Pretty Simple G-chem question....

[jdpaul14]

Hey guys,

So this question is Topscore Exam #3, question 47.

Which of the following has an octet on the central atom?

a) ICl4+
b) SeF4
c) PBr+4
d) IF3
e) BrF4-

How would I go about solving this problem? I know I am supposed to look at the valence of each element but then what?

Thanks a lot,

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[Sublimation]

draw out the structures of each. Then determine the formal charge of the central atom. the one that gives u a formal charge of 0 has a full oct.

 

[jdpaul14]

Ok so I did what you said...but I am getting a formal charge of +2 for PBr4+. When I calculate the formal charge of P I get +1 and then you add the +1 of the molecule and I get +2. Am I doing this right?

PBr4+ is the answer by the way....

Thanks again!

 

[Dentite99]

Here's how I'd solve it:

Phosphorus has 5 valence e-
Br has 7 valence e-, but with a +4 it has 4 e- taken away= 3 valence e-

P's 5e- + Br 3e-= 8=octet.

 

[jdpaul14]

Hey Dentite,

The answer is (PBr4)+....its a +1 for the entire PBr molecule, would this make a difference?

 

[Dentite99]

Hmm, I'm not sure, but my thinking is that P needs 3 e- to fill it's shell and would only need 3Br. I think the + on the molecule makes the P is missing an e- and therefore need 4 e- and thus 4 Br's to give it a full octet.
I liked it better when I thought it was PBr4+! ha

 

[Sublimation]

Hey Dentite,

The answer is (PBr4)+....its a +1 for the entire PBr molecule, would this make a difference?

Valence e- of Br is 7X4= 28 + 5P = 33 -1(due to the postive charge)=32

Now draw out the molecule. Yes P has a formal charge of +1 however remember u can drag one of the lone pairs of the Br down to fill in the octet giving the Br the +1 charge. Now the P has the full octet. The best way to do this is get fast when u draw them out. If you cant pick it out right off the bat go and draw them all out. Ull see that the only one that will let you get a full octet on the P is PBr4+ (dont forget resonance Hope this clears it up

 

[Sublimation]

Hmm, I'm not sure, but my thinking is that P needs 3 e- to fill it's shell and would only need 3Br. I think the + on the molecule makes the P is missing an e- and therefore need 4 e- and thus 4 Br's to give it a full octet.
I liked it better when I thought it was PBr4+! ha

P can make 5 bonds.

 

[PooyaH]

Just draw the compound: PBr4

P has 5 valence electrons and 4 of them are sharing electrons with Br. each bonds represents 2 electrons. You need to realize since 4 of the valence electrons are forming a bond with 4 Br's, P is left out with one unshared electron! This would give P 9 electrons 8 (4 bonding) + 1 = 9!
This gives P a negative charge and since the comound is (PBr4)+, this would cancel that negative charge and would give the central atom a zero charge. I hope you can see it, it's hard to explain it this way, and it's a lot easier to draw it!

 

[Dentite99]

Nice explanation.

 

[PooyaH]

...or you can say that PBr4+ has 32 valence electrons and after you draw the compound you would see that there are 24 unshared electrons on 4 Br's (6 on each). Thus, 32 valence electrons - 24 unshared electrons = 8 shared electrons which satisfies the octet rule!

 

[Sublimation]

If you ever form a positive formal charge on an inner atom beyond the second row, you will always shift electrons to make double or triple bonds to minimize the formal charge. or just remember that an octet is when you have a set of 8 valence electrons associated with one atom.

 

[jdpaul14]

Thanks a lot guys, this really helped

 

[Shinpe]

draw out the structures of each. Then determine the formal charge of the central atom. the one that gives u a formal charge of 0 has a full oct.

Are you sure about this? I thought a full octet means a total 8 electrons around the atom (half of the shared, all of the lone pairs). I don't think the formal charge has to necessarily be 0 for it to have a full octet

Here's an example of where what you're saying doesn't work:
SO4(2-)
Sulfur has a formal charge of +2 when it has the full octet (4 single bonds with each O) but a formal charge of 0 when it has an expanded octet (2 double bonds and 2 single bonds).

Ain't that right?

 

[GiTsticker]

I agree, the only atoms with full octets AND formal charges of 0 are the noble gases. Formal charges are a moot point for this problem, which explains the previous confusion.

 

[Shinpe]

I agree, the only atoms with full octets AND formal charges of 0 are the noble gases. Formal charges are a moot point for this problem, which explains the previous confusion.

I wouldn't say that either. CH4, NH3, H20 all have zero formal charges and full octets.

 

[GiTsticker]

Heh, ya and those too...

 

[Shinpe]

Heh, ya and those too...

And SO2 and PBr3 and ........... ;p