lodine is the only halogen that occurs in a positive oxidation state, in impurities within Chile saltpeter, ${\mathbf{\text{NaNO}}}_{\mathbf{\text{3}}}$.

(a) Is this mode of occurrence consistent with iodine's location in the periodic table? Explain.

(b) In the production of ${\mathbf{\text{I}}}_{\mathbf{\text{2}}}\mathbf{\text{,I}}{{\mathbf{\text{O}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{-}}}$reacts with $\mathbf{\text{HS}}{{\mathbf{\text{O}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{-}}}$:

$\mathbf{\text{I}}{{\mathbf{\text{O}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{-}}}\mathbf{\text{(aq) + HS}}{{\mathbf{\text{O}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{-}}}\mathbf{\text{(aq)}}\mathbf{\to }\mathbf{\text{HS}}{{\mathbf{\text{O}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{-}}}\mathbf{\text{(aq) + S}}{{\mathbf{\text{O}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{2 -}}}{\mathbf{\text{(aq) +H}}}_{\mathbf{\text{2}}}{\mathbf{\text{O(I) +I}}}_{\mathbf{\text{2}}}\mathbf{\text{(s)}}\mathbf{}$

[unbalanced] Identify the oxidizing and reducing agents.

(c) If $\mathbf{0}\mathbf{.}\mathbf{78}\mathbf{}\mathbf{mol}\mathbf{\%}$ of an ${\mathbf{\text{NaNO}}}_{\mathbf{\text{3}}}$ deposit is${\mathbf{\text{NaIO}}}_{\mathbf{\text{3}}}$, how much${\mathbf{\text{I}}}_{\mathbf{\text{2}}}$ (in g) can be obtained from $\mathbf{\text{1.000}}$ton (2000. Ib) of the deposit?

Electronegativity: An atom's chemical behavior in which it attracts the shared electron pair to itself is known as electronegativity. As the number of energy levels grows larger, electronegativity decreases.

The element iodine can be found near the bottom of the periodic table. As a result of its low electronegativity, it can adopt positive oxidation states. Therefore, iodine's position in the periodic table corresponds to its mode of occurrence.

The balanced chemical equation is given by

$2{\text{IO}}_3^{-}\left(\text{aq}\right)+5{\text{HSO}}_3^{-}\text{(aq)}\to 3{\text{HSO}}_4^{{-}_{\left(aq\right)}}+2\text{S}{{\text{O}}_4}^{2-}\left(\text{aq}\right)+{\text{H}}_2{\text{O}}_{\left(\text{I}\right)}+{\text{I}}_{2(\text{s})}$

In this reaction $\text{2I}{{\text{O}}_{\text{3}}}^{\text{-}}$converts to solid iodine, in which reduction occurs.

Thus, $\text{I}{{\text{O}}_{\text{3}}}^{\text{-}}$is a oxidizing agent. $\text{HS}{{\text{O}}_{\text{3}}}^{\text{-}}$acts as a reducing agent because $\text{HS}{{\text{O}}_{\text{3}}}^{\text{-}}$converts to $\text{HS}{{\text{O}}_{\text{4}}}^{\text{-}}$in which oxidation occurs.

The mol % of sodium nitrate is given by

$0.78\mathrm{mol}\%{\mathrm{NaIO}}_3$leaves ( $\text{100.00 - 0.78}$)$\text{mol} {\text{NaNO}}_{\text{3}}{\text{= 99.22mol\%NaNO}}_{\text{3}}$

The mass of solid iodine is calculated as,

Mass (g) of iodine:

Therefore, the required value is $1.1\times {10}^4\mathrm{g}{\mathrm{I}}_2$.