JTD, Td and SP notes,for chemistry higher level

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1. TETRAHEDRAL COMPLEXES The energy of the t 2g orbitals is raised most because they are closest to the ligands. This crystal field splitting is the opposite way round to that in octahedral complexes

2. t2 e

3. The magnitude of the crystal field splitting ∆ t tetrahedral complexes is considerably less than in octahedral fields. There are two reasons for this: CFSE in tetrahedral complexes is quite small and it is always less than the pairing energy . Due to this reason pairing of electrons is energetically unfavourable . Thus all the tetrahedral Complexes are high-spin Complexes. d , d 5 and d 10 arrangements the CFSE is zero in both octahedral and tetrahedral complexes. The octahedral CFSE is greater than the tetrahedral CFSE It follows that octahedral complexes are generally more stable and more common than tetrahedral complexes. 1. There are only four ligands instead of six, so the ligand field is only two-thirds the size : hence the ligand field splitting is also two-thirds the size 2. The direction of the orbitals does not coincide with the direction of the ligands . This reduces the crystal field splitting by roughly a further two thirds . Tetrahedral crystal field splitting roughly 2/3 × 2/3 = 4/9 of the octahedral crystal field splitting. Tetrahedral CFSE scaled for comparison with octahedral values, assuming ∆ t = 4/9 ∆ o

4. The value of crystal field splitting parameter in [CoCl 6 ] 4– ion is 18000 cm –1 . Calculate the value of CFSE in [CoCl 4 ] 2–

5. Any non-linear molecular system in a degenerate electronic state will be unstable and will undergo some sort of distortion to lower its energy and remove the degeneracy. More simply, molecules or complexes which have an unequally filled set of orbitals (either t 2g or e g ), will be distorted. Jahn–Teller distortion Strong distortions resulting from uneven filling of the e g orbitals

6. Among the following compounds (MnO 2 , Mn 2 O 3 , MnO , KMnO 4 ) Jahn-Teller distortion will be highest in: a) MnO b) Mn 2 O 3 c) MnO 2 d) KMnO 4 Jahn teller effect is not observed in high spin complexes of: A) Mn 2+ B) Cr 2+ C) Cu 2+ D) Fe 3+

7. TETRAGONAL DISTORTION OF OCTAHEDRAL COMPLEXES (JAHN-TELLER DISTORTION) The shapes of transition metal complexes are affected by whether the d orbitals are symmetrically or asymmetrically filled. If the d electrons are symmetrically arranged, they will repel all six ligands equally. Thus the structure will be a completely regular octahedron

8. If the d electrons are asymmetrically arranged, they will repel some ligands in the complex more than others. Thus the structure is distorted because some ligands are prevented from approaching the metal as closely as others. This causes significant distortion of the octahedral shape. Distortion caused by the asymmetric filling of the t 2g orbitals is usually too small to measure. If the dz 2 orbital contains one more electron than the dx 2 −y 2 orbital then the ligands approaching along +z and −z will encounter greater repulsion than the other four ligands. The repulsion and distortion result in the elongation of the octahedron along the z axis. This is called tetragonal distortion If the d x 2− y 2 orbital contains the extra electron, then elongation will occur along the x and y axis. Thus there will be four long bonds and two short bonds. This is equivalent to compressing the octahedron along the z axis and is called tetragonal compression .

9. Z-out Z-in

10. Crystal field stabilization energy for z-in distortion? Crystal field stabilization energy for z-out distortion of d 9 ? It is easier to weaken two bonds rather than stretching four metal-ligand bonds

11. Ti 3+ octahedral complexes prefer to undergo z-in due to the greater value of the crystal field stabilization energy. CFSE due to z-in distortion is –2δ 2 /3. CFSE due to z-out distortion is –δ 2 /3.

12. Coordinatively labile nature of [Cr(H 2 O) 6 ] 2+ : The [Cr(H 2 O) 6 ] 2+ undergoes substitution easily since the Cr(II) ion is a high spin d 4 system with one electron in the eg orbital. Hence it is electronically degenerate and shows Jahn-Teller distortion. Hence the hydrated Cr(II) ion is coordinatively labile. [Cu( en ) 3 ] 2+ is unstable since JT distortion brings strain into the ethylene diamine molecule that is added along the z-axis. Hence only [Cu( en ) 2 (H 2 O) 2 ] 2+ is formed. Au(II) ion is less stable and undergoes disproportionation to Au(I) and Au(III) even though the Cu(II) and Ag(II) ions are comparatively more stable. Why? The Δ value increase down the group. Hence, in Au(II) ion, it reaches a maximum, which causes high destabilization of the last electron, which is now occupying the d x2-y2 . This makes Au(II) reactive, which may undergo either oxidation to Au(III), a d 8 system or reduction to Au(I), a d 10 system.

13. Jahn-Teller distortion on [Cu( en )3]2+

14. Jahn-Teller distortion on [Cu( en )2(H2O)2]2+

15. Identify the correct statement about [Ni(H₂O) 6 ] +2 and [Cu(H₂O) 6 ] +2 (b) Ni-O(equatorial) and Cu-O(equatorial) (c) All Ni-O bond lengths are equal whereas Cu-O (equatorial) bonds are shorter than Cu- O(axial) bonds (d) All Cu-O bond lengths are equal whereas Ni-O(equatorial) bonds are shorter than Ni- O(axial)bonds