Question

Compound $YB{a_2}C{u_3}{O_7}$ is a superconductor. The oxidation number of copper in the compound will be:
[ oxidation number of $Y = + 3$ ]
A.$+ \dfrac{7}{3}$
B.$- \dfrac{7}{3}$
C.$+ \dfrac{2}{3}$
D.$+ \dfrac{1}{5}$

Answer 1

We know that superconductors are those materials which can conduct energy through it without resistance. It means that superconductors can carry a current indefinitely without the loss of energy. We will determine the oxidation state of each element present in the compound to find the oxidation state of copper.

Complete step-by-step answer: Oxidation number is basically the number of electrons gained or lost by the element in formation of the compound.
So here we will first determine the oxidation state of each element present in the compound.
Oxidation state of $B{a_2} = ( + 2) \times 2 = 4$
Oxidation state of $O = ( - 2) \times 7 = - 14$
Oxidation state of $Y = + 3$ (already given in question)
As we know this is a neutral compound so all the oxidation states will be equal to zero.
Let the oxidation state of copper be $x$
As there are three atoms of copper combined together so we will take the oxidation state of copper as $3x$
So,
$3 + 4 + 3x - 14 = 0 \\\ 7 + 3x - 14 = 0 \\\ 7 + 3x = 14 \\\ 3x = + 14 - 7 \\\ 3x = + 7 \\\ x = + \dfrac{7}{3} \\\$
So, the oxidation state of copper in $YB{a_2}C{u_3}{O_7}$ is $+ \dfrac{7}{3}$

Hence the correct answer is Option A.

Note: $YB{a_2}C{u_3}{O_7}$ is a stoichiometric compound which possesses a deficiency of electrons stabilized by the crystal structure in which there are holes present making the copper appear to have an average oxidation state of $+ \dfrac{7}{3}$ So, the superconductivity of it would be due to these positive holes being able to move freely in the solid when an external potential difference is applied across it.