What is the name of #Ag_2O#?
1 Answer
May 12, 2016
Here's my thought process:
- From formula:
#"Ag"# #-># silver, just from looking at the periodic table. #"O"# : second column from neon, so the oxidation state is#""^(-2)# since it would want to gain two electrons as a free element to get an octet valency.- This is a combination of a metal (
#"Ag"# ) and nonmetal (#"O"# )#-># ionic compound. - Ionic compound
#-># oxidation state of#"O"# #=# actual charge. Thus, the charge on oxygen is#""^(2-)# .#-># anion #"O"# : anion stem = "ox"- Anion ending in an ionic compound: "-ide"
- There are two equivalents of silver, due to the
#color(blue)(""_2)# subscript. - The charge on silver for the neutral
#"Ag"_2"O"# is gotten from solving#color(blue)(2)stackrel("Ag")overbrace((""^(?+))) + stackrel("O")overbrace((""^(2-))) = 0# for#"?"# . - Thus, the charge on silver is
#color(blue)(""^(1+))# .
Knowing the charge, and recalling the roman numeral for
Name:
cation name + roman numeral in parentheses + anion stem + anion ending
- Cation name: silver
- Roman numeral:
#\mathbf("I")# - Anion stem: "ox"
- Anion ending: "-ide"
Silver + (I) + ox + ide
#-># Silver(I) oxide