To find the oxidation state of copper in the superconducting compound $YBa_{2}Cu_{3}O_{7}$, we start by assuming the oxidation states of each element based on common oxidation states and the necessary neutrality of the compound. In this compound:

• Yttrium ($Y$) is in its usual oxidation state of +3.
• Barium ($Ba$) typically has an oxidation state of +2.
• Oxygen ($O$) typically has an oxidation state of -2.

The formula for the compound is given as $YBa_{2}Cu_{3}O_{7}$. The overall charge of the compound must be zero. Setting up the equation based on the oxidation states and the stoichiometry of the compound, we get:

$+3 + 2(+2) + 3(x) + 7(-2) = 0$

$3 + 4 + 3x - 14 = 0$

$3x - 7 = 0$

$3x = 7$

$x = \frac{7}{3}$

So, the oxidation state of copper (Cu) in $YBa_{2}Cu_{3}O_{7}$ is $\frac{7}{3}$, or more conventionally expressed as +$\frac{7}{3}$ or +2.33 when averaged over the three copper atoms. This mixed valency is a characteristic of the copper oxide layers in high-temperature superconducting materials, allowing for the unique electronic properties that enable superconductivity.