odyssey question

[rmm30]

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what is formal charge of P in P(CH3)4. Answer is +1.
for this to be true would the species in question have to be charged i.e.
P(CH3)4 +

 

[recyrb]

I would think so, eliminating the 1 extra electron, P(5 valence)-4(bonds) making a plus 1 formal charge.


with that said, can a central atom have only one excess electron as oppose to the a pair?

 

[invictusx]

I would think so, eliminating the 1 extra electron, P(5 valence)-4(bonds) making a plus 1 formal charge.


with that said, can a central atom have only one excess electron as oppose to the a pair?

As a radical, yes.

P(CH3)4 would indeed have a formal charge of +1 on phosphorous. I think they're not putting the formal charge for a reason just so you could figure it out (or else why ask that question at all?)

 

[rmm30]

As a radical, yes.

P(CH3)4 would indeed have a formal charge of +1 on phosphorous. I think they're not putting the formal charge for a reason just so you could figure it out (or else why ask that question at all?)

you mean P(CH3)4 + would have a formal charge or +1 on P right? because as you said you could have a radical species where the formal charge on P would be 0.

why ask the question at all? just because you have a charged species does not automatically put the formal charge on the central atom. See the sulfate ion lewis. Likewise you can have a neutral species with formal charges that cancel each other out. See carbon monoxide lewis.


Could somebody please confirm that this question from destroyer is wrong / right so this can be resolved????

 

[invictusx]

you mean P(CH3)4 + would have a formal charge or +1 on P right? because as you said you could have a radical species where the formal charge on P would be 0.

why ask the question at all? just because you have a charged species does not automatically put the formal charge on the central atom. See the sulfate ion lewis. Likewise you can have a neutral species with formal charges that cancel each other out. See carbon monoxide lewis.


Could somebody please confirm that this question from destroyer is wrong / right so this can be resolved????

Oh oops, I mean P(CH3)4+

 

[NewBee12]

Formal charge is Group # - ( 1/2 # shared electrons- # unshared electrons). Thus,.,,,, 5- ( 1/2 (8) - 0 )= 1 Thus molecule has no formal charge.
i hope this helps

 

[rmm30]

correct, so this is an error in answer key?

 

[AmpedUp]

Formal charge is Group # - ( 1/2 # shared electrons- # unshared electrons). Thus,.,,,, 5- ( 1/2 (8) - 0 )= 1 Thus molecule has no formal charge.
i hope this helps

But that's for the entire molecule. We want the formal charge of Phosphorus.

P wants 5 electrons, but there's only 4 electrons bonded to it. Therefore, 5-4 = +1.

You can also think of P(CH3)4 as PH4...the same number of groups are bonded to P. They're both tetrahedral. (Side note: P exists as a tetrahedron - P4H10).

Anyway, forget the formal charge equation. This is the general equation I use:

What an element "wants" (valence electrons) minus the actual number of electrons bonded to element = formal charge.

For example, what is the formal charge of N in NH3?

N wants 5 electrons and has 5 electrons (2 electrons from the lone-pair + 3 electrons from the N-H bond). Therefore, 5-5 = 0.

What about ammonium NH4?

N wants 5 electrons and has 4 (there's no lone-pair electrons on NH4). Therefore, 5-4 = +1.

By the way, N and P are mostly similar...except that P can expand with d-orbital (like most 3rd period elements).


I hope that clears it up!

 

[rmm30]

But that's for the entire molecule. We want the formal charge of Phosphorus.

P wants 5 electrons, but there's only 4 electrons bonded to it. Therefore, 5-4 = +1.

By the way, forget the formal charge equation. This is the general equation I use:

What an element "wants" (valence electrons) minus the actual number of electrons bonded to element = formal charge.

For example what is the formal charge of N in NH3?

N wants 5 electrons and has 5 electrons (2 electrons from the lone-pair + 3 electrons from the N-H bond). Therefore, 5-5 = 0.

What about ammonium NH4?

N wants 5 electrons and has 4 (there's no lone-pair electrons on NH4). Therefore, 5-4 = +1.

I hope that clears it up!

I really do appreciate your help but no its not cleared up. Ammonium without the +1 charge? I think you understand that there is no such thing as NH4 without the +1 charge. It is a violation of the octet rule. the reason the Phosphorous compound can exist as a neutral species is bc it can have expanded octets.

I will just assume destroyer made a typo and forgot to put in the +1 charge in the question.

 

[AmpedUp]

I really do appreciate your help but no its not cleared up. Ammonium without the +1 charge? I think you understand that there is no such thing as NH4 without the +1 charge. It is a violation of the octet rule. the reason the Phosphorous compound can exist as a neutral species is bc it can have expanded octets.

