Solve Mystery of Titanium Trichloride's Electron Structure

[Dual Op Amp]

I have gone to www.webelements.com and have looked at a substance known as titanium trichloride. The electron structure is odd, a D orbital has only one electron. This doesn't make sense, the D orbital isn't filled, and the atom should be unstable. Can someone help.

 

[Astronuc]

Electron configuration of Ti is

$$1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^2\,4s^2$$

Ionization potentials 1) 6.83 eV, 2) 13.6 eV, 3) 27.6 eV, 4) 44.66 eV. One can see the relatively large difference between 3rd and 4th potentials. The difference between the 3rd and 4th valence states for Zr for instance is ~ 9.2 eV, and not surprisingly Zr forms compounds in the tetravalent state.

It is not surprising that Ti forms di- and tetra-valent compounds, however it may surprise some that it does form tri-valent compounds, and one of the most useful is TiCl3. However the compound is very reactive with air and water.

Some caution is needed - "Because of its sensitivity toward oxygen and water, anhydrous titanium trichloride should always be handled under an inert atmosphere. The submitters report that titanium trichloride in bottles that have been opened and resealed undergoes a slow deterioration that causes erratic results . . . . This decomposition is frequently detectable by the evolution of white fumes from the titanium trichloride during transfer. If a number of small-scale reactions are to be performed, the use of a Schlenk tube is advisable to extend the useful life of the titanium trichloride."

 

[Dual Op Amp]

That doesn't make sense, it would either fill the entire fourth shell, or delete it totally. I don't understand. An atom is more stable when the entire valence shell is filled, so titanium should try to either fill or empty the third and fourth shell, right?

 

[Gokul43201]

You'll find this wackiness in most compounds involving transition metals.

The energy levels with d-orbitals (within Eg and t2g) being so close to each other allows d-block elements (transition metals) to exhibit multiple valencies. So, while +4 (or[Ar]) is the preferred oxidation state of Ti, the +2 and +3 states are not uncommon. Geometric (steric, or other) factors are also responsible for determining the stablest composition.

 

[Lyuokdea]

Usually the 4s energy level has a slightly lower energy associated with it than the 3d enerfgy level. Thus electrons will fall into the 4s before the 3d, the only common exception to this is that if the atom can form a half full or completely full d shell by moving one electron from the 4s orbital to the 3d orbital, then it will usually do that. For instance, Chromium will have a
[Ar]3d^5 4s^1 configuration instead of [Ar]3d^4 4s^2

~Lyuokdea

 

[Astronuc]

An atom is more stable when the entire valence shell is filled, so titanium should try to either fill or empty the third and fourth shell, right?

Yes that is right, and that is why one will find bi-valent or tetravalent compounds, and the tetravalent compounds are much certainly more stable. For examples, TiO2, a common white pigment in paints, is very stable.

Both dihalides and trihalides are produced by reducing their respective tetrahalide. Remember, TiCl3 is very reactive, readily reacting with air and moisture, and is an effective reducing agent. TiCl2 is even more reactive.

 

[Dual Op Amp]

I'm still confused, it would have to bond with a whole bunch of atoms in order to fill or empty it's valence shell.

 

[Dual Op Amp]

Okay, now I'm really confused.

 

[Astronuc]

Here is a reference:

http://www.millenniumchem.com/Products+and+Services/Products+by+Type/Titanium+Dioxide+-+Paint+and+Coatings/r_TiO2+Fundamentals/Titanium+Ores_EN.htm

In nature, Ti tends to form oxides with other metals, and its most natural state is Ti(4+).

In the industrial chemical realm, one can 'force' Ti to 2+ and 3+ states depending on the stoichiometry, so one can make TiCl2, TiCl3 and TiCl4. Of these, TiCl4 is most stable. As previously states, TiCl3 is more reactive (i.e. less stable than TiCl4) and TiCl2 is even more reactive than TiCl3.

Also remember that the halides, of which F and Cl are the most reactive, draw electrons from metals, particularly those from the left side of the periodic table. So Ti certainly would not fill the 'outer shell'.

 

[Dual Op Amp]

But either way it doesn't add up to a complete valence shell, so it should try to find a way to add up to a complete valence shell. But, it doesn't. The entire third shell should be completely filled or emptied, but that doesn't happen. I'm just not getting this, I must of missed something.

 

[Gokul43201]

What you're missing is that while TiCl3 exists, it is not very stable - because of the partially filled outer shell.

"But it doesn't" ? No...it does. That's what it means to say that it is very reactive...it tries to reach a stabler configuration "if it can". But it exists in something like a metastable state, which, if slightly perturbed will fall into a more stable state.

 

[Dual Op Amp]

So, that titanium is not completely stable. So, why doesn't it bond with another atom to get even more stable?

 

[Dual Op Amp]

The titanium atom is trying to fill up the entire valence shell, the only way to do that is to either fill the entire third shell or completely empty the D,P, and S orbitals in the third shell, but it doesn't do that. In order to do that, it would have to bond with fourteen atoms!

 

[Astronuc]

Metals very rarely exist in the environment in their 'free' or elemental state. Most metal is founds in compounds of metal oxides, metal sulphides, carbonates, silicates, sulphates, etc.

