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 #""^(1+)# is #\mathbf("I")#, we now have the name.

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