Consequences Of Jahn-Teller Distortion - Coordination Chemistry | Inorganic Chemistry PDF Download

Consequences of Jahn-Teller Distortion 

• Stability of Cu²⁺ Complexes: For a given ligand, the relative stabilities of complexes of dipositive metal ions of first transition series is in order
  
  Ba²⁺ < Sr²⁺ < Ca²⁺ < Mg²⁺ < Mn²⁺ < Fe²⁺ < Co²⁺ < Ni²⁺ < Cu²⁺ > Zn²⁺
  
  This is called Irving-William Series. The extra stability of Cu²⁺ complexes is due to the Jahn-Teller distortion. During distortion two electrons are lowered in energy while one is raised an equal amount of energy.
• The complex [Cu(en)₃]²⁺ is unstable because Jahn-Teller distortion causes strain into ethylenediamine molecule attached along z-axis.
  
  
  Similarly, trans- [Cu(en)₂(H₂O)₂] ²⁺ is more stable than the cis-[Cu(en)₂(H₂O)₂]²⁺ because Jahn-Teller distortion causes strain in ethylenediamine molecule along z-axis in the later.
  
    
•  In the cr ystalline KCuF₃, the two Cu-F distances on z-axis are 196 pm and the other four Cu-F distances in xy plane are 207 pm. This is due to the Jahn-Teller distortion (compression).
• Complexes of Au²⁺ (d9 configuration) ion are unstable and undergo disproportionation to Au (I) and Au (III) whereas the complexes of Cu²⁺ (d⁹) Ag²⁺ (d⁹) are comparatively more stable. We know that crystal field splitting value increases on moving down the group. The high value of ∆ causes the high destabilization of d_{x² – y²} orbital. Therefore, Au²⁺ becomes extremely reactive and undergoes oxidation to Au³⁺, a d⁸ system or reduction to Au⁺, a d¹⁰ system. Au³⁺ forms a square planar complex while Au⁺ forms a linear complex.
  
  Applications of CFSE 
   
• Ionic Radii of Transition Metal Cations: Ionic radii of dipositive or tripositive metal ions are expected to decrease linearly from Ca²⁺ to Zn²⁺ because of increasing effective nuclear charge.  But some ions show discrepancies from this regular trend of decreasing ionic radii.

The ionic radii is determined b y how closely a ligand can approach the metal ion. If both the t_2g and e_g orbitals are symmetrical filled the ionic radii will not be deviated. For example ionic radii of Ca²⁺, Mn²⁺ and Zn²⁺ fall on curve because these ions are screened by a spherical charge distribution. In d₁ syst em the electron will be in t_2g orbital which does not point towards ligands, therefore, the ionic radii is decreased from d₁ to d₃ system. When we go to d₄ high spin the radii increase because the coming electron would be in eg orbital and hence ligands shall repel farther. This trend will continue and the above curve is obtained. On the basis of above statements we can also explain the trend of ionic radii obtained in the case of tripositive metal ions.

• Hydration Energy of Metal Ions: Hydration energy is the energy released when gaseous metal cations get hydrated.

Mⁿ⁺(g) + 6H₂O → [M(H₂O)₆]ⁿ⁺ + Hydration energy

The hydration energy of metal ions should increase linearly from Ca²⁺ to Zn²⁺ because the charge density is increase but it is not observed. This is because when the metal cations of 3d -series transition metals get hydrated, a high spin aqua complex is formed. Therefore, energy released is not only hydration energy but also CFSE. Thus:

Experimental Hydration Energy    =     Theoretical Hydration Energy      +    CFSE

If the experimental hydration energies for dipositive metal ions of the 3d transition elements are plotted as a function of atomic no., a double humped curved is formed.

The two humps in such a curve due to contribution of CFSE to the hydration energy.  Since the CFSE of Ca²⁺, Mn²⁺, and Zn²⁺ in high spin Oh complexes is zero. Therefore, the hydration energies of these ions lie on the straight line and does not deviate from the experimental value. 

• Lattice Energy: Lattice energy is the change in energy when one mole of an ionic crystal is formed from its constituent gaseous ions.

M+ (g) + X– (g) → MX (s) + Lattice energy

According to Born Landes’s equation:

Lattice energy (U) =  

i. e. Lattice energy = 

Where N = Avogadro’s no., A = Madelung Constant which depends upon the geometry of the crystal, Z+ and Z^– are the charges on the cat ion and anion respectively, r⁰ = interionic distance, n = Born exponent.

