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Periodic Table. So we now know how multielectron atoms behave How do multiatom molecules behave?. Ionic Bonds. Between elements on left (electropositive) and right (electronegative) of periodic table. Na. -5eV. Na.
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Periodic Table So we now know how multielectron atoms behave How do multiatom molecules behave?
Ionic Bonds Between elements on left (electropositive) and right (electronegative) of periodic table
Na -5eV Na % -------------------------------------------------------------------------------- clear all %Constants (all MKS, except energy which is in eV) hbar=1.055e-34;m=9.110e-31;epsil=8.854e-12;q=1.602e-19; %Lattice Np=200;a=(10e-10/Np);R=a*[1:1:Np];t0=(hbar^2)/(2*m*(a^2))/q; %Hamiltonian,H = Kinetic,T + Potential,U + Ul + Uscf T=(2*t0*diag(ones(1,Np)))-(t0*diag(ones(1,Np-1),1))-(t0*diag(ones(1,Np-1),-1)); UN=(-q*11/(4*pi*epsil))./R;% Z=11 for sodium l=1;Ul=(l*(l+1)*hbar*hbar/(2*m*q))./(R.*R); Uscf=zeros(1,Np);change=1; while change>0.05 [V,D]=eig(T+diag(UN+Uscf));D=diag(D);[DD,ind]=sort(D); E1s=D(ind(1));psi=V(:,ind(1));P1s=psi.*conj(psi);P1s=P1s'; E2s=D(ind(2));psi=V(:,ind(2));P2s=psi.*conj(psi);P2s=P2s'; E3s=D(ind(3));psi=V(:,ind(3));P3s=psi.*conj(psi);P3s=P3s'; [V,D]=eig(T+diag(UN+Ul+Uscf));D=diag(D);[DD,ind]=sort(D); E2p=D(ind(1));psi=V(:,ind(1));P2p=psi.*conj(psi);P2p=P2p'; n0=(2*(P1s+P2s)+P3s)+(6*P2p); %config 1s2 2s2 2p6 3s1 n=n0*(10/11); % 11 electrons Unew=(q/(4*pi*epsil))*((sum(n./R)-cumsum(n./R))+(cumsum(n)./R)); %Uex=(-q/(4*pi*epsil))*((n./(4*pi*a*R.*R)).^(1/3));%Unew=Unew+Uex; change=sum(abs(Unew-Uscf))/Np,Uscf=0.5*Uscf+0.5*Unew; end [E1s E2s E2p E3s] + SCF loop ans = -735.2999 -45.6916 -33.6766 -5.7165 (-5.7 is Ionization potential ie,energy for converting Na Na+)
Cl -12.3eV Cl % -------------------------------------------------------------------------------- clear all %Constants (all MKS, except energy which is in eV) hbar=1.055e-34;m=9.110e-31;epsil=8.854e-12;q=1.602e-19; %Lattice Np=200;a=(10e-10/Np);R=a*[1:1:Np];t0=(hbar^2)/(2*m*(a^2))/q; %Hamiltonian,H = Kinetic,T + Potential,U + Ul + Uscf T=(2*t0*diag(ones(1,Np)))-(t0*diag(ones(1,Np-1),1))-(t0*diag(ones(1,Np-1),-1)); UN=(-q*17/(4*pi*epsil))./R;% Z=17 for chlorine l=1;Ul=(l*(l+1)*hbar*hbar/(2*m*q))./(R.*R); Uscf=zeros(1,Np);change=1; while change>0.05 [V,D]=eig(T+diag(UN+Uscf));D=diag(D);[DD,ind]=sort(D); E1s=D(ind(1));psi=V(:,ind(1));P1s=psi.*conj(psi);P1s=P1s'; E2s=D(ind(2));psi=V(:,ind(2));P2s=psi.*conj(psi);P2s=P2s'; E3s=D(ind(3));psi=V(:,ind(3));P3s=psi.*conj(psi);P3s=P3s'; [V1,D1]=eig(T+diag(UN)+diag(Uscf)/2);D1=diag(D1);[DD1,ind1]=sort(D1); [V,D]=eig(T+diag(UN+Ul+Uscf));D=diag(D);[DD,ind]=sort(D); E2p=D(ind(1));psi=V(:,ind(1));P2p=psi.*conj(psi);P2p=P2p'; E3p=D(ind(2));psi=V(:,ind(2));P3p=psi.*conj(psi);P3p=P3p'; [V2,D2]=eig(T+diag(UN)+diag(Uscf)/2);D2=diag(D2);[DD2,ind2]=sort(D2); n0=2*(P1s+P2s+P3s)+(6*P2p)+(5*P3p); %config 1s2 2s2 2p6 3s2 3p5 n=n0*(16/17); %17 electrons Unew=(q/(4*pi*epsil))*((sum(n./R)-cumsum(n./R))+(cumsum(n)./R)); %Uex=(-q/(4*pi*epsil))*((n./(4*pi*a*R.*R)).^(1/3));%Unew=Unew+Uex; change=sum(abs(Unew-Uscf))/Np,Uscf=0.5*Uscf+0.5*Unew; End [E1s E2s E2p E3s E3p] E=2*(DD1(ind(1))+DD1(ind(2))+DD1(ind(3)))+6*DD2(ind(1))+5*DD2(ind(2)) E3p = -12.7622 But need electron affinity, ie, Cl Cl- Keeping track of E below
