Problem 2

Copper(I) oxide

\ce{Cu2O}

is a semiconductor used in solid-state electronics and solar cells. The cubic unit cell has lattice constant

a = 427.0

pm, with oxygen atoms (A) forming a body-centered cubic sub-lattice and copper atoms (B) occupying positions forming a face-centered cubic sub-lattice. (a) What are the coordination numbers of Cu and O? (b) Calculate the shortest

\ce{O-O}

\ce{Cu-O}

, and

\ce{Cu-Cu}

distances. (c) Calculate the theoretical density of stoichiometric

\ce{Cu2O}

in g cm

^{-3}

. (d) A non-stoichiometric sample has

0.2\%

of all Cu atoms oxidized to

\ce{Cu^2+}

. What percentage of Cu sites are vacant, and what is

x

\ce{Cu_{2-x}O}

? (e) Write balanced equations for the reaction of

\ce{Cu2O}

with: (1) atmospheric

\ce{O2}

in humid air; (2) dilute sulfuric acid; (3) hot concentrated sulfuric acid.

Step 2 of 5: Interatomic distances

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d(\ce{O-O}) = \tfrac{1}{2}a\sqrt{3} = 369.8\ \text{pm};\quad d(\ce{Cu-O}) = \tfrac{1}{4}a\sqrt{3} = 184.9\ \text{pm};\quad d(\ce{Cu-Cu}) = \tfrac{1}{2}a\sqrt{2} = 301.9\ \text{pm}

Analysis

Oxygen atoms occupy corners and the center of the cube, so the closest $\ce{O-O}$ distance is half the body diagonal: $\frac{1}{2} \times 427.0 \times \sqrt{3} = 369.8$ pm. Cu atoms sit on tetrahedral positions along the body diagonal, giving $d(\ce{Cu-O}) = 184.9$ pm. The closest Cu atoms lie on adjacent faces separated by half the face diagonal: $\frac{1}{2} \times 427.0 \times \sqrt{2} = 301.9$ pm.