Question

1) trans-FeCl2(dppe)2 is a green colored compound. Upon reduction of this coordination compound with NaBH4, and orange colored trans-FeH2(dppe)2 results. Explain the origin in the difference of colors by drawing a d-orbital splitting diagram for each complex.

inorganic chemistry

Answer 1

Trans-Fe complexes of both the compounds are given below.

Ppha ろ ? ce Fee (Green) Oxidation state of Fe in trans- Fell, dppe), = +2 PhP PPh2 Fe H (Orange) H Phap PPh2 Oxidation state of Fe in trans-FeH, (dppe)2 =+2

To know the type of splitting we need to calculate the coordination number (CN) of central metal (Fe) for the above mentioned complexes.

Fe (At. No.) = 26 Valence orbital 22 configuration 9 = 3d 462 Feet Valence electronic 2 de configuration 141/0/110 C.N of Feet in trans. FeCl₂ (depe), = 6 C.N of Feet in trans-Fetta (appe), = 6 Coctahedral complex Coctahedral complex)

As we can see, the CN of both the complexes is 6. This means octahedral splitting occurs due to crystal field. Both Cl^{​​​​​-}​​​​ and H^{​​​​​​-} ions are weak field ligands. but H^{​​​​​​-} is even weaker than Cl^{​​​​​-} .  Hence the splitting and electronic distribution is as follows.

CASE-1:01 CASE-2 H Weg 3d6 3d 1/1/1/11 141 1/11 Lal1111tag

Both the complexes have the same profile as far as oxidation number, coordination number, pattern of splitting and electronic distribution is concerned. But, varies in Crystal Field Splitting Energy(CFSE or ∆_o)

In CASE - 1 and CASE - 2 as described above, ∆^{​​​​​​}_o > ∆'_o

Hence CFSE in the case of Cl_{​​​​​2} complex is greater than that of hydridecomplex. This causes the difference in colour as seen in the complexes.