I will just assume destroyer made a typo and forgot to put in the +1 charge in the question.

I just edited my post. I accidentally hit "submit reply."

It's not a typo. Destroyer is absolutely correct. Look up the Lewis Dot Structure of Tetramethylphosphonium Chloride if you have any doubts. It's a tetrahedral structure. The extra electron on P gives it a positive charge.

 

[rmm30]

I just edited my post. I accidentally hit "submit reply."

It's not a typo. Destroyer is absolutely correct. Look up the Lewis Dot Structure of Tetramethylphosphonium Chloride if you have any doubts. It's a tetrahedral structure. The extra electron on P gives it a positive charge.

Please please please go to the top of the thread and view the original question straight out of destroyer. Where do you see a Chlorine atom in the structure? The compound you are refering to is an IONIC compound formed from the Pch3 4 cation and a chlorine anion. The compound destroyer is referring to is neutral. For your own benefit you have to know why this is a mistake. P ch3 4 as a neutral species exists as a radical. In which case the charge on P is 0 and destroyer is wrong.

 

[AmpedUp]

Please please please go to the top of the thread and view the original question straight out of destroyer. Where do you see a Chlorine atom in the structure? The compound you are refering to is an IONIC compound formed from the Pch3 4 cation and a chlorine anion. The compound destroyer is referring to is neutral. For your own benefit you have to know why this is a mistake. P ch3 4 as a neutral species exists as a radical. In which case the charge on P is 0 and destroyer is wrong.

Here you go buddy!

http://www.chemicalbook.com/ProductChemicalPropertiesCB6385341_EN.htm

That's why it's a +1 charge.

 

[rmm30]

Here you go buddy!

http://www.chemicalbook.com/ProductChemicalPropertiesCB6385341_EN.htm

That's why it's a +1 charge.

There is just no getting through to you. Are you even reading these posts? The compound you are showing me is tetramethyl phospONIUM. It is a CATION. Like H30+ (i.e. HydrONIUM) or NH4+ (i.e. ammONIUM). The question originally asks for the charge on P for P(CH3)4 or Tetramethyl Phosphorous. The charge on P here is 0 and it exists as a radical. nevermind.

 

[AmpedUp]

There is just no getting through to you. Are you even reading these posts? The compound you are showing me is tetramethyl phospONIUM. It is a CATION. Like H30+ (i.e. HydrONIUM) or NH4+ (i.e. ammONIUM). The question originally asks for the charge on P for P(CH3)4 or Tetramethyl Phosphorous. The charge on P here is 0 and it exists as a radical. nevermind.

LOL same thing. The 1 electron on P (the radical) is the same thing as having a + charge. This is a CHARGED species in nature (bound to preferably chlorine or bromine). I'm reading your posts, but I'm simply trying to refute the claim that this is a typo. It's not.

 

[rmm30]

Do you understand that P(CH3)4 and P(CH3)4 + are NOT the same thing. And that the formal charges on P are 0 and +1 respectively. If this question appears on your exam (or one just like it) you will choose the wrong answer.

Can we at least agree that P(CH3)4 and P(CH3)4 + are not the same compound?

 

[AmpedUp]

Do you understand that P(CH3)4 and P(CH3)4 + are NOT the same thing. And that the formal charges on P are 0 and +1 respectively. If this question appears on your exam (or one just like it) you will choose the wrong answer.

Can we at least agree that P(CH3)4 and P(CH3)4 + are not the same compound?

Ok. you really don't have to use that harsh tone with me. It doesn't help anything. You probably think I'm dumb, but whatever...it's just a forum.

So I did some more reading about this...And yes, if it was a radical, it would be a neutral species (which we are assuming P(CH3)4 is in this question). I used the formal charge equation: #Valence Electrons - 1/2(Bonding electrons) - Nonbonding Electrons. So for P in this molecule, it's 5 - 1/2(8) - 1 = 0. So I guess you're correct. I was thinking about the overall charge of the molecule, which if you took the radical away would make it an overall charge of +1.

P(CH3)4+ SHOULD be in the question instead of P(CH3)4, because that's what I assumed when i first looked at the question...I was thinking overall charge when I should've been focusing on P itself. My bad.

 

[rmm30]

not trying to be harsh at all. Just trying to get to the bottom of it and sometimes that requires some thought and discussion/disagreement. Lewis structures are important. they are fundamental to gc and orgo and it is something that will definitely be on the dat. I had a question about the accuracy of a destroyer problem and it threw me off. Took some time to think it through and this thread solidified this topic so that when it does show up ill be ready and hope you will to. And no I dont think you are dumb at all. Just engaging this forum proves that you are educated and motivated to succeed. good luck w your pursuits. moving on...