Question

The compound $YB{{a}_{2}}C{{u}_{3}}{{O}_{7}}$which shows superconductivity have copper in oxidation state of $+\dfrac{7}{3}$.
Assume that the rare earth element ytterbium is in its usual $+3$oxidation state. If true enter 1 and for false enter 0.

Answer 1

Hint:The oxidation number can be defined as the number of electrons present in an atom of a molecule which the atom can share, lose or gain during chemical bond formation with different elements.
Oxidation number is also called oxidation state.

Complete step by step answer:
To check whether the above statement is true or false we first need to calculate the oxidation state of copper.
Total charge on the compound $YB{{a}_{2}}C{{u}_{3}}{{O}_{7}}$is zero.
Oxidation state of $C{{u}_{{}}}$= to be calculate so let it be x
Oxidation state of barium (Ba) = +2
Oxidation state of $B{{a}_{2}}=(+2)\times 2$
Oxidation state oxygen = $-2$
Oxidation state of ${{O}_{7}}=(-2)\times 7=-14$
Oxidation state of Y = +3
To find oxidation state of Cu we have to add oxidation state of all atoms of the compound \[YB{{a}_{2}}C{{u}_{3}}{{O}_{7}}\] and equate that to zero as the total charge on the compound is zero.
$\begin{align}
  & (3\times 1)+(2\times 2)+(3\times x)+(-14)=0 \\
 & 3+4+3x-14=0 \\
 & x=\dfrac{7}{3} \\
\end{align}$ …… (i)
On solving equation (i) we will get the value of x as $\dfrac{7}{3}$
We got oxidation state of $C{{u}_{3}}=\dfrac{7}{3}$so the statement in the question i.e. “The compound $YB{{a}_{2}}C{{u}_{3}}{{O}_{7}}$which shows superconductivity have cupper in oxidation state of $+\dfrac{7}{3}$ .
Assume that the rare earth element ytterbium is in its usual $+3$oxidation state.” is true.

Hence, the answer is 1.

Note:
Oxidation number and oxidation state both are the same. Don’t get confused about these two terms.
Some general rules to find oxidation number (O.N.):
O.N. of free element = 0 (always)
O.N. of monatomic ion = charge of the ion
O.N. of hydrogen = -1 and +1 when combined with less electronegative elements.
O.N. of oxygen = -2 and -1 in peroxides
O.N. of group 1 = +1
O.N. of group 2 = +2
O.N. of group 17 in binary compounds = -1
Total O.N. of neutral compound = 0