What is the most likely way in which an atom of magnesium (mg) will satisfy the octet rule when it forms bonds?
by losing two electrons
by losing four electrons
by gaining six electrons
by gaining eight electrons

               By losing two electrons

                       Magnesium belong to alkali earth metals (Group II) (s-Block) in periodic table with electronic configuration as given below,

                                     Mg    =    1s², 2s², 2p⁶, 3s²

As shown above Mg contains two electrons in its valence shell. Hence, in order to react and attain noble gas configuration Mg must loose two electrons as follow,

                                         Mg    →    Mg⁺²  +  2 e⁻

Note that after loosing two electrons the electronic configuration becomes as follow,

                               Mg⁺²  =      1s², 2s², 2p⁶   or   [Ne]

Hence, in order to satisfy the octet rule (presence of eight electrons in valence shell) Mg must donate or loose two electrons because it is far more difficult to gain 6 electrons to complete octet of third shell respectively.

it should be p1 v1 = p2 v2