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Concept 06 / 07Chemical Bonding and Molecular Structure

Valence Bond Theory, Hybridisation and Sigma and Pi Bonds

Bond formation by orbital overlap, sigma and pi bonds and their relative strengths, the idea of hybridisation, sp, sp2, sp3, sp3d, sp3d2 and sp3d3 with examples, determining hybridisation from steric number, s character and its effects, and multiple bonds in ethene and ethyne.

Figure 1
Valence bond theory (Heitler, London, Pauling) explains a covalent bond as the overlap of half-filled atomic orbitals on two atoms, each contributing one electron, with the two electrons paired with opposite spins. Greater overlap means a stronger bond. The energy falls as the atoms approach until nuclear repulsion balances the attraction at the equilibrium bond length. Sigma and pi bonds. A sigma (σ\sigma) bond forms by head-on (axial) overlap along the internuclear axis: s-s, s-p or p-p end-on, or with hybrid orbitals. Electron density is concentrated between the nuclei and the bond is cylindrically symmetric, so rotation about it is free. A pi (π\pi) bond forms by sideways overlap of parallel p orbitals, with electron density above and below the axis and a nodal plane along it. Sideways overlap is less effective, so a π\pi bond is weaker than a σ\sigma bond between the same atoms, and it prevents rotation (the origin of cis-trans isomerism). A single bond is one σ\sigma; a double bond is σ+π\sigma + \pi; a triple bond is σ+2π\sigma + 2\pi. A π\pi bond never forms alone. Hybridisation. Methane has four equivalent C-H bonds at 109.5∘109.5^{\circ}, which pure 2s2s and 2p2p orbitals cannot explain. Pauling proposed that orbitals of similar energy on the same atom mix (hybridise) to form an equal number of equivalent hybrid orbitals, directed so as to minimise repulsion. Hybrid orbitals form only σ\sigma bonds or hold lone pairs; unhybridised p orbitals form π\pi bonds. The number of hybrid orbitals equals the steric number from VSEPR. sp (SN 2): linear, 180∘180^{\circ}; BeCl2\mathrm{BeCl_{2}}, C2H2\mathrm{C_{2}H_{2}}, CO2\mathrm{CO_{2}} carbon. sp2^{2} (SN 3): trigonal planar, 120∘120^{\circ}; BF3\mathrm{BF_{3}}, C2H4\mathrm{C_{2}H_{4}}, graphite. sp3^{3} (SN 4): tetrahedral, 109.5∘109.5^{\circ}; CH4\mathrm{CH_{4}}, NH3\mathrm{NH_{3}}, H2O\mathrm{H_{2}O}, diamond. sp3^{3}d (SN 5): trigonal bipyramidal; PCl5\mathrm{PCl_{5}}, SF4\mathrm{SF_{4}}, XeF2\mathrm{XeF_{2}}. sp3^{3}d2^{2} (SN 6): octahedral; SF6\mathrm{SF_{6}}, XeF4\mathrm{XeF_{4}}, [CoF6]3−[\mathrm{CoF_{6}}]^{3-}. sp3^{3}d3^{3} (SN 7): pentagonal bipyramidal; IF7\mathrm{IF_{7}}. dsp2^{2} (square planar) occurs in transition metal complexes such as [Ni(CN)4]2−[\mathrm{Ni(CN)_{4}}]^{2-} (see Coordination Compounds). Ethene and ethyne. In ethene each carbon is sp2^{2}: three σ\sigma bonds (two C-H, one C-C) in a plane at 120∘120^{\circ}; the unhybridised 2p2p orbitals overlap sideways to form the π\pi bond perpendicular to the plane. In ethyne each carbon is sp: one C-C σ\sigma and one C-H σ\sigma in a line, plus two mutually perpendicular π\pi bonds from the two remaining p orbitals on each carbon. Counting: ethyne has 3σ3\sigma and 2π2\pi bonds; benzene 12σ12\sigma and 3π3\pi. s character. sp (50%50\% s), sp2^{2} (33%33\%), sp3^{3} (25%25\%). More s character means the orbital is held closer to the nucleus: shorter, stronger bonds (C-H in ethyne < ethene < ethane in length), greater electronegativity of the carbon, and greater acidity of the attached H (ethyne's C-H is acidic enough to react with sodium). Bond angles also widen with s character (180∘180^{\circ}, 120∘120^{\circ}, 109.5∘109.5^{\circ}). Key Takeaways: - σ\sigma: head-on overlap, free rotation; π\pi: sideways overlap, weaker, restricts rotation. - Double bond =1σ+1π= 1\sigma + 1\pi; triple =1σ+2π= 1\sigma + 2\pi. - Number of hybrid orbitals == steric number: 2 sp, 3 sp2^{2}, 4 sp3^{3}, 5 sp3^{3}d, 6 sp3^{3}d2^{2}, 7 sp3^{3}d3^{3}. - Hybrid orbitals form σ\sigma bonds and hold lone pairs; π\pi bonds use unhybridised p orbitals. - Higher s character: shorter bond, higher electronegativity, more acidic C-H. - Pitfall: counting a lone pair as not requiring a hybrid orbital. Brain-Teaser 1: Give the hybridisation of each carbon in CH2=CH−C≡N\mathrm{CH_{2}=CH-C\equiv N} and count the σ\sigma and π\pi bonds. Solution: CH2\mathrm{CH_{2}} carbon: sp2^{2}; CH carbon: sp2^{2}; nitrile carbon: sp. Bonds: C-H 3σ3\sigma; C=C 1σ+1π1\sigma + 1\pi; C-C 1σ1\sigma; C≡N 1σ+2π1\sigma + 2\pi. Total 6σ6\sigma and 3π3\pi. Brain-Teaser 2: What is the hybridisation of the central atom in NH4+\mathrm{NH_{4}^{+}}, SO3\mathrm{SO_{3}}, XeO3\mathrm{XeO_{3}} and BrF5\mathrm{BrF_{5}}? Solution: NH4+\mathrm{NH_{4}^{+}}: 44 bonds, 00 lp: sp3^{3} (tetrahedral). SO3\mathrm{SO_{3}}: 33 σ\sigma domains, 00 lp (double bonds count once): sp2^{2}, planar. XeO3\mathrm{XeO_{3}}: 33 σ\sigma bonds and 11 lp: sp3^{3}, pyramidal. BrF5\mathrm{BrF_{5}}: 55 bonds and 11 lp: sp3^{3}d2^{2}, square pyramidal. Counting only σ\sigma bonds and lone pairs gives the steric number directly. Brain-Teaser 3: Why can rotation occur about the C-C bond of ethane but not about the C=C bond of ethene at room temperature? Solution: The C-C σ\sigma bond is cylindrically symmetric, so rotating one methyl group does not change the overlap. The π\pi bond in ethene requires the two p orbitals to stay parallel; rotating one carbon by 90∘90^{\circ} would destroy the overlap and break the π\pi bond, which costs about 250250 kJ/mol, far more than thermal energy at room temperature. Hence cis and trans alkenes are distinct compounds.

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Valence Bond Theory, Hybridisation and Sigma and Pi Bonds — DPP

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