[Ari Ben Canaan]

Page 7 Part (iv)

The mark scheme stays F is Fe²⁺/FeSO₄

My answer would have been [Fe(H₂O)₆]²⁺

Do you think that would have earned the mark ?

[Ari Ben Canaan]

Also, I have a question regarding d-orbital splitting.

Lets say an octahedral complex is going to be formed. The d_z² and the d_x²-y² would be repelled to the higher energy level, correct ?

Now lets say we have a square planar couple being formed...

Would only the d_x²-y² be at the higher energy level ?

Also, what would be the splitting in a tetrahedral complex ?

[illusionz]

(Original post by Ari Ben Canaan)
Also, I have a question regarding d-orbital splitting.

Lets say an octahedral complex is going to be formed. The d_z² and the d_x²-y² would be repelled to the higher energy level, correct ? yes

Now lets say we have a square planar couple being formed...

Would only the d_x²-y² be at the higher energy level ?

Also, what would be the splitting in a tetrahedral complex ?

How the splitting works in a square planar complex depends on where you take your axes to lie! Generally we assume that the ligands occupy the xy plane and so the z axis has no ligands along it.

The dx2-y2 is indeed the highest energy orbital, and the splitting pattern:



The reason the dz2 is higher than the dxz/dyz is that the dz2 has a region of electron density in the xy plane, whereas the other two orbitals have nodal planes which are the xy plane, hence meaning that the dz2 experiences more repulsion from ligands than the xz/yz. The orbitals in the xy plane are repulsed even more, with the xy having lobes between orbitals and so has less repulsion than the x2-y2 which sticks along the same axes as the ligands.

http://en.wikipedia.org/wiki/Crystal...tting_diagrams

[Ari Ben Canaan]

(Original post by illusionz)
x

Interesting....

I have another question... Let's consider [CuCl4]2- ... It is tetrahedral.

Since the dz2 orbital lies along the same direction as one of the Cl ions is attached to... It would be at a higher energy level than the dxy dzy and dxz orbitals in an octahedral complex, correct ?

I can also see that the dyz and dxz experience some repulsion even though they don't lie exactly along the same path as the Cl ligands in that area.

However, I cannot see how the dxy orbital experience so much repulsion that it is higher than dz2.

[charco]

(Original post by Ari Ben Canaan)
Interesting....

I have another question... Let's consider [CuCl4]2- ... It is tetrahedral.

Since the dz2 orbital lies along the same direction as one of the Cl ions is attached to... It would be at a higher energy level than the dxy dzy and dxz orbitals in an octahedral complex, correct ?

I can also see that the dyz and dxz experience some repulsion even though they don't lie exactly along the same path as the Cl ligands in that area.

However, I cannot see how the dxy orbital experience so much repulsion that it is higher than dz2.

Yes, crystal field splitting is effectively reversed in tetrahedral complexes such as tetrachlorocuprate(II), with the eg orbitals being of lower energy than the dxy, dxz, dyz orbitals.

[illusionz]

(Original post by charco)
x

Misread something ignore this quote!

(Original post by Ari Ben Canaan)
Interesting....

I have another question... Let's consider [CuCl4]2- ... It is tetrahedral.

Since the dz2 orbital lies along the same direction as one of the Cl ions is attached to... It would be at a higher energy level than the dxy dzy and dxz orbitals in an octahedral complex, correct ? Yes. Delta t (ie delta for a tetrahedral complex) is smaller than Delta o (octahedral complex) - it's roughly 4/9 the size. So the lower energy level of the tetrahedral splitting is higher than the lower energy level for the octahedral splitting.

I can also see that the dyz and dxz experience some repulsion even though they don't lie exactly along the same path as the Cl ligands in that area.

However, I cannot see how the dxy orbital experience so much repulsion that it is higher than dz2.

Your mistake here is thinking that the ligands have to lie along one of the principle axes. Axes are all arbritrary, they are simply three orthogonal lines which we can take to be anywhere.

The simplest way to visualise a tetrahedron like this is to draw a cube, with its centre on the origin. The points of a tetrahedron make up 4 points in the cube as shown here:



You can see that none of the M-L bonds lie along the axes.

However, the xy,xz and yz orbitals are closer to the ligands than the x2-y2 and z2. The xz/xy/yz orbitals point towards the centre of the edges of the cube but the x2-y2 and z2 point towards the centre of the faces of the cube. Hopefully you can see that the former is closer to the ligands than the latter.

[Ari Ben Canaan]

(Original post by charco)
Yes, crystal field splitting is effectively reversed in tetrahedral complexes such as tetrachlorocuprate(II), with the eg orbitals being of lower energy than the dxy, dxz, dyz orbitals.

Could you have a look at the very first post... Its a really quick question.

[charco]

(Original post by Ari Ben Canaan)
Could you have a look at the very first post... Its a really quick question.

I can't open it ..

[Ari Ben Canaan]

(Original post by charco)
I can't open it ..

Page 7 Part (iv)

See attached.

[charco]

(Original post by Ari Ben Canaan)
Page 7 Part (iv)

See attached.

If ever you see the words "pink solid" always go for copper.

The pale green solution is iron(II) sulphate

[Ari Ben Canaan]

(Original post by charco)
If ever you see the words "pink solid" always go for copper.

The pale green solution is iron(II) sulphate

Yup, I knew about the pink solid but my question is would they have accepted [Fe(H2O)6]2+ ions in lieu of Fe2+ or FeSO4.

The mark scheme just says Fe2+/FeSO4

[charco]

(Original post by Ari Ben Canaan)
Yup, I knew about the pink solid but my question is would they have accepted [Fe(H2O)6]2+ ions in lieu of Fe2+ or FeSO4.

The mark scheme just says Fe2+/FeSO4

yes, they would