
The equivalent weight of \[{\text{Mn}}{{\text{O}}_2}\] is equal to molecular weight when it is converted to:
A.\[{\text{M}}{{\text{n}}_2}{{\text{O}}_3}\]
B.\[{\text{MnO}}\]
C.\[{\text{MnO}}_5^{2 - }\]
D.\[{\text{MnO}}_4^ - \]
Answer
570.6k+ views
Hint: Equivalent becomes equal to the molecular weight when numbers of mole are one. Hence we need to check in the above question that when \[{\text{Mn}}{{\text{O}}_2}\] is converted to one of the above options, oxidation state of manganese changes by one.
Formula used:
\[\dfrac{{{\text{Molecular weight}}}}{{{\text{n}} - {\text{factor}}}} = {\text{Equivalent weight}}\]
Complete step by step answer:
Equivalent weight as the name suggests is the mass of 1 equivalent. It is also known as the gram equivalent they both are the same thing. It is a mass of a given substance that combines with fixed mass of a fixed quantity of another substance.
For Elements equivalent weight of an element is the mass which combines what is placed is one gram of hydrogen. We get the value of equivalent weight by dividing the atomic weight with valence.
For a redox reaction n-factor is calculated as the change in oxidation state of the element which undergoes the redox change. In the above question we need to check where the change in oxidation state of metal is 1 because then only equivalent weight and molecular weight will be equal (according to formula). Let us consider when it converts to option 1.
\[{\text{Mn}}{{\text{O}}_2} \to {\text{M}}{{\text{n}}_2}{{\text{O}}_3}\]
Let us calculate oxidation state of central element,
Oxidation state of manganese in \[{\text{Mn}}{{\text{O}}_2}\]\[ + 2 \times - 2 = 0\]
\[{\text{oxidation state of Mn}} = 4\]
\[{\text{2}} \times {\text{oxidation state of Mn in M}}{{\text{n}}_2}{{\text{O}}_3} + 3 \times - 2 = 0\]
\[{\text{oxidation state of Mn}} = \dfrac{6}{2} = 3\]
Hence the change in oxidation state is 1 that is from 4 to 3. So equivalent weight will be the same as molecular weight.
The correct option is option A.
Note:
Molecular weight is the mass of a molecule that is calculated using carbon 12 isotope as a reference. It is also known as atomic mass. The unit of molecular mass is \[{\text{g mo}}{{\text{l}}^{ - 1}}\] .
Formula used:
\[\dfrac{{{\text{Molecular weight}}}}{{{\text{n}} - {\text{factor}}}} = {\text{Equivalent weight}}\]
Complete step by step answer:
Equivalent weight as the name suggests is the mass of 1 equivalent. It is also known as the gram equivalent they both are the same thing. It is a mass of a given substance that combines with fixed mass of a fixed quantity of another substance.
For Elements equivalent weight of an element is the mass which combines what is placed is one gram of hydrogen. We get the value of equivalent weight by dividing the atomic weight with valence.
For a redox reaction n-factor is calculated as the change in oxidation state of the element which undergoes the redox change. In the above question we need to check where the change in oxidation state of metal is 1 because then only equivalent weight and molecular weight will be equal (according to formula). Let us consider when it converts to option 1.
\[{\text{Mn}}{{\text{O}}_2} \to {\text{M}}{{\text{n}}_2}{{\text{O}}_3}\]
Let us calculate oxidation state of central element,
Oxidation state of manganese in \[{\text{Mn}}{{\text{O}}_2}\]\[ + 2 \times - 2 = 0\]
\[{\text{oxidation state of Mn}} = 4\]
\[{\text{2}} \times {\text{oxidation state of Mn in M}}{{\text{n}}_2}{{\text{O}}_3} + 3 \times - 2 = 0\]
\[{\text{oxidation state of Mn}} = \dfrac{6}{2} = 3\]
Hence the change in oxidation state is 1 that is from 4 to 3. So equivalent weight will be the same as molecular weight.
The correct option is option A.
Note:
Molecular weight is the mass of a molecule that is calculated using carbon 12 isotope as a reference. It is also known as atomic mass. The unit of molecular mass is \[{\text{g mo}}{{\text{l}}^{ - 1}}\] .
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