Shape of Complexes

[maihunpagal23]

Can anyone tell me how can we determine shape of complexes??
Say for instance when H2O is added do Cu 2+ ion, it forms 6 bonds with it but when NH3 is added to it, it forms a tetrahedral shape?? Why is this??


And secondly tell me how can we draw the shape of 3d orbitals??

 

[maihunpagal23]

Also refer to JUne 07, p4 of chemistry ques 4, part a
how will we determine its shape when it gets split???

 

[kahynickel]

FOR SHAPES OF COMPLEXES

These are recall questions. Follow the following rules.

1- Recognize the type of ligand such as monodentate (donating one electron pair/molecule such as H2O, NH3, Cl-,Br-, OH-, CN- etc), bidentate (donating two electron pairs/ molecule
NH2CH2CH2NH2

2- If six monodentate ligands are given then the shape would be OCTAHEDRAL.

Examples: [Cu(H2O)6]2+, [Fe(H2O)6]3+, [Ni(NH3)6]2+, [Fe(CN)6]3-
In all these examples i have included monodentate ligand of one type only. Now let's treat some different monodentate ligands bur the total number of ligands will be six so the shape will be OCTAHEDRAL.

[Cu(NH3)4(H2O)2]2+ both ligands are monodentate shape is octahedral

[Cu(Cl)4(H2O)2]2- both ligands are monodentate shape is octahedral

[Cr(H2O)3(OH)3] both ligands are monodentate shape is octahedral

Now you were asking how we know the shape of the complexes. The answer is understand the reactions of Cu2+ in syllabus.

You should be aware that when a salt of copper(II) like CuSO4 is dissolved in water it produces a pale blue solution containing [Cu(H2O)6]2+ ions as complex ions- this entity is often written as Cu2+(aq) so be cautious it means [Cu(H2O)6]2+. Now remember that whenever any transition metal salt is poured into water it forms [M(H2O)6]n+ where M can be Fe3+, Ni2+, Co2+, Cr3+, etc [Do J10/41 Q.2] It is therefore a strict rule that always six water molecules are attached to a metal ions. [ but Why six molecules of water and why not 4 or 5 molecules is another discussion- leave that- So just remember this]

Now the pale blue solution is mixed with dil ammonia it initially forms a pale blue ppt which is Cu(OH)2. On adding excess ammonia it gives a dark blue complex [Cu(NH3)4]2+ or [CU(NH3)4(H2O)2]2+. you should remember these reactions and the formulas.

3- If FOUR ligands are given then two shapes are possible SQUARE PLANAR or TETRAHEDRAL. But how to decide between square planar and tetrahedral- the answer is here.

It depends upon the LIGANDS. If a strong ligand is given such as NH3, CN- then the shape would be square planar.
For example [Cu(NH3)4]2+, [Ni(CN)4]4- both complexes contain strong field ligands so the shape is square planar.

In case of weak field ligands such as Cl-, F-, Br- etc a tetrahedral arrangement is preferred. [CuCl4]2-, [FeCl4]2-, etc. Remember that H2O is a weak field ligand but it always form six coordinate covalent bond to a metal ion so i did not mention it here)

But what about [Pt(NH3)2(Cl)2]????? the answer is square planar since NH3 are much stronger ligands than Cl- so their effect is more pronounced than Cl.

 

[kahynickel]

FOR SHAPES AND SPLITTING PATTERN OF d-ORBITALS

This is the area of transition metals that was only asked in J7- the first paper of the new format after N6. So I think this deserves a part of question in June 2011.

The question raised by maihunpagal23 is that how do we know that the shape of the d-orbital is changed after splitting is WRONG. Clear your mind that the shape of all the d-orbitals do not change. It is the energy of these orbitals which change.

First the shapes: (Memorize this detail iff u can't understand)

there are 5 d-orbitals dxy, dxz, dyz, dx2-y2, dz2. The first three are in between the axes i.e X, Y ans Z so you can't draw them on lines. The last 2 orbitals are along the axes.