Mankind devotes enormous quantities of energy to refine and purify metals. The we develop special alloys (e.g. stainless steels, superalloys, etc) and compounds (e.g. TiCl3) for specific purposes. These alloys and compounds do not exist in nature.

TiCl3 is manufactured in a very controlled environment, which is believe involves the reduction of TiCl4. At the moment, I cannot find a process in the public domain.

It must stored in a sealed container which is inert so that it will not react with air or water (moisture).

In the presence of O and the halides, Ti gives up electrons, so it is 'emptying' its valence electrons. Its most stable state is to give up 4 electrons 3d² and 4s². The particular valence depends on the chemical environment and the availability or lack of anionic species that will allow it to settle to its most stable form.

TiCl2 is a 'vigorous' reducing agent, and TiCl3 is an 'effective' reducing agent, i.e. TiCl3 is slightly less reactive than TiCl2.

 

[Dual Op Amp]

Yes, this has gone past titanium trichloride. As I read www.webelements.com , I found more elements that the valence shells don't add up. The titanium would empty the 3D and 4S shell, but what about the 3P and the 3S shells? In order to become stable, it would need to empty those.

 

[Dual Op Amp]

When I look online at the valence orbitals of titanium, I find that the valence orbitals are a 4S^2 and a 3D^2, but what about the valence orbitals 3P^6 and 3S^2?

 

[Astronuc]

Valence orbitals 3P⁶ and 3S² have much higher binding energies (ionization potentials). If those electrons where available, one would find TiO₆ (using 3P⁶, 3S², 4S², 3D²), but we see such a compound.

In theory, one could see TiCl12, but only 6 Cl could fit around a Ti atom and the other 6 would be excluded. However, Ti4 is the most stable form of the Ti chlorides.

Valence electrons are the outer most electrons - think of element Groups from the Periodic Table.

Bonds form to minimize energy. Think of heat of formation.

 

[Dual Op Amp]

So, you're saying that Titanium bonds with the most stable configuration there is, besides filling the entire shell, because it can't fit all of the desired atom onto itself?

 

[Dual Op Amp]

Is that what he's saying?

 

[Dual Op Amp]

Does anyone here know the answer to what I justed asked?

 

[Dual Op Amp]

Is anyone out there?
There...
There...
There...

 

[Astronuc]

So, you're saying that Titanium bonds with the most stable configuration there is, besides filling the entire shell, because it can't fit all of the desired atom onto itself?

Not quite - and perhaps my example was somewhat misleading.

Titanium, like other metals on the left side and middle of the periodic table, surrender electrons to the more electronegative elements on the right side of the periodic table. So Titanium, with its (4S², 3D²) has several options with regard to sharing (rather than loosing) electrons in a chemical bond. I may share 2, 3, or 4 valence electrons. The most stable state arises when Ti shares 4 electrons, as in TiCl₄ and TiO₂.

The other states, Ti²⁺ and Ti³⁺ do exist, but only under 'special conditions', and compounds such as TiCl₂ and TiCl₃ are consequently very reactive, and they tend to form Ti⁴⁺ if the environment allows.

The ionization energies for Ti are:

Ionisation Energies (kJ /mol)
M -> M⁺ 658
M⁺ -> M²⁺ , 1310
M²⁺ -> M³⁺ , 2652
M³⁺ -> M⁴⁺ , 4175
M⁴⁺ -> M⁵⁺ , 9573
M⁵⁺ -> M⁶⁺ , 11516

TiCl₆ is not feasible because the binding energies of the '3p'-electrons are too high.

In addition, TiCl₄ has a tetrahedral stucture - see - http://www.webelements.com/webelements/compounds/text/Ti/Cl4Ti1-7550450.html - and this structure would preclude two additional Cl atoms.

 

[Dual Op Amp]

Alright, for half a year I researched that when an entire valence shell is filled the atom is stable. For example, oxygen, with six valence electrons, doulble bonds with another oxygen, in order to fill the entire valence shell, count it up. 2S^2,2P^4 + 2 = 8. This is the octec rule. Now, I assumed, as the atom got larger, it would just need to share more atoms, in order to COMPLETELY fill or empty it's valence shell. Sodium and chloride do that, sodium:valence = 1,chloride:valence=7. They bond, both of them follow the octec rule. Okay, but as I got into larger atoms, I noticed one in particular, titanium. This atom has three shells, and fills up the fourth. Now, it bonds with four, in order to become stable, but that would leave 3P and 3S. The atom would still be unstable. 3P and 3S are still in the picture. According to what I've learned, the ENTIRE valence shell must be emptied or filled, in order to reach a stable atom. But, titanium won't do this. So, I must have missed something. Where's that something?

 

[Astronuc]

Editorial correction in a previous post

. . . one would find TiO₆ (using 3P⁶, 3S², 4S², 3D²), but we see such a compound.

should read - . . . one would find TiO₆ (using 3P⁶, 3S², 4S², 3D²), but we do not see such a compound.

I noticed other threads about valence electrons and atomic/ molecular bonds. Allow to review those, and then respond with a sufficiently comprehensive answer.

