Question

Which of the following after reacting with KI does not remove iodine?
A \[CaS{{O}_{4}}\]
B \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\]
C \[HN{{O}_{3}}\]
D \[HCl\]

Answer 1

Hint: Any good oxidising agent can remove iodine from KI. The oxidising agent is defined as the agent which tends to gain electrons itself (reduce itself) and make others oxidise by removing electrons from it. In other words, the oxidising agent is one that tends to increase the oxidation number of other compounds reacting with it.

Complete Step by Step Solution:
Now greater the oxidation number of central metal, the greater the tendency to reduce (neutral) itself and oxidise other compounds, and is termed a good oxidising agent.

In \[CaS{{O}_{4}}\], Sulphur (S) is the central metal and its oxidation number in this compound is +6. The charge on Ca is +2, the charge on oxygen is -2, and completely compound is neutral such as
$ 2\text{ }+\text{ }x\text{ }+\text{ }4\left( -2 \right)=\text{ }0 \\$
$ 2\text{ }+\text{ }x\text{ }\text{ }8\text{ }=\text{ }0 \\$
$ x\text{ }=\text{ }+6 \\$
$ 2\text{ }+\text{ }x\text{ }+\text{ }4\left( -2 \right)=\text{ }0 \\$
$ 2\text{ }+\text{ }x\text{ }\text{ }8\text{ }=\text{ }0 \\$
$ x\text{ }=\text{ }+6 \\$

In \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\], the central metal is Cr and its oxidation number in this complex is +6. The charge on potassium (K) is +1, the charge on oxygen is -2 and the compound is completely neutral such as
$ 2\left( +1 \right)\text{ }+\text{ }2x\text{ }+\text{ }7\left( -2 \right)\text{ }=\text{ }0 \\$
$ 2\text{ }+\text{ }2x\text{ }-14\text{ }=\text{ }0 \\$
$ 2x\text{ }-\text{ }12\text{ }=\text{ }0 \\$
$ x\text{ }=\text{ }+6 \\$
$ 2\left( +1 \right)\text{ }+\text{ }2x\text{ }+\text{ }7\left( -2 \right)\text{ }=\text{ }0 \\$
$ 2\text{ }+\text{ }2x\text{ }-14\text{ }=\text{ }0 \\$
$ 2x\text{ }-\text{ }12\text{ }=\text{ }0 \\$
$ x\text{ }=\text{ }+6 \\$

In \[HN{{O}_{3}}\], central metal is nitrogen and its oxidation state in this complex is +5. The charge on hydrogen is +1, the charge on oxygen is -2, and the compound completely is neutral such as
$ 1\text{ }+\text{ }x\text{ }+\text{ }3\left( -2 \right)\text{ }=\text{ }0 \\$
$ 1\text{ }+\text{ }x\text{ }\text{ }6\text{ }=\text{ }0 \\$
$ x\text{ }\text{ }5\text{ }=\text{ }0 \\$
$ x\text{ }=\text{ }+5 \\$
$ 1\text{ }+\text{ }x\text{ }+\text{ }3\left( -2 \right)\text{ }=\text{ }0 \\$
$ 1\text{ }+\text{ }x\text{ }\text{ }6\text{ }=\text{ }0 \\$
$ x\text{ }\text{ }5\text{ }=\text{ }0 \\$
$ x\text{ }=\text{ }+5 \\$

Now in \[HCl\], hydrogen has a +1 oxidation state and thus, it is not a good oxidising agent. So \[HCl\]will not release iodine after reaction with KI.
Thus, the correct option is D.

Note: All the compounds, \[CaS{{O}_{4}}\], \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\], and \[HN{{O}_{3}}\] are good oxidising agents as compared to \[HCl\]. Thus, all three compounds tend to get reduced (reduce their oxidation number through gaining of the electron) and oxidise KI (increase the oxidation number of K with the release of iodine from it).