Question

What are the products of the dissociation of magnesium iodide? (select all correct answers): $ \sim 2I + (aq) \sim Mg2 - (aq) \sim I2 - (aq) \sim 2I - (aq) \sim Mg(s) \sim Mg2 + (aq)$

Answer 1

Hint : In order to answer this question, to know the products of the dissociation of magnesium iodide, first we will discuss which type of compound, magnesium iodide is. And then we will write the actual reaction of dissociation of magnesium iodide and will discuss more about that dissociation reaction.

Complete Step By Step Answer:
Magnesium iodide is a soluble ionic compound that dissociates completely when dissolved in water to produce magnesium cations and iodide anions in aqueous solution.
Now, magnesium is located in group 2 of the Periodic Table, which implies that magnesium atoms tend to lose 2 electrons and become cations that carry a $2 + $ charge.
 Iodine, on the other hand, is located in group 17 of the Periodic Table, which implies that iodine atoms tend to gain 1 electron and become anions that carry a $1 - $ charge.
As you know, ionic compounds must be neutral, i.e. the overall positive charge carried by the cations must be balanced by the overall negative charge carried by the anions.
In your case, you need 2 iodide anions in order to balance the $2 + $ charge of the magnesium cation.
Therefore, the chemical formula for magnesium iodide is:
$[M{g^{2 + }}] + 2[{I^ - }] \to Mg{I_2}$
We can now say that when magnesium iodide dissolved in water, it produces $M{g^{2 + }}$ and ${I^ - }$ ions.
$Mg{I_{2(aq)}} \to M{g^{2 + }}_{(aq)} + 2{I^ - }_{(aq)}$

Note :
Magnesium iodide is a soluble ionic compound that dissociates completely when dissolved in water to produce magnesium cations and iodide anions in aqueous solution. Hence, magnesium iodide is soluble in water.