The correct answer is ##Mg####I_2##.

The correct name of this compound is magnesium iodide. Magnesium has +2 and +1 oxidation states , with the +2 being the most common.

Let us see how we got the answer; Look at the electronic arrangement of Mg and I atoms.

Mg ( Z= 12) has 12 electrons with following electronic configuration. 1##s^2##2##s^2##2##p^6##3##s^2##

It loses two electron in its 3s subshell to achieve stability and forms ion ##Mg^(2+)##.

##Mg^(2+)## = 1##s^2##2##s^2##2##p^6##

I on the other hand has five electrons in its 5p sub shell and wants to gain an extra electron to achieve stable noble gas configuration.Iodine atom on gaining one electron forms negative Iodide ion, ##I^-## ion.

I = [ Kr } 4##d^10## 5##s^2## 5##p^5##

##I^(-)## = [ Kr } 4##d^10## 5##s^2## 5##p^6##

Two Iodine atoms gains one electron each from one Mg atoms, each Mg atom loses two electrons to two Iodine atom , in this process Mg atom becomes ##Mg^(2+)## ion and each Iodine atom after gaining one electron becomes ##I^(-)## ion.

Mg^(2+)## and 2 ##I^(-)

or the formula is ##Mg## ##I_2##

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