Chemistry - Ligands and Splitting Confusion

Basically, I have the idea in my head that ligands donate a lone pair to create a coordinate bond with empty orbitals (I think d-orbitals, but I'm not 100%). But for example, how can 6 ligands form if there are only 5 orbitals in the sub-shell. And when they split, How are there any free orbitals for the electrons to move to. I know my understanding is flawed, so I would really appreciate a clear explanation as I've been looking everywhere to find one! Thank you :smile:

Reply 1

Kian Stevens

16

Original post by emily380

Basically, I have the idea in my head that ligands donate a lone pair to create a coordinate bond with empty orbitals (I think d-orbitals, but I'm not 100%). But for example, how can 6 ligands form if there are only 5 orbitals in the sub-shell. And when they split, How are there any free orbitals for the electrons to move to. I know my understanding is flawed, so I would really appreciate a clear explanation as I've been looking everywhere to find one! Thank you :smile:

Your idea of ligands donating a lone pair of electrons to a central metal ion, to form coordinate bonds with it, is good because that's what a ligand is.

There may be 5 orbitals in the d subshell, but an orbital is a space which can accommodate a maximum of 2 electrons (with opposite spins). Hence, 10 electrons can occupy the d subshell in total.

And what do you mean by 'split'? What 'splits'?

Reply 2

emily380

OP

14

Original post by Kian Stevens

Your idea of ligands donating a lone pair of electrons to a central metal ion, to form coordinate bonds with it, is good because that's what a ligand is.

There may be 5 orbitals in the d subshell, but an orbital is a space which can accommodate a maximum of 2 electrons (with opposite spins). Hence, 10 electrons can occupy the d subshell in total.

And what do you mean by 'split'? What 'splits'?

For example xopper has a full 3d subshell, so how can it form complexes with ligands donating an electron pair if the d orbital is full.

I mean the d-orbitals becoming energetically different, i.e. not degenerate due to the effect of the ligand repulsion.

(Also thank you so much for your reply!!)

(edited 5 years ago)

Reply 3

BTAnonymous

22

Original post by emily380

For example xopper has a full 3d subshell, so how can it form complexes with ligands donating an electron pair if the d orbital is full.

I mean the d-orbitals becoming energetically different, i.e. not degenerate due to the effect of the ligand repulsion.

(Also thank you so much for your reply!!)

You mean that the d orbital splits and so the electrons are able to move between these split orbitals? I believe that is called d-d transition.

Reply 4

emily380

OP

14

Original post by BTAnonymous

You mean that the d orbital splits and so the electrons are able to move between these split orbitals? I believe that is called d-d transition.

Yep, that's what I mean :smile:

Reply 5

Kian Stevens

16

Original post by emily380

For example xopper has a full 3d subshell, so how can it form complexes with ligands donating an electron pair if the d orbital is full.

I mean the d-orbitals becoming energetically different, i.e. not degenerate due to the effect of the ligand repulsion.

(Also thank you so much for your reply!!)

Complex ions will form with transition metal ions, not elements.

It's like asking 'how can you form NaCl with elemental sodium and chlorine?'. You can't, you'd need to turn them into ions first.

Reply 6

BTAnonymous

22

Original post by emily380

Yep, that's what I mean :smile:

Remember a transition metal has an incomplete d sub shell so there will always be an orbital which can accept a pair of electrons, hence when this orbital splits the electrons can absorb a photon of visible light which excites these electrons to the other part of the d orbital which has slightly more energy.

Reply 7

emily380

OP

14

Original post by BTAnonymous

Remember a transition metal has an incomplete d sub shell so there will always be an orbital which can accept a pair of electrons, hence when this orbital splits the electrons can absorb a photon of visible light which excites these electrons to the other part of the d orbital which has slightly more energy.

Thank you. I'm still unsure on how it has incomplete d orbitals after coordinate bonding with up to 6 ligands?

Reply 8

username3718068

11

Original post by emily380

Basically, I have the idea in my head that ligands donate a lone pair to create a coordinate bond with empty orbitals (I think d-orbitals, but I'm not 100%). But for example, how can 6 ligands form if there are only 5 orbitals in the sub-shell. And when they split, How are there any free orbitals for the electrons to move to. I know my understanding is flawed, so I would really appreciate a clear explanation as I've been looking everywhere to find one! Thank you :smile:

The 6 ligands form coordinate bonds on the next energy level. So for example, the first row transition metal series the ligands will form coordiante bonds on the 4th energy level not on the d orbitals of the 3rd energy level. As for how 6 ligands form... the one s, 3 p's and 2 d's orbitals will hybridise to form 6 lots of sp3d2 hybrid orbitals, and since each orbitals holds a maximum of 2 electrons, thats how you get 12 electrons. The electrons in the 3d orbitals remain unchanged, which is the reason why you get coloured compounds (i.e. the electron being promoted from the dxy,dyz,dxz to dx2-y2 and dz2). The exception to this is Al3+ though which has its 6 ligands on the 3rd energy rather than the 4th since Al3+ has no electrons in the 3rd energy level so you will get 6 x 3sp3d2 hybrid orbitals on the Al3+ ion instead of 4sp3d2. this is undergraduate chem pretty much.12th standard chemistry in essence. Wish they made a levels as difficult as 12th standard chemistry. :biggrin:

(edited 5 years ago)

Reply 9

BTAnonymous

22

Original post by emily380

Thank you. I'm still unsure on how it has incomplete d orbitals after coordinate bonding with up to 6 ligands?

That's a good question and is beyond my knowledge and is also beyond a level. However, I found this image:

You can see that the 4p shell is filled yet the 3d one still remains incomplete.

Why? I'm not sure to be completely honest but google will have the answer somewhere :smile:

Reply 10

BTAnonymous

22

https://www.chemguide.co.uk/atoms/properties/3d4sproblem.html

this explains some of the orbital filling issues which are (fortunately!) beyond our syllabus but might provide a satisfactory understanding to not leave you worrying.

It was a very good question, I've never thought about that before.

Original post by BTAnonymous

That's a good question and is beyond my knowledge and is also beyond a level. However, I found this image:

You can see that the 4p shell is filled yet the 3d one still remains incomplete.

Why? I'm not sure to be completely honest but google will have the answer somewhere :smile:

I don't believe that the picture is correct.

It is generally thought that the orbitals used are those of the 4 level, i.e. 4s, 4p & 4d

If you want to use the hybridisation model then an octahedral field would use 4s¹ 4p³ 4d² hybridised orbitals, leaving the 3d orbitals untouched.

i.e. six sp3d2 orbitals

Reply 12

BTAnonymous

22

Original post by charco

I don't believe that the picture is correct.

It is generally thought that the orbitals used are those of the 4 level, i.e. 4s, 4p & 4d

If you want to use the hybridisation model then an octahedral field would use 4s¹ 4p³ 4d² hybridised orbitals, leaving the 3d orbitals untouched.

i.e. six sp3d2 orbitals

well hopefully none of that is on my exam on Tuesday lol