In the reaction, \(\mathrm{I}_{2}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow 2 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2}\), equivalent weight of iodine will be equal to: (a) Twice its molecular weight (b) Its molecular weight (c) \(\frac{1}{2}\) its moleculear weight (d) \(\frac{1}{4}\) its molecular weight

Understanding how to balance a chemical reaction is crucial when it comes to studying chemistry. A balanced chemical reaction shows that matter is conserved, which means the number of atoms of each element is the same on both sides of the equation. In the exercise given, balancing is already done:
\(\textrm{I}_2 + 2 \textrm{S}_2\textrm{O}_3^{2-} \longrightarrow 2 \textrm{I}^- + \textrm{S}_4\textrm{O}_6^{2-}\).
Here, we have 2 iodine atoms reacting with 2 thiosulfate molecules to give 2 iodide ions and 1 tetrathionate ion. Notice the coefficients 2 for thiosulfate and iodide to ensure the number of sulfur and iodine atoms are balanced on both sides. This also ensures that the charge is balanced, with a total charge of -4 on both sides of the equation.

Determining oxidation states of elements in a molecule or an ion is a fundamental step in understanding redox reactions. In the context of our exercise, iodine undergoes a change in oxidation state from 0 in \(\textrm{I}_2\) to -1 in \(\textrm{I}^-\).
Oxidation state signifies the hypothetical charge an atom would have if all bonds to atoms of different elements were completely ionic. To calculate the change in oxidation state, we need to check the charges before and after the reaction. In this case, Iodine starts with no charge as a molecule (I_2) and ends up with a -1 charge as an iodide ion (I^-). This is a reduction reaction for iodine as its oxidation state decreases.
Determining the correct oxidation states is often made easier with some rules such as hydrogen generally being +1, oxygen typically being -2, and the overall charge of ions indicating the cumulative oxidation state.

The valence factor, often known as the 'n factor', is an integer that represents the change in oxidation state per atom or the number of electrons exchanged per molecule or ion in a redox reaction. In the reaction we have, iodine changes its oxidation state by one unit per atom. However, since each \(\textrm{I}_2\) molecule contains two I atoms, the total change for the molecule is two units, thus giving us a valence factor of 2.
This concept of the valence factor is particularly helpful when we want to find equivalent weight, as it's involved in the formula \(E =\frac{M}{n}\), where \(E\) is equivalent weight, \(M\) is molecular weight, and \(n\) is the valence factor.

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It allows us to calculate the amounts of substances consumed and produced, based on the balanced chemical equation. For the given reaction, stoichiometry tells us that one molecule of \(\textrm{I}_2\) reacts with two molecules of thiosulfate (\(\textrm{S}_2\textrm{O}_3^{2-}\)) to produce two iodide ions (\(\textrm{I}^-\)) and one tetrathionate ion (\(\textrm{S}_4\textrm{O}_6^{2-}\)).
In terms of equivalent weight calculation, stoichiometry is integral because, without the correct molar ratios as derived from a balanced equation, it would be impossible to calculate equivalent weights accurately. This harmonious interplay between balanced equations, stoichiometry, and equivalent weights is what allows chemists to quantify reactions and determine the correct amounts of reactants and products.