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
6 cs, p
7 ba, n
8 al, s
9 al, i
10 rb, n
11 cs, s
12 ca, o
13 al, f
14 fr, f

Explanation:

Step1: Determine ion - charges

Use the group number of the elements in the periodic table to find their common ionic charges. Metals usually lose electrons and form cations, non - metals usually gain electrons and form anions.

Step2: Write ionic formula

Combine the ions in a ratio such that the overall charge of the compound is zero.

Step3: Name the compound

Name the metal first, then the non - metal with the ending changed to “-ide”.

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
6	Cs, P	Cs$^{+}$, P$^{3 -}$	Cs$_{3}$P	Cesium phosphide
7	Ba, N	Ba$^{2+}$, N$^{3 -}$	Ba$_{3}$N$_{2}$	Barium nitride
8	Al, S	Al$^{3+}$, S$^{2 -}$	Al$_{2}$S$_{3}$	Aluminum sulfide
9	Al, I	Al$^{3+}$, I$^{-}$	AlI$_{3}$	Aluminum iodide
10	Rb, N	Rb$^{+}$, N$^{3 -}$	Rb$_{3}$N	Rubidium nitride
11	Cs, S	Cs$^{+}$, S$^{2 -}$	Cs$_{2}$S	Cesium sulfide
12	Ca, O	Ca$^{2+}$, O$^{2 -}$	CaO	Calcium oxide
13	Al, F	Al$^{3+}$, F$^{-}$	AlF$_{3}$	Aluminum fluoride
14	Fr, F	Fr$^{+}$, F$^{-}$	FrF	Francium fluoride