Question

Write the formula of each of the following set of compounds :
Copper (I) oxide and Copper (II) oxide :
A. $C{{u}_{2}}O\text{ }and\text{ }CuO$
B. $Cu\text{ }and\text{ }C{{u}_{2}}$
C. $Cu{{(O)}_{2}}\text{ }and\text{ }CuO$
D. None of the above

Answer 1

The oxidation number of oxygen in compounds is usually -2, and in peroxides it is -1.
For oxygen the oxidation number is -2, because oxygen is a divalent compound (double bonded) but in case of peroxides the oxidation number of oxygen is -1.

Complete step by step answer:
-To find the oxidation number of a metal in the compounds we should know the oxidation number of the remaining elements in the compound other than the metal.
-We have to find Copper (I) oxide and Copper (II) oxide among the given options.
- Coming to given options, option A, $C{{u}_{2}}O\text{ }and\text{ }CuO$.
- Oxidation of number of copper in $C{{u}_{2}}O$
2x + (-2) =0
2x = 2
x = 1
- Oxidation number of copper in $C{{u}_{2}}O$ is +1.
- Oxidation of number of copper in$CuO$.
x + (-2) =0
x – 2=0
x = 2
Oxidation number of copper in $CuO$ is +2.
-Coming to option B, $Cu\text{ }and\text{ }C{{u}_{2}}$. We know that the pure metal has an oxidation number of zero in its pure form. So, in $Cu\text{ }and\text{ }C{{u}_{2}}$, copper has an oxidation number of zero. So, option B is wrong.
- Coming to option C, $Cu{{(O)}_{2}}\text{ }and\text{ }CuO$.
-Oxidation number of copper in $Cu{{(O)}_{2}}$
x + 2(-2) =0
x - 4 = 0
x = 4
- Oxidation number of copper in $CuO$
x – 2 =0
x = 2
-The oxidation number of copper in $Cu{{(O)}_{2}}\text{ }and\text{ }CuO$ are +4and +2 are respectively.
- Copper (I) oxide and Copper (II) oxide is going to be exhibited by$C{{u}_{2}}O\text{ }and\text{ }CuO$.

- So the correct option is A.

Note: Oxidation state or oxidation number of an atom or ion is the number of electrons gained or lost by an atom or ion compared to the neutral atom present in the same chemical compound. The oxidation number of the metals is going to change with the composition of the chemicals.