Question

Sodium chloride and sodium iodide react with concentrated sulfuric acid.
Which statements are correct?
1. Sodium chloride is not oxidized by concentrated sulfuric acid.
2. No color change is seen when concentrated sulfuric acid is added to sodium chloride
3. Sodium iodide is oxidized by concentrated sulfuric acid.
(A) 1, 2 and 3 are correct
(B) 1 and 2 only are correct
(C) 2 and 3 only are correct
(D) 1 only is correct

Answer 1

Hint: We must know that sulphuric acid acts as acid as well as an oxidizing agent in case of halide ions but this ability varies down the group. Sulphuric acid acts as an acid with all the halide ions but it is not strong enough to oxidize fluoride and chloride ions. Hence we will solve this by considering the fact that how its properties varies in case of halide ions.

Complete answer:
When a concentrated sulphuric acid reacts with a halide ion, it gives hydrogen ion to produce hydrogen halide. These hydrogen halides are in gaseous state and immediately escape from the system but we can see steamy fumes if they are exposed to moist air. This property is the same for all halide ions.
-Concentrated sulphuric acid can also act as an oxidizing agent other than being an acid as it oxidizes the halide ions. But this behavior is not valid in cases of fluoride and chloride ions hence we only get fumes as a product in these cases.
-Reaction of sodium chloride with concentrated sulphuric acid:-
$NaCl+{{H}_{2}}S{{O}_{4}}\to HCl+NaHS{{O}_{4}}$
In case of sodium chloride we only get steamy fumes of HCl as sulphuric acid is not strong enough to oxidize chloride ions. Since only fumes are produced in this case, hence we would not see any color change in the solution.
- Reaction of sodium iodide with concentrated sulphuric acid:-
$\begin{align}
& -NaI+{{H}_{2}}S{{O}_{4}}\to HI+NaHS{{O}_{4}} \\
& -{{H}_{2}}S{{O}_{4}}+8{{H}^{+}}+8{{I}^{-}}\to 4{{I}_{2}}+{{H}_{2}}S+4{{H}_{2}}O \\
\end{align}$
As we can see, sodium iodide reacts with sulphuric acid to give fumes of hydrogen iodide along with the major production of iodine due to which the color of the solution changes to red. In this case sulphuric acid is strong enough to oxidize the iodide ions.
So from the above data we may see that all the given statements in the question are right.

Hence the correct option is: (A) 1, 2 and 3 are correct.

Note:
Remember that red color was formed because when ${{I}_{2}}$ produced, it got in contact with sodium iodide and gave ${{I}_{3}}^{-}$ ion which is the main reason behind it.
-The hydrogen sulphide (${{H}_{2}}S$) gas produced cannot be seen but it can be observed due to its pungent smell (like bad egg) and also we shouldn’t inhale it much due to its toxic properties.