Chapter 10, Problem 79E

### Chemistry

10th Edition
Steven S. Zumdahl + 2 others
ISBN: 9781305957404

Chapter
Section

### Chemistry

10th Edition
Steven S. Zumdahl + 2 others
ISBN: 9781305957404
Textbook Problem

# The unit cell of MgO is shown belowlDoes MgO have a structure like that of NaCl or ZnS? If the density of MgO is 3.58 g/cm3, estimate the radius (in centimeters) of the O2− anions and the Mg2+ cations.

Interpretation Introduction

Interpretation: The radius of Magnesium and its structure has to be estimated.

Concept introduction:

Unit cell: In a crystal, the smallest repeating units of lattice are known as unit cell.

In packing of atoms in a crystal structure, the atoms are imagined as spheres and closely packed in a regular pattern. The two major types of close packing of the spheres in the crystal are – hexagonal close packing and cubic close packing. Cubic close packing structure has face-centered cubic (FCC) unit cell.

In face-centered cubic unit cell, each of the six corners is occupied by every single atom. Each face of the cube is occupied by one atom.

Each atom in the corner is shared by eight unit cells and each atom in the face is shared by two unit cells. Thus the number of atoms per unit cell in FCC unit cell is,

8×18atomsincorners+6×12atomsinfaces=1+3=4atoms       The side length of one unit cell is given bya=2R2where  a=side lenghth of unit cellR=radiusofatom

Explanation

Explanation

To determine the structure of MgO

The structure of MgO resembles NaCl

From the given figure, the structure of MgO is similar to NaCl because it has 4â€‰Mg+ and 4â€‰O2- ions per face centered cubic cell.

To calculate the mass of MgO

Molar mass of MgO = 40.31â€‰g

Mass of MgO = 4â€‰MgOâ€‰formulaâ€‰unitsÃ—1â€‰molâ€‰MgO6.022Ã—1023atomsÃ—40.31â€‰gâ€‰MgO1â€‰molâ€‰MgO

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  = 2.678Ã—10-22â€‰gâ€‰MgO

The mass of MgO is calculated by using the formula units of MgO and Avogadro number and molar mass. The mass of MgO is found to be 2.678Ã—10-22â€‰gâ€‰ .

To calculate the volume and cube edge length

Mass of MgO = 2.678Ã—10-22â€‰gâ€‰

Volume of unit cell = 2.678Ã—10-22â€‰gâ€‰MgOÃ—1â€‰cm33.58â€‰g

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  = 7.48Ã—10-23â€‰cm3

Volume of unit cell = l3

l= cube edge length

l = (7

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