Question

The compound ${YbBa}_2{Cu}_3{O}_7$, which shows superconductivity has copper in oxidation state $x$, then $x$ is :
(Assume that the rare earth element ytterbium is in the usual $+3$ oxidation state)

• A. $+\frac{7}{3}$
• B. $-\frac{7}{3}$
• C. $+2$
• D. $+\frac{5}{3}$

Answer 1

The correct option is A $+\frac{7}{3}$

The oxidation states of $Yb,Ba,Cu$ and $O$ are $+3,+2,x$ and $-2$ respectively.

$+3+2(+2)+3x+7(-2)=03x=7x=+\frac{7}{3}$

Hence, the oxidation state of copper is $+\frac{7}{3}$.