Chemistry 401

(a) The two structure types for six-cordinate complexes are octahedral and trigonal prismatic, shown below:

(b) The trigonal prism is the rare structure

The hint concerning the color of [Co(en)₂Cl₂]⁺ is useful because the enantiomerically resolvable form of this ion must be the cis isomer (the trans isomer is not optically active). This suggests that the violet salt, CoCl₃·4NH₃, must have a cis structure, as well, shown below.

The air oxidation of CoCO₃ gives the pink cis-chelate complex:

Addition of HCl causes an acid/base reaction with the carbonate ligand, releasing carbon dioxide and maintaining the cis geometry (the violet complex alluded to by the hint):

The final green product is the correct color for a trans complex, so the concentrated HCl must be inducing a cis-trans isomerization:

As the number of en ligands increases the step-wise formation constant decreases for each ion, as expected based on entropy considerations (and perhaps somewhat on steric grounds). As the atomic number of the metal increases the step-wise formation increases for K₁ and K₂, which is primarily a Z* effect and in complete agreement with the Irving-Williams series. K₃ for Cu²⁺ is out of line because Cu²⁺ does not easily form six-coordinate complexes because of the extreme Jahn-Teller distortion found in these complexes.

Consider the two possibilities for the yellow complex. If the complex is mer, then the possible products are either cis or trans:

If the yellow complex is fac, then only a cis product is possible:

Since a trans product was found, the original yellow complex must have been mer since this is the only one that allows a trans product.

The assumption in this line of reasoning is that the substitutions on the cobalt by water and chloride are at the same positions as the nitrite, i.e., there is no scrambling of the ligands during the reaction processes.

First, draw the structures of each complex:

W(CO)₆ is an 18 electron complex, so will be rather unreactive to either associative or dissociative routes. Since the complex is six-coordinate, the likely reaction mechanism will be via a dissociative route:

IrCl(PPh₃)₂(CO) is a four-coordinate, 16 electron complex. Addition of a ¹³CO forms an 18 electron intermediate, which should be relatively favorable:

Thus, the iridium complex is predicted to exchange CO much more rapidly than the tungsten complex.

Two different perspectives of each structure is shown. In the octahedral structure there are 2 different sets methyl groups (bold and not bold) that are inequivalent, i.e., there is no symmetry operation that relates the two sets of methyl groups (within each set, a C₂ axis exchanges the each pair, so each pair is equivalent). In the trigonal prism structure all four methyl groups are equivalent: front and back are related by mirrors, right and left are related by mirrors, and across the front-back diagonal there is a C₂ axis.

Use the electron configuration and/or spectrochemical series to answer the questions asked.

(a) [Cr(OH₂)₆]²⁺ or [Mn(OH₂)₆]²⁺

Both complexes have the same ligands, water, which is a weak ligand, so the LFSE is determined by the electron configuration. Cr²⁺ is d⁴, or t_2g³e_g¹ with LFSE = –6Dq while Mn²⁺ is d⁵, or t_2g³e_g² with LFSE = 0. [Cr(OH₂)₆]²⁺ has the larger LFSE (more stabilization).

(b) [Mn(OH₂)₆]²⁺ or [Fe(OH₂)₆]³⁺

Both complexes have the same ligands, water, which is a weak ligand, and both are d⁵ or t_2g³e_g² so the LFSE = 0 for both complexes.

(c) [Fe(OH₂)₆]³⁺ or [Fe(CN)₆]^3–

Both complexes have the same metal in the same oxidation state, Fe³⁺, which is d⁵. Water is a weak field ligand (high spin) so the electron configuration is t_2g³e_g² with LFSE = 0. Cyanide is a strong field ligand (low spin) so the electron configuration is t_2g⁵ with LFSE = –20Dq + 2P. [Fe(CN)₆]^3– has the larger LFSE (more stabilization).

(d) [Fe(CN)₆]^3– or [Ru(CN)₆]^3–

Both complexes have the same ligands, CN^–, which is a strong field (low spin) ligand and the electron configurations for both metals are d⁵ so the LFSE = –20Dq + 2P. Ru³⁺ is higher on the Irving-Williams series (larger Z*) for metals than Fe³⁺ so the ruthenium complex will have the larger LFSE.

(e) tetrahedral [FeCl₄]^2– or tetrahedral [CoCl₄]^2–

Both complexes have the same ligands, chloride, which is a weak ligand (although this usually does not matter in a tetrahedral environment), so the LFSE is determined by the electron configuration. Fe²⁺ is d⁶, or e³t₂³ with LFSE = –6Dq_t while Co²⁺ is d⁷, or e⁴t₂³ with LFSE = –12Dq_t. Tetrahedral [CoCl₄]^2– has the larger LFSE (more stabilization).

