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The CsCl structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 69\()\). Given that the density of cesium chloride is \(3.97 \mathrm{~g} / \mathrm{cm}^{3}\), and assuming that the chloride and cesium ions touch along the body diagonal of the cubic unit cell, calculate the distance between the centers of adjacent \(\mathrm{Cs}^{+}\) and \(\mathrm{Cl}^{-}\) ions in the solid. Compare this value with the expected distance based on the sizes of the ions. The ionic radius of \(\mathrm{Cs}^{+}\) is \(169 \mathrm{pm}\), and the ionic radius of \(\mathrm{Cl}^{-}\) is \(181 \mathrm{pm}\).

Short Answer

Expert verified
The distance between the centers of adjacent Cs鈦 and Cl鈦 ions in the CsCl structure is calculated to be \(3.59 脳 10^{-8} cm\). This value is close to the expected distance based on the sizes of the ions (\(3.50 脳 10^{-8} cm\)), confirming that the ions touch along the body diagonal of the cubic unit cell.

Step by step solution

01

Formula for the body diagonal of the cubic unit cell

In a cubic unit cell, the body diagonal connects two opposite corners of the cube, passing through the center of the cube. According to Pythagorean theorem, the length of the body diagonal (d) can be calculated using the formula: \(d = \sqrt{a^2 + a^2 + a^2}\) where "a" is the length of the unit cell's edge.
02

Calculate the length of the unit cell's edge

Knowing the density of CsCl (\(3.97 \frac{g}{cm^3}\)), we can find the mass of one unit cell: mass = density 脳 volume of the unit cell \(mass = (3.97 \frac{g}{cm^3}) 脳 (a^3)\) Since one CsCl unit cell consists of 1 Cs鈦 ion and 1 Cl鈦 ion, we can get the molar mass of CsCl: molar mass = mass of Cs鈦 + mass of Cl鈦 = (132.9 g/mol) + (35.45 g/mol) = \(168.35 \frac{g}{mol}\) Using Avogadro's number, we can calculate the mass of one CsCl formula unit: mass of 1 CsCl formula unit = \(\frac{168.35 \frac{g}{mol}}{6.022 脳 10^{23} formula units/mol} = 2.797 脳 10^{-22} g\) Now, we can find the volume of the unit cell and the length of its edge (a): \(a^3 = \frac{mass}{density} = \frac{2.797 脳 10^{-22} g}{3.97 \frac{g}{cm^3}} = 7.045 脳 10^{-23} cm^3\) a = \(\sqrt[3]{7.045 脳 10^{-23} cm^3} = 4.15 脳 10^{-8} cm\)
03

Calculate the distance between the centers of adjacent Cs鈦 and Cl鈦 ions

Using the formula for the body diagonal (d) from step 1 and the calculated length of the unit cell's edge (a): \(d = \sqrt{a^2 + a^2 + a^2}\) \(d = \sqrt{3 脳 (4.15 脳 10^{-8} cm)^2} = 7.19 脳 10^{-8} cm\) The distance between the centers of adjacent Cs鈦 and Cl鈦 ions is half of the body diagonal (since it starts at the center), so: distance = \(\frac{1}{2} 脳 7.19 脳 10^{-8} cm = 3.59 脳 10^{-8} cm\)
04

Compare the calculated distance with the expected distance based on ionic radii

The expected distance between the centers of adjacent Cs鈦 and Cl鈦 ions can be calculated by summing their ionic radii: expected distance = ionic radius of Cs鈦 + ionic radius of Cl鈦 expected distance = \(169 pm + 181 pm = 350 pm = 3.50 脳 10^{-8} cm\) Comparing the calculated distance (3.59 脳 10鈦烩伕 cm) with the expected distance (3.50 脳 10鈦烩伕 cm), we can see that they are very close to each other, which is consistent with the assumption that the ions touch along the body diagonal of the cubic unit cell.

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Key Concepts

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

Cubic Unit Cell
Imagine a salt crystal on your kitchen table, except it's not your typical table salt. We're talking about cesium chloride (CsCl), and it's arranged in a structure known as a cubic unit cell. In this formation, every ion's position is methodically placed within a cube-shaped space. These cubes are like the building blocks of the crystal, stacking snugly together to form the overall structure of the solid.

