Chemical Bonding II & III, Orbital Hybridization, Molecular Orbitals PDF

CHEM 10021: Bonding and Molecular Structure Part 5: Chemical bonding II; orbital hybridisation (carbon, nitrogen and oxygen). Learning Outcomes ▪ Understand what sigma and pi bonds are ▪ Describe orbital hybridisation as valence bond theory. ▪ Understand hybridization of atomic orbitals:...

CHEM 10021: Bonding and Molecular Structure Part 5: Chemical bonding II; orbital hybridisation (carbon, nitrogen and oxygen). Learning Outcomes ▪ Understand what sigma and pi bonds are ▪ Describe orbital hybridisation as valence bond theory. ▪ Understand hybridization of atomic orbitals: o sp3 hybridization o sp2 hybridization o sp hybridization Valence Bond Theory ▪ Valence bond theory considers the interaction of atomic orbitals on separate atoms as they are brought together to form molecules. ▪ Bonds result from the pairing of unpaired electrons in atomic orbitals. ▪ The wavefunction of an electron pair is formed by superimposing the wavefunctions of the fragments of the molecule. Sigma (σ) and pi (π) Bonds ▪ σ-bond: cylindrically symmetrical about the bond axis with no nodal plane across the bond axis. Top: σ bond formed between two s orbitals; Middle: σ orbital formed between an s and a p orbital; Bottom: σ bond formed from two P orbitals. ▪ π-bond: electron density in 2 lobes with a single nodal plane along the bond axis ▪ A single bond is a σ-bond ▪ A double bond is a σ-bond plus one π-bond ▪ A triple bond is a σ-bond plus two π-bonds  1. sp3 Hybridisation of Atomic Orbitals Methane CH4 Let’s look at carbon as an example – C has 4 unpaired electrons and therefore makes 4 covalent bonds. First look at its electronic structure (1s2 2s2 2p2) as an energy diagram showing the valence electrons only (i.e. those in the higher energy levels, 2s and 2p). We can see that since 2s has 2 and two of the p orbitals have one each that 4 bonds could not form here and so what happens? One of the 2s electrons gets promoted up into the one empty p orbital and now the C atom can form 4 bonds as it has 4 electrons in separate orbitals ready to pair. Hybridisation Now, let’s hybridize them – this means that we combine the existing orbitals that contain the 4 valence electrons, in this example that is the 2s and the three p orbitals The hybrid atomic orbitals will have some characteristics of the separate orbitals that were mixed (hybridized) to make them. This means that all of the hybrid orbitals will have some s and some p character. sp3 hybridisation: formed from one s orbital and three p orbitals – 2sp3 The hybrid atomic orbitals do not have different shapes – they are just orientated differently in space. In methane, four H atoms bond to the C by electron pairing of the H single electron and each electron in the hybrid orbitals- each bond is a sigma bond and methane is tetrahedral, so sp3 gives this shape. σ(C2sp3, H1s) Where did the energy come from to promote that electron in the 2s orbital of carbon up to a p orbital in the first place? - the answer is it came from the formation of the bonds which is exergonic. Ethane Two types of bonds in ethane – σ between the two carbons (C2sp3, C2sp3) and σ bonds between the carbon and the hydrogens (C2sp3, H1s) Nitrogen This time there is no promotion of an electron into a higher energy orbital (like there was for carbon) because it would not increase the number of unpaired electrons. The orbitals hybridize to form four 2sp3 orbitals, one of which will contain 2 electrons and that is a lone pair Ammonia NH3: trigonal pyramidal with a bond σ(N2sp3, H1s) Oxygen Like for nitrogen, there is no promotion of an electron into a higher energy orbital (like there was for carbon) because it would not increase the number of unpaired electrons. Water: bent geometry σ(O2sp3, H1s) Now with two lone pairs the angel is even smaller as they are pushing the bonds down through repulsion (bent geometry)  2. sp2 Hybridisation of Atomic Orbitals: formed from one s orbital and two p Carbon can also form sp2 hybrid orbitals such as in ethylene (C2H4) (remember one electron is promoted up from the 2s into one of the 2p orbitals to give four unpaired electrons The 2py is coming out of the page towards us – away from the trigonal planar geometry. Ethylene: the C=C bond has one σ and one π orbital σ(C2sp2, C2sp2) π(C2py, C2py) σ(C2sp2, H1s) Rotation is disallowed around the double bond 3. sp hybridization of atomic orbitals sp hybrid orbitals: the combination of one s orbital and one p orbital Carbon – C can do this type of hybridization as well Acetylene C2H2 π(C2py, C2py) π(C2px, C2px) σ(C2sp,C2sp) CHEM 10021: Bonding and Molecular Structure Part 6: Chemical bonding III; molecular orbitals. Learning Outcomes ▪ Understand the difference between homonuclear molecules and heteronuclear molecules. ▪ Recognise homonuclear molecules with MOs originating from s orbitals. ▪ Recognise homonuclear molecules with MOs originating from s and p orbitals. ▪ Recognise heteronuclear diatomic molecules. Molecular Orbital (MO) Theory Remember orbitals are wavefunctions – remember electron probability density. ▪ In MO theory, valence electrons are delocalized over the entire molecule- they are not confined to individual atoms or bonds. ▪ MOs (wavefunctions) arise from adding together/superimposing, atomic orbitals or wavefunctions. ▪ A linear combination of atomic orbitals (LCAO) creates MOs (bonding and antibonding orbitals). ▪ N MOs can be constructed by N atomic orbitals Homonuclear molecules with MOs originating from s orbitals Bonding Orbitals Orbitals are wavefunctions and bonding orbitals arise from a LCAO – constructive interference of the wavefunctions (orbitals).  