transition metal question
Please can someone explain the attached Q. The answer is D.
HI -
Is this to do with ligands? (F-)
Is it something to do with the fact that Ti is only stable is the following Oxidation states +2,+3,+4
K? Some to do with K+?
electron promotion?
Can someone map out the various bonds and why D does not work.
[A level question]
Thanks
(edited 9 years ago)
It is related to the oxidation state of titanium, not the type of ligands present.
When oxidised, the s electrons are lost first, followed by the d electrons.
You need to be able to calculate the oxidation state of titanium in each complex, this is critical.
Titanium will only oxidise to +4 as this gives a closed shell electron configuration. Electrons in the p molecular orbitals will not be lost (in constrast to d orbitals, they are held much closer to the nucleus and are less shielded / experience a greater effective nuclear charge).
When oxidised, the s electrons are lost first, followed by the d electrons.
You need to be able to calculate the oxidation state of titanium in each complex, this is critical.
Titanium will only oxidise to +4 as this gives a closed shell electron configuration. Electrons in the p molecular orbitals will not be lost (in constrast to d orbitals, they are held much closer to the nucleus and are less shielded / experience a greater effective nuclear charge).
Original post by ChemistryBud
It is related to the oxidation state of titanium, not the type of ligands present.
Thanks for the reply.
Im not sure if I have done this right.
I have simply used O.N to calculate what Ti would have to be:
K being +1,
F being -1
O-being -2
for
A: Ti would be +3
B: +4
C: +4
D: +6
As D is +6 this is not a stable so is unlikely to exist.
Is that what they were after? This would require memorizing the stable states of the transition metals? Or can you work out the stable states from the electronic config they give?
Original post by RMIM
Thanks for the reply.
Im not sure if I have done this right.
I have simply used O.N to calculate what Ti would have to be:
K being +1,
F being -1
O-being -2
for
A: Ti would be +3
B: +4
C: +4
D: +6
As D is +6 this is not a stable so is unlikely to exist.
Is that what they were after? This would require memorizing the stable states of the transition metals? Or can you work out the stable states from the electronic config they give?
Im not sure if I have done this right.
I have simply used O.N to calculate what Ti would have to be:
K being +1,
F being -1
O-being -2
for
A: Ti would be +3
B: +4
C: +4
D: +6
As D is +6 this is not a stable so is unlikely to exist.
Is that what they were after? This would require memorizing the stable states of the transition metals? Or can you work out the stable states from the electronic config they give?
Correct, you can determine which oxidation states will occur from the electronic configuration. Its not that a +6 O.S is unstable, it is more that after +4 the ionisation energy will be too large.
Edit: good job working out the oxidation state for titanium in each of them complexes
(edited 9 years ago)
Original post by ChemistryBud
Correct, you can determine which oxidation states will occur from the electronic configuration. Its not that a +6 O.S is unstable, it is more that after +4 the ionisation energy will be too large.
Edit: good job working out the oxidation state for titanium in each of them complexes
Edit: good job working out the oxidation state for titanium in each of them complexes
Ah I see. Thanks for the help ChemistryBud.
But what is the actual bonding?
Take the first one
K3TiF6
are we talking about ligands here with lone pairs going into empty orbitals for the F (F-)? Or am I mixing the next chapter with this one?
(edited 9 years ago)
Original post by RMIM
Ah I see. Thanks for the help ChemistryBud.
But what is the actual bonding?
Take the first one
K3TiF6
are we talking about ligands here with lone pairs going into empty orbitals for the F (F-)? Or am I mixing the next chapter with this one?
But what is the actual bonding?
Take the first one
K3TiF6
are we talking about ligands here with lone pairs going into empty orbitals for the F (F-)? Or am I mixing the next chapter with this one?
Its great that your asking this type of question. I'm not too sure what the next chapter entails but I'll take a guess and if this approach is too advanced let me know.
Ok, so the first complex will actually exist as [TiF6]K3
The potassium ions (K+) are not directly bonded to titanium, whilst the six fluoride ligands are. Without potassium present it would exist as [TiF6]3-. Check that you can work out the oxidation state for this complex (that is, when a complex is charged), the answer you get should remain the same.
In the case of potassium, we say that it is ionically bonded to the titanium hexafluoride complex (in which the fluoride ligands are covalently bonded to titanium). Now, how do we know the fluoride ligands are covalently bonded?
