Answer to Question #160785 in Inorganic Chemistry for KAARTIKA A/