Question

complete this table.

name	empirical formula
lithium fluoride

Explanation:

Step1: Determine charge of magnesium ion

Magnesium (Mg) in group 2 loses 2 electrons to form $Mg^{2 + }$.

Step2: Determine charge of bromide ion

Bromine (Br) in group 17 gains 1 electron to form $Br^{-}$.

Step3: Balance charges for magnesium bromide

To balance the +2 charge of Mg with -1 charge of Br, we need 2 Br⁻ ions. So empirical formula is $MgBr_{2}$.

Step4: Determine charge of lithium ion

Lithium (Li) in group 1 loses 1 electron to form $Li^{+}$.

Step5: Determine charge of fluoride ion

Fluorine (F) in group 17 gains 1 electron to form $F^{-}$.

Step6: Balance charges for lithium fluoride

The +1 charge of Li and -1 charge of F balance in a 1:1 ratio. So empirical formula is $LiF$.

Answer:

name	empirical formula
lithium fluoride	$LiF$