# Iodine solution is prepared by dissolving iodine in A) NaOHB) Na$_2$CO$_3$C) H$_2$OD) KI

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Hint: The preparation of iodine solution can be done in consideration with the polarity of the compound. In this question we will look after the polar nature of given compounds, and the dissociation of compounds in water.

Now, we have to prepare the iodine solution. So, first, let us know about the iodine molecule.
> Thus, the iodine molecule is covalent in nature, as it exists in diatomic nature, and the most important it is a nonpolar molecule. In other terms, we can say that it doesn’t ionize in the water.
> Now, if we talk about the given options. The first option is sodium hydroxide.
As we know that NaOH is a type of alkali metal hydroxide, and there is presence of Na-OH bond. So, it is the most polar compound, and highly soluble in water. On dissociation, there will be formation of sodium ions, and the hydroxide ions.
> The second option is sodium carbonate. It is an ionic bond.
As we know, it is a salt, composed of sodium ions, and the carbonate ions.
> We have the third option, i.e. water. It cannot be considered, as the iodine molecule is insoluble in the polar solvent, as mentioned.
> Talking about the fourth option, i.e. KI. It is an ionic polar compound.
As we know, on dissociation it will ionize into the potassium ions, and the iodide ions.
So, we can say that the iodine ion obtained from the KI will react with the iodine, and the iodine solution will be prepared.

Hence, the correct option is (D).

Note: Don’t get confused as all the given options are polar, so, we considered the dissociation of the compounds in water. As mentioned, iodine will not ionize in water. We can also use sodium iodide instead of the potassium iodide.