Ionic Bonds (NaCl) Na+ Cl- GS wavefn -5eV -12.3eV Na Cl + -IP(Na) + IP(Cl) = -7.3 eV U = 8.9 eV EA(Cl) = -IP(Cl) + U -IP(Na) + EA(Cl) = 1.6 eV This is the energy to create ions. Now, need to include electrostatic energy to create NaCl crystal
From Ions to crystal For each Na+ 6 Cl- at dist a (face centers) 12 Na+ at dist a √2 (edge centers) 8 Cl- at dist a √3 (corners) 6 Na+ at dist 2a 24 Cl- at dist a √5, … etc E(NaCl) – E(Na+) – E(Cl-) = -6.q2/4pe0a + 12.q2/4pe0a√2 - … (a = 0.28 nm) = -7.0 eV So net ionic bonding energy ≈ 7.0-1.6 = 5.4 eV
From Ions to crystal In water, large dielectric constant ≈ 80 This decreases each term in electrostatic contribution New bonding energy ≈ 7/80 – 1.6 = -0.7 eV (Negative!) So NaCl spontaneously dissociates in water (dissolves)
Covalent Bonds Between elements of comparable electronegativity
Covalent Bonds H2 molecule H = ∑iTi + ∑aTNa + ∑iaUia + ∑abUab + ∑ijUij Nucl- Nucl El- El El KE Nucl KE Nucl- El uL(r) uR(r) • ≈ fLuL(r) + fRuR(r) Choose a basis set Two hydrogen 1s orbitals Coefficients {fL, fR} Analogy: Grid of points as our basis set Coefficients {f1,f2,.... ,fN}
Covalent Bonds H2 molecule H S Hmn = ∫um*Hun dV s = ∫uL*uR dV uL(r) uR(r) f≈ fLuL + fRuR HF = EF (left multiply by uL,R* and integrate over dv) HLL HLRfL 1 s fL = E HRL HRRfR s 1 fR
Covalent Bonds a b b a H = 1 s s 1 S = a b fL 1 s fL = E b a fR s 1 fR H S a = ∫uL(r)HuL(r)dV = ∫uR(r)HuR(r)dV b = ∫uL(r)HuR(r)dV s = ∫uL(r)uR(r)dV For 1s orbitals of H, b < 0 (recall off-diag terms on a grid were -t) The off-diagonal term reduces E
Finding Eigenmodes E-a Es-b Es-b E-a ES-H = HF = ESF Eigenvalues of S-1H (ES-H)F = 0 det(ES-H) = 0 EB = (a+b)/(1+s) EA = (a-b)/(1-s) (EB- a)fL + (EBs-b)fR = 0 fL = fR = A F = (uL+uR)A ∫|F|2dV = 1 A = 1/2(1+s)
uLuR uA = (uL-uR)/√2(1-s) ANTIBONDING EA uB = (uL+uR)/√2(1+s) BONDING EB EB = (a+b)/(1+s) EA = (a-b)/(1-s)
Covalent Bonds • -4.5 eV • 0.074 nm • Net decrease in curvature of wavefunction by sharing els • This decreases electronic energy as atoms come closer • Must trade off with increase in nuclear and el repulsion
Various kinds of bonds • Ionic bond due to electron transfer between dissimilar ions (weak, stabilized by long-ranged electrostatics, a = 0.2nm) • Covalent bond due to electron sharing between similar ions (strong, directional, short-ranged, a = 0.07nm stabilized by quantum kinetic energy/curvature) Other kinds of bonds exist – • Metallic bonds involve one electron shared by entire lattice • Van-der-Waals involves higher order dipole-dipole interaction • H-bonding is weak interaction involving one H atom, as in ice We will soon see how these bonds can extend to form a 3-D network and create a solid