 

[Dual Op Amp]

Okay, but what about what I just posted.

 

[Astronuc]

Titanium has a completed 4s shell, like Calcium. In addition, the 3S and 3P shells are completed. However, the 3d shell is partially complete. The two 4S electrons and the two 3d electrons are available for forming bonds with other elements - or compound ions.

So the Ti⁴⁺ ion is quite stable. Before I give a complete answer, I want to confirm some information. Please be patient.

 

[Dual Op Amp]

But, aren't those subshells? I thought the atom had to fill it's SHELL, not it's subshell. The atom of carbon fills the entire second shell. Not just the subshell, right? The 4S is completed, yet it is used.
1s, that's the first shell.
2S,P that's the second shell.
3S,P,D that's the third shell.
Up until now, I've learned that the entire SHELL must be filled, not a subshell.
Oh shoot, didn't read the please be patient.

 

[Astronuc]

RE: Subshells - yes the s, p, d, f, g . . . are subshells, but they are most 'stable' when filled. BTW, I don't like to use the term 'shells' and 'subshells' but that's sought of an unavoidable convention.

Titanium is a transitional metal, and that is what I would like to address with the right information. Along the way, I would like to be to address some of the other posts and threads with the same answer.

Thanks for your patience.

 

[Dual Op Amp]

Yes, but isn't it the entire SHELL must be filled in order to stabilize the entire atom?

 

[Coolblueflame]

umm just a thought ..how bout a titanium cloride bucky ball?

 

[Dual Op Amp]

It seems we're not going anywhere with this. If only sub-shells were filled to stabilize an atom, then why does the octec rule work? If that were true, then carbon could simply make two bonds, and then it would be stable. All it would have to do is empty it's P subshells. That's not what happens, instead it tries to fill the entire second shell. So, why it titanium so different. Like I said, I've learned that an entire SHELL, not sub-shell, but SHELL has to be filled, in order for it to stabilize. So, where am I wrong? Do you have any sites to where I can learn some of this?

 

[Astronuc]

Welcome to the world of Transitional Metal Chemistry. Unfortunately, I have not found a website that goes into sufficient detail, but I did find a number of university chemistry courses that a uniquely devoted to transitional metals and their chemistry.

The notion that a full shell provides stability is supported by the fact that the nobel gases, e.g. He, Ne, Ar, Kr, Xe are stable with completely full primary shells, and full p subshells - in fact they are monatomic gases - unlike diatomic N₂ and O₂, or F₂ or Cl₂.

Carbon cannot be compared to the transitional metals, since it does not have any 'd' orbitals. It is the presence of the 'd' and 'f' suborbitals (subshells) that allows for the multi-valent ions of the transitional metals.

In forming compounds, it is not just the cation (+ ion) that affects the structure of the molecule, but also the anion (- ion), and in complex molecules or coordination compounds, it is the type of ligand (atoms or molecules surrounding a central metal ion) that affects the metal ion valence.

I will see if I can find some appropriate references, and I will get back to you on carbon.

 

[Dual Op Amp]

No wait, let me rephrase my question, how is argon stable, when it still has the d orbital to fill. If I know this, I would be able to understand Ti.
argon is Titanium without the 4s and 3d.
According to what I've learned, an entire shell must be filled. Argon avoids this rule too.

 

[Dual Op Amp]

AHAA!
Argon can be stable, because the 4S orbital interupts the 3D orbital. This, for some weird reason, allows argon to be stable. Which allows titanium tetrachloride to be stable.
But, what I don't completely understand it how does a lower 4S make it stable?
Hmm...

 

[chem_tr]

Hello

Single electrons in a whole orbital tend to be thrown out to give a more stable electronic configuration; that is why potassium with electron configuration [Ar]4s¹ throws one electron to reach the most stable configuration, [Ar].

Titanium(III) chloride is a classical d¹ ion, thus very eager to give away its single (and somewhat troublesome) electron, to reach titanium(IV) ion, again [Ar].

So why is argon so stable, thus not reactive at all? Its basic electronic shells are COMPLETELY filled, with no half-filled orbitals. If you manage to bombard with an electron, you'll end up argon again, after a system interconversion, $Ar^-\longrightarrow Ar+e^-$.

Let me show the other alternative, namely, pulling out an electron; in this case, a cation like Ar⁺ will be formed, but this process is extremely energy consuming, so the negligible amount of ionization energy is very eagerly satisfied by the system, to give finally the most stable form, Ar: $Ar^++e^-\longrightarrow Ar$

You are okay to think that 3d orbital can be filled, and if you manage to do this (as this will be VERY hard, due to 3d orbitals of extremely higher level than that of 4s and 3p), with enough number of electrons, the super anion (probably be 7-) might be somewhat stable, but another problem occurs here; argon is argon, with its small radius will be very unstable with this number of extraordinary electrons; an unusual value of $$\displaystyle \frac {charge}{radius}$$ will not possibly cause the anion to be stable. You must keep this anion at extremely low temperatures to avoid its decomposition (here, ionization energies will be very low to give away these extra electrons, finally meeting at the usual place, [Ar])

Hope this helps.