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Question: The eq. wt. of iodine of mass M in, \({I_2} + 2{S_2}{O_3}^{2 - } \to 2{I^ - } + {S_4}{O_6}^{2 - }\) ...

The eq. wt. of iodine of mass M in, I2+2S2O322I+S4O62{I_2} + 2{S_2}{O_3}^{2 - } \to 2{I^ - } + {S_4}{O_6}^{2 - } is:
A. MM
B. M2\dfrac{M}{2}
C. M4\dfrac{M}{4}
D. none of these

Explanation

Solution

As we know that the equivalent weight of a compound refers to the mass of a substance that reacts exactly with an arbitrarily fixed amount of any other compound. Thus, the equivalent weight of a compound can be represented as:
Equivalent weight=molecular weightnumber of electrons gained or lostEquivalent{\text{ }}weight = \dfrac{{molecular{\text{ }}weight}}{{number{\text{ }}of{\text{ }}electrons{\text{ }}gained{\text{ }}or{\text{ }}lost}}

Complete step by step answer:
When we want to determine the equivalent weight of iodine in the given reaction, we need to first find out the number of electrons gained by iodine. The change in the oxidation state of I in the reactant and product gives the number of electrons gained by I2{I_2}.
As we know that the oxidation state is the charge obtained by the atom of an element when it loses or gains electrons while going from its free state to a combined state with atoms of other elements. The oxidation state of atoms in their elementary state is assumed as zero. Therefore, the oxidation state of Iodine in I2{I_2} is 0.
Therefore, the oxidation state of iodine in I{I^ - } is 1 - 1.
Here, the change in the oxidation state will make it 11. So, the number of electrons gained by iodine is 11.
One molecule of I2{I_2} has two iodine atoms. Then, the total number of electrons gained by two iodine atoms is 11. As the molecular weight of I2{I_2} is M. Since, two electrons are gained per mole of I2{I_2} so the equivalent weight of iodine can be derived by dividing it by 2 i.e., M2\dfrac{M}{2}.

So, the correct answer is Option B.

Note: We need to understand the concept of equivalent mass of a substance is defined as the number of parts by mass of it which combine with or displace 1.00781.0078 parts by mass of hydrogen, 88 parts by mass of oxygen and 35.535.5 parts by mass of chlorine. The atoms combine together to form chemical compounds, such that the elements are always present in definite proportions by mass and this property can be used to make chemically different molecules with different mass, equal.