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Atoms are assumed to touch in closest packed structures, yet every closest packed unit cell contains a significant amount of empty space. Why?

Short Answer

Expert verified
The presence of empty space in closest packed structures, such as face-centered cubic (fcc) and hexagonal closest packed (hcp) arrangements, is due to the spherical shape of atoms. Even when atoms are packed as closely as possible, their spherical nature results in empty spaces called interstitial spaces between them. These unoccupied spaces account for approximately 26% of the total volume in closest packed structures, making the packing efficiency about 74%. It is impossible to fill these spaces with additional atoms without expanding the lattice or changing the arrangement of existing atoms.

Step by step solution

01

Closest packed structures of atoms

Closest packed structures are arrangements of atoms in crystal lattice where the atoms are packed as closely as possible. There are two main types of closest packed structures: hexagonal closest packed (hcp) and face-centered cubic (fcc) structures. In both these structures, the atoms occupy a certain pattern which minimizes the interstitial spaces between them.
02

Properties of closest packed structures

In closest packed structures, each atom has 12 nearest neighbors (in fcc and hcp). The packing efficiency, which is the ratio of the volume occupied by the atoms to the total volume of the crystal, is approximately 74% for both fcc and hcp structures. This means that approximately 26% of the crystal volume is empty space.
03

Reason for empty space in closest packed structures

The reason for the existence of empty space in closest packed structures is due to the fact that atoms are spherical in shape. When arranging spheres in the most compact way possible (face-centered cubic or hexagonal close-packed arrangement), there will still be some leftover space, called interstitial space, between the spheres. It is impossible to completely fill this space with additional atoms without expanding the lattice or altering the arrangement of the existing atoms. So, even in the densest arrangement, the closest packed structures will always contain some amount of empty space.

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Most popular questions from this chapter

Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of \(197 \mathrm{pm}\), calculate the density of solid calcium.

MnO has either the \(\mathrm{NaCl}\) type structure or the \(\mathrm{CsCl}\) type structure (see Exercise 67). The edge length of the \(\mathrm{MnO}\) unit cell is \(4.47 \times 10^{-8} \mathrm{~cm}\) and the density of \(\mathrm{MnO}\) is \(5.28 \mathrm{~g} / \mathrm{cm}^{3}\) a. Does \(\mathrm{Mn} \mathrm{O}\) crystallize in the \(\mathrm{NaCl}\) or the \(\mathrm{CsCl}\) type structure? b. Assuming that the ionic radius of oxygen is \(140 . \mathrm{pm}\), estimate the ionic radius of manganese.

The molar heat of fusion of sodium metal is \(2.60 \mathrm{~kJ} / \mathrm{mol}\), whereas its heat of vaporization is \(97.0 \mathrm{~kJ} / \mathrm{mol}\). a. Why is the heat of vaporization so much larger than the heat of fusion? b. What quantity of heat would be needed to melt \(1.00 \mathrm{~g}\) sodium at its normal melting point? c. What quantity of heat would be needed to vaporize \(1.00 \mathrm{~g}\) sodium at its normal boiling point? d. What quantity of heat would be evolved if \(1.00 \mathrm{~g}\) sodium vapor condensed at its normal boiling point?

What quantity of energy does it take to convert \(0.500 \mathrm{~kg}\) ice at \(-20 .{ }^{\circ} \mathrm{C}\) to steam at \(250 .{ }^{\circ} \mathrm{C} ?\) Specific heat capacities: ice, \(2.03 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} ;\) liquid, \(4.2 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} ;\) steam, \(2.0 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} ; \Delta H_{\mathrm{vap}}=\) \(40.7 \mathrm{~kJ} / \mathrm{mol} ; \Delta H_{\mathrm{fus}}=6.02 \mathrm{~kJ} / \mathrm{mol}\)

In each of the following groups of substances, pick the one that has the given property. Justify your answer. a. highest boiling point: \(\mathrm{HBr}, \mathrm{Kr}\), or \(\mathrm{Cl}_{2}\) b. highest freezing point: \(\mathrm{H}_{2} \mathrm{O}, \mathrm{NaCl}\), or HF c. lowest vapor pressure at \(25^{\circ} \mathrm{C}: \mathrm{Cl}_{2}, \mathrm{Br}_{2}\), or \(\mathrm{I}_{2}\) d. lowest freezing point: \(\mathrm{N}_{2}, \mathrm{CO}\), or \(\mathrm{CO}_{2}\) e. lowest boiling point: \(\mathrm{CH}_{4}, \mathrm{CH}_{3} \mathrm{CH}_{3}\), or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) f. highest boiling point: HF, HCl, or HBr CC=O g. lowest vapor pressure at \(25^{\circ} \mathrm{C}: \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CCH}_{3}\), or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\)

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