Question

part 2: determine the products formed by each of the following pairs of elements after an ionic reaction.

pair of elements charges ionic formula of the compound formed name of compound

ex k, i k+1 i-1 ki potassium iodide
1 be, f
2 al, br
3 na, cl
4 mg, s
5 al, n

Explanation:

Step1: Determine charges of Be and F

Beryllium (Be) loses 2 electrons to form $Be^{2 + }$, fluorine (F) gains 1 electron to form $F^{-}$.

Step2: Write ionic formula

To balance charges, we need 2 F⁻ for 1 $Be^{2+}$, so formula is $BeF_{2}$.

Step3: Name the compound

It is Beryllium fluoride.

Step4: Determine charges of Al and Br

Aluminum (Al) loses 3 electrons to form $Al^{3+}$, bromine (Br) gains 1 electron to form $Br^{-}$.

Step5: Write ionic formula

To balance charges, we need 3 $Br^{-}$ for 1 $Al^{3+}$, so formula is $AlBr_{3}$.

Step6: Name the compound

It is Aluminum bromide.

Step7: Determine charges of Na and Cl

Sodium (Na) loses 1 electron to form $Na^{+}$, chlorine (Cl) gains 1 electron to form $Cl^{-}$.

Step8: Write ionic formula

The formula is $NaCl$.

Step9: Name the compound

It is Sodium chloride.

Step10: Determine charges of Mg and S

Magnesium (Mg) loses 2 electrons to form $Mg^{2+}$, sulfur (S) gains 2 electrons to form $S^{2 - }$.

Step11: Write ionic formula

The formula is $MgS$.

Step12: Name the compound

It is Magnesium sulfide.

Step13: Determine charges of Al and N

Aluminum (Al) loses 3 electrons to form $Al^{3+}$, nitrogen (N) gains 3 electrons to form $N^{3 - }$.

Step14: Write ionic formula

The formula is $AlN$.

Step15: Name the compound

It is Aluminum nitride.

Answer:

#	Pair of Elements	Charges	Ionic Formula of the Compound Formed	Name of Compound
2	Al, Br	$Al^{3+}, Br^{-}$	$AlBr_{3}$	Aluminum bromide
3	Na, Cl	$Na^{+}, Cl^{-}$	$NaCl$	Sodium chloride
4	Mg, S	$Mg^{2+}, S^{2 - }$	$MgS$	Magnesium sulfide
5	Al, N	$Al^{3+}, N^{3 - }$	$AlN$	Aluminum nitride