[tashkent]

Hello everyone!

I have a query regarding determining the empty orbitals used by the following metal ions in forming complexes with water:  Cd+2, Ag+, Ce+3, Fe+2, Mn+2 and Zn+2.  I was asked to use the Crystal Field Theory.  I'm not quite sure with my answers:

For Cd+2: 4f, since its last filled orbital is 4d (4d10).
For Ag+: 4d, since its last filled orbital is 4d9.
For Ce+3: 4f, since its last filled orbital is 4f12.
For Fe+2: 3d, since its last filled orbital is 3d6.
For Mn+2: 3d, since its last filled orbital is 3d5.
For Zn+2: 4s, since its last filled orbital is 3d10.

I have not deducted the electrons of each metal ions with their corresponding charges.  Am I on the right track?

Merry Christmas and Happy New Year to all.

Regards,
Tashkent

[Donaldson Tan]

please don't double post in future. thank you.

[tashkent]

Hello.

Are there any inorganic chemistry wizards willing to reply to my query?

Thank you.

[Mitch]

I have no clue what you are doing. I recommend you actually read about crystal field theory first.

[tashkent]

Ok, mitch.  This is my second attempt on this conundrum:

a)   For Cd+2, there are 46 electrons (Cd has an atomic number of 48).

Electronic configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d8

Since Cd has a +2 charge, its coordination number is 4.  Therefore, it can accommodate 4 electron pairs.  The empty orbitals are one 4d orbital and three 5p orbitals.


b)   For Ag+, there are 46 electrons (Ag has an atomic number of 47).

Electronic configuration: 1s2 2s2 2p6 3s2 2p6 4s2 3d10 4p6 5s2 4d8

Since Ag has a +1 charge, its coordination number is 2.  Therefore, it can accommodate 2 electron pairs.  The empty orbitals are one 4d orbital and one 5p orbital.


c)   For Ce+3, there are 55 electrons (Ce has an atomic number of 58).

Electronic configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s1

Since Ce has a +3 charge, its coordination number is 6.  Therefore, it can accommodate 6 electron pairs.  The empty orbitals are six 4f orbitals.


d)   For Fe+2, there are 24 electrons (Fe has an atomic number of 26).

Electronic configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d4

Since Fe has a +2 charge, its coordination number is 4.  Therefore, it can accommodate 4 electron pairs.  The empty orbitals are three 3d orbitals and one 4p orbital.


e)   For Mn+2, there are 23 electrons (Mn has an atomic number of 25).

Electronic configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d3

Since Mn has a +2 charge, its coordination number is 4.  Therefore, it can accommodate 4 electron pairs.  The empty orbitals are three 3d orbitals and one 4p orbital.


f)   For Zn+2, there are 28 electrons (Zn has an atomic number of 30).

Electronic configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d8

Since Zn has a +2 charge, its coordination number is 4.  Therefore, it can accommodate 4 electron pairs.  The empty orbitals are one 3d orbital and three 4p orbitals.


I would appreciate any replies from you guys.  Am I on the right track?  Thank you.

Sincerely,
Tashkent

[Demotivator]

A little better but still all wrong.

Water is a weak field ligand and will therefore produce high spin states due to small crystal field splitting of d orbitals. This has consequences on which empty d orbitals are available for bonding, eg 3d or 4d?

For Fe2+ coord number is 6.
electronic config is 3d6, high spin. That means all 3d orbitals are occupied. That leaves 4s, three 4p, and two 4d orbitals as empty for bonding. If it was low spin because of large crystal field splitting, two 3d orbitals would have been empty, instead.

Mn2+ is also six coordinate octahedral (high spin).

Ag+ : all 4d orbitals are filled (d10, not d8). That leaves empty 5s and 5p (linear sp hybridization).

Zn2+ is tetrahedral. It's d orbitals are filled (d10, not d8). That leaves only s and p orbitals (as in tetrahedral sp3 hybridization).

I think Cd2+ is also tetrahedral.

[tashkent]

Hello, Demotivator!  Thanks for the reply.

I've read from Petrucci (Gen.Chem) that Fe+2 usually assumes a coordination number of 6, while Zn+2 assumes a coordination number of 4.  Is there a reason for this?  Is there a strategy or technique in knowing the coordination number of a transition metal by just knowing the metal ion?  According to my prof, the coordination number can be determined by multiplying the charge by 2, but this doesn't work for Fe+2 and Mn+2 since they have six coordinate octahedral geometry.  Why is that?

And for Ag+, the electronic configuration is [Kr]4d9 5s1, isn't it?  I think the empty orbitals for this metal ion are two 5p orbitals.  Please correct me if I'm wrong.

And how about Ce+3?  The electronic configuration is [Xe]4f1.  The empty orbitals will be six 4f orbitals, I think.  Is it right?

I would really appreciate your comments on this conundrum.  Thanks so much.

Sincerely,
Tashkent

[Mitch]

All 2nd and 3rd row transition metals will be octahedrally compled, it's the first row transitions that can be tricky.

[Demotivator]

The rule from your prof has so many exceptions it doesn't deserve to be called a rule.  It probably works best if the charge is 3, or if 1, 2 or 3 provided the transition metal is on the far right hand side like Ni cu zn.

For Ag+ it's [Kr]4d10 5s0.  It prefers a filled d  and empty s orbital.

True, For Ce3+ [Xe]4f1.  However, the 4f orbitals of the lanthanide elements are "shielded" and do not participate in sigma bonding as a rule. For an octahedral complex, only 2 of the five d orbitals, as well as s and three p point in the correct directions. I don't know what the Ce3+ coordination geometry really is, though. You might assume six coord.