According to Born Lande’s equation, the lattice energy of an ionic crystal increases with increase in product of Z⁺ and Z^– and decreases with interionic distance (r⁰). Consequently, the lattice energy for halides of the dipositive metal ions of the 3d -elements should increase from Ca²⁺ to Zn²⁺. But a humped curve is obtained. This is because of contribution of CFSE to the lattice energy. 

• Structure of Spinels: Spinels are mixed metal oxide having formula AB₂O₄ or AO.B₂O₃ where A can be group 2 metal or a transition metal in +2 oxidation state and B can be a group 13 metal or transition metal in +3 oxidation state. The oxide anions form a cubic closed packed (ccp) or face centred cubic (fcp) lattice with four tetrahedral (Td) voids and eight octahedral (Oh) voids per AB₂O₄ unit.

No. of tetrahedral voids = 2× no. of octahedral voids.

No. of octahedral voids = No. of lattice points.

 Spinels are of two types: 

(a) Normal Spinels (b) Inverse Spinels

a. Normal Spinels : In normal spinels A²⁺ ions occup y one eight of the tetrahedral voids and B3+ ions occupy half of the octahedral voids. This is written as [AII][BIII]O₄.
For example – MgAl₂O₄, MgCr₂O₄, NiCr₂O₄, Mn₃O₄, Co₃O₄ etc.

b. Inverse Spinels: In inverse spinel, half of the B³⁺ ions have exchanged places with all the A²⁺ ions i.e. the A²⁺ occupy 1/4 of the octahedral voids along with half of the B³⁺ ions, whereas half of the B³⁺ ions occup y 1/8 of the Oh voids. It is represented as

 [B][AB]O₄. Examples are CrFe₂O₄, Fe₃O₄, NiFe₂O₄, MgFe₂O₄, NiAl₂O₄ etc. In general, the spinels having Fe³⁺ ions are inverse spinels.

Predicting Structures of Spinels: Structures of spinels are determined by the CFSE of the cations occupying the tetrahedral and octahedral voids.
The metal cations in Td and Oh voids are surrounded b y four and six oxide ion respectively. Thus, metal ions are considered to form Td and Oh complexes respectively. The oxide ion because as weak ligand. Structure is determined by followed way:

• Compare CFSE in Oh and Td field for both A²⁺ and B³⁺ ions
• If CFSE in Oh voids is greater than B³⁺ in Td and A²⁺ in both Td and Oh, the spinel will be normal.
• If CFSE for A²⁺ ions in Oh void is greater than A²⁺ in Td and B³⁺ in both Td and Oh, that will be inverse spinel.

Examples: 

• Mn3O₄ →Mn^IIMn₂^IIIO₄

Since, Mn³⁺ in Oh filed has the highest CFSE, therefore it will be stabilized in Oh voids. Consequently, Mn³⁺ will occupy the Oh voids and the Mn²⁺ ion will occupy the tetrahedral voids. Hence Mn₃O₄ is a normal spinel.

Caution: Calculate all the CFSE in terms of ∆₀ for comparing. The ∆t can be converted into ∆₀ by 
multiplying . For comparing only see the magnitude and not the sign.

•  Fe₃O₄ → Fe^IIFe₂^IIIO₄

Since CFSE for Fe²⁺ in octahedral field is highest . Therefore, Fe²⁺ ions will occup y Oh voids and Fe³⁺ ions will occupy both Td and Oh voids. Thus Fe₃O₄ in an inverse spinel.

• NiFe₂O₄ → Ni^IIFe₂^IIIO₄

Since CFSE for Ni²⁺ is highest in Oh field. Therefore, Ni²⁺ ions will occupy Oh voids and Fe³⁺ ions will occupy both Td and Oh voids. Thus Ni2Fe₂O₄ in an inverse spinel.
Note- Spinel having Fe³⁺ ions are inverse spinel because Fe³⁺ (d⁵) ion has CFSE equal to zero in both octahedral Td voids.

• Co₃O₄ → Co^IICo₂^IIIO₄
  For Co³⁺, oxide ion behaves as strong field ligand hence low spin complex is formed.

Since CFSE for Co³⁺ is highest in Oh field, Co³⁺ ions will occupy octahedral and Co²⁺ ions will occupy tetrahedral voids. Hence Co₃O₄ is a normal spinel.

Shortcut → Calculate CFSE for both A²⁺ and B³⁺ ions in Oh filed. If CFSE of B³⁺ is higher than A²⁺ ions, the spinel will be normal spinel. If CFSE of A²⁺ is higher than that of B³⁺, then the spinel will be inverse spinel.
 

Limitations of CFT: 

• Considers only d orbitals and s and p orbitals are ignored.
• Does not consider the covalent character in transition metal complexes.
• Can not explain the Spectrochemical series.