The equivalent weight of $\text{C}{{\text{u}}_{2}}\text{S}$(mol. Wt. = M) in the following reaction is:
$\text{C}{{\text{u}}_{2}}\text{S + 4Mn}{{\text{O}}_{4}}\text{ }\to \text{ C}{{\text{u}}^{2+}}\text{ + S}{{\text{O}}_{2}}\text{ + M}{{\text{n}}^{2+}}$
A. M/2
B. M/6
C. M/8
D. M/4
Answer
668.1k+ views
Hint: Equivalent of the substance is defined as the ratio of the molecular weight or mass of the compound to the n - factor of the compound or the acidity or basicity. N - factor can be calculated by determining the change in the oxidation state.
Complete Step-by-Step Answer:
- In the given question, we have to calculate the equivalent weight of the copper sulfide from the given reaction.
- It is given in the question that the molecular weight of the copper sulfide is M.
- Now, we have to calculate the n - factor. Firstly, we have to calculate the change in the oxidation state of the copper and sulphur.
- So, change in the oxidation state of copper is:
$\begin{align}
& \text{C}{{\text{u}}_{2}}\text{S C}{{\text{u}}^{2+}} \\
& 2\text{x - 2 = 0 x = +2} \\
& \text{x = +1} \\
\end{align}$
- So, here the oxidation of copper takes place from +1 to +2 and the change in the oxidation state is (2-1) = 1.
- Now, change in the oxidation state of the sulphur will be:
$\begin{align}
& \text{C}{{\text{u}}_{2}}\text{S S}{{\text{O}}_{2}} \\
& \text{1 }\times \text{ 2 + x = 0 x - 2 }\times \text{ 2 = 0} \\
& \text{2 + x = 0 x - 4 = 0} \\
& \text{x = -2 x = 4} \\
\end{align}$
- So, the Oxidation of the sulphur also take place from -2 to +4 and the change in the oxidation state of sulphur is (4-(-2)) = 4+2 = 6.
- So, the n - factor will be 6 + 2(1) = 8.
- The equivalent weight of the copper (I) sulphide is:
$\text{Equivalent weight = }\dfrac{\text{Molecular weight}}{\text{n - factor}}$
$\text{Equivalent weight = }\dfrac{\text{M}}{8}$
Therefore, option C is the correct answer.
Note: The oxidation state of the single element is zero but if it has any charge present on it then it is considered as n - factor. Whereas the basicity is defined for the acidic substance and the acidity is defined for the basic substance.
Complete Step-by-Step Answer:
- In the given question, we have to calculate the equivalent weight of the copper sulfide from the given reaction.
- It is given in the question that the molecular weight of the copper sulfide is M.
- Now, we have to calculate the n - factor. Firstly, we have to calculate the change in the oxidation state of the copper and sulphur.
- So, change in the oxidation state of copper is:
$\begin{align}
& \text{C}{{\text{u}}_{2}}\text{S C}{{\text{u}}^{2+}} \\
& 2\text{x - 2 = 0 x = +2} \\
& \text{x = +1} \\
\end{align}$
- So, here the oxidation of copper takes place from +1 to +2 and the change in the oxidation state is (2-1) = 1.
- Now, change in the oxidation state of the sulphur will be:
$\begin{align}
& \text{C}{{\text{u}}_{2}}\text{S S}{{\text{O}}_{2}} \\
& \text{1 }\times \text{ 2 + x = 0 x - 2 }\times \text{ 2 = 0} \\
& \text{2 + x = 0 x - 4 = 0} \\
& \text{x = -2 x = 4} \\
\end{align}$
- So, the Oxidation of the sulphur also take place from -2 to +4 and the change in the oxidation state of sulphur is (4-(-2)) = 4+2 = 6.
- So, the n - factor will be 6 + 2(1) = 8.
- The equivalent weight of the copper (I) sulphide is:
$\text{Equivalent weight = }\dfrac{\text{Molecular weight}}{\text{n - factor}}$
$\text{Equivalent weight = }\dfrac{\text{M}}{8}$
Therefore, option C is the correct answer.
Note: The oxidation state of the single element is zero but if it has any charge present on it then it is considered as n - factor. Whereas the basicity is defined for the acidic substance and the acidity is defined for the basic substance.
Recently Updated Pages
Which will be the least stable resonating structure class 11 chemistry CBSE

How many 5 digit telephone numbers can be construc-class-11-maths-CBSE

How do you find the angle of the resultant vector class 11 physics CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

What is the need and importance of classification class 11 biology CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

10 examples of friction in our daily life

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference between physical and chemical change class 11 chemistry CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

