Question

Inorganic Chemistry Tutorial: Isomerism and Chirality of Coordination Compounds

Answer 1

\({\bf{Definitions:}}\)
linkage isomerism: occurs when a ligand links to the same atom, but using different atoms
ionization isomerism: occurs when the ligand and counterion swap places
special case: hydrate isomerism, in which one such ligand is water
coordination isomerism: occurs when different ions form from the same molecular formula


ex of a coordination isomer in which the complex cation and anion swap places

Answer 2

\({\bf{Tetrahedral~Complexes:}}\)
two main cases:
1. 4 different ligands [MABCD] enantiomers, which look just like their non-coordinated counterpart
2. 2 nonsymmetric bidendate ligands

special note: you have to take into account the ligand bonding patterns, you can't just look at the ligands themselves when you're determining chirality

Answer 3

\({\bf{Square~Planar~Complexes:}}\)
Consider MA2B2. the 2 pairs of identical ligands can be paired such that matching ligands are adjacent (cis, C2v) or across from each other (trans, D2h)



Now consider [MA2BC], with two identical ligands. This also produces 2 isomers (cis, where the 2 identical ligands are adjacent, and trans, where the 2 identical ligands are across from each other)

(this is the best diagram I could find, just consider the 4 planar atoms for now)



Third case to consider: 2 bidendate ligands [M(AB)2]. This is basically just the first case, except with bonding considerations.

Answer 4

\({\bf{Pyramidal~Complexes:}}\)
2 types to consider for ML5: trigonal bipyramidal and square pyramidal
Often, these conformations are close in energy so their isomers are not isolable.



They undergo Berry pseudorotation in which a trigonal bipyramidal complex distorts into a square pyramidal complex by "tucking in" two of its equatorial ligands to get a square plane, and "pushing out" the two axial ligands to become the new equatorial ligands (easier to see in an illustration)

Answer 5

\({\bf{Octahedral~Complexes:}}\)

First case: [MA4B2] where the B ligands can be adjacent (cis) or on opposite sides of the plane



Second case: [MA3B3], which has 2 isomers: mer and fac
> mer: all the A ligands are arranged in one plane, and the B ligands are in the perpendicular plane (named after meridian, since you can think of the ligands as occupying flat sides of perpendicular hemispheres) C2v.

> fac: all the A ligands occupy one face of the octahedron and the B ligands are on the other side

Third case: [MA2B2C2]. 5 isomers:
1. all-trans. all ligands have their matching ligand on the opposite side of the compound
2. all-cis. all ligands have their matching ligand adjacent to them.



3-5: one trans pair of ligands, and two cis pairs of ligands. three diff. arrangments

Answer 6

Special cases: [MA3B2C] has 3 arrangments: fac, mer-trans, and mer-cis
- in the fac arrangement, the 3 A ligands are fac
- in the mer-trans arrangement, the 3 Aligands are mer while the B ligands are trans, and the C ligand occupies the same mer plane as the A ligands
- in the mer-cis arrangement, the 3 A ligands are mer and the B ligands are cis (basically, take the mer trans arrangement and swap a B ligand with a C ligand)

Answer 7

\({\bf{Chirality:}}\) (reminder: for polydendate cases you must also consider bonding)

Some considerations:
1. ML3 for bidendate ligand L is chiral (ML3 doesn't have axis of symmetry due to the 90 degree staggering of the bonds)





2. introducing [MA2B2] with cis ligands B is chiral under the same argument as case 1
3. changing case 2 to a trans B ligand scenario introduces a mirror plane, and thus makes the compound achiral

Answer 8

\({\bf{Chiral~Octahedral~Complexes:}}\)
Some notation to introduce: imagine looking down the C3 axis such that the compound resembles a helix. Now consider the direction of the bond from the "upper ligand" (the ligand closer to you, or the one that sticks upward out of the page) to the "lower ligand")

if the direction is clockwise, designate with Δ, and if counterclockwise designate with Λ



**note** the CW and CCW designations here have nothing to do with how the compounds rotate polarized light. That is instead designated w/ d(+) which rotates clockwise and l(-) which rotates counterclockwise.

Answer 9

Source material is Chapter 7.6-7.10 of Inorganic Chemistry, 7th edition, Weller, et. al.

As a side note I really appreciate how I can spend several hours reading a chapter, taking notes, summarizing them, finding relevant illustrations and examples, and not even get 1 medal, while useless, low-effort spam posts get 5-10+

Answer 10

oh woowow

Answer 11

why did you close it??
are you doing this for medal? or do you really want someone to benefit from it?