Hybridization of SnCl3 -

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I just had my exam today, so I had this question that is, what is the type of hybridization which occurs in SnCl3 - , my answer is sp2 hybridization because I thought that the last electron left wouldn't be hybridized. Am I correct?
 

Answers and Replies

  • #2
DrDu
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Hybridization is not an observable property of an atom, so this question makes little sense.
More appropriate would be to ask with which kind of assumed hybridization would best fit the observed bond properties. Normally it makes little sense to consider hybridization beyond second row atoms as the size of s and p orbitals is becomes too different and the energy from bond formation doesn't make up for the propagation. So the best description would probably one where there is no hybridization, the free electron pair fills the s orbital and the bonding is due to the p orbitals, only.
 
  • #3
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hmmm, let me quote the whole question to avoid any miscommunication.
Tin (Sn) is an element of Group 14. What type of hybrid orbitals would tin use in SnCl3 (-1)

A. sp
B. sp2
C. sp3
D. sp3d

I'm not sure if this changes anything but I hope there's an answer to this.

PS: one more question
For the hydrogen atom, the electron transition with the longest wavelength is

A. n5 to n4
B. n7 to n 6
C. n6 to n1
D. n2 to n1

My friends say the answer should be B because the energy released when the electron moves from n=7 to n=6 is the smallest, thus the frequency is the smallest, and the wavelength is the longest.

However I did a quick check on the internet, it seems most the answer given were D.

Am I overlooking something important?
 
  • #4
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I think it should be sp3. 2 paired electrons in the s orbital and 3 unpaired electron in p orbital used for bonding
 
  • #5
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wouldn't that be 5 valence electrons when there's only 3 chlorine attached and the charge is negative 1
 
  • #6
DrDu
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Answer to the second question is D as you corretly said. The first question doesn't exactly rise my confidence in the school system of your country.
 
  • #7
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Answer to the second question is D as you corretly said. The first question doesn't exactly rise my confidence in the school system of your country.
Yes, and I would like to add that the majority of my country's people would agree with you. The country is going through brain drain, there's no helping it.

Anyhow, can you give a brief explanation as to why the answer is D? Because I remember the difference of frequency between two energy levels decreases as it goes higher up, so the frequency produced for answer B should much lower than D.
 
  • #8
Borek
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Strange, I would expect B to be correct as well (for the hydrogen question).
 
  • #9
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Hybridization is not an observable property of an atom, so this question makes little sense.
More appropriate would be to ask with which kind of assumed hybridization would best fit the observed bond properties. Normally it makes little sense to consider hybridization beyond second row atoms as the size of s and p orbitals is becomes too different and the energy from bond formation doesn't make up for the propagation. So the best description would probably one where there is no hybridization, the free electron pair fills the s orbital and the bonding is due to the p orbitals, only.
Taking a look at a model of SnCl3- (http://ce.sysu.edu.cn/echemi/inocbx/ic3/Sn/SnCl3-.html) it looks like it can be described with an hybridization. If the free electron filled the s orbital and the bonding were due to the p orbitals, wouldn't the bond angles be closer to 90 degrees?
 
  • #10
DrDu
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Yes, and I would like to add that the majority of my country's people would agree with you. The country is going through brain drain, there's no helping it.

Anyhow, can you give a brief explanation as to why the answer is D? Because I remember the difference of frequency between two energy levels decreases as it goes higher up, so the frequency produced for answer B should much lower than D.
I am deeply sorry. I wanted to write "B is correct". Confounded the letters.
 
  • #11
DrDu
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If the free electron filled the s orbital and the bonding were due to the p orbitals, wouldn't the bond angles be closer to 90 degrees?
According to crystal structures of salts containing the anion (e.g. K[SnCl_3].KCl.H2O), the bond angle found is indeed very close to 90 degree, e.g. 87.7 and 90.8 degrees for the salt mentioned.
See: http://dx.doi.org/10.1016/0022-1902(62)80247-4 [Broken]
Similar conclusions are found from IR or Raman spectroscopy in the solution:
http://www.springerlink.com/content/xw4733261885q755/
 
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  • #12
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Is there any way to find this hybridization without any actual spectroscopy or such experiments? I mean, can something like the electronic configuration be used? After all, the OP was asked this for in a theoretical manner.
 