(a) d-block portion of the Periodic Table with the ions that form square-planar complexes highlighted in yellow:

Sc	Ti	V	Cr	Mn	Fe	Co	Ni Ni2+	Cu Cu2+	Zn
Y	Zr	Nb	Mo	Tc	Ru	Rh Rh+	Pd Pd2+	Ag Ag3+	Cd
La	Hf	Ta	W	Re	Os	Ir Ir+	Pt Pt2+	Au Au3+	Hg

(b) Examples of square-planar complexes:

PtCl₄^2–, Pt(NH₃)₄²⁺, Ni(CN)₄^2–, AuCl₄^–, Cu(NH₃)₄²⁺, PdCl₄^2–

There are optical isomers when the nitrito groups are cis, geometric isomers when the nitrito groups are trans, and the nitrito groups are ambidentate, able to bond either through the N atom or an O atom.

a) [Cr(NH₃)₆]³⁺

Hexaamminechromium(III) ion

The oxidation state of Cr is 3+ so the electron configuration is d³ (t_2g³) so LFSE = –12Dq

b) [Cu(NH₃)₄(OH₂)₂]²⁺

trans–Tetraamminediaquacopper(II) ion or cis–tetraamminediaquacopper(II) ion

The oxidation state of Cu is 2+ so the electron configuration is d⁹ (t_2g⁶e_g³) so LFSE = –6Dq

c) [Ti(OH₂)₆]³⁺

Hexaaquatitanium(III) ion

The oxidation state of Ti is 3+ so the electron configuration is d¹ (t_2g¹) so LFSE = –4Dq

d) [Co(CN)₆]^3– (low spin)

Hexacyanocobaltate(III) ion

The oxidation state of Co is 3+ so the electron configuration is d⁶ (t_2g⁶) so LFSE = –24Dq + 2P

e) [Ni(NH₃)₄Cl₂]

cis–Tetraamminedichloronickel(II) or trans–tetraamminedichloronickel(II)

The oxidation state of Ni is 2+ so the electron configuration is d⁸ (t_2g⁶e_g²) so LFSE = –12Dq

a) [Cr(NH₃)₆]³⁺: Cr³⁺, d³, Not Jahn-Teller active.

b) [Cu(NH₃)₄(OH₂)₂]²⁺: Cu²⁺, d⁹, Jahn-Teller active, probably will axially elongate.

c) [Ti(OH₂)₆]³⁺: Ti³⁺, d¹, Jahn-Teller active, will axially compress.

d) [Co(CN)₆]^3–: Co³⁺, d⁶, low spin (cyanide is a strong ligand), Not Jahn-Teller active

e) [Ni(NH₃)₄Cl₂]: Ni²⁺, d⁸, Not Jahn-Teller active

a) [Mo(CO)₆]

Metal/Ligande^– countTotal

 

Mo⁰d⁶6

 

CO2×612

 

Total e^–18, stable

 

b) [(η⁶–C₆H₆)₂Cr]

Metal/Ligande^– countTotal

 

Cr⁰d⁶6

 

η⁶–C₆H₆2×612

 

Total e^–18, stable

 

c) [(η³–C₃H₅)Co(CO)₂]

Metal/Ligande^– countTotal

 

Co⁺d⁸8

 

CO2×24

 

η³–C₃H₅4×14

 

Total e^–16, likely stable

 

d) [Fe(diphos)(CO)₃] (diphos = Ph₂PCH₂CH₂PPh₂, Ph = C₆H₅)

Metal/Ligande^– countTotal

 

Fe⁰d⁸8

 

CO2×36

 

diphos4×14

 

Total e^–18, stable

 

All of these complexes are six-coordinate so we can treat them as octahedral. The LFSE is determined by the electron configuration and the ligand field strength of the ligands. The number of unpaired spins, n, found from the electron configuration is used to find the spin-only moment μ = [n(n+2)]^½ μ_B

a) cis–diaqua–cis–dichloro–cis–difluorocobaltate(II) ion

Co²⁺ is d⁷ and these are all weak field ligands. therefore:

The electron occupation is t_2g⁵e_g² so the LFSE = 8Dq

The number of unpaired electrons is 3, so μ = [3(3+2)]^½ = 3.87 μ_B

b) trans–dicyanobis(ethylenediamine)iron(III) ion

Fe³⁺ is d⁵ and these are all strong field ligands. Therefore:

The electron occupation is t_2g⁵e_g⁰ so the LFSE = 20Dq – 2P

The number of unpaired electrons is 1, so μ = [1(1+2)]^½ = 1.73 μ_B

c) trans–dichlorotetrakis(triphenylphosphine)nickel(II)

Ni²⁺ is d⁸ so the field strength of the ligands does not matter. Therefore:

The electron occupation is t_2g⁶e_g² so the LFSE = 12Dq

The number of unpaired electrons is 2, so μ = [2(2+2)]^½ = 2.83 μ_B

d) tris(bipyridine)ruthenium(II) ion

Ru²⁺ is d⁶ and these are strong field ligands. Therefore:

