Química Geral · Bonding theories
Teoria dos orbitais moleculares
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O que você vai aprender
- Combine s and p atomic orbitals into σ and π bonding and antibonding molecular orbitals and describe their nodes and energies.
- Fill the molecular orbital diagram of a period 1 or period 2 homonuclear diatomic, using the σ2p/π2p order switch between N₂ and O₂, and calculate its bond order.
- Predict magnetism, bond length and bond strength for diatomics and their ions, such as O₂⁺, O₂⁻, O₂²⁻ and N₂⁺.
- Describe heteronuclear diatomics such as CO, CN⁻ and NO qualitatively from electron counts, isoelectronic partners and atomic orbital energies.
- Distinguish localized σ bonds from delocalized π systems in benzene and ozone by combining hybridization with molecular orbital theory.
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Bonding and antibonding orbitals
When two atomic orbitals overlap they combine into two molecular orbitals (MOs); n atomic orbitals always give n MOs. The in-phase combination builds up electron density between the nuclei: this bonding MO lies lower in energy than the atomic orbitals. The out-of-phase combination has a nodal plane between the nuclei: this antibonding MO, marked *, lies higher, and it is raised slightly more than the bonding MO is lowered. Take z as the internuclear axis. End-on overlap (s with s, s with p_z, p_z with p_z) gives σ MOs, which are cylindrically symmetric about the axis. Side-on overlap of p_x with p_x, or p_y with p_y, gives π MOs; each π MO has a nodal plane that contains the axis, and a π* MO has a second nodal plane between the nuclei. Orbitals with zero net overlap, such as an s orbital on one atom facing a p_x orbital on the other, do not combine. Each MO holds at most two electrons of opposite spin. Bond order = ½ × (electrons in bonding MOs − electrons in antibonding MOs). H₂ is σ1s², bond order 1. H₂⁺ has one electron, bond order ½: a real but weaker bond. He₂ is σ1s² σ*1s², bond order 0, and because the antibonding MO is raised more than the bonding MO is lowered, two helium atoms repel rather than form a stable molecule.
Second-row homonuclear diatomics
For Li₂ to Ne₂ only the valence MOs need be drawn: σ1s and σ*1s are both full and cancel. Each atom brings one 2s and three 2p orbitals, so eight valence MOs form: σ2s, σ*2s, σ2p, σ*2p, two degenerate π2p and two degenerate π*2p. For O₂, F₂ and Ne₂ the energy order is σ2s < σ*2s < σ2p < π2p < π*2p < σ*2p. For Li₂ to N₂ the 2s and 2p energies are close, so the σ2s and σ2p orbitals mix (s–p mixing); this pushes σ2p above π2p, giving σ2s < σ*2s < π2p < σ2p < π*2p < σ*2p. Fill from the bottom and place one electron in each degenerate orbital, spins parallel, before pairing (Hund's rule). N₂ (drawn) has 10 valence electrons: σ2s² σ*2s² π2p⁴ σ2p², so its bond order is (8 − 2)/2 = 3. Across the period the bond orders are Li₂ 1, Be₂ 0, B₂ 1, C₂ 2, N₂ 3, O₂ 2, F₂ 1, Ne₂ 0. Between atoms of similar size, a higher bond order means a shorter, stronger bond: N₂ (110 pm, about 945 kJ/mol), O₂ (121 pm, about 498 kJ/mol), F₂ (141 pm, about 159 kJ/mol). B₂ is paramagnetic because its two π2p electrons sit in separate orbitals; with σ2p below π2p they would pair in σ2p instead, so this observation supports the switched order.
Magnetism and diatomic ions
A species with unpaired electrons is paramagnetic and is drawn into a magnetic field; a species with all electrons paired is diamagnetic and is weakly pushed out. O₂ has 12 valence electrons: σ2s² σ*2s² σ2p² π2p⁴ π*2p². The last two electrons occupy the two degenerate π* orbitals singly with parallel spins, so O₂ has bond order 2 and two unpaired electrons. Liquid oxygen is held between the poles of a strong magnet, while liquid nitrogen pours straight through. The Lewis structure O=O (drawn) pairs every electron and cannot explain this. When an ion forms, ask which orbital the electron leaves or enters. Removing an electron from O₂ empties part of π*, so O₂⁺ has bond order 2.5 and a shorter, stronger bond. Adding electrons fills π*: superoxide, O₂⁻, has bond order 1.5 and one unpaired electron; peroxide, O₂²⁻, has bond order 1 and is diamagnetic. Bond lengths follow: O₂⁺ about 112 pm, O₂ 121 pm, O₂⁻ about 130 pm, O₂²⁻ about 149 pm. In N₂ the highest occupied MO is the bonding σ2p, so ionization weakens the bond: N₂⁺ has bond order 2.5, with its one unpaired electron in σ2p.
