Question

ionic match - maker
objective: create your \singles ion ad\ and make 5 matches - find that special ion
*complete the ionic bond - write the formula and name of your compound
*explain/show how bond (formula) is created using oxidation numbers (charges), & models with dot structures
element (your identity - symbol & name) rb
your ad (tell us about yourself)

• name, family, group, valence electrons, metal/nonmetal, what you do with your electrons

rubidium group 1 metal 1 valence i give electrons to

• what are you looking for in an ionic match?

non - metal
your matches
*for each match: give the formula, name, and show/explain how bond is created (use back of sheet to show models & charges)
match #1 formula rb3p name: rubidium phosphide
*show/explain how bond created
match #2 formula rb3n2 name: rubidium nitride
*show/explain how bond created
match #3 formula rbc name: rubidium carbide
*show/explain how bond created
match #4 formula
*show/explain how bond created
match #5 formula
*show/explain how bond created
find someone who is not a match for you & explain why it didnt work.
*not a match (list element symbol / name)
*explain why an ionic bond is not possible

Explanation:

Step1: Identify properties of Rubidium (Rb)

Rubidium is in Group 1, has 1 valence - electron, is a metal, and loses 1 electron to form a +1 ion ($Rb^+$).

Step2: Match #1 - with Phosphorus (P)

Phosphorus is a non - metal in Group 15, gains 3 electrons to form $P^{3 - }$. To balance charges, the formula is $Rb_3P$ (rubidium phosphide). The bond is created when 3 $Rb^+$ ions bond with 1 $P^{3 - }$ ion due to electrostatic attraction between oppositely charged ions.

Step3: Match #2 - with Nitrogen (N)

Nitrogen is a non - metal in Group 15, gains 3 electrons to form $N^{3 - }$. The formula is $Rb_3N$ (rubidium nitride). Three $Rb^+$ ions bond with one $N^{3 - }$ ion through electrostatic forces.

Step4: Match #3 - with Carbon (C)

Carbon can form various compounds. For an ionic compound with Rb, considering carbon as carbide ($C^{4 - }$), the formula is $Rb_4C$ (rubidium carbide). Four $Rb^+$ ions bond with one $C^{4 - }$ ion.

Step5: Match #4 - with Oxygen (O)

Oxygen is a non - metal in Group 16, gains 2 electrons to form $O^{2 - }$. The formula is $Rb_2O$ (rubidium oxide). Two $Rb^+$ ions bond with one $O^{2 - }$ ion.

Step6: Match #5 - with Sulfur (S)

Sulfur is a non - metal in Group 16, gains 2 electrons to form $S^{2 - }$. The formula is $Rb_2S$ (rubidium sulfide). Two $Rb^+$ ions bond with one $S^{2 - }$ ion.

Step7: Non - match example

Helium (He) is a noble gas. Noble gases have a full outer electron shell and are very stable. They do not gain, lose, or share electrons easily. So, an ionic bond between Rb and He is not possible as He does not form ions under normal conditions.

Answer:

Match #1 Formula: $Rb_3P$, Name: Rubidium phosphide, Bond creation: 3 $Rb^+$ ions bond with 1 $P^{3 - }$ ion due to electrostatic attraction.
Match #2 Formula: $Rb_3N$, Name: Rubidium nitride, Bond creation: 3 $Rb^+$ ions bond with 1 $N^{3 - }$ ion due to electrostatic attraction.
Match #3 Formula: $Rb_4C$, Name: Rubidium carbide, Bond creation: 4 $Rb^+$ ions bond with 1 $C^{4 - }$ ion due to electrostatic attraction.
Match #4 Formula: $Rb_2O$, Name: Rubidium oxide, Bond creation: 2 $Rb^+$ ions bond with 1 $O^{2 - }$ ion due to electrostatic attraction.
Match #5 Formula: $Rb_2S$, Name: Rubidium sulfide, Bond creation: 2 $Rb^+$ ions bond with 1 $S^{2 - }$ ion due to electrostatic attraction.
NOT A MATCH: He (Helium), Explanation: Helium is a noble gas with a full outer - shell and does not form ions under normal conditions, so an ionic bond with Rb is not possible.