The electron occupation is t_2g⁶e_g⁰ so the LFSE = 24Dq – 2P

The number of unpaired electrons is 0, so μ = [0(0+2)]^½ = 0 μ_B

e) cis–dicyanobis(oxalato)manganate(II) ion

Mn²⁺ is d⁵ and cyanide is a strong field ligand while oxalate is near the border. This suggests that the net will give a low-spin complex. Therefore:

The electron occupation is t_2g⁵e_g⁰ so the LFSE = 20Dq – 2P

The number of unpaired electrons is 1, so μ = [1(1+2)]^½ = 1.73 μ_B

Use the EAN rule to determine the stability of each of these complexes.

a) hexacarbonylvanadium(0)

V(0) is d⁵ in this complex.

Each CO contributes 2 electrons: total = 6(2) = 12

Total electron count = 5 + 12 = 17, predicted to be unstable. (In fact, V(CO)₆ is stable – there is a disproportionation into [V(CO)₆⁺ ][V(CO)₆^– ], giving 16 and 18 electron complexes.)

b) bis(η⁵–cyclopentadienyl)bis(η¹–cyclopentadienyl)titanium(IV)

Ti(IV) is d⁰ in this complex.

Each η⁵–cp contributes 6 electrons: total = 2(6) = 12

Each η¹–cp contributes 2 electrons: total = 2(2) = 4

Total electron count = 12 + 4 = 16, predicted to be stable.

c) bis(η⁴–cyclooctadiene)nickel(0)

Ni(0) is d¹⁰ in this complex.

Each η⁴–COD contributes 4 electrons: total = 2(4) = 8

Total electron count = 10 + 8 = 18, predicted to be stable.

d) hexacyanoferrate(III)

Fe(III) is d⁵ in this complex.

Each CN^– contributes 2 electrons: total = 6(2) = 12

Total electron count = 5 + 12 = 17, predicted to be unstable. (It is stable, however; the EAN rule fails in this case.)

e) bis(η⁶–benzene)chromium(0)

Cr(0) is d⁶ in this complex.

Each benzene contributes 6 electrons: total = 2(6) = 12

Total electron count = 6 + 12 = 18, predicted to be stable.

Although the chemical formula might suggest that waters in transition metal compounds are hydrates, often some of the waters act as ligands that chemically bond, in some cases fairly strongly, to the metal ion. In the case of CoCl₂·6H₂O, heating transforms the compound into the tetrahedral complex [CoCl₂(H₂O)₂]. The bond strength between the Co²⁺ and the O in H₂O is strong enough that 150 °C is not enough energy to break the bond.

a) tris(oxalato)chromate(III) ion

[Cr(ox)₃]^3– has Cr in the +3 oxidation state, which is d³, and approximately an octahedral crystal field so LFSE = –12Dq.

b) hexacyanoferrate(II) ion

[Fe(CN)₆]^4– has Fe in the +2 oxidation state, which is d⁶, and a strong octahedral crystal field so LFSE = –24Dq + 2P.

c) tetrachloro–η²–etheneplatinate(II) ion

[Pt(CH₂=CH₂)Cl₄]^2– has Pt in the +2 oxidation state, which is d⁸, and approximately an octahedral crystal field so LFSE = –12Dq.

a) bis(η⁵–cyclopentadienyl)cobalt(II)

Using the EAN rule, Co(cp)₂:

Co²⁺ is d⁷: 7 e^–

cp^– contributes 6 e^– × 2: 12 e^–

Total: 19 e^–, not stable.

b) tetracarbonylnickel(0)

Using the EAN rule, Ni(CO)₄:

Ni⁰ is d¹⁰: 10 e^–

CO contributes 2 e^– × 4: 8 e^–

Total: 18 e^–, stable.

c) tricarbonyl(η⁵–cyclopentadienyl)manganese(I)

Using the EAN rule, Mn(cp)(CO)₃:

Mn⁺ is d⁶: 6 e^–

cp^– contributes 6 e^– × 1: 6 e^–

CO contributes 2 e^– × 3: 6 e^–

Total: 18 e^–, stable.

Cr³⁺ is d³ so that the LFSE is –12Dq independent of the strength of the ligand, which is largely thermodyamically favorable. Any reduction in the number of ligands will reduce the LFSE (for example, in tetrahedral coordination the LFSE drops by more than 50%), which is not favorable. Z* for Cr³⁺ will not be too different from Z (Cr is on the leftward side of the Periodic Table), which implies that the ionic radius will be large enough that the presence of six ligands does not bring about too much steric repulsion. Yet, since Cr is a first row transition metal, the ionic radius is still small enough that the coordination sphere can not expand to 7-coordinate (or higher) without introducing sizable steric repulsions. These combination of factors (and probably others) lead to the high incidence of 6-coordinate Cr³⁺ complexes.