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Periodic Table

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

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  1. Periodic Table So we now know how multielectron atoms behave How do multiatom molecules behave?

  2. Ionic Bonds Between elements on left (electropositive) and right (electronegative) of periodic table

  3. 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+)

  4. 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

  5. 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

  6. 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

  7. 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)

  8. Covalent Bonds Between elements of comparable electronegativity

  9. 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}

  10. 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

  11. 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

  12. 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)

  13. 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)

  14. 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

  15. 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

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