Heteronuclear diatomics
When the two atoms differ, the atomic orbitals of the more electronegative atom lie lower in energy. A bonding MO then has more character of the lower-energy, more electronegative atom, and an antibonding MO has more character of the higher-energy atom. Carbon monoxide (drawn) has 4 + 6 = 10 valence electrons, the same as N₂: the two are isoelectronic, so CO also has bond order 3 and is diamagnetic, and its π* orbitals lie mainly on carbon. CN⁻ and NO⁺ are isoelectronic with N₂ too, each with bond order 3. Nitric oxide, NO, has 5 + 6 = 11 valence electrons; the eleventh occupies a π* orbital, so NO has bond order 2.5 and one unpaired electron, and it is paramagnetic. Because that electron is antibonding, NO is easily ionized (first ionization energy 9.26 eV, against 15.58 eV for N₂) and NO⁺ has the shorter bond (106 pm, against 115 pm in NO). Adding an electron instead gives NO⁻, which is isoelectronic with O₂: bond order 2 and two unpaired electrons.
Localized and delocalized bonds
Hybridization and MO theory work together. In benzene (drawn) each carbon is sp²: its three sp² orbitals make localized C–C and C–H σ bonds in a flat hexagon, and one unhybridized 2p orbital per carbon is left perpendicular to the ring. Those six 2p orbitals combine into six π MOs, three bonding and three antibonding. The six π electrons fill the three bonding MOs, so the π bonding is spread over the whole ring rather than fixed in three double bonds. Each C–C bond gets a π bond order of 3/6 = ½ on top of its σ bond, a total of 1.5, and all six bonds are 140 pm long, between ethene's C=C (134 pm) and ethane's C–C (154 pm). In ozone the central oxygen has three electron domains (two σ bonds and a lone pair) and is sp². One 2p orbital from each oxygen, perpendicular to the molecular plane, gives three π MOs: bonding, nonbonding and antibonding. Of ozone's 18 valence electrons, 4 are in the two σ bonds and 10 are in five in-plane lone pairs, which leaves 4 π electrons to fill the bonding and nonbonding π MOs. Only the bonding pair adds π bonding, shared over two O–O links, so each O–O bond order is 1 + ½ = 1.5 and both bonds are 128 pm. The π bond of ethene or of a single C=C in cyclohexene, by contrast, is localized between two atoms.
What to carry forward
Count the valence electrons, fill the right diagram (π2p below σ2p up to N₂; σ2p below π2p from O₂ on), and use Hund's rule for degenerate orbitals. Bond order = ½ × (bonding − antibonding electrons) predicts relative bond strength and length, and unpaired electrons predict paramagnetism. For an ion, ask which orbital the electron leaves or enters. For a heteronuclear diatomic, count electrons and find an isoelectronic partner. In benzene and ozone, sp² hybrids build the localized σ framework and the leftover p orbitals form delocalized π MOs.
Problemas de prática (40)
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Heteronuclear diatomics
- Bond order and magnetism of COPredict the bond order and magnetism of carbon monoxide by comparison with its isoelectronic partner N₂.
Magnetism and diatomic ions
- Predict the bonding in Li₂Predict the bond order and magnetism of Li₂ from its valence MO configuration.
- What happens when F₂ is ionized?Predict how removing an electron from F₂ changes its bond length and magnetism.
Bonding and antibonding orbitals
- Can an s orbital combine with a pₓ orbital?Predict from orbital symmetry that an s orbital and a perpendicular p orbital on a neighbouring atom have zero net overlap.
Classificar 5
Delocalized π systems
- Which ring has delocalized π electrons?Classify six-membered rings by whether their π electrons are delocalized over more than two atoms.
Heteronuclear diatomics
- Find the species isoelectronic with N₂Classify diatomic species as isoelectronic with N₂ by counting valence electrons.
Magnetism and diatomic ions
- Which dioxygen species is diamagnetic?Classify dioxygen species as paramagnetic or diamagnetic from their π* occupancy.
Bonding and antibonding orbitals
- Side-on overlap of 2p orbitalsClassify the molecular orbitals formed by side-on overlap of parallel p orbitals as π orbitals.
- Describe the σ*1s orbitalIdentify the features of an antibonding σ orbital formed from two 1s orbitals.
Comparar 4
Second-row diatomics and bond order
- Rank six species by bond orderRank homonuclear and heteronuclear diatomic species by bond order calculated from their MO configurations.
