CH2740
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CH2740
SECTION 1
Compulsory question
1 Answer ALL parts.
(a) For an infinite linear chain of hydrogen atoms with equidistant atoms (lattice constant a):
(i) Draw a simple band structure E(k). Use correct labelling.
(ii) Explain how to increase or reduce the width of the band.
(iii) Explain why the energy at k=0 is smaller than at k>0.
(iv) Is this material a metal or a semiconductor?
[25%]
SECTION 2
Answer ONE question from Question 2 and 3
The D2h character table is appended to this paper after SECTION 2
2 Answer ALL parts.
(a) Using Hückel theory, we can describe the 几 electrons in conjugated hydrocarbons and aromatic systems. However, we can also use Hückel theory to describe G bonds between electrons in H3, using the 1s orbitals at each H atom as the atomic orbital basis:
(i) Set up the secular determinants for linear and cyclic H3 .
(ii) Find the roots of the secular determinant for both systems in terms of the Coulomb integrals a and
the resonance integrals F .
(Hints: Use x = (a − E)/F. One of the roots in the triangular case is x = −2.)
(iii) Draw the molecular orbital energy levels of both systems.
(Remember that the resonance integral F has a negative value.)
(iv) Determine which state, linear or triangular, is more stable for the following molecules/ions:
(i) H3+ ,(ii) H3 , (iii) and H3- .
[50%]
MODEL ANSWERS
(1) Model Answers (a)
[covered in lectures]
inbetween them (= higher energy).
(iv)This is a metal.
(ii) The width of the band depends on the interatomic distance a. The smaller the distance, the stronger the orbital overlap and the wider the band.
(iii) The energy at k=0 is smaller because all Hydrogen s orbitals are in phase with each other and form a totally symmetric delocalised electronic state (0 nodes). At high k, the s orbitals have oscillating sign and the delocalised state shows nodal planes
(2) Model Answers
[very similar to problems seen in lecture and practised in workshop]
(a)
(i) The secular determinants in terms of a and F are Linear:
E a
F 0
F
a E
F
F | = 0
Triangular:
| F a E F | = 0
(ii) We resolve the secular equations as follows:
Linear:
X =
|1 X 1 | = 0
X ⋅ (X2 − 2) = 0
X1,2 = ±√2, X3 = 0
E1 = a + (2F
E2 = a
E3 = a (2F
Triangular:
X 1 1
1 1 X
X1 = −2
3
X
2X + 1 =
(iii)
(iv)The energy expressions for the triangular cation, anion, and neutral H3 are:
EH = 2a + 4F
EH3 = 3a + 3F
EH3 = 4a + 2F
The energy expressions for the linear cation, anion, and neutral H3 are:
EH = 2a + 2√2F
EH3 = 3a + 2√2F
EH3 = 4a + 2√2F
Therefore H3+ is triangular, H3 is triangular, and H3- is linear.
2022-09-14