University Physics IV · topic UPIV-12.01
Covalent bonding and molecular orbitals
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Covalent bonding and molecular orbitals: predict a finite change
A molecular orbital is a one-electron state that spans every nucleus in the molecule; in the LCAO approximation it is a weighted sum of atomic orbitals, with the weights fixed by minimising the energy. Use the supplied topic model Y_out = 54.422*((z*z-z-(1-exp(-2*z*r)*(1+z*r))/r)*exp(-z*r)*(1+z*r+z*z*r*r/3)-z*(z-2)*exp(-z*r)*(1+z*r))/(1-exp(-2*z*r)*(1+z*r+z*z*r*r/3)^2), where Y_out is σg–σu* GAP (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. r is Nuclear separation R, measured in a₀. Keep z (Orbital exponent ζ) = 1 fixed. For this exercise use only 2.6375 a₀ ≤ r ≤ 7.3625 a₀ on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at r = 3.677 a₀ and 4.3385 a₀. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δr across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
σg–σu* GAP against r
Predictions from the supplied Covalent bonding and molecular orbitals model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): σg–σu* GAP / eV
Horizontal axis (x): r: Nuclear separation R / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Nuclear separation R / a₀ | σg–σu* GAP / eV |
|---|---|
| 2.6375 | 6.760930580278907 |
| 3.81875 | 2.8654448163440684 |
| 5 | 1.160076560980908 |
| 6.18125 | 0.4463045184677629 |
| 7.3625 | 0.1648737517547915 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of r separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At r = 3.677 a₀, substitute into Y_out = 54.422*((z*z-z-(1-exp(-2*z*r)*(1+z*r))/r)*exp(-z*r)*(1+z*r+z*z*r*r/3)-z*(z-2)*exp(-z*r)*(1+z*r))/(1-exp(-2*z*r)*(1+z*r+z*z*r*r/3)^2): Y₀ = 3.18268 eV.
- At r = 4.3385 a₀, the same fixed-parameter model gives Y₁ = 1.93761 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.24507 eV; the input change is Δr = 0.6615 a₀.
- Average sensitivity is ΔY_out/Δr = -1.88219 eV per a₀. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 3.18268 eV, Y₁ = 1.93761 eV; ΔY_out = -1.24507 eV; average sensitivity = -1.88219 eV per a₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Covalent bonding and molecular orbitals: infer an input with uncertainty
A molecular orbital is a one-electron state that spans every nucleus in the molecule; in the LCAO approximation it is a weighted sum of atomic orbitals, with the weights fixed by minimising the energy. Use the supplied topic model Y_out = 54.422*((z*z-z-(1-exp(-2*z*r)*(1+z*r))/r)*exp(-z*r)*(1+z*r+z*z*r*r/3)-z*(z-2)*exp(-z*r)*(1+z*r))/(1-exp(-2*z*r)*(1+z*r+z*z*r*r/3)^2), where Y_out is σg–σu* GAP (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. r is Nuclear separation R, measured in a₀. Keep z (Orbital exponent ζ) = 1 fixed. For this exercise use only 2.6375 a₀ ≤ r ≤ 7.3625 a₀ on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.76641231 eV, with a bounded reading uncertainty ±0.090231 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0004725 a₀, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
σg–σu* GAP against r
Predictions from the supplied Covalent bonding and molecular orbitals model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): σg–σu* GAP / eV
Horizontal axis (x): r: Nuclear separation R / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Nuclear separation R / a₀ | σg–σu* GAP / eV |
|---|---|
| 2.6375 | 6.760930580278907 |
| 3.81875 | 2.8654448163440684 |
| 5 | 1.160076560980908 |
| 6.18125 | 0.4463045184677629 |
| 7.3625 | 0.1648737517547915 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(r) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(2.6375) = 6.76093 eV and Y_out(7.3625) = 0.164874 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 2.6375 + (0.766412 − 6.76093)(7.3625 − 2.6375)/(0.164874 − 6.76093) = 6.9316 a₀. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives r = 5.51975 a₀; substituting back gives Y_out = 0.76641231 eV.
- The interpolation error in the input is |r_line − r| = 1.41185 a₀. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.67618131 eV and 0.85664331 eV. Sort the roots even if the response decreases: 5.38137 a₀ ≤ r ≤ 5.67452 a₀.
- The lower and upper input deviations are 0.13838 a₀ and 0.154774 a₀. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer r ≈ 5.51975 a₀; compatible interval [5.38137, 5.67452] a₀. Endpoint interpolation gives 6.9316 a₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.