Question

The structures of another class of high-temperature ceramic superconductors are shown below.

(a) Determine the formula of each of these four superconductors.

(b) One of the structural features that appears to be essential for high-temperature superconductivity is the presence of planar sheets of copper and oxygen atoms. As the number of sheets in each unit cell increases, the temperature for the onset of superconductivity increases. Order the four structures from the lowest to highest superconducting temperature.

(c) Assign oxidation states to Cu in each structure assuming that Tl exists as T1³⁺. The oxidation states of Ca, Ba, and O are assumed to be +2, +2, and -2, respectively.

(d) It also appears that copper must display a mixture of oxidation states for a material to exhibit superconductivity. Explain how this occurs in these materials as well as in the superconductor in Exercise 79.

Short answer

(a) The formula for the superconductors is $\mathrm{T}1{\mathrm{Ba}}_2{\mathrm{CuO}}_5,\mathrm{T}1{\mathrm{Ba}}_2{\mathrm{CaCu}}_2{\mathrm{O}}_7,\mathrm{T}1{\mathrm{Ba}}_2{\mathrm{Ca}}_2{\mathrm{Cu}}_3{\mathrm{O}}_9\mathrm{and}\mathrm{T}1{\mathrm{Ba}}_2{\mathrm{Ca}}_2{\mathrm{Cu}}_3{\mathrm{O}}_9$for structure A, B, C and D respectively.

(b)$a<b<c<d$

(c) Cu in structure (a) will be Cu³⁺, in structure (b) will be 1 Cu²⁺ and 1Cu^3+, in structure (c) will be 2 Cu²⁺ and 1Cu³⁺, in structure (d) will be 3 Cu²⁺ and 1Cu³⁺

(d) In case of Exercise 16.73 the superconductor material achieves variable copper oxidation states. It can be obtained by omitting oxygen at various sites in the lattice.

Step-by-step solution

Step 1(a): Determine formula for the structures

$$\begin{gathered} \mathrm{Ba}: 2 \mathrm{inside} \mathrm{unit} \mathrm{cell} \\ \mathrm{T}1:8 \mathrm{corners}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.\mathrm{T}1}}{\mathrm{corner}}=1\mathrm{T}1 \\ \mathrm{Cu}: 4 \mathrm{edges}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\mathrm{Cu}}}{\mathrm{edge}}=1\mathrm{Cu} \\ \mathrm{O}:6 \mathrm{faces}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\mathrm{O}}}{\mathrm{faces}}+8 \mathrm{edges}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\mathrm{O}}}{\mathrm{edges}}=5\mathrm{O} \end{gathered}$$

Therefore, for structure (a) the formula for the superconductor will be $\mathrm{T}1{\mathrm{Ba}}_2{\mathrm{CuO}}_5$

Step 2(b): Determine formula for the structures

$$\begin{gathered} \mathrm{Ba}: 2 \mathrm{inside} \mathrm{unit} \mathrm{cell} \\ \mathrm{T}1:8 \mathrm{corners}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.\mathrm{T}1}}{\mathrm{corner}}=1\mathrm{T}1 \\ \mathrm{Cu}: 8 \mathrm{edges}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\mathrm{Cu}}}{\mathrm{edges}}=2\mathrm{Cu} \\ \mathrm{O}:10 \mathrm{faces}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\mathrm{O}}}{\mathrm{faces}}+8 \mathrm{edges}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\mathrm{O}}}{\mathrm{edges}}=7\mathrm{O} \\ \mathrm{Ca}: 1 \mathrm{inside} \mathrm{unit} \mathrm{cell} \end{gathered}$$

Therefore, for structure (b) the formula for the superconductor will be $\mathrm{T}1{\mathrm{Ba}}_2{\mathrm{CaCu}}_2{\mathrm{O}}_7$