Atomic orbitals Molecular orbital 1sa + 1sb = σ1s which is a bonding MO and also a wavefunction σ1s: designates a molecular orbital that is cylindrically symmetric about the bond axis (with no nodal plane along the bond axis). Now consider the molecule on the right as a wavefunction: when waves interfere constructively, the amplitude increases where they overlap. Increased amplitude in the region between the nuclei translates to an enhanced probability density (ψ2) between the nuclei. And any electron in a bonding MO will be attracted to BOTH nuclei and will be lower in energy (more stable because of the attraction – i.e. harder to remove) This lower energy of the bonding orbital is decreased compared to the atomic orbitals and can be represented as an energy diagram If this was H + H -> H2 then H2 is much more stable than the H atoms Antibonding Orbitals Antibonding orbitals arise from a LCAO – though destructive interference of the wavefunctions (orbitals). Atomic orbitals Antibonding orbitals 1sa + 1sb = σ1s* which is an antibonding MO and also a wavefunction σ1s* designates an antibonding molecular orbital When wavefunctions interfere destructively, the amplitude decreases where they overlap. Decreased amplitude in the internuclear region translates to a diminished probability density (ψ2) between the nuclei and a node between the two nuclei. An electron in this antibonding orbital is essentially excluded from the internuclear region, and thus has a higher energy than if it was in an atomic orbital. This higher energy of the antibonding orbital is increased compared to the atomic orbitals and can also be represented as an energy diagram An antibonding orbital is raised in energy by approximately the same amount that the bonding orbital is lowered in energy. Remember that N molecular orbitals can be constructed by N atomic orbitals- therefore 2 atomic orbitals generate 2 molecular orbitals (one bonding and one antibonding, one lower in energy and one higher in energy). Examples: 1. Hydrogen (H2) 2. Helium (He2) There are 2 at lower energy and 2 at higher energy so there is no net loss or gain – and so does He2 exist? Bond order can help us answer this question Bond order = ½ (no. of bonding electrons – no. of antibonding electrons) So, for He2 it would be: ½(2-2)= 0 this means “no bond” – so He2 does not really exist (small amount does and it is the weakest chemical bond known) Comparing this to H2: ½( 2-0) = 1 this means “1 bond” – so H2 exists These examples have been 1s so now let’s consider 2s orbitals (which are bigger). Using Li as an example (Li 1s2 2s1): (σ1s)2 (σ1s*)2 (σ12s)2 So, for Li2 it would be: ½(4-2)= 1 which means there is one bond Example: Be (σ1s)2 (σ1s*)2 (σ2s)2 (σ2s*)2 So, using all electrons for Be2 bond order = ½(4-4)= 0 which means there is no bond Or if we just use the valence electrons, bond order = ½(2-2)= 0 which means there is no bond Homonuclear molecules with MOs originating from s AND p orbitals Here we’re going to talk about the 2px and 2py but we haven’t done 2pz for now to keep it simple. Bonding MOs formed by LCAO of 2px and 2py as a result of constructive interference 2pxa + 2pxb = π2px Or 2pya + 2pyb = π2py This new π orbital is a molecular wavefunction with a nodal plane along the bond axis Antibonding MOs formed by LCAO of 2px and 2py as a result of destructive interference  2pxa + 2pxb = π2px* Or 2pya + 2pyb = π2py* π*-orbitals: Molecular wavefunction (molecular orbital) with TWO nodal planes. One nodal plane is through the bonding axis and the other is between nuclei. Boron and carbon – energy diagrams just with valence electrons so the 1s orbital is left out for simplicity B2: (σ2s)2 (σ2s*)2 π2px π2py bond order = ½(4-2) = 1 C2: (σ2s)2 (σ2s*)2 (π2px)2 (π2py)2 bond order = ½(6-2) = 2 B2 has 4 electrons in lower energy bonding orbital and 2 electrons in higher energy antibonding orbital and a bond order of 1 C2 has 6 electrons in lower energy bonding orbital and 2 electrons in higher energy antibonding orbital and a bond order or 2 Bonding MOs formed by LCAO of 2pz orbitals 2pza + 2pzb = σ 2pz Constructive interference results in a region of increased amplitude between nuclei, and thus an increased probability density between nuclei (therefore lower energy MO); nodes pass through nuclei, but no nodes along the bond axis. Antibonding MOs formed by LCAO of 2pz orbitals 2pza + 2pzb = σ 2pz* Destructive interference results in a nodal plane between the nuclei, and decreased probability density between nuclei (therefore higher energy MO); nodes pass through and between the nuclei, but no nodes along the bond axis Both are σ: cylindrically symmetric with no nodal plane about the bond axis If nuclear charge (Z) < 8 then π lower in energy than σ2pz but if nuclear charge >/= 8 then σ2pz is lower in energy. The antibonding orbitals’ energies do not change with Z O2 – double bond but also biradical so highly reactive For HETERONUCLEAR diatomic molecules: The relative E ordering is (π2px) and (π2py) < (σ2pz) if Z < 8 for both atoms. You do not need to predict the energy level ordering if either one of the atoms has Z = or > 8.