Delocalized π systems
- Rank C–C bond lengths, benzene includedPlace benzene's delocalized C–C bonds in a bond-length ranking with localized single, double and triple bonds.
Magnetism and diatomic ions
- Why ionization weakens N₂ but strengthens O₂Compare the effect of ionization on the bonds of N₂ and O₂ from the character of each molecule's highest occupied orbital.
- Rank O–O bond lengths in dioxygen speciesRank O₂⁺, O₂, O₂⁻ and O₂²⁻ by bond length using bond orders from the MO diagram.
Calcular 11
Second-row diatomics and bond order
- Bond order of the C₂²⁻ ionCalculate the bond order of the dicarbide (acetylide) ion and connect it to an isoelectronic diatomic.
- Antibonding electrons in F₂Count the electrons in antibonding valence orbitals of F₂ and relate the count to its single bond.
- Bond order of N₂ from its MO diagramCalculate the bond order of N₂ from its valence molecular orbital configuration.
Delocalized π systems
- Occupied π orbitals of benzeneCount the occupied π molecular orbitals of benzene from its p orbitals and π electrons.
- Count ozone's π electronsCount the electrons in the delocalized π system of ozone by subtracting σ-bond and in-plane lone-pair electrons from the valence total.
Bonding and antibonding orbitals
- How many valence MOs in N₂?Apply conservation of orbitals to count the molecular orbitals formed from the valence orbitals of two atoms.
- Bond order of H₂⁺Calculate the bond order of a one-electron diatomic ion from its MO configuration.
Magnetism and diatomic ions
- Bond order of N₂⁺Calculate the bond order of N₂⁺ by removing an electron from the highest occupied orbital of N₂.
- Bond order of superoxideCalculate the bond order of the superoxide ion by adding an electron to the O₂ diagram.
Heteronuclear diatomics
- Unpaired electrons in NO⁻Count the unpaired electrons of the NO⁻ ion by applying Hund's rule to its degenerate π* orbitals.
- Bond order of nitric oxideCalculate the bond order of nitric oxide from its valence electron count.
Encontre o erro 4
Bonding and antibonding orbitals
- What went wrong for Be₂?Diagnose an electron-counting error that gives the wrong bond order for Be₂.
- Find the flaw in a He₂ argumentDiagnose why filled bonding and antibonding orbitals together make two atoms repel rather than merely not bond.
Second-row diatomics and bond order
- Spot the error in an O₂ diagramDiagnose a Hund's-rule error in filling the degenerate π* orbitals of O₂.
Delocalized π systems
- Diagnose a hybridization error for ozoneDiagnose an incorrect hybridization assignment that would rule out ozone's delocalized π system.
Interprete as evidências 4
Second-row diatomics and bond order
- What B₂'s magnetism revealsUse the observed paramagnetism of B₂ to decide between the two possible orderings of σ2p and π2p.
- What kind of double bond does C₂ have?Interpret the MO configuration of C₂ to identify which orbitals supply its bond order.
- Order the valence MOs of N₂Arrange the valence molecular orbitals of N₂ by energy, placing π2p below σ2p.
Bonding and antibonding orbitals
- Name the orbital from its nodesIdentify a diatomic molecular orbital from the nodal planes described for it.
Trabalhe de trás para frente 2
Second-row diatomics and bond order
- Which molecule has this configuration?Work back from a valence MO configuration to the neutral homonuclear diatomic it describes.
Magnetism and diatomic ions
- Identify a dinitrogen ion from its MO cluesWork back from a bond order and the character of the singly occupied orbital to the dinitrogen ion they describe.
Planeje uma síntese 1
- Test the MO prediction for O₂Choose an experiment that tests whether a substance has unpaired electrons.
Explique por quê 5
Delocalized π systems
- Hybridization plus MOs in benzeneExplain benzene's equal C–C bonds by combining sp² hybridization for the σ framework with delocalized π molecular orbitals.
Heteronuclear diatomics
- Where the π* orbitals of CO lieExplain why the antibonding π* orbitals of carbon monoxide are concentrated on carbon.
- Why NO ionizes so easilyExplain the low ionization energy of NO and the shorter bond of NO⁺ from the antibonding character of the electron removed.
Magnetism and diatomic ions
- Why liquid oxygen sticks to a magnetExplain the paramagnetism of O₂ from the occupancy of its π* orbitals.
Second-row diatomics and bond order
- Why σ2p and π2p swap at O₂Explain the change in σ2p/π2p ordering between N₂ and O₂ in terms of s–p mixing and the 2s–2p energy gap.
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