[arvin_infinity]

So just wanted to check my understanding: Pls correct me if am wrong

1) If all 4 ligands are the same then it's a Tetrahedral
2) If elements from d8 groups are present then it's a Square planar
except: and PtCl4, NiCl4


+rep

[EierVonSatan]

1) This is not a factor in determining geometry
2) d⁸ metal ions tend to be square planar

[nevetstreblig]

Guys, what is a d⁸ metal ion? I'm doing chem A level but have never come across that term o.O What determines the shape? (In slightly more Laymans terms)

[a.partridge]

the number of electrons in the outer d orbital - you should be able to do that at A level.

otherwise your definition is OK but will not do in the 'real world' beyond A level : )

[EierVonSatan]

(Original post by nevetstreblig)
Guys, what is a d⁸ metal ion? I'm doing chem A level but have never come across that term o.O What determines the shape? (In slightly more Laymans terms)

I'm not sure it's a term you'll see at A-level or not, it's just what I've learned to call them. It's simply a metal ion with 8 d electrons. I don't even think you need to know how to determine shape, other than learning a few examples

[JMaydom]

Predicting the geometry is rather advanced for A-level, but if you can follow, this is why.

It can be predicted/rationalised by crystal/ligand field theory. In this, the d orbitals become non degenerate due to ligand environment. The ligands split the d orbitals into a higher and lower set (slightly more complex for square planar) and the splitting is determined by geometry.

If the d electron count is right, the electrons can all enter the lower states, making the system lower in energy compared to degenerate d orbitals. d8 is a configuration that benefits from such a distortion to D4h as 8 electrons can fill all the lower energy states in D4h.

Electronically all d8 metals should be square planar (D4h), but this is where sterics plays a role and the extent of the splitting of the d orbitals. (see the spectrochemical series and factors determining delta)
The larger the splitting, the greater the driving force for D4h. Splitting tends to be greater for the heavier metals, hence Pt shows a greater tendency for D4h than Ni.
Also, the D4h geometry places the ligands closer together, so bulky ligands will prefer tetrahedral. If the metal is larger, sterics is less of an issue. This also helps to explain Pt's tendency for D4h.


Hope that explains it!

[arvin_infinity]

(Original post by EierVonSatan)
1) This is not a factor in determining geometry
2) d⁸ metal ions tend to be square planar

I guess what I should have said is

d⁸ metal ions =square planar (with their exception)

and anything else Tetrahedral


Yh am only doing this for A-level

[EierVonSatan]

(Original post by arvin_infinity)
I guess what I should have said is

d⁸ metal ions =square planar (with their exception)

and anything else Tetrahedral


Yh am only doing this for A-level

Yeah, I was agreeing with you

(there are always exceptions in chemistry )

[arvin_infinity]

(Original post by a.partridge)

otherwise your definition is OK but will not do in the 'real world' beyond A level : )

yh this is only for A-level

(Original post by JMaydom)
x

If I had seen this a bit earlier like a month ago I would have tried to learn it - for now don't really follow

[arvin_infinity]

(Original post by EierVonSatan)
Yeah, I was agreeing with you

(there are always exceptions in chemistry )

Haha
We love those exceptions!

What if I was given this and asked to state the shape:
[Co(H2NCH2CH2NH2)₂Cl₂]

Its square planar shape isn't it!

[EierVonSatan]

(Original post by arvin_infinity)
Haha
We love those exceptions!

What if I was given this and asked to state the shape:
[Co(H2NCH2CH2NH2)₂Cl₂]

Its square planar shape isn't it!

You have 2 bidentate ligands and 2 monodentate ligands, so it would be 6-coordinate

[arvin_infinity]

(Original post by EierVonSatan)
You have 2 bidentate ligands and 2 monodentate ligands, so it would be 6-coordinate

my brain is fried!
Of courssseee !

[arvin_infinity]

(Original post by EierVonSatan)
You have 2 bidentate ligands and 2 monodentate ligands, so it would be 6-coordinate

Found another one : cis-Diammine(1,1-cyclobutanedicarboxylato)platinu m(II)
Is this even a complex ion?

there are only 2 coordinate bonds (according to the diagram I see)+ 2 ordinary covalent bonds

[EierVonSatan]

(Original post by arvin_infinity)
Found another one : cis-Diammine(1,1-cyclobutanedicarboxylato)platinu m(II)
Is this even a complex ion?

there are only 2 coordinate bonds (according to the diagram I see)+ 2 ordinary covalent bonds

It's better known as carboplatin and yes it's a complex ion. As it's d⁸ it's likely (well is in this case) to be square planar

[arvin_infinity]

(Original post by EierVonSatan)
It's better known as carboplatin and yes it's a complex ion. As it's d⁸ it's likely (well is in this case) to be square planar

I was only not sure cuz according to this image there are only 2 coordinate bonds..and my understanding is that all ligands must be bonded by coordinated bonds


am I missing something!

Also for square planar there should be 4 coordinate bonds

[thegodofgod]

(Original post by arvin_infinity)
I was only not sure cuz according to this image there are only 2 coordinate bonds..and my understanding is that all ligands must be bonded by coordinated bonds


am I missing something!

Also for square planar there should be 4 coordinate bonds

The Pt-O bonds are co-ordinate bonds too, not sure why they haven't got arrows on them though

[EierVonSatan]

(Original post by arvin_infinity)
I was only not sure cuz according to this image there are only 2 coordinate bonds..and my understanding is that all ligands must be bonded by coordinated bonds


am I missing something!

Also for square planar there should be 4 coordinate bonds

It's unusual to draw arrows to represent coordinate bonds in complexes. Here's a picture without them for carboplatin:

Spoiler:

Show

[img]http://spider.science.strath.ac.uk/p...arboplatin.gif[/img]

Same sort of problem with cis-platin:

Spoiler:

Show

[img]http://upload.wikimedia.org/wikipedi...tin-stereo.svg[/img]

Take your pick

edit: what's wrong with the img tags

[arvin_infinity]

(Original post by EierVonSatan)
It's unusual to draw arrows to represent coordinate bonds in complexes. Here's a picture without them for carboplatin:

Spoiler:

Show

[img]http://spider.science.strath.ac.uk/p...arboplatin.gif[/img]

Same sort of problem with cis-platin:

Spoiler:

Show

[img]http://upload.wikimedia.org/wikipedi...tin-stereo.svg[/img]

Take your pick

edit: what's wrong with the img tags

yh it's forbidden to look at!
Oh..interesting..actually I remember my teacher saying it's not that common to put the arrow even ..apparently it's not usual to include the "pair of e" neither


Probably my last question on this:
How would I know carboplatin can form cis/trans!
According to textbook conditions required for Cis/trans sterioisomerisms are:
(Carboplatin structure is in the textbook btw)

• 4 coordinate complex with a square planar shape- 2 different ligands 2 of one and 2 of another
• Octahedral (6 coordinate complex) - 2 different ligands - 4 of one and 2 of another one
• Octahedral (6 coordinate complex ) 3 bidentate or 2 bidentate +2 mono dentate or 1 bidentate + 4 monodentate

[EierVonSatan]

(Original post by arvin_infinity)
Probably my last question on this:
How would I know carboplatin can form cis/trans!

It can't

The cycle is one ligand

[arvin_infinity]

(Original post by EierVonSatan)
[/B]
It can't

The cycle is one ligand

Phew..that means no exception so far! yeeey
It's a bidentate isn't it !
Then why do we name it like cis-Diammine(1,1-cyclobutane..