[benjster85]

The formula is SnO₂.

Is there any way to know which name to give it?? Or more helpfully, when is a peroxide is a possibility?

[Hunter2]

It is tin-IV-oxide. Look in which group of elements you can find tin in the perodic system of elements. Peroxide is more like, if the element has only one oxidation number like these elements in first and second group. BaO₂ for example  is Bareiumperoxide.

[sjb]

It is tin-IV-oxide. Look in which group of elements you can find tin in the perodic system of elements. Peroxide is more like, if the element has only one oxidation number like these elements in first and second group. BaO₂ for example  is Bareiumperoxide.

Not really, though this is a possible simplification.

Hydrogen has more than one oxidation number, but exists as a regular oxide (water) and hydrogen peroxide [and admittedly, the oxidation thing doesn't play a part here].

Peroxides will have O-O bonds.

[Hunter2]

Hydrogen is also in the first group. In compounds to oxygen it can only be +1. So water and hydrogen peroxide is possible. What I want to say is that the oxidationnumber of hydrogen in H₂O₂ can not be +2, if oxygen has -2. Its only -1 for oxygen.

[benjster85]

Your explanation was not super helpful. If I saw BaO₂ I would know that compound was a peroxide. But that is what I'm saying. Based on balance of charges, it's not obvious which ion of tin is present!

Looking at the group number is not helpful for tin, since it has two possible oxidation states. You are using your prior knowledge that it must be an oxide, but that is not obvious if you start with the goal of balancing the charges.

Anyone else understand what I'm asking?

[magician4]

from charge balance alone, you can't decide this question (if there is a multitude of options): you'll always need to take an additional look at for example X-ray structures, IR spectra , chemical behaviour  and thatlike if you're in doubt.

on the other hand: if there is a multitude of options, molecules always would tend to form the most stable variety - if they can (which is the moment where kinetics come into play, too)

now, Sn^(+II) isn't very hard to oxidize ( E⁰ Sn^(+II) /  Sn^(+IV) + 2 e^- - 0.15 V), and peroxide is a very powerfull oxidator (E⁰ H₂O₂ + 2 H⁺ + 2 e^- / 2 H₂O + 1,78 V ) , and even if there was a polymorphism (Sn^+II)(O₂^2-) / (Sn^+IV)(O^2-)₂ , my expectation would be that the respective internal redox reaction towards  (Sn^+IV)(O^2-)₂ would take place in a split second.

So: I'm not aware of any metal in low oxidation state (when there are higher oxidation states easily available), that would form a stable peroxide, as something like that tends to self-destruct in a New York minute.
But who knows, somewhere in deep space where it's very cold...
... a "New York minute" might become long enough to last for eons, and those species might be found:
There's a whole zoo of unlikely molecules*⁾ , nevertheless "stable" at 3 K, to be found out there...


regards

Ingo

*⁾
for example cyan-isocyan, di-isocyan ... just to name two of them

[Hunter2]

It is at least learning and knowledge that SnO₂ is a dioxide and not a peroxide. And again you already know yourself tin has at least more oxidation numbers +2 and +4. Why should SnO₂ a tin-II-peroxide, if peroxides it self are strong oxidisers and would form tin-IV easily.

at magican4: What is a New York minute?

[magician4]

@ Hunter2:

http://www.urbandictionary.com/define.php?term=New%20York%20Minute


... and I thought that this expression wasn't hopelessly outdated by now.
(If so, pls. tell me, as I'm no native English speaker and always am eager to learn)

regards

Ingo

[Hunter2]

Cannot tell because I am also not native speaker. Cu in studentencity.de

[mjpam]

Would the simplest way to figure this out be to apply the atomic orbital formalism?

Atomic tin has an atomic ground state configuration of [Kr] 4d¹⁰5s²5p², making its dipositive ion Sn(II) stannous ion especially stable due to a completely full 4d orbital and completely empty 5p; and atomic oxygen has an atomic ground state configuration of [He] 2s⁶, making its dinegative ion O(VIII) oxide ion especially stable due to a completely full 2p orbital. Niether the tetrapositive Sn(IV) stannic not the mononegative O(VII) peroxide ion have a ground state atomic orbital configuration in which the atom's highest-energy orbital is completely empty, making them higher energy with respect to their respective counterparts. Therefore, to the first approximation that is necessary in introductory chemistry courses at any level, tin(II) oxide is more stable than tin(IV) perxoide.