Step 3(c): Determine formula for the structures

$$\begin{gathered} \mathrm{Ba}: 2 \mathrm{inside} \mathrm{unit} \mathrm{cell} \\ \mathrm{T}1:8 \mathrm{corners}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.\mathrm{T}1}}{\mathrm{corner}}=1\mathrm{T}1 \\ \mathrm{Cu}: 12 \mathrm{edges}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\mathrm{Cu}}}{\mathrm{edge}}=3\mathrm{Cu} \\ \mathrm{O}:14 \mathrm{faces}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\mathrm{O}}}{\mathrm{faces}}+8 \mathrm{edges}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\mathrm{O}}}{\mathrm{edges}}=9\mathrm{O} \\ \mathrm{Ca}: 2 \mathrm{inside} \mathrm{unit} \mathrm{cell} \end{gathered}$$

Therefore, for structure (c) the formula for the superconductor will be $\mathrm{T}1{\mathrm{Ba}}_2{\mathrm{Ca}}_2{\mathrm{Cu}}_3{\mathrm{O}}_9$

Step 4(d): Determine formula for the structures

$$\begin{gathered} \mathrm{Ba}: 2 \mathrm{inside} \mathrm{unit} \mathrm{cell} \\ \mathrm{T}1:8 \mathrm{corners}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.\mathrm{T}1}}{\mathrm{corner}}=1\mathrm{T}1 \\ \mathrm{Cu}: 16 \mathrm{edges}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\mathrm{Cu}}}{\mathrm{edge}}=4\mathrm{Cu} \\ \mathrm{O}:18 \mathrm{faces}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\mathrm{O}}}{\mathrm{faces}}+8 \mathrm{edges}\times \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\mathrm{O}}}{\mathrm{edges}}=11\mathrm{O} \\ \mathrm{Ca}: 3 \mathrm{inside} \mathrm{unit} \mathrm{cell} \end{gathered}$$

Therefore, for structure (d) the formula for the superconductor will be $\mathrm{T}1{\mathrm{Ba}}_2{\mathrm{Ca}}_3{\mathrm{Cu}}_4{\mathrm{O}}_{11}$

Determine order of superconductivity

Structure a has one planar sheet of Cu and O atoms and the number increases by one for each of the remaining structures. The order of superconductivity temperature from lowest to highest temperature is a<b<c<d.

Determine oxidation states of Cu

We need to assign oxidation states to Cu in each structure assuming that T1 exists as T1³⁺. The oxidation states of Ca, Ba, and O are assumed to be +2, +2, and -2, respectively.

Let the oxidation state of Cu be x.

$$\begin{gathered} \mathrm{T}1{\mathrm{Ba}}_2{\mathrm{CuO}}_5 \\ 3+2\left(2\right)+\mathrm{x}+5\left(-2\right)=0 \\ \mathrm{x}=+3 \end{gathered}$$

Therefore, Cu in structure (a) will be Cu³⁺

$$\begin{gathered} \mathrm{T}1{\mathrm{Ba}}_2{\mathrm{CaCu}}_2{\mathrm{O}}_7 \\ 3+2\left(2\right)+2+2\mathrm{x}+7\left(-2\right)=0 \\ \mathrm{x}=+\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \end{gathered}$$

Therefore, Cu in structure (b) will be 1 Cu²⁺ and 1Cu³⁺

$$\begin{gathered} T1B{a}_{2}C{a}_{2}C{u}_3{O}_9 \\ 3+2\left(2\right)+2\left(2\right)+3x+9\left(-2\right)=0 \\ x=+\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \end{gathered}$$

Therefore, Cu in structure (c) will be 2 Cu²⁺ and 1Cu³⁺

$$\begin{gathered} \mathrm{T}1{\mathrm{Ba}}_2{\mathrm{Ca}}_3{\mathrm{Cu}}_4{\mathrm{O}}_{11} \\ 3+2\left(2\right)+3\left(2\right)+4\mathrm{x}+11\left(-2\right)=0 \\ \mathrm{x}=+\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$4$}\right. \end{gathered}$$

Therefore, Cu in structure (d) will be 3 Cu²⁺ and1Cu³⁺

Explanation for part (d)

Different oxidation states of copper can be achieved for this superconductor material by varying the numbers of Ca, Cu, and O in each unit cell. In the previous step the mixtures of copper oxidation states have been calculated. In case of Exercise 16.73 the superconductor material achieves variable copper oxidation states. It can be obtained by omitting oxygen at various sites in the lattice.