  • #13
DrDu
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Is there any way to find this hybridization without any actual spectroscopy or such experiments? I mean, can something like the electronic configuration be used? After all, the OP was asked this for in a theoretical manner.
My general advice is never to invoke hybridization when you are not forced to do so, especially not in main group compounds beyond the second row.
Keep it simple!
I gave a theoretical justification in post #2.
Even in compound like water and NH3 little is gained in assuming hybridized orbitals.
Also bond angles aren't a good argument in general. After all, there is no physical law why orbitals should line up 100% for optimal bonding.
 
  • #14
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From post #2...

Normally it makes little sense to consider hybridization beyond second row atoms as the size of s and p orbitals is becomes too different and the energy from bond formation doesn't make up for the propagation.
I'm not sure I understand that properly. You say, the energy from bond formation doesn't make up for the propagation, which propagation is that?

My general advice is never to invoke hybridization when you are not forced to do so, especially not in main group compounds beyond the second row.
Keep it simple!
I gave a theoretical justification in post #2.
Even in compound like water and NH3 little is gained in assuming hybridized orbitals.
Also bond angles aren't a good argument in general. After all, there is no physical law why orbitals should line up 100% for optimal bonding.
Thanks for the reply :smile:

But the last statement puts me into confusion again. Bond angles do indicate the structure where the molecule is in maximum stability which every molecule is trying to attain, right? So orbitals should pretty much line up like that, I believe. Wasn't hybridization brought in to explain this, in the first place?
 
  • #15
DrDu
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I'm not sure I understand that properly. You say, the energy from bond formation doesn't make up for the propagation, which propagation is that?
I mean the energy necessary to lift e.g. an s electron to a p electron before hybridization. I confused promotion with propagation, sorry.
 
  • #16
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I mean the energy necessary to lift e.g. an s electron to a p electron before hybridization. I confused promotion with propagation, sorry.
Aha! Did some more reading into it, and I get it now. Thanks :smile:
 
  • #17
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Is there any way to find this hybridization without any actual spectroscopy or such experiments? I mean, can something like the electronic configuration be used? After all, the OP was asked this for in a theoretical manner.
Yes, there is a way within a theoretical framework. Count all of the electrons, etc..., create your Lewis structure, and then count the number of electron domains, from there you'd get your molecular geometry. Within this case, there is a lone pair of electrons on the tin atom, but it is also a tetrahedral molecule (electron-pair geometry), however, its molecular geometry is trigonal pyramidal, so the hybridization for tetrahedral generally is sp^3.

Sometimes, for instance in general chemistry, or organic chemistry, you can get the hybridization on that alone, however, there are instances where a molecules geometry won't exactly match up to its hybridization so knowing how to draw and deduce the sigma/pi - bonds, and knowings vesper theory would be more paramount than simply going on the "status-quo" so to speak. But within his course and this question, it is simply trying to test him on a more basic level of understanding.
 
  • #18
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Yes, there is a way within a theoretical framework. Count all of the electrons, etc..., create your Lewis structure, and then count the number of electron domains, from there you'd get your molecular geometry. Within this case, there is a lone pair of electrons on the tin atom, but it is also a tetrahedral molecule (electron-pair geometry), however, its molecular geometry is trigonal pyramidal, so the hybridization for tetrahedral generally is sp^3.

Sometimes, for instance in general chemistry, or organic chemistry, you can get the hybridization on that alone, however, there are instances where a molecules geometry won't exactly match up to its hybridization so knowing how to draw and deduce the sigma/pi - bonds, and knowings vesper theory would be more paramount than simply going on the "status-quo" so to speak. But within his course and this question, it is simply trying to test him on a more basic level of understanding.
Gotcha. Thanks for the explanation!
 
