   Chapter 12, Problem 45GQ

Chapter
Section
Textbook Problem

The very dense metal iridium has a face-centered cubic unit cell and a density of 22.56 g/cm3. Use this information to calculate the radius of an atom of the element.

Interpretation Introduction

Interpretation:

The radius of the iridium atom has to be identified.

Concept introduction:

• An ionic radii are the radius of an atom's ion in ionic crystals structure.
• An ionic solid is made up cations and anions held together by electrostatic forces in a rigid array or lattice.
• Positive charge ions are cations and negative charge ions are anions.
• Lattice Energy is mainly depends on the charge on the ion and radius or size of the ion.
• Ionic radius increases from top to bottom on the periodic table.

Ionic radius decreases from left to right the periodic table.

Explanation

The density of the calcium atom is given below,

density=22.56g/cm3

Determining the mass of four atoms of Ca in a unit cell

192.217 g6.022 × 1023atoms/mol  = 31.913 × 10¯23g/atom31.913 × 10¯23g/atom  × 4 atoms = 127.655 × 10¯22g

Determining the volume of the unit cell:

Massdensity = volume

127.655 × 10-22g22.56g/cm3 = volumevolume =5.66×10-22cm3

Determining the edge length of the unit cell:

cube root of diameter is d =5

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