The reaction involves: .
Here, each molecule of donates 1 electron, leading to a total of 2 electrons being transferred per molecule of .
Oxidation states of sulfur: In , the average oxidation state of sulfur is +2. During the reaction, sulfur in is partially oxidized to , where the average oxidation state becomes +2.5.
Definition of equivalent weight: The equivalent weight of a substance is given by: Here, n represents the number of electrons transferred per molecule in the reaction.
Calculation: For , , because each molecule transfers 1 electron. Therefore, the equivalent weight of is equal to its molecular weight, M. Thus, the equivalent weight of in this reaction is M.
Correct Answer:
Option 2: M
Explanation:
1. Analyze the Reaction:
I2 + 2Na2S2O3 → 2NaI + Na2S4O6
2. Determine the Change in Oxidation State:
In Na2S2O3, the average oxidation state of sulfur is +2.
In Na2S4O6, the average oxidation state of sulfur is +2.5.
3. Calculate the Change per Molecule:
Each S atom in Na2S2O3 changes its oxidation state from +2 to +2.5, a change of 0.5.
There are two sulfur atoms in Na2S2O3, so the total change in oxidation state per molecule of Na2S2O3 is 2 * 0.5 = 1.
4. Determine the n-factor:
The n-factor (number of electrons transferred per molecule) is equal to the total change in oxidation state, which is 1.
5. Calculate the Equivalent Weight:
Equivalent weight = Molecular weight / n-factor
Equivalent weight = M / 1 = M
Given below are two statements:
Statement (I): The first ionization energy of Pb is greater than that of Sn.
Statement (II): The first ionization energy of Ge is greater than that of Si.
In light of the above statements, choose the correct answer from the options given below:
Identify the ion having 4f electronic configuration.