I need to find the correct representation formulas from 4 different options.
So I mark in red Circle what I sure is incorrect and in blue Circle what I think is correct for the 2 others options I need help :) (I think option 3 is also incorrect.)
http://p1cture.me/upload.php


Thanks.

Pop, I don't see any pictures. You might want to try again.

i need to find the correct representation formulas from 4 different options.
so i mark in red Circle what i sure is incorrect and in blue Circle what i think is correct for the 2 others options i need help :) (i think option 3 is also incorrect.)
http://p1cture.me/images/37238349284060897651.jpg


thanks.

Now, that's better :)

You're right that 3 is wrong, this is because N has 5 outermost electrons, three are used in the single bonds and there should be 2 left.

As for the 4th, Al has 3 outermost electrons, three of them are in the single bonds, how many are left, and how many are there in the picture? Is that the same? :)

Well about the 4th I think is also incorrect 3 electrons are in the single bonds and 2 others not take part in the bonds. So to exist the Octet rule Al need to make 5 bonds but here we have only 3 bonds yes?


Thank you.

Sorry I got a problem :)

Hm... I don't quite follow your reasoning...

Al has a total of 3 outermost electrons.

In each single bond, there is one electron involved/used.

Hence, since there are 3 single bonds, electrons are used.

This means that all the 3 outermost electrons of Al are used in bonds, there are none left behind. In 4, you have 2 electrons not in any bonds, this is not true! There should not be any electrons there.

As you can see, Al does not as such have 8 outermost electrons after bonding, this is why it tends to form something else through a process called dimerisation, where you have:

AlF_3 + AlF_3 \rightarrow Al_2F_6

:)

Oh OK after I read your answer 4 times I got where I did mistake.
So if we conclude this question, I can say that number 1 (where I mark in blue) is the only true here?

Thank you :)

As you said at the start, both structures I and II are correct. II is in the same family as water and has the Lewis structure you show. Often AlF3 is shown as an Al atom with three bonded fluorine atoms; the F atoms all have an octet of electrons but Al has an "open sextet" which is why it is a very strong Lewis acid (if you have gotten to that sort of thing)

I mean II is incorrect this why I mark it in red. H has 1 outermost electrons, three are used in the single bonds and there should be no left any other electrons but in the picture is show more 2 electrons this why II is incorrect.
Only NO I is correct in this picture this what I think :)

Thank you.

I finally noticed the extra pair of electrons on H. Correct, that's wrong/