The cubic unit cell is special because it's the simplest form out there, with sides of equal length and angles of 90 degrees. Picture a six-sided dice, and that's the basic shape we're dealing with. In the case of CsCl, each corner of this dice represents the location of a chloride ion (Cl鈦), while smack-dab in the center lies a cesium ion (Cs鈦). This arrangement results in a particularly elegant and straightforward layout to analyze. When students grasp this concept, understanding the rest of the crystal's properties becomes a walk in the park.
Ionic Radius
Our journey into the microcosmic realms brings us to the fascinating world of ionic radius. This term describes the size of an ion -- an atom that has lost or gained electrons and thus carries a charge. It's a fundamental property in understanding not just our CsCl crystal, but ionic compounds in general. The ionic radius is like a personal bubble for each ion, a space that it effectively 'occupies'.

In our CsCl structure, the cesium ion (Cs鈦) has given away an electron and, as a result, shrinks down a bit from its original, neutral state. On the flip side, the chloride ion (Cl鈦) has snagged an extra electron, puffing it up larger than its neutral counterpart. This 'size change' is crucial when trying to puzzle out how these ions will pack together and what distance will separate them at the center of their bonding embrace. Imagine you and a friend trying to stand as close as possible while holding balloons 鈥 the size of the balloons (like the ions' radii) will determine how close you can actually get.
Body Diagonal Calculation
Next up is the body diagonal calculation, the hero of this numerical narrative that provides answers to the intimate distances between our ionic couple, Cs鈦 and Cl鈦. Picture again the cubic unit cell, in all its three-dimensional glory. If we draw a line connecting two opposite corners, slicing right through the heart of the cube, what we鈥檝e traced out with our finger is the body diagonal. To calculate its length, we turn to the Pythagorean theorem, that old algebraic ally found in most math textbooks.

For a cubic unit cell with an edge length 'a', the diagonal is given through an elegant equation: \(d = \sqrt{3a^{2}}\). This simple yet powerful formula unravels the mystery of the unit cell's internal geometry, revealing the spatial secrets held within. And why does this matter? Because it helps students determine the edge-to-edge length of cubes that represent the real distance between our ion pairs within the crystal lattice. This melding of geometry and chemistry is what makes the study of crystal structures such an intriguing and visually stimulating experience.

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Most popular questions from this chapter

General Zod has sold Lex Luthor what Zod claims to be a new copper-colored form of kryptonite, the only substance that can harm Superman. Lex, not believing in honor among thieves, decided to carry out some tests on the supposed kryptonite. From previous tests, Lex knew that kryptonite is a metal having a specific heat capacity of \(0.082 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and a density of \(9.2 \mathrm{~g} / \mathrm{cm}^{3}\) Lex Luthor's first experiment was an attempt to find the specific heat capacity of kryptonite. He dropped a \(10 \mathrm{~g} \pm 3 \mathrm{~g}\) sample of the metal into a boiling water bath at a temperature of \(100.0^{\circ} \mathrm{C} \pm 0.2^{\circ} \mathrm{C}\). He waited until the metal had reached the bath temperature and then quickly transferred it to \(100 \mathrm{~g} \pm\) \(3 \mathrm{~g}\) of water that was contained in a calorimeter at an initial temperature of \(25.0^{\circ} \mathrm{C} \pm 0.2^{\circ} \mathrm{C}\). The final temperature of the metal and water was \(25.2^{\circ} \mathrm{C}\). Based on these results, is it possible to distinguish between copper and kryptonite? Explain. When Lex found that his results from the first experiment were inconclusive, he decided to determine the density of the sample. He managed to steal a better balance and determined the mass of another portion of the purported kryptonite to be \(4 \mathrm{~g} \pm 1 \mathrm{~g}\). He dropped this sample into water contained in a \(25-\mathrm{mL}\) graduated cylinder and found that it displaced a volume of \(0.42 \mathrm{~mL} \pm 0.02 \mathrm{~mL}\). Is the metal copper or kryptonite? Explain. Lex was finally forced to determine the crystal structure of the metal General Zod had given him. He found that the cubic unit cell contained four atoms and had an edge length of 600\. pm. Explain how this information enabled Lex to identify the metal as copper or kryptonite. Will Lex be going after Superman with the kryptonite or seeking revenge on General Zod? What improvements could he have made in his experimental techniques to avoid performing the crystal structure determination?

You are asked to help set up a historical display in the park by stacking some cannonballs next to a Revolutionary War cannon. You are told to stack them by starting with a triangle in which each side is composed of four touching cannonballs. You are to continue stacking them until you have a single ball on the top centered over the middle of the triangular base. a. How many cannonballs do you need? b. What type of closest packing is displayed by the cannonballs? c. The four corners of the pyramid of cannonballs form the corners of what type of regular geometric solid?

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