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Which of He, F, P or N can not be the central atom in a lewis structure? Its a multiple choice question with only one answer. I have seen somewhere that He, F and H cannot serve as central atoms so I don't know how to go about this.

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  • $\begingroup$ See chemistry.stackexchange.com/questions/7800/… for some insight. $\endgroup$ – Todd Minehardt Sep 14 '16 at 1:34
  • $\begingroup$ "Central" seems a poor constraint given the atoms. Diatomic molecules don't have a "central" atom. I think the question would be as Which of He, F, P or N can not be an atom in a polyatomic Lewis structure? $\endgroup$ – MaxW Sep 14 '16 at 2:26
  • $\begingroup$ @MaxW Isn't a diatomic molecule already polyatomic? $\endgroup$ – Martin - マーチン Sep 14 '16 at 4:47
  • $\begingroup$ @Martin - マーチン - Yes but I was trying to reword the question to avoid the He monoatomic molecule. I can draw a Lewis structure for a helium atom/molecule. Since it is the only atom in the molecule it is also the "central" atom. I also avoid the use of "central" since a diatomic molecule doesn't have a "central" atom. $\ce{N2}$ and $\ce{F2}$ of course have valid Lewis dot structures. $\endgroup$ – MaxW Sep 14 '16 at 4:52
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    $\begingroup$ @Max I agree, the question is very ambiguous and the concept of "central atom" is also not very well defined. There is enough to go wrong here. I am not even sure if the original author of the question understands the very restricted nature of his choice of Lewis structure. $\endgroup$ – Martin - マーチン Sep 14 '16 at 4:58
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The way I interpret the term "central atom" is that in a binary compound or ion $\ce{EX_$n$^{$z\pm$}}$ (with $\ce{E} \neq \ce{X}$), $\ce{E}$ is the central atom. This interpretation is obviously up for debate, especially with regard to diatomic molecules (where $\ce{E} = \ce{X}$), since the term is not well-defined.

What I want to point out is that fluorine and hydrogen can both act as central atoms, even if you neglect the case of $\ce{F2}$, which (according to "my definition") does not contain a central atom per se.

In liquid hydrogen fluoride1 the following autoionisation equilibrium ($K \approx 2 \times 10^{-12}$ at $273~\mathrm{K}$) is established2,3

$$\ce{3 HF <=> H2F+ + HF2-}$$

$\ce{H2F+}$ (the fluoronium ion) has fluorine as the central atom, is isoelectronic with $\ce{H2O}$, and is therefore bent.

$\ce{HF2-}$ (the bifluoride ion) has hydrogen as the central atom, with a 3-centre-4-electron bond, and is linear. (Some discussion can be found here.)

enter image description here

The only answer I would accept is therefore $\ce{He}$.


1 Please do not make liquid hydrogen fluoride at home. Unless you absolutely have to do so, please do not make it in the lab either.

2 Housecroft & Sharpe, Inorganic Chemistry, 4th ed., p 277

3 Weller et al., Inorganic Chemistry, 6th ed., p 147

n.b. for our German friends, I am sure Holleman-Wiberg describes it somewhere too but I don't have access to the book now.

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    $\begingroup$ @MaxW IUPAC defines "molecule" as an electrically neutral entity consisting of more than one atom. $\endgroup$ – orthocresol Sep 14 '16 at 5:40
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    $\begingroup$ @MaxW I don't know where it says that the ideal gas law only applies to molecules. No assumption is made of the type of particles in an ideal gas, because it doesn't affect the model. To be honest, I do not really know what the answer is intended to be, nor am I interested in guessing the intent of the question author. I merely wanted to point out the existence of counter-examples to the statement that "F and H cannot be central atoms". $\endgroup$ – orthocresol Sep 14 '16 at 5:45
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    $\begingroup$ @MaxW Such is life, when you get poorly-worded, ambiguous questions in high school... :) $\endgroup$ – orthocresol Sep 14 '16 at 5:51
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    $\begingroup$ Interestingly there is quite a lot now about helium containing ions, but in my first skimming, I could only find examples where helium would act as a terminal compound. That is if you neglect $\ce{He_{(1;2)}@C_{60;70}}$. $\endgroup$ – Martin - マーチン Sep 14 '16 at 6:06
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    $\begingroup$ Ho-Wi: Lehrb. d. Anorg. Chem. (102. Auflage), Kap. XII 2., p. 449. (german edition). "Das Ionenproduct des in reinem flüssigen Fluorwasserstoff vorliegenden Autoprotolysegleichgewichts $$\ce{ 3HF <--> H2F+ + HF2- }$$ beträgt ca. $10^{-10.7}$ bei $0~^\circ\mathrm{C}$ [...]." Roughly translated: The ion product of the auto-ionisation equilibrium in pure liquid hydrogen fluoride (...) is about $10^{-10.7}$ at $0~^\circ\mathrm{C}$ [...]." $\endgroup$ – Martin - マーチン Sep 14 '16 at 6:32
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I'm not sure what the official definition for a central atom is, but as I understand, it is a device used to figure out how to draw a Lewis structure. Whichever is the most electropositive atom that can form enough bonds is, in general, a central atom.

In that case, $\ce{F}$ would be a "central atom" in the molecule $\ce{F2}$; $\ce{P}$ would be in something like $\ce{H3PO4}$; and $\ce{N}$ would be in $\ce{NH3}$.

However, since $\ce{He}$ is a noble gas in the first row, it never participates in covalent bonding and is thus never a "central atom." A

If the question is a multiple-choice question that comes from that kind of a trick-question place, then it might be possible. If it is some kind of super-advanced chemistry, I am sorry that I am not qualified to give you a better answer.

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