Question

write the formula for each simple ionic compound shown below:
name
rubidium sulfide
sodium bromide
beryllium fluoride
lithium sulfide
aluminum phosphide
silver nitride
chemical formula
rb2s

Explanation:

Step1: Determine the ions

• For Rubidium sulfide ($\text{Rb}_2\text{S}$): Rubidium ($\text{Rb}$) forms $\text{Rb}^+$ ion (since it is in Group 1, loses 1 electron), sulfur ($\text{S}$) forms $\text{S}^{2 - }$ ion (gains 2 electrons to complete octet). Using criss - cross rule (swap the magnitudes of charges), we get $\text{Rb}_2\text{S}$.
• For Sodium bromide: Sodium ($\text{Na}$, Group 1) forms $\text{Na}^+$ ion, bromine ($\text{Br}$, Group 17) forms $\text{Br}^-$ ion. Using criss - cross rule (charges 1 and 1), the formula is $\text{NaBr}$.
• For Beryllium fluoride: Beryllium ($\text{Be}$, Group 2) forms $\text{Be}^{2+}$ ion, fluorine ($\text{F}$, Group 17) forms $\text{F}^-$ ion. Using criss - cross rule (swap charges 2 and 1), the formula is $\text{BeF}_2$.
• For Lithium sulfide: Lithium ($\text{Li}$, Group 1) forms $\text{Li}^+$ ion, sulfur ($\text{S}$) forms $\text{S}^{2 - }$ ion. Using criss - cross rule (swap charges 1 and 2), the formula is $\text{Li}_2\text{S}$.
• For Aluminum phosphide: Aluminum ($\text{Al}$, Group 13) forms $\text{Al}^{3+}$ ion, phosphorus ($\text{P}$, Group 15) forms $\text{P}^{3 - }$ ion. Using criss - cross rule (charges 3 and 3, which cancel out), the formula is $\text{AlP}$.
• For Silver nitride: Silver ($\text{Ag}$, typically + 1 oxidation state in simple ionic compounds) forms $\text{Ag}^+$ ion, nitrogen ($\text{N}$, Group 15) forms $\text{N}^{3 - }$ ion. Using criss - cross rule (swap charges 1 and 3), the formula is $\text{Ag}_3\text{N}$.

Answer:

Name	Chemical Formula
Sodium bromide	$\text{NaBr}$
Beryllium fluoride	$\text{BeF}_2$
Lithium sulfide	$\text{Li}_2\text{S}$
Aluminum phosphide	$\text{AlP}$
Silver nitride	$\text{Ag}_3\text{N}$