Question

From experimental data, it was found that the complexes [NiCl (PPh.),] 5. a and (PdCL,(PPh,] are paramagnetic and diamagnetic, respectively Propose the geometry of the complexes based on this information. Justify your answers. [6 marks] AN 20 b. Propose a possible structure for each of the following compounds: AM P JAN 20 [Co2(CO) i. [2 marks] [(n'-CHJCr(CO) ii. EXAM [2 marks] Use the 18-electron rule to determine if the proposed structures in C. part(b) are correct. XAM 20 14 marks] d. Explain the mean ing of a-donor and n-acceptor as applied to organometallic compounds with CO ligands. 2 0 [3 marks] Describe and explain the effect of lanthanoid contraction to the atomic radii of d-block metals. JAN 20 13 marks]

Answer 1

Geometry of complex depends on the magnetic behaviour of complexes. Further structure of carbonyl complexes are determined on the basis of 18 electron rule.
solution. (a) [Ni cl,(PPhy),] Ni2t 3d8 4 go Ground state-1 11/ 14 14 111 O Excited state - Mi til 1411 0 0 3d 45 | 901 e de PPh PPh 3 853 -> Tetrahedral [Pdcho (PPh3), ] (Paramag hetic) Pd2t4d8 550 Ground State - 11111111 Excited State - M116 ♡ CD ů a pphy PPhz diamagnetic) dope & Square Plomar. geometry

(b) (i) [Co. (CO)g ] 0 | 08C-G CG-C=O → Due to small size of Cobalt, two Carbonyl groups can easily overlap with two two Cobalt atoms & behal as bridged carbonyl groups. (li) [(ME-GGHG) Gr (CO), ?????????????????? nonnnnn (©) [Co, (20)g] į terminal co= 12é 2 bridged co=4 é 2 Cobalt atom = 18e 1 co-co= 2 é - Total = 36€ é fer Co = 18 e. [(28- CHo) Cr (co) Cratom = bé 3 terminal.coz 6e- e from benzene=66- (d) co ligand act as both 6-donos & re-acceptos It has lone pair of electron in the non- bonding molecular orbital & hence, act as .. G-dortor while it has Bi-antibonding bi orbital. That's why also act as r- acceptor lig and as it re-antibonding orbital is vacant. 1be- UU o

e) Due to lanthanide contraction, d-block elements a after Lanthanide series faces decrease in ionic madii at greater rate than expected. It is due to poor shielding of nuclear charge by 4f electrons & hence the 6S elections are drawn towards the nucleus this resulting in smaller atomic radius of d-block metals which results in similar size of 4d & 5d elements.