Question #82730

When using an independent particle model, an MO ψ is expressed in LCAO form, i.e. as a linear combination of atomic orbitals {ϕ_1, ϕ_2, . . . , ϕ_K}, (1)
and the expansion coefficients are determined from secular equation: the Hamiltonian is the one electron operator h and the “orbital energy” ε. Find the MOs and orbital energies for a linear
polyene (Cn Hn+2), taking ϕ_µ to be 2p_z AOs on C and including only nearest-neighbor matrix
element, with
(ϕ_µ|h|ϕ_µ) = α, (ϕ_µ|h|ϕ_µ±1) = β (2)
and with the neglect of the overlap.
Give explicitly the MOs and orbital energies for N = 3 and N = 4.
1

Expert's answer

2018-12-27T07:30:24-0500

Answer on Question #82730 - Chemistry - Physical Chemistry

Question:

When using an independent particle model, an MO ψ\psi is expressed in LCAO form, i.e. as a linear combination of atomic orbitals {φ1,φ2,,φK}\{\varphi_{-}1,\varphi_{-}2,\dots ,\varphi_{-}K\}, (1)

and the expansion coefficients are determined from secular equation: the Hamiltonian is the one electron operator hh and the "orbital energy" ε\varepsilon. Find the MOs and orbital energies for a linear

polyene (Cn Hn+2), taking φμ\varphi_{-}\mu to be 2p_z AOs on C and including only nearest-neighbor matrix element, with


(φμhφμ)=α,(φμhφμ±1)=β(2)(\varphi_ {-} \mu | h | \varphi_ {-} \mu) = \alpha , (\varphi_ {-} \mu | h | \varphi_ {-} \mu \pm 1) = \beta (2)


and with the neglect of the overlap.

Give explicitly the MOs and orbital energies for N=3N = 3 and N=4N = 4.

Solution:


φiHφi=αi\left\langle \varphi_ {i} \right| \mathbf {H} \left| \varphi_ {i} \right\rangle = \alpha_ {i}φiHφi=βij\left\langle \varphi_ {i} \right| \mathbf {H} \left| \varphi_ {i} \right\rangle = \beta_ {i j}φiφj=Sij\left\langle \varphi_ {i} \right| \varphi_ {j} \rangle = S _ {i j}


where αi\alpha_{i} is termed the Coulomb integral, βij\beta_{ij} the resonance integral and SijS_{ij} the overlap integral. We are using normalized AOs, so SiiS_{ii} . Furthermore, the two atoms are identical, so


α1=α2;\alpha 1 = \alpha 2;β12=β21;\beta 1 2 = \beta 2 1;(c12+c22)α+2c1c2βE(c12+c22+2c1c2S)=0(c _ {1} ^ {2} + c _ {2} ^ {2}) \alpha + 2 c _ {1} c _ {2} \beta - E (c _ {1} ^ {2} + c _ {2} ^ {2} + 2 c _ {1} c _ {2} S) = 0Ec1=Ec2=0\frac {\partial E}{\partial c _ {1}} = \frac {\partial E}{\partial c _ {2}} = 0(αE)c1+(βES)c2=0(\alpha - E) c _ {1} + (\beta - E S) c _ {2} = 0(βES)c1+(αE)c2=0(\beta - E S) c _ {1} + (\alpha - E) c _ {2} = 0αEβESβESαE=(αE)2(βES)2=0\left| \begin{array}{c c} \alpha - E & \beta - E S \\ \beta - E S & \alpha - E \end{array} \right| = (\alpha - E) ^ {2} - (\beta - E S) ^ {2} = 0E1=α+β1+SandE2=αβ1SE _ {1} = \frac {\alpha + \beta}{1 + S} \quad \text{and} \quad E _ {2} = \frac {\alpha - \beta}{1 - S}ΨiΨi=ci12+ci22+2ci1ci2S=1\left\langle \Psi_ {i} \right| \Psi_ {i} \rangle = c _ {i 1} ^ {2} + c _ {i 2} ^ {2} + 2 c _ {i 1} c _ {i 2} S = 1Ψ1=12(1+S)(φ1+φ2)andΨ2=12(1+S)(φ1φ2)\Psi_ {1} = \frac {1}{\sqrt {2 (1 + S)}} \left(\varphi_ {1} + \varphi_ {2}\right) \quad \text{and} \quad \Psi_ {2} = \frac {1}{\sqrt {2 (1 + S)}} \left(\varphi_ {1} - \varphi_ {2}\right)


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