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Without doing calculations, indicate how you would expect the lattice energies of \(\mathrm{LiCl}(\mathrm{s}), \mathrm{KCl}(\mathrm{s}), \mathrm{RbCl}(\mathrm{s}),\) and \(\mathrm{CsCl}(\mathrm{s})\) to compare with the value of \(-787 \mathrm{kJmol}^{-1}\) determined for \(\mathrm{NaCl}(\mathrm{s})\) on page \(543 .\) [ Hint: Assume that the enthalpies of sublimation of the alkali metals are comparable in value. What atomic properties from Chapter 9 should you compare?

Short Answer

Expert verified
The expected order of lattice energies based on atomic radii is: LiCl > NaCl > KCl > RbCl > CsCl.

Step by step solution

01

Consider atomic properties

Chapter 9 discusses the periodic trends in atomic properties. One such property that affects lattice energy is atomic radius. As you go down the group from Lithium to Cesium in the alkali metals, atomic radius increases. This means that Li is smallest and Cs is largest.
02

Relate atomic radius to lattice energy

Lattice energy depends inversely on the distance between the ions in the lattice, that is, the larger the distance, the smaller the lattice energy, and vice versa. Since Li is smaller than Na, we would expect LiCl to have a higher lattice energy than NaCl. Conversely, KCl, RbCl, and CsCl should have lower lattice energies than NaCl because K, Rb, and Cs are larger than Na.
03

Order lattice energies based on atomic radius

Based on the assessment of atomic radius in relation to lattice energy, the lattice energies should follow the order: LiCl > NaCl > KCl > RbCl > CsCl.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Atomic Radius
The concept of atomic radius is fundamental when discussing lattice energies. The atomic radius is essentially the distance from the center of an atom's nucleus to its outermost electrons. As we move down a group in the periodic table, each element has an additional electron shell compared to the one above it. This increases the distance of the valence electrons from the nucleus, thus increasing the atomic radius. Additionally, as the number of electron shells increases, the shielding effect becomes greater, reducing the attraction between the electrons and the nucleus and leading to a larger atomic radius.

In the context of lattice energy, the atomic radius plays an influential role. A smaller atomic radius usually means that the ions in a solid can pack closer together, leading to a relatively high lattice energy. Conversely, a larger atomic radius can result in lower lattice energy, as the ions are further apart and their interactions are weaker. This understanding helps us predict trends in lattice energies among compounds as we look across different elements in the periodic table.
Periodic Trends and Lattice Energy
Periodic trends refer to patterns in the properties of elements across the periodic table. Lattice energy's relationship with atomic radius is a classic example of these trends. Since lattice energy is the energy required to separate a compound's ions to an infinitely apart state, it is affected by the size of the ions involved.

In a group, as the atomic number increases, so does the size of the ion, resulting in lower lattice energy. This is because the effective nuclear charge experienced by the outer electrons decreases, making it easier to separate the ions.

Factors Affecting Lattice Energy

Other factors to consider include charge density and ionic charge. Higher charge and smaller ionic radius lead to stronger attraction between ions, thus increasing lattice energy. In a period, from left to right, the lattice energy typically increases because ions generally become smaller and more charged. However, when comparing ions with different charges, the one with the higher charge will usually have a higher lattice energy, assuming similar ionic sizes.
Alkali Metals and Their Lattice Energies
Alkali metals, found in Group 1 of the periodic table, have distinctive properties that affect their lattice energies. These elements, including Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), and Cesium (Cs), possess a single electron in their outermost shell, which they can easily lose to form cations.

Due to their increasing atomic radii down the group, the lattice energies of alkali metal compounds decrease. This is why LiCl will have a higher lattice energy compared to NaCl, as lithium has a smaller radius allowing for tighter packing of ions in the crystal lattice.

Exploring Lattice Energies in Alkali Metal Halides

The comparison of lattice energies in compounds like NaCl, KCl, RbCl, and CsCl is particularly illustrative. Each successive alkali metal ion is larger than the previous one, implying a corresponding decrease in lattice energy. For instance, CsCl has significantly lower lattice energy than NaCl due to the much larger size of the Cs+ ion compared to Na+. These trends are consistent with the principles of periodicity and offer great insights into the periodic properties of elements.

