[NTP]

Hi,


Heres the Question:
Oxides of the second Period Include:
BO3
N2O5
N2O3
CO2
Li2O


I checked the answer and it says that the incorrect of all these is B03 . I don't know why this is so since in the question it wasn't defined if the oxide of the second period is acidic or basic. Could be cos Boron is a semi-metal but I can't understand why BO3 is the incoreect one and not the others. Can someone kindly please explain this to me.
Thank you in advance!
NTP

[billnotgatez]

Balance

This is a trick question



Li +1 -7
Be +2 -6
B  +3 -5
C  +4 -4
N  -3 +5
O  -2 +6
F  -1 +7
Ne   0


N₂O₃ is perplexing

[NTP]

But why BO3 is not an oxide then??

[billnotgatez]

Which of these are most likely
BO
BO₃
BO₂
B₃O₂
B₂O₃
B₃O₃
B₂O₂
B₂O

[NTP]

BO2 maybe not sure but

[billnotgatez]

BO2 maybe not sure but

Nope

Why is it H₂O and not H₁₀O ?

Why did I put that list of elements down and numbers beside them?

There are certain concepts on how compounds are formed. At one time I learned the octet rule which may be passe today. Certainly the N₂O₃ compound is not likely to be explained by the octet rule. So the original question was double tricky.

[Donaldson Tan]

But why BO3 is not an oxide then??

Balance

This is a trick question

BO_{3 don't exist, that's why it is an invalid choice.}

[wereworm73]

There's no way you can have a neutral, uncharged molecule for "BO₃".  Each oxygen atom wants to take a pair of electrons in order to fill its electron shell, and since you're dealing with 3 oxygens, this would be a total of 6 electrons.  However, one boron atom can only give up 3 valence electrons.  So, you're going to need 2 boron atoms for every 3 oxygens, and the resulting molecule would be B₂O₃.

[billnotgatez]

Somebody want to explain N₂O₃

[wereworm73]

Alrighty then.  Nitrogen has 5 valence electrons, though it doesn't necessarily give away all of them.  With N₂O₃, the nitrogen atoms only give away 3 electrons each.
     ..     ..
O=N-O-N=O

This compound isn't very stable, though, and the nitrogens would actually prefer to completely empty their shells (in the presence of highly electronegative atoms like oxygen or fluorine) than partially empty them.  If N₂O₃ is dissolved in water long enough, a nitrogen can give away its remaining two electrons to an oxygen to form a nitrate ion (as nitric acid), along with some nitric oxide gas (though not in just one step--the N₂O₃ breaks down into NO & NO₂ first).

The nitric oxide, incidentally, also has some valence electrons left and will lend them to oxygens when given the chance.   

[NTP]

Thanks all for the help 


I have a similar question to this one which says this:

Oxides of the Third Period include:
Na2O
MgO2
Al2O3
SiO2
Cl2O7


So in this case it is MgO2 is the incorrect one because Mg = +2 and O = -2 therefore should be Mg2O2 right?
While the others are correct since:
Na = +1, O = -2  therefore Na2O
Al = +3,  O = -2  therefore Al2O3
Cl = -1, +1, -3, -5, -7, 0 = -2, therefore Cl2O7

SiO2 is a bit hard because Si has a valency of 4 since it is a element in group 4 so how come it is SiO2 and Not Si2O4?
So which one is wrong SiO2 or MgO2? I say MgO2 but not sure why

[Albert]

Yes, MgO2 is wrong. It's just a simple maths problem, up to a certain point: SiO2 is correct because, given that O = -2 and Si = +4 (in this case), considering 1 atom of silicon, you'll need 2 atoms of oxygen for balancing the charges.

On the other hand, MgO2 is incorrect, as you pointed out. However, it should be MgO: 2 positives charges v. 2 negative charges. That's it. Who asked for four ones? 

[billnotgatez]

Li +1 -7
Be +2 -6
B  +3 -5
C  +4 -4
N  -3 +5
O  -2 +6
F  -1 +7
Ne   0
Na +1 -7
Mg +2 -6
Al +3 -5
Si +4 -4
P  -3 +5
S  -2 +6
Cl -1 +7
Ar   0

[NTP]

Thanks all for your help 

[billnotgatez]

Cl = -1, +1, -3, -5, -7, 0 = -2, therefore Cl2O7

I am not sure your reasoning is correct, but Cl₂O₇ seems to be a valid compound