Q22.36P - Chemistry: Molecular Nature Of Matter And Change

Question

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?

Step-by-Step Solution

Answer

(a) Iodine's position in the periodic table corresponds to its mode of occurrence.

(b) $\text{I}{{\text{O}}_{\text{3}}}^{\text{ - }}$ is an oxidizing agent. $\text{HS}{{\text{O}}_{\text{3}}}^{\text{ - }}$acts as a reducing agent because $\text{HS}{{\text{O}}_{\text{3}}}^{\text{ - }}$converts to $\text{HS}{{\text{O}}_{\text{4}}}^{\text{ - }}$.

(c) The mass of solid iodine is $1.1\times {10}^4 \mathrm{g}$.

Step 1: Electronegativity

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.

Step 2:Subpart (a)

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.

Step 3: Subpart (b)

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.

Step 4: Subpart (c)

The mol % of sodium nitrate is given by

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

The mass of solid iodine is calculated as,

Mass (g) of iodine:

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