The most common approach used in organometallic chemistry for determining the of type bonding present is the 'covalent bond classification method'.
http://en.wikipedia.org/wiki/Covalent_Bond_Classification
I can't emphasise enough how useful it is to read the first few paragraphs- it only takes 5 minutes and for A-level I reckon you can stop as soon as it mentions Z-type ligands. If you find it easy to grasp, you may even wish to learn the equations that follow (they offer a simple way of determing the oxidation state e.t.c).
Let me know how you get on/ if you would prefer a different answer that follows the methods taught in your syllabus (im not sure what these are
) the above approach gets examined every year at university (just in case you think its a load of rubbish!)To test your understanding, see if you can tell me the type of bonding present in Ti(OH)4, Ti(NMe2)4, and finally the anticancer drug cisplatin Pt(Cl)2(NH3)2. If you can Ill be impressed!
(edited 9 years ago)
Original post by ChemistryBud
Its great that your asking this type of question. I'm not too sure what the next chapter entails but I'll take a guess and if this approach is too advanced let me know.
Ok, so the first complex will actually exist as [TiF6]K3
The potassium ions (K+) are not directly bonded to titanium, whilst the six fluoride ligands are. Without potassium present it would exist as [TiF6]3-. Check that you can work out the oxidation state for this complex (that is, when a complex is charged), the answer you get should remain the same.
In the case of potassium, we say that it is ionically bonded to the titanium hexafluoride complex (in which the fluoride ligands are covalently bonded to titanium). Now, how do we know the fluoride ligands are covalently bonded?
The most common approach used in organometallic chemistry for determining the of type bonding present is the 'covalent bond classification method'.
http://en.wikipedia.org/wiki/Covalent_Bond_Classification
I can't emphasise enough how useful it is to read the first few paragraphs- it only takes 5 minutes and for A-level I reckon you can stop as soon as it mentions Z-type ligands. If you find it easy to grasp, you may even wish to learn the equations that follow (they offer a simple way of determing the oxidation state e.t.c).
Let me know how you get on/ if you would prefer a different answer that follows the methods taught in your syllabus (im not sure what these are) the above approach gets examined every year at university (just in case you think its a load of rubbish!)
To test your understanding, see if you can tell me the type of bonding present in Ti(OH)4, Ti(NMe2)4, and finally the anticancer drug cisplatin Pt(Cl)2(NH2)2. If you can Ill be impressed!
Ok, so the first complex will actually exist as [TiF6]K3
The potassium ions (K+) are not directly bonded to titanium, whilst the six fluoride ligands are. Without potassium present it would exist as [TiF6]3-. Check that you can work out the oxidation state for this complex (that is, when a complex is charged), the answer you get should remain the same.
In the case of potassium, we say that it is ionically bonded to the titanium hexafluoride complex (in which the fluoride ligands are covalently bonded to titanium). Now, how do we know the fluoride ligands are covalently bonded?
The most common approach used in organometallic chemistry for determining the of type bonding present is the 'covalent bond classification method'.
http://en.wikipedia.org/wiki/Covalent_Bond_Classification
I can't emphasise enough how useful it is to read the first few paragraphs- it only takes 5 minutes and for A-level I reckon you can stop as soon as it mentions Z-type ligands. If you find it easy to grasp, you may even wish to learn the equations that follow (they offer a simple way of determing the oxidation state e.t.c).
Let me know how you get on/ if you would prefer a different answer that follows the methods taught in your syllabus (im not sure what these are
) the above approach gets examined every year at university (just in case you think its a load of rubbish!)To test your understanding, see if you can tell me the type of bonding present in Ti(OH)4, Ti(NMe2)4, and finally the anticancer drug cisplatin Pt(Cl)2(NH2)2. If you can Ill be impressed!
You have the formula for cisplatin incorrect.
PtCl2(NH3)2
Haha thanks charco :-) I was half asleep.
Original post by charco
You have the formula for cisplatin incorrect.
PtCl2(NH3)2
PtCl2(NH3)2
I take it it just the NH3 part that is wrong and nothing to do with the bracket too? - as the displayed formula would be the same for Cl2 or (Cl)2 ?
I think the bracket is normally left out for Cl2 but I tend to put it in for clarity to show that each Cl- ligand is bonded directly to Pt and that the chloride ligands are not bonded to one another. If you google the structure it might be a bit clearer what mean by this. it certainly isn't something thay you would ever lose a mark for in the exam.
Original post by ChemistryBud
I think the bracket is normally left out for Cl2 but I tend to put it in for clarity to show that each Cl- ligand is bonded directly to Pt and that the chloride ligands are not bonded to one another. If you google the structure it might be a bit clearer what mean by this. it certainly isn't something thay you would ever lose a mark for in the exam.
yes I think it's more clear with the brackets.
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