Question

1. Write the metal d" configuration and identify the number of valence electrons around the metal atom in each of the following molecules [Rh(PMe3)4]* [(n®-C6H6)Cr(CO)3] [Rh(bipy)2Cl] * [(n-C3H5)Fe(CO)2]: co a co RhRh coc1 co [Ir(CO)CI(PPhz)2]

Answer 1

For all the complexes, the valence electron counting is done using neutral atom method.

[Rh(PMe₃)₄]⁺

The oxidation state of the metal atom is +1. Hence it is a d⁸ ion.

Now, valence electron counting around the metal

Rh………..9e-

4PMe₃……….2x4 = 8e-

1+ charge……………= -1e-

Total valence electron = 16e-

The electron counting is done for all other complexes similarly.

[(n⁶-C₆H₆)Cr(CO)₃]

The chromium atom is in zero oxidation state. Hence, it is a d⁶ ion.

Cr……………6e-

(n⁶-C₆H₆)……….6e-

3CO………….6e-

Total valence electron = 18e-

[Rh(bipy)₂Cl]⁺

The oxidation state of the Rh is +2 and hence it is a d⁷ ion.

Rh……………9e-

2bipy……….8e-

Cl……………1e-

2+ charge……….-2e-

Total valence electron = 16e-

[(n⁵-C₅H₅)Fe(CO)₂]^-

The oxidation state of Fe is zero, hence it is a d⁸ ion.

(n⁵-C₅H₅)…………..5e-

Fe……………………8e-

2CO………………..4e-

1- charge…………..1e-

Total valence electron = 18e-

[Rh₂(CO)₄Cl₂]

The oxidation state of Rh is +1, hence it is a d⁸ ion.

2Rh……………..18e-

4CO……………….8e-

2Cl(bridging)……..6e-

Total valence electon 32e-, i.e. 16e- for each metal atom.

[Ir(CO)Cl(PPh₃)₂]

The oxidation state of Ir is +1, hence it is a d⁸ ion.

Ir…………….9e-

CO……………2e-

Cl……………..1e-

2PPh₃…………..4e-

Total valence electrons 16e-