character. 2s 2 2p 1 B Three sp 2 hybrid orbitals unused 2p suborbital ENERGY

This figure illustrates the 3 hybrid orbitals combined with the unused 2p orbital, which is perpendicular to the hybrid orbitals. sp 2 orbitals The result of adding one s and two p orbitals together is a trigonal planar arrangement of electron domains

H + empty 2p orbital H + H + B H σ bond H H sp 2 Geometry and Bonding

sp Hybridization Imagine BeH 2 (the Be-H bond is covalent), with Be having the electron configuration: [He]2s 2 Here, we have a situation where no bonding electrons are available. To make two Be-H bonds, Be must create two hybrid orbitals by mixing two atomic orbitals (the 2s orbital and one of the 2p orbitals). This yields sp hybrid orbitals (50% s, 50% p) 2s 2 ENERGY 2p 0 1s 1 Be 2 H X

2s 2 ENERGY 2p 0 Be Two sp hybrid orbitals unused 2p suborbitals sp Hybridization

Hybridization of Lone Electron Pairs Ex. What is the hybridization of Oxygen in H 2 O? The valence electron configuration of O is [He]2s 2 2p 4 2p 4 2s 2 As you see, there are two unpaired O electrons. Does this mean that these two p-suborbitals can overlap with the two Hydrogen 1s orbitals without hybridizing?? ENERGY O 1s 1 2 H

No!! The reason is that we now have two sets of lone pairs of electrons that are substantially different in energy (2s and 2p). The orbitals will hybridize to form degenerate (equal energy) sets of electrons. Lone pair must always be equal in energy with each other, and with bonding electrons. 2p 4 2s 2 ENERGY O 1s 1 2 H X H O H 2s electrons 2p electrons Hybridization of Lone Electron Pairs BAD!!

Water has sp 3 hybridization 2p 4 2s 2 1s 1 ENERGY σ bonds H2OH2O H O H sp 3 electrons O 2 H Four sp 3 hybrid orbitals Lone pairBonding electrons

So What Do We Know So Far? Total Electron Domains Around Atom (Bond + LP) Hybridization 2 sp 3 sp 2 4sp 3

Double and Triple Bonding How can orbital overlap be used to explain double and triple bonds? What kind of interactions are these? Lets look at ethene, C 2 H 4 C C H H H H The hybridization of each carbon is sp 2 because each is surrounded by three electron domains. The geometry around each C is trigonal planar. sp 2

2s 2 2p 2 sp2 hybrid orbitals unhybridized p-electron C C H H H H Forming Double Bonds C We can see that for each carbon atom, we need three sp 2 orbitals and three unpaired electrons to make three sigma bonds. But how is the double bond formed?

All double bonds consist of 1 σ-bond and 1 π-bond Double Bonds formed by simultaneous σ and π interaction The remaining p-electrons form a π bond. This bond forms due to attraction between the parallel p-orbitals. The like-phase regions are drawn toward one another and overlap. H + H + H + H +

Triple Bonds formed by 1 σ -bond and 2 π -bonds. Ex. HCN HC N sp Can you draw the orbital diagram for this molecule? π π

Examples How many σ and π bonds are in each of the following molecules? Give the hybridization of each carbon. CH 3 CH 2 CHCHCH 3 CH 3 CCCHCH 2

sp 3 d and sp 3 d 2 hybridization Atoms like S, Se, I, Xe … etc. can exceed an octet because of sp 3 d and sp 3 d 2 hybridization (combination of ns, np, and nd orbitals where n>3). This results in either trigonal bipyramidal or octahedral skeletal geometry sp 3 d sp 3 d 2

Exceeding an Octet. Example: SF 6 3p 4 3s 2 Energy 3d 0 sp 3 d 2 hybrid orbitals sp 3 6 F S SF 6 Fluorine lone pair

S Exceeding an Octet. Example: SF 6 F 3 lone pair unpaired electron overlap x 6

Look Familiar ???

Examples: What is the hybridization of the central atom? CO 2 H 2 CO CH 3 CCH IF 5 PCl 5 SeOF 4