Question

In 1984, the first organometallic compounds with H₂ as an ?² ligand were synthesized: M(CO)₃(PR₃)₂(H₂) with M = Mo or W and R = cyclohexyl or isopropyl. These complexes illustrate back-bonding with H₂ as a ligand.

a. Draw a structure for this molecule (just use R for the substituents on the P atoms). Because the R groups are very bulky, assume that the two PR₃ ligands are in a trans configuration. Only one isomer is then possible. Show that this compound satisfies the 18-electron rule.

b. Draw the diagram shown in lecture showing how the two 1s atomic orbitals in the H atoms give rise to σ and σ* molecular orbitals in H₂, with the σ orbital being full.

c. Draw diagrams showing how L→M σ bonding occurs as well as M→L back-bonding into the H₂ σ* orbital. If the metal is particularly electron-rich, why would the H₂ bond be considerably weakened when coordinating with the metal?

d. What metal could we replace Mo or W with such that the bonding in the complex could be better described as two hydrides (H)₂ instead of a dihydrogen (H₂)? Explain your choice.

Answer 1

This complex fellow 18 electron rule because H2 atom donate two electron to metal, structure and calculation of 18 electron rule are shown in image. This image also show answers of rest ef the questions.

9. Structure of molecule CO co - -/- PR3 -co R₂PF- - HH la e rule - Mo-4d ss'=6e- 3 co provide = 3x2=6 e- B 2 PR3 provide = 2x2 = 46- H₂ provide = 20 Total = lde- b. Molecular orbital of H₂ Со H-atom H-atom Molecular orbital ML back bonding C. LAM o bonding C D metal filled orbital CAD/11) orbital H₂M Ligand la empty filled e orbital Ligand so empty orbital ... chromium can replace mo or w when two H. atom present as a ligand instead H₂ because in case of co the back bonding is stronger a due to high electon density on Cr, thus of orbital of H₂ More filled thus H-4 single bond break down.