[Shadow]

I can't understand how to solve a problem. I need to determine the type of hybridisation in [FeF6]^4-. Here how I started:
Fe²⁺ is the central ion. The valent electron configuration would be: 3s² 3p⁶ 3d⁶. F^- makes a weak ligand field so the d orbital of Fe should be filled (the numbers determine the electrons): 2d_xy, 1d_yz, 1d_zx, 1_dx²-y², 1d_z². So there are no empty d orbitals and the 6F^- ions have to fill the 4s, 4p_x, 4p_y, 4p_z. So far it is sp³, but there are 2F^- ions left. Where to add them? 5s? Then I get a type of hybridisation that doesn't exist.

[ZuraKotaro]

Ah yes, hybridization, I remember having a headache over this back in highschool, it's actually not as complicated as you think.

Let's go through this, step by step, suppose you have a carbon monoxide molecule (CO), what is the bond between the two atoms? A double bond. How many regions of electron density (i.e. how many different bond groups or lone pairs) are around each atom? The answer is three each (2 lone pairs + 1 double bond), therefore the hybridization around each atom is sp², because one s + two p gives three hybrid orbitals.

Now suppose you have a methane molecule (CH₄), how many regions of electron density is around the carbon atom? 4. Therefore sp³ hybridized (1s + 3p = 4 hybrid orbitals).

Now let's consider your case, FeF₆^4-, how many regions of electron density is around the Fe atom? 6 (6 bonds). Therefore sp³d² hybridized (1s + 3p + 2d = 6 hybrid orbitals).

That wasn't so hard wasn't it? And if you wanna go real detailed and talk about which s, p, and d orbitals are used, then they're just the outermost orbitals, in this case 3s, 3p, and 2d (remember that d orbitals are one energy level below it's corresponding s and p orbitals).

Hope this helps!

[Borek]

they're just the outermost orbitals, in this case 3s, 3p, and 2d

No such thing as a 2d orbital.

[Shadow]

Ah yes, hybridization, I remember having a headache over this back in highschool, it's actually not as complicated as you think.

Let's go through this, step by step, suppose you have a carbon monoxide molecule (CO), what is the bond between the two atoms? A double bond. How many regions of electron density (i.e. how many different bond groups or lone pairs) are around each atom? The answer is three each (2 lone pairs + 1 double bond), therefore the hybridization around each atom is sp², because one s + two p gives three hybrid orbitals.

Now suppose you have a methane molecule (CH₄), how many regions of electron density is around the carbon atom? 4. Therefore sp³ hybridized (1s + 3p = 4 hybrid orbitals).

Now let's consider your case, FeF₆^4-, how many regions of electron density is around the Fe atom? 6 (6 bonds). Therefore sp³d² hybridized (1s + 3p + 2d = 6 hybrid orbitals).

That wasn't so hard wasn't it? And if you wanna go real detailed and talk about which s, p, and d orbitals are used, then they're just the outermost orbitals, in this case 3s, 3p, and 2d (remember that d orbitals are one energy level below it's corresponding s and p orbitals).

Hope this helps!

Thanks for the answer. I knew that way of determining the type of hybridisation. I didn't know that it can be used on complex ions (because of the ligand field, I know that ,depending on its strength, it makes changes when filling the d orbirals with electrons). There is a different explanation in my book, and I wanted to solve this problem according to it, but the way you wrote is less complicate so I will stick to it. By the way, the answer is right. I hope that you thought of 4s, 4p, and 3d at the end.

[Shadow]

Tried to determine the type of hybridisation in [Ni(CN)₄]^- using the traditionaly way (sp³), but that is not the right answer. dsp² is the right one because CN^- makes a very strong ligand field.

[hairygorillaz]

Let's go through this, step by step, suppose you have a carbon monoxide molecule (CO), what is the bond between the two atoms? A double bond. How many regions of electron density (i.e. how many different bond groups or lone pairs) are around each atom? The answer is three each (2 lone pairs + 1 double bond), therefore the hybridization around each atom is sp², because one s + two p gives three hybrid orbitals.

Carbon monoxide is a triple bond with one lone pair on carbon. So there are two regions of electron density and its sp hybridized

[cheese (MSW)]

The currently accepted description of bonding in TM cmplxs (ligand field theory) is usually not encountered until the third year of a BSc program in chemistry.   Hybridization is not part of LFT.  Applying  LFT  to [FeF6]^4-:  Fe(II) [Ar] 3d^6; F^- is a weak field ligand hence the Fe(II) is high spin: t2g (↑↓) (↑)(↑)→Δoct→(↑)(↑)eg  with four unpaired electrons.  The d electrons in TM cmplxs play no part in dictating the stereochemistry of the cmplx; they are said to be stereochemically inactive. 
So what you (and your teacher) need to know:  in TM cmplxs ignore the d e⁻s and just go with number of ligands; [FeF6]^4- is six coordinate or has octahedral coordination.  If you must have hybridization then it is sp^3d^2 (nsnp^3 (n-1)dx^2-y^2 dz^2).  But as I said hybridization is not part of LFT: the six bonding MOs in say [FeF6]^4- comprise of a singly degenerate a1, a triply degenerate t1u, and a doubly degenerate eg set of MOs not six equivalent MOs if hybridization were the correct picture.  If I were your teacher I would stick with main group cmpds.
As for four-coordinate TM cmplxs, even experts have problems predicting whether an ML4 cmplx will be tetrahedral of square planar (unless it’s Pt(II) which is always sq plnr) so it is unfair to ask a high school student what the geometry will be.  In tetrahedral cmplxs (e.g., [MCl4]^n- M = Fe(III), Co(III), Ni(II)) the AOs used to form the bonding MOs are nsp3 mixed with ns(n-1)d^3; in sq planar (e.g., [Ni(CN)4]^2-; [PtCl4]^2-) the AOs used in bonding are the ns npxpy and (n-1)dx^2-y^2 AOs.  Recall that there are no ligands along the z axis and hence npz and (n-1)dz^2 cannot be part of the bonding MOs in sq cmplxs.