Chapter 10: Problem 24
What are some of the properties of fullerenes that make them potentially so useful?
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 10: Problem 24
What are some of the properties of fullerenes that make them potentially so useful?
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeIt takes less enetgy to dissociate a diatomic fluorine mole. cule than a diatomic oxygen molecule (in fact. less than one-third as much). Why is it easier to dissociate fluorine?
In Section 10.2. we discussed two-lobed \(p_{\text {re }} p_{\text {s }}\) and \(p_{z}\) states and four-lohed hybrid \(s p^{3}\) states. Another kind of hybrid state that sticks out in just one direction is the sp. formed from a single \(p\) state and an \(s\) state. Consider an urbitrary combination of the \(2 s\) state with the \(2 p_{x}\) state. Let us represent this by \(\cos \tau \psi_{2.0 .0}+\sin T \psi_{2.1 .0}\) (The trig factors ensure norinalization. In carrying out the integral, crow terms integrate to \(0 .\) leaving \(\cos ^{2}+\int\left|\psi_{2,0,0}\right|^{2} d b^{\prime}+\sin ^{2} \tau \int\left|\psi_{21,0}\right|^{2} d V\). which is \(\left.1 .\right)\) (a) Calculate the probability that an electron in such a state would be in the \(+z\) -hemisphere. (Nore: Here. the cross tems do nor integrate to \(0 .\) ) (b) What value of \(r\) leads to the maximum probability. what is the value of the maximum probability, and what is the cotresponding ratio of \(\psi_{2,0.0}\) to \(\phi_{2,1,0} ?\) (c) Using a computer, make a density (shading) plot of the probability density - density versus \(r\) and \(\theta\) for the \(r\) -value found in part (b).
Section 10.2 discusses \(\sigma\) -bonds and \(\pi\) -bonds for \(p\) -states and \(\sigma\) -bonds for s-states. but not \(\pi\) -bonds for s-states. Why not?
In a concise yet fairly comprehensive way, explain why doped semiconductors are so pervasive in modem technology.
The bond length of the \(\mathrm{N}_{2}\) molecule is \(0.1 \mathrm{nm},\) and its effective spring constant is \(2.3 \times 10^{3} \mathrm{~N} / \mathrm{m}\). (a) From the size of the energy jumps for rotation and vibration. deterinine whether either of these modes of energy storage should be active at \(300 \mathrm{~K}\). (b) According to the equipattivion theorem, the heat capacity of a diatomic molecule storing energy in rotations but not vibrations should be \(\frac{5}{2} R(3\) translational +2 rotational degrees of freedom). If it is also storing energy in vibrations. it should be \(\frac{7}{2} R\) (adding 2 vibrational degrees). Nitrogen's molar heat capacity is \(20.8 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\) at \(300 \mathrm{~K}\). Does this agree with your findings in part (a)?
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