Chemical Bonding
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Overview
Chemical Bond
The nature of the chemical bond was elucidated through quantum mechanics in the 1920s and 1930s. Here we explore the chemical bond by computing the following: (1) the potential energy curve of a diatomic molecule (2) the polarity of the bond for a series of diatomic molecules
We load the QuantumChemistry package with Maple's with command
withQuantumChemistry;
AOLabels,ActiveSpaceCI,ActiveSpaceSCF,AtomicData,BondAngles,BondDistances,Charges,ChargesPlot,CorrelationEnergy,CoupledCluster,DensityFunctional,DensityPlot3D,Dipole,DipolePlot,Energy,ExcitationEnergies,ExcitationSpectra,ExcitationSpectraPlot,ExcitedStateEnergies,ExcitedStateSpins,FullCI,GeometryOptimization,HartreeFock,Interactive,Isotopes,MOCoefficients,MODiagram,MOEnergies,MOIntegrals,MOOccupations,MOOccupationsPlot,MOSymmetries,MP2,MolecularData,MolecularGeometry,NuclearEnergy,NuclearGradient,OscillatorStrengths,Parametric2RDM,PlotMolecule,Populations,RDM1,RDM2,RTM1,ReadXYZ,Restore,Save,SaveXYZ,SearchBasisSets,SearchFunctionals,SkeletalStructure,Thermodynamics,TransitionDipolePlot,TransitionDipoles,TransitionOrbitalPlot,TransitionOrbitals,Variational2RDM,VibrationalModeAnimation,VibrationalModes,Video
Energy of the chemical bond
Define the hydrogen fluoride (HF) molecule as a function of its internuclear distance R
HF ≔ H,0,0,0,F,0,0,R;
HF≔H,0,0,0,F,0,0,R
Setting R to 1.0 Angstrom (1 angstrom = 10-10 m) with the subs command, we have
HF1 ≔ subsR=1.0, HF;
HF1≔H,0,0,0,F,0,0,1.00000000
With the Energy command we can compute the energy of HF at R = 1 angstrom
E1 ≔ EnergyHF1, method='Parametric2RDM', basis=dz;
E1≔−100.15637329
We can also compute the Energy as a list of R values (calculation may take a minute or so)
E_ls ≔ seqR,EnergyHF, method='Parametric2RDM', basis=dz, R=0.7,0.8,0.9, 1.0, 1.1, 1.2,1.6,1.8,2.2,2.6;
E_ls≔0.70000000,−100.04258858,0.80000000,−100.12769696,0.90000000,−100.15593391,1.00000000,−100.15637329,1.10000000,−100.14341514,1.20000000,−100.12470651,1.60000000,−100.04862055,1.80000000,−100.02176025,2.20000000,−99.99225294,2.60000000,−99.98235671
We can make a plot of the energy as a function of R known as the potential energy curve
energycurve ≔ splineE_ls, 'R',2:plotenergycurve, R=0.6..2.6, axes=boxed,labels=R (angstroms),"E hartrees",labeldirections=horizontal,vertical;
(a) From the plot of the energy for HF, estimate the internuclear distance R in angstroms at which the energy reaches its minimum? (b) Convert your result in (a) to meters. (c) From the plot of the energy for HF, estimate the bond dissociation energy, the energy to dissociate the HF molecule from its energy minimum? (d) Convert the energy in (c) from hartrees (energy unit in atomic units) to kJ/mol (kilojoules/mole). Hint: We can easily convert units in Maple. For example, if the dissociation energy is 1.5 hartrees (or denoted as 1.5 E0), we can convert it to kJ/mol
Edissociation ≔ 1.5⋅UnitsUnit'hartrees';NA ≔ evalfScientificConstantsConstant'NA'⋅UnitsUnit'1mol' ; Edissociation ≔ convertEdissociation,units, 'kJ'⋅NA;
Edissociation≔1.50000000E0
NA≔6.0221408610231mol
Edissociation≔3938.24945900kJ1mol
Polarity of the chemical bond
A chemical bond is described as polar if there exists a separation of partial negative and partial positive charges. For a diatomic molecule the polarity of the bond can be estimated in at least two ways: (1) the difference in the electronegativities of the two atoms connected by the bond and (2) the dipole moment of the molecule. We can compute and plot the dipole moment of a molecule with the Dipole and DipolePlot commands of the Quantum Chemistry Toolbox. Note that the Toolbox uses the standard convention that the direction of the dipole moment is from the negative partial charge to the positive partial charge, which differs from some general chemistry textbooks.
The dipole moment of HF is
HF ≔ H,0,0,0,F,0,0,0.917; dipole ≔ DipoleHF, basis=dzp; total_dipole ≔ sqrtadddipolei,22,i=1..3;
HF≔H,0,0,0,F,0,0,0.91700000
dipole≔X0.Y0.Z−2.02931607
total_dipole≔2.02931607
The plot of the dipole moment of HF is (click on the plot to rotate)
DipolePlotHF, basis=dzp;
The dipole moment of HCl is
HCl ≔ H,0,0,0,Cl,0,0,1.274; dipole ≔ DipoleHCl, basis=dzp; total_dipole ≔ sqrtadddipolei,22,i=1..3;
HCl≔H,0,0,0,Cl,0,0,1.27400000
dipole≔X0.Y0.Z−1.45323422
total_dipole≔1.45323422
The plot of the dipole moment of HCl is (click on the plot to rotate)
DipolePlotHCl, basis=dzp;
The dipole moment of HBr is
HBr ≔ H,0,0,0,Br,0,0,1.414; dipole ≔ DipoleHBr, basis=dzp; total_dipole ≔ sqrtadddipolei,22,i=1..3;
HBr≔H,0,0,0,Br,0,0,1.41400000
dipole≔X0.Y0.Z−1.40698952
total_dipole≔1.40698952
The plot of the dipole moment of HBr is (click on the plot to rotate)
DipolePlotHBr, basis=dzp;
The dipole moment of HI is
HI ≔ H,0,0,0,I,0,0,1.609; dipole ≔ DipoleHI, basis=dzp; total_dipole ≔ sqrtadddipolei,22,i=1..3;
HI≔H,0,0,0,I,0,0,1.60900000
dipole≔X0.Y0.Z−0.99485228
total_dipole≔0.99485228
The plot of the dipole moment of HI is (click on the plot to rotate)
DipolePlotHI, basis=dzp;
(e) Use the total (or net) dipole moments to arrange HF, HCl, HBr, and HI from lowest to highest bond polarity. (f) Use the differences in electronegativity of A and B in AB to arrange HF, HCl, HBr, and HI from lowest to highest bond polarity.
(g) Are the results in (e) consistent with the results in (f)?
Hint: For part (f) the electronegtaivities of atoms can be directly obtained in the Quantum Chemistry Toolbox with the AtomicData command
AtomicDataFelectronegativity;
3.98000000
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