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Chemical Bonding

Copyright (c) RDMCHEM LLC 2020

 

 

Overview

Chemical Bond

Overview

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

Chemical Bond

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

(2.1)

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

(2.1.1)

 

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

(2.1.2)

 

With the Energy command we can compute the energy of HF at R = 1 angstrom

> 

E1 ≔ EnergyHF1, method='Parametric2RDM', basis=dz;

E1≔−100.15637329

(2.1.3)

 

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

(2.1.4)

 

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≔1.50000000⁢E0

NA≔6.02214086⁢1023⁢1mol

Edissociation≔3938.24945900⁢kJ⁢1mol

(2.1.5)

 

> 

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

(2.2.1)

 

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

(2.2.2)

 

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

(2.2.3)

 

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

(2.2.4)

 

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

(2.2.5)

 

> 


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