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Most popular questions from this chapter

Some vapor pressure data for Freon- \(12, \mathrm{CCl}_{2} \mathrm{F}_{2}\), once a common refrigerant, are \(-12.2^{\circ} \mathrm{C}, 2.0 \mathrm{atm} ; 16.1^{\circ} \mathrm{C}\) 5.0 atm; \(42.4^{\circ} \mathrm{C}, 10.0\) atm; \(74.0^{\circ} \mathrm{C}, 20.0\) atm. Also, \(\mathrm{bp}=-29.8^{\circ} \mathrm{C}, \quad T_{\mathrm{c}}=111.5^{\circ} \mathrm{C}, \quad P_{\mathrm{c}}=39.6 \mathrm{atm} .\) Use these data to plot the vapor pressure curve of Freon-12 What approximate pressure would be required in the compressor of a refrigeration system to convert Freon- 12 vapor to liquid at \(25.0^{\circ} \mathrm{C}\) ?

In your own words, define or explain the following terms or symbols: (a) \(\Delta H_{\text {vap }} ;\) (b) \(T_{c} ;\) (c) instantaneous dipole; (d) coordination number; (e) unit cell.

We have learned that the enthalpy of vaporization of a liquid is generally a function of temperature. If we wish to take this temperature variation into account, we cannot use the Clausius-Clapeyron equation in the form given in the text (that is, equation 12.2 ). Instead, we must go back to the differential equation upon which the Clausius-Clapeyron equation is based and reintegrate it into a new expression. Our starting point is the following equation describing the rate of change of vapor pressure with temperature in terms of the enthalpy of vaporization, the difference in molar volumes of the vapor \(\left(V_{g}\right),\) and liquid \(\left(V_{1}\right),\) and the temperature. $$\frac{d P}{d T}=\frac{\Delta H_{\mathrm{vap}}}{T\left(V_{\mathrm{g}}-V_{1}\right)}$$ Because in most cases the volume of one mole of vapor greatly exceeds the molar volume of liquid, we can treat the \(V_{1}\) term as if it were zero. Also, unless the vapor pressure is unusually high, we can treat the vapor as if it were an ideal gas; that is, for one mole of vapor, \(P V=R T\). Make appropriate substitutions into the above expression, and separate the \(P\) and \(d P\) terms from the \(T\) and \(d T\) terms. The appropriate substitution for \(\Delta H_{\text {vap }}\) means expressing it as a function of temperature. Finally, integrate the two sides of the equation between the limits \(P_{1}\) and \(P_{2}\) on one side and \(T_{1}\) and \(T_{2}\) on the other. (a) Derive an equation for the vapor pressure of \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{l})\) as a function of temperature, if \(\Delta H_{\mathrm{vap}}=\) \(15,971+14.55 T-0.160 T^{2}\left(\text { in } J m o l^{-1}\right)\) (b) Use the equation derived in (a), together with the fact that the vapor pressure of \(\mathrm{C}_{2} \mathrm{H}_{4}(1)\) at \(120 \mathrm{K}\) is 10.16 Torr, to determine the normal boiling point of ethylene.

Can \(\mathrm{SO}_{2}\) be maintained as a liquid under a pressure of 100 atm at \(0^{\circ} \mathrm{C} ?\) Can liquid methane be obtained under the same conditions?

The normal melting point of copper is \(1357 \mathrm{K}\), and \(\Delta \mathrm{H}_{\text {fus }}\) of \(\mathrm{Cu}\) is \(13.05 \mathrm{kJ} \mathrm{mol}^{-1}\). (a) How much heat, in kilojoules, is evolved when a \(3.78 \mathrm{kg}\) sample of molten Cu freezes? (b) How much heat, in kilojoules, must be absorbed at 1357 K to melt a bar of copper that is \(75 \mathrm{cm} \times\) \(15 \mathrm{cm} \times 12 \mathrm{cm} ?\) (Assume \(d=8.92 \mathrm{g} / \mathrm{cm}^{3}\) for \(\mathrm{Cu}\).)

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