Among the following oxides in order amphoteric, Basic, Acidic, Neutral and mixed oxides respectively:

$\ce{B2O3, SnO2, CO, Pb3O4, CaO}$

I know metals are basic and non metals are acidic . But how can we compare??

  • $\begingroup$ Compare to what? $\endgroup$ – Nilay Ghosh Mar 5 '17 at 15:44
  • $\begingroup$ @NilayGhosh comparing among themselves $\endgroup$ – hey Mar 5 '17 at 16:32

Well starting off with the easiest:

  • $\ce{CaO}$ is basic, because simple alkali and alkali earth oxides form Hydroxides in water, so they deprotonate the $\ce{H2O}$ to form $\ce{OH-}$ in solution which is a base then. If you follow the rows in the PSE the oxides will change from basic to acidic like most none-metal oxides will be acidic or better they will form acids like $\ce{SO3}$ would form $\ce{H2SO4}$ with water which then behaves like an acid. For the earlier elements in the rows so alkali and alkali earth due to the big difference in electronegativity it's an ionic compound so we are talking about Calcium-cations and Oxide-anions here. But the strongest base in water is $\ce{OH-}$ also called a solvobase (which is the strongest base there is in a solvent). Oxide is a deprotonated $\ce{OH-}$ so it would be stronger than $\ce{OH-}$ thus it will deprotonate the water every time. So all of these ionic oxides tend to be basic.

  • $\ce{CO}$ is a neutral one. $\ce{CO}$ could be seen as the anhydride of formic acid, $\ce{HCOOH}$ , take one $\ce{H2O}$ away and you will end with $\ce{CO}$ still $\ce{CO}$ and $\ce{H2O}$ will not form formic acid without a catalysts and it has no reaction at all with water thus being neutral.

  • $\ce{Pb3O4}$ is a mixed oxide you can easily see this if you compare the oxidation states. If this was a regular oxide $\ce{Pb}$ would be +IV and Oxide would be -III but Oxide only exists as -II so this has to be a different case. We could double it to yield $\ce{Pb6O8}$ then we could have 2 $\ce{Pb(+IV)}$. This isn't the case however and its a mixture of a plumbate which exists as one unit $\ce{[PbO4]{4-}}$ with a $\ce{Pb(+IV)}$ in the center and the counter ions are 2 $\ce{Pb(+II)}$, so a $\ce{Pb2[PbO4]}$ which is $\ce{Pb3O4}$.

  • $\ce{B2O3}$: Boron oxide is pretty much the same as with boric acid, although we can write it as a $\ce{B(OH)3}$ it is an acid as we are approaching the non metals now.

  • $\ce{SnO2}$: had to look that one up but it is amphoteric forming either stannic acids or stannates.

I hope I explained most of that stuff correctly though.


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