Question

ionic bonding practice
follow all directions for credit

draw the lewis dot diagrams of magnesium and iodine.

based on the dot diagram, does magnesium want to gain or lose electrons, and how many?

based on the dot diagram, does iodine want to gain or lose electrons, and how many?

how many iodine atoms and magnesium atoms will be in this molecule, based on how many electrons need to be gained/lost?

in the space below, draw the lewis dot diagrams of all the elements that will be in the ionic compound. if there is more than one of an atom draw its dot structure multiple times. draw arrows to show where the electrons being transferred will move.

write dot structures, and charges, of the new ions formed in bracket notation for the ionic compound made with magnesium and iodine.

using the above steps, write the bracket notation for the ionic compound created by combing aluminum and chlorine.

Explanation:

Step1: Determine electron transfer for Mg and I

Magnesium (Mg) has 2 valence electrons. It loses 2 electrons to achieve a stable octet. Iodine (I) has 7 valence electrons. It gains 1 electron to achieve a stable octet. So, 1 Mg atom will transfer 2 electrons to 2 I atoms.

Step2: Draw Lewis dot diagram for electron transfer

For Mg: $\cdot\!Mg\cdot$ loses 2 electrons. For I: $\cdot\!\ddot{I}\cdot$ gains 1 electron each. Arrows show electron transfer from Mg to 2 I atoms.

Step3: Write bracket notation for ions

Mg loses 2 electrons to form $Mg^{2 + }$: $[\text{Mg}]^{2 + }$. Each I gains 1 electron to form $I^{-}$: $[\cdot\!\ddot{I}\cdot]^{-}$. The ionic compound is $MgI_{2}$, so the bracket notation is $[\text{Mg}]^{2 + }[\cdot\!\ddot{I}\cdot]^{-}[\cdot\!\ddot{I}\cdot]^{-}$.

For Aluminum (Al) and Chlorine (Cl):

Step1: Determine electron transfer for Al and Cl

Aluminum (Al) has 3 valence electrons. It loses 3 electrons. Chlorine (Cl) has 7 valence electrons. It gains 1 electron. So, 1 Al atom will transfer 3 electrons to 3 Cl atoms.

Step2: Write bracket notation for ions

Al loses 3 electrons to form $Al^{3 + }$: $[\text{Al}]^{3 + }$. Each Cl gains 1 electron to form $Cl^{-}$: $[\cdot\!\ddot{Cl}\cdot]^{-}$. The ionic compound is $AlCl_{3}$, so the bracket notation is $[\text{Al}]^{3 + }[\cdot\!\ddot{Cl}\cdot]^{-}[\cdot\!\ddot{Cl}\cdot]^{-}[\cdot\!\ddot{Cl}\cdot]^{-}$.

Answer:

For Mg and I: $[\text{Mg}]^{2 + }[\cdot\!\ddot{I}\cdot]^{-}[\cdot\!\ddot{I}\cdot]^{-}$
For Al and Cl: $[\text{Al}]^{3 + }[\cdot\!\ddot{Cl}\cdot]^{-}[\cdot\!\ddot{Cl}\cdot]^{-}[\cdot\!\ddot{Cl}\cdot]^{-}$