Question

Yttrium oxide, Barium carbonate and copper oxide react to form a super conductor $YBa_2Cu_3O_7$. The average oxidation state of copper in the compound is

• A. +2
• B. +2.33
• C. +2.67
• D. +3

Answer 1

Answer: (B)

+2.33

Answer 2

The compound given is $YBa_2Cu_3O_7$. To find the average oxidation state of copper, we need to know the oxidation states of the other elements in the compound.

1. Oxidation state of Yttrium (Y): Yttrium is a rare earth element and typically exhibits an oxidation state of +3.
2. Oxidation state of Barium (Ba): Barium is an alkaline earth metal (Group 2) and always exhibits an oxidation state of +2 in its compounds.
3. Oxidation state of Oxygen (O): In most compounds, especially oxides, oxygen exhibits an oxidation state of -2.

Let the average oxidation state of copper (Cu) be 'x'.

The sum of the oxidation states of all atoms in a neutral compound is zero. For $YBa_2Cu_3O_7$: $(1 \times +3) + (2 \times +2) + (3 \times x) + (7 \times -2) = 0$

Substitute the known values: $(1 \times +3) + (2 \times +2) + (3 \times x) + (7 \times -2) = 0$

Now, solve for x: $3 + 4 + 3x - 14 = 0$ $7 + 3x - 14 = 0$ $3x - 7 = 0$ $3x = 7$ $x = \frac{7}{3}$

Convert the fraction to a decimal: $x = 2.333...$

Rounding to two decimal places, the average oxidation state of copper is +2.33.