  • #19
DrDu
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Yes, there is a way within a theoretical framework. Count all of the electrons, etc..., create your Lewis structure, and then count the number of electron domains, from there you'd get your molecular geometry. Within this case, there is a lone pair of electrons on the tin atom, but it is also a tetrahedral molecule (electron-pair geometry), however, its molecular geometry is trigonal pyramidal,
up to here, that is plain VSEPR theory.

so the hybridization for tetrahedral generally is sp^3.
This has not the least scientific basis. Can you give a serious reference (I don't mean a high school text book) for this claim?
 
  • #21
cgk
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What do you mean it is not scientifically based? As for a reference, am I not good enough?!
There is no proof by authority in science.

But in any-case, here you go:

http://www.seas.upenn.edu/~ese111/Hybridization.pdf [Broken]

...

This is basic general chemistry 101...
From the first link you post:
sp HYBRIDIZATION

This involves making linear combination of s and p atomic orbital to form molecular orbitals that are directed along certain directions. This leads to the so called covalent bonding.
ouch... it hurts.

No, but seriously: Don't take those introductory texts too seriously. It is not uncommon that they are written by people who themselves do not have a deep understanding of theory (just for reference: That does not mean that they are not good scientists; just that you cannot assume that a typical organics professor has any concept of what a "Slater determiant", a "molecular orbital" or a "covalent bond" actually is, and should take their theoretical explanations with a grain of salt).

Regarding "tetrahedral implies sp3": For example, tetrahedral coordination is also found in coordination complexes and these are most definitely not sp3 hybridized. And even in covalent complexes, tetrahedral coordination can result for any number of reasons; but expecting a hybridization to occur in later than 2nd row atoms is normally not called for, for the reasons DrDu already explained earlier in this thread.

To get a clearer picture of those issues, you might want to read Kutzeniggs "Chemical bonding in higher main group elements" (http://dx.doi.org/10.1002/anie.198402721 ).
 
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  • #22
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Don't take those introductory texts too seriously
It isn't that I am taking it too seriously, it is just that I am telling the other poster, for his general chemistry course what to expect. I know it is just a mathematical model trying to understand how certain molecules would interact/bond, but my point here isn't being solely focused on the uselessness of going past sp, or up to d orbitals, rather, as said before, if posed the question of it (previous post stated "theoretical framework") then you could use a simple way of getting what it may look like based on the mathematical framework developed by Pauling.

As noted before: "This is basic general chemistry 101"..., not an overly advanced course. And, in the U.S. this is taught the exact way those links state, we aren't taught or some of us aren't told that going beyond the second row is useless, usually we are given a bunch of question of a bunch of chemical molecules and are told to figure out its hybridization... However, at least in organic chemistry it is rather different. The professor I took organic chemistry with only stressed that the sp2 hybridization was only valid and we didn't deal with anything beyond that. However, we did touch on sp3 of carbon-hydrogen bonds though, but that isn't what I am trying to discuss here.

Thanks for the link though. I appreciate being more knowledgeable of chemistry. But, one thing, I can only view the abstract...
 
  • #23
DrDu
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As noted before: "This is basic general chemistry 101"...
Kurokari doesn't seem to be from US. So I think it is quite irrelevant what may be taught in a specific class, what counts is a correct understanding of science.
Having done my thesis in molecular physics I share the folling complaints with many theoretical chemists:

1. Why are introductory chemistry students treated with mutilated theories of bonding they don't have the mathematical means to understand instead of being taught chemistry?

2. Why do introductory texts perpetuate even those concepts from Paulings books (which are still apedagogical masterpieces and a source of scientific insight) from the first half of the 20th century which have been proven wrong or seriously modified since then when quantitative checks of the predictions became possible with the introduction of computers?

I think the dominance of Pauling in introductory chemistry is - not too surprisingly - especially an US problem.

As far as the article by Kutzelnigg (who wrote many pedagogical articles trying to correct some of the most serious theoretical misconceptions among chemists) I fear you either have to buy it or get it from your library.
 

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