There are 4 lobes of dx2-y2 orbital and 2 lobes of dz2 orbital i.e a total of 6 lobes are along the axes.In a metal with no ligands all the d-orbitals are of EQUAL ENERGY. Now imagine six ligands around a transition metal ion will always influence the orbitals which are in the direction of these ligands.Ligands have electrons in their respective orbitals which directly influence the six lobes of dx2-y2 and dz2 orbitals which causes repulsions and thus the energy of these two orbitals increases- In this way dxy, dxz, dyz are not directly influenced by the ligands since they are not in the direction of these ligands. now the 5 d-orbitals are not of equal energy they split in to two sets of orbitals. dxy, dxz, dyz are called t2g set at lower energy and dx2-y2, dz2 are called eg set at higher energy.

IMPORTANT

If the question asks " how the d-orbitals split into two sets of orbitals in case of an octahedral complex" like in J7 Q.4 then the above explanation is required.

But if the questions ask " why the 5 d-orbitals split into two sets of orbitals having different energy " then You have give them the point that Ligands have electrons in their respective orbitals which directly influence the six lobes of dx2-y2 and dz2 orbitals which causes repulsions and thus the energy of these two orbitals increases- In this way dxy, dxz, dyz are not directly influenced by the ligands since they are not in the direction of these ligands.

SPLITTING PATTERN

You may be asked to draw the splitting pattern of two complexes 1) having a weak ligand 2) a strong ligand

Can you do it!!!

 

[maihunpagal23]

awesome explanation, thanks a lot..
but wait a minute, tell me how will I answer the last question put forward by you???

 

[kahynickel]

draw the splitting pattern and then think.................. you will find the answer.

 

[maihunpagal23]

a stronger ligand will cause more repulsion and will create a larger energy gap b/w the two orbitals, but i think the shape will remain the same

 

[kahynickel]

you got it right. The energy gap between t2g and eg would be larger in case of a strong ligand and small in case of a weak ligand. Good work. Now draw the splitting pattern.

The shape will remain same as octahedral.

 

[maihunpagal23]

how??? Wont they split in the same pattern???

 

[maihunpagal23]

please reply yar

 

[kahynickel]

Listen dude! You are confusing the things. Remember that:

1- The shape of any d-orbitals do not CHANGE during the splitting. It remain same.
2- I am talkiing about the ENERGIES of d-orbitals that change.
3- Before splitting all the d-orbitals have same energies. After the splitting the dxy, dxz and dyz are at low energy while both dx2-y2 and dz2 orbitals are at high energy.

The spliiting pattern remain same means that the dx2-y2 and dz2 will be at higher energy and dxy, dxz and dyz are at low energy. BUT the energy gap would be larger in case of strong ligand, while small incase fo a weak ligand.

Now tell me clearly what is popping in ur mind??

 

[Xam]

THANKS A LOT NICKEL!
A PAPI(KISS) ON UR CHEEK FRM MY SIDE
STAY BLESSD. N THNX ALOT.

 

[maihunpagal23]

Listen dude! You are confusing the things. Remember that:

1- The shape of any d-orbitals do not CHANGE during the splitting. It remain same.
2- I am talkiing about the ENERGIES of d-orbitals that change.
3- Before splitting all the d-orbitals have same energies. After the splitting the dxy, dxz and dyz are at low energy while both dx2-y2 and dz2 orbitals are at high energy.

The spliiting pattern remain same means that the dx2-y2 and dz2 will be at higher energy and dxy, dxz and dyz are at low energy. BUT the energy gap would be larger in case of strong ligand, while small incase fo a weak ligand.

Now tell me clearly what is popping in ur mind??

kuttay k bachay
mjhe ye btaaaa agar mai is ki diagram bnaun axises pe to wahe bnengy na jo pehle bnay thay???

 

[kahynickel]

Read my post again. It has the all the answer.

By the way improve your moral character. No one has been able to provide such an explanation as I have given.
So cautious next time.

 

[WellWIshER]

Read my post again. It has the all the answer.

By the way improve your moral character. No one has been able to provide such an explanation as I have given.
So cautious next time.

did u read his USERNAME

HE IS MENTALY RETARDED :crazy: :crazy: :%) :%) :Yahoo!:

 

[maihunpagal23]

Read my post again. It has the all the answer.

By the way improve your moral character. No one has been able to provide such an explanation as I have given.
So cautious next time.

did u read his USERNAME

HE IS MENTALY RETARDED :crazy: :crazy: :%) :%) :Yahoo!:

L mera