Answer to Question #143827 in Physical Chemistry for Booker

Question #143827
Consider the carbonate ion, CO32-
i) Setup and solve the Huckel secular equations for the pi-electrons.
ii) Express the energies in terms of the Coulomb integrals]
iii) Determine the delocalization energy.
1
Expert's answer
2020-11-12T07:09:30-0500

i) the molecular orbitals |ψi⟩ can be described as a linear combination of the 2pz atomic orbital at carbon with corresponding c1 coefficient and 2pz atomic orbitals at oxygen with corresponding c2, c3 and c4 coefficients

c1(H11-ES11)+c2(H12-ES12)+c3(H13-ES13)+c4(H14-ES14)=0

c1(H21-ES21)+c2(H22-ES22)+c3(H23-ES23)+c4(H24-ES24)=0

c1(H31-ES31)+c2(H32-ES32)+c3(H33-ES33)+c4(H34-ES34)=0

c1(H41-ES41)+c2(H42-ES42)+c3(H43-ES43)+c4(H44-ES44)=0

"\\alpha"i = Hii

"\\beta" ij = Hij; "\\beta" ij = "\\beta" ji

Sij = 0

Sii = 1

only adjecent orbitals have interactions

c1("\\alpha"1-E)+c2("\\beta" 12)+c3("\\beta" 13)+c4("\\beta" 14)=0

c1("\\beta" 21)+c2("\\alpha"2-E)+0+0=0

c1("\\beta" 31)+0+c3("\\alpha"3-E)+0=0

c1("\\beta" 41)+0+0+c4("\\alpha"4-E)=0

"\\beta" 12 = "\\beta" 13="\\beta" 14="\\beta"

"\\alpha"2 = "\\alpha"3 = "\\alpha"4

"\\begin{vmatrix}\n \u03b11-E & \u03b2 & \u03b2 & \u03b2\\\\\n \u03b2 & \u03b12-E & 0 & 0\\\\\n\u03b2& 0 & \u03b12-E & 0\\\\\n\u03b2& 0 & 0 & \u03b12-E\\\\\n\\end{vmatrix}" = 0

"\\alpha"1"\\alpha"23-3"\\alpha"22"\\beta"2-3"\\alpha"1"\\alpha"22E-"\\alpha"23E+6"\\alpha"2"\\beta" 2E+3"\\alpha"1"\\alpha"2E2+3"\\alpha"22E2-3"\\beta"2E2-"\\alpha"1E3-3"\\alpha"2E3+E4=0

Solving this system we obtain:

E1=0.5("\\alpha"1+"\\alpha"2-("\\alpha"12-2"\\alpha"1"\\alpha"2+"\\alpha"22+12"\\beta"2)0.5)

E2=E3="\\alpha"2

E4=0.5("\\alpha"1+"\\alpha"2+("\\alpha"12-2"\\alpha"1"\\alpha"2+"\\alpha"22+12"\\beta"2)0.5)

ii) E1=0.5("\\alpha"1+"\\alpha"2-("\\alpha"12-2"\\alpha"1"\\alpha"2+"\\alpha"22+12"\\beta"2)0.5)

E2=E3="\\alpha"2

E4=0.5("\\alpha"1+"\\alpha"2+("\\alpha"12-2"\\alpha"1"\\alpha"2+"\\alpha"22+12"\\beta"2)0.5)

iii) E(delocalized) = 0.5("\\alpha"1+"\\alpha"2-("\\alpha"12-2"\\alpha"1"\\alpha"2+"\\alpha"22+12"\\beta"2)0.5)

Let determine E(localized)

1-carbon, 2-oxygen with double bond

c1(H11-ES11)+c2(H12-ES12)=0

c1(H21-ES21)+c2(H22-ES22)=0


c1("\\alpha"1-E)+c2("\\beta" 12)=0

c1("\\beta" 21)+c2("\\alpha"2-E)=0


"\\begin{vmatrix}\n \u03b11-E & \u03b2 \\\\\n \u03b2 & \u03b12-E \\\\\n\\end{vmatrix}" =0


"\\alpha"1"\\alpha"2-"\\beta"2-"\\alpha"1E-"\\alpha"2E+E2 = 0

E1 = 0.5("\\alpha"1+"\\alpha"2-("\\alpha"12-2"\\alpha"1"\\alpha"2+"\\alpha"22+4"\\beta"2)0.5)

E2 = 0.5("\\alpha"1+"\\alpha"2+("\\alpha"12-2"\\alpha"1"\\alpha"2+"\\alpha"22+4"\\beta"2)0.5)

E(delocalization) = 0.5("\\alpha"1+"\\alpha"2+("\\alpha"12-2"\\alpha"1"\\alpha"2+"\\alpha"22+4"\\beta"2)0.5)-0.5("\\alpha"1+"\\alpha"2-("\\alpha"12-2"\\alpha"1"\\alpha"2+"\\alpha"22+ 12"\\beta"2)0.5) = 0.5(-("\\alpha"12-2"\\alpha"1"\\alpha"2+"\\alpha"22+4"\\beta"2)0.5+("\\alpha"12-2"\\alpha"1"\\alpha"2+"\\alpha"22+ 12"\\beta"2)0.5)


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