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Chapter 37: Problem 4

Does it make sense to distinguish individual NaCl molecules in a salt crystal? What about individual \(\mathrm{H}_{2} \mathrm{O}\) molecules in an ice crystal? Explain.

Short Answer

Expert verified
No, individual NaCl molecules cannot be distinguished in a salt crystal due to the ionic nature of the compound and the resulting lattice structure of ions. However, individual H2O molecules can be distinguished in an ice crystal because they maintain their molecular identity in the solid state despite being bonded to each other via hydrogen bonds.

Step by step solution

01

Introduction to Ionic and Covalent Bonding

First, it is important to understand what Ionic and Covalent bonding are. Ionic bonding occurs when valence electrons are transferred from one atom to another, creating positively and negatively charged ions. These ions are attracted to each other due to their opposite charges, forming ionic compounds. NaCl is an example of an ionic compound. Covalent bonding, on the other hand, involves the sharing of valence electrons between atoms, leading to the formation of molecules. H2O is an example of a molecule formed through covalent bonding.
02

Explain the Structure of a Salt Crystal (NaCl)

In a crystal of NaCl, one cannot distinguish separate NaCl molecules because they do not exist. This is due to the ionic nature of the compound - each sodium atom loses an electron to become a Na+ ion and each chlorine atom gains an electron to become a Cl- ion. These ions form a repeating pattern in a lattice structure rather than existing as individual NaCl molecules.
03

Explain the Structure of an Ice Crystal (H2O)

Conversely, it is possible to distinguish individual H2O molecules in an ice crystal. The water molecules are bonded to each other through a type of covalent bonding known as hydrogen bonding. Despite the molecules being fixed in a lattice structure in the solid ice state, they still maintain their individual molecular identity, which is different than ionic compounds.

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Most popular questions from this chapter

Photovoltaic (PV) cells convert sunlight energy directly into electricity, with no moving parts (recall Fig. 37.20 ). In a PV cell, photons incident on a semiconductor \(P N\) junction promote electrons to the conduction band, producing electron-hole pairs and driving current through an external circuit (Fig. 37.25 ). Commercially available PV cells are \(15-20 \%\) efficient, meaning they convert this fraction of incident sunlight into electrical energy; the theoretical maximum efficiency is around \(33 \%\) for silicon-based PV cells. An important limitation on PV efficiency is the relation between the solar spectrum (IMAGE CANNOT COPY) and PV cells' semiconductor band-gap energy. For silicon, the band gap is \(1.14 \mathrm{eV}\); photons with less energy can't promote electrons to the conduction zone and are thus unavailable for the PV energy conversion. Conversely, photons with more than the band-gap energy give up their excess energy as heat, also reducing PV efficiency. One way to improve PV efficiency is to make multi-layer cells with several \(P N\) junctions using semiconductors with different band gaps. For a multi-layer PV cell to be effective, a. the junction with the largest band gap should be closest to the top of the PV cell. b. the junction with the largest band gap should be closest to the bottom of the PV cell. c. the largest band gap should correspond to infrared wavelengths. d. the smallest band gap should correspond to ultraviolet wavelengths.

The lower-energy states in a covalently bound diatomic molecule can be found approximately from the so-called Morse potential \(U(r)=U_{0}\left(e^{2\left(r-r_{0}\right) / a}-e^{-2\left(r-r_{0}\right) / a}\right),\) where \(r\) is the atomic separation and \(U_{0}, r_{0},\) and \(a\) are constants determined from experimental data. Calculate \(d U / d r\) and \(d^{2} U / d r^{2}\) to show that \(U\) has a minimum, and find expressions for (a) \(U_{\min }\) and (b) the separation \(r_{\min }\) at the minimum energy.

Integrating Equation 37.5 over all energies gives the total number of states per unit volume in a metal. Therefore, integrating from \(E=0\) to \(E=E_{\mathrm{F}}\) - that is, over the occupied states only-gives the number of conduction electrons per unit volume. Carry out this integration to show that the electron number density is given by $$n=\left(\frac{2^{9 / 2} \pi m^{3 / 2}}{3 h^{2}}\right) E_{\mathrm{F}}^{3 / 2}$$

How do type I and type II superconductors differ?

The transition from the ground state to the first rotational excited state in diatomic oxygen \(\left(\mathrm{O}_{2}\right)\) requires \(356 \mu \mathrm{eV}\). At what temperature would the thermal energy \(k T\) be sufficient to set diatomic oxygen into rotation? Would you ever find diatomic oxygen exhibiting the specific heat of a monatomic gas at normal pressure?
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