Question

What is the oxidation state of copper in the compound, $Y{ Ba }_{ 2 }{ Cu }_{ 3 }{ O }_{ 7 }$ which shows superconductivity? Assume that the rare earth element yttrium is in its usual +3 oxidation state.
(a) $\dfrac { 7 }{ 3 } $
(b) 3
(c) 2
(d) 5

Answer 1

Hint: The oxidation number of an atom in a species is the total number of electrons it gains or loses while forming a bond with another atom present in the species. We can find out the oxidation number of an atom in a species using two methods.

Complete step by step solution:
There are two ways of solving this question since they have not mentioned what type of oxidation number. There are two types of oxidation numbers. (a) Calculated or average oxidation number and (b) observed oxidation number. Sometimes the values for both the calculated oxidation number and the observed oxidation number are the same. But they are different for some species. For such species we have to find the actual oxidation number by using the chemical structure of the species.
Calculated oxidation number:
For finding the calculated oxidation number of a particular atom in a species, we have to assume the oxidation number of all other atoms present in the species. The valence number of the atoms with their sign is taken as their assumed oxidation number. For example, if we have to find the oxidation number of hydrogen in ${ H }_{ 2 }O$, we will assume the oxidation number of oxygen as -2 where 2 is the valency of O atom. Therefore the oxidation number of H will be:
$\begin{matrix} Sum\quad of\quad all\quad the\quad atoms \\ present\quad in\quad the\quad species \end{matrix}=\begin{matrix} Overall\quad charge \\ present\quad on\quad the\quad species \end{matrix}$
Let the oxidation number of H be x,
$2x-2=0$ (since there are two atoms of hydrogen present in the molecule, therefore x is multiplied by 2.Also the overall charge on the molecule is zero)
$\Rightarrow x=+1$
Now for the species $Y{ Ba }_{ 2 }{ Cu }_{ 3 }{ O }_{ 7 }$, let us assume the oxidation number of copper be x. The oxidation number of barium, yttrium, and oxygen will be +2, +3 and -2 respectively.
$3+(2\times 2)+3x+[7\times (-2)]=0$
$\Rightarrow x=\dfrac { 7 }{ 3 } $
Thus the oxidation state of copper is $\dfrac { 7 }{ 3 } $.

Hence the correct answer is (a) $\dfrac { 7 }{ 3 } $.

Note: In this case we cannot calculate the oxidation state of copper using the other method since for calculating the observed oxidation number we need to know the structure of the compound and how the bonded electrons are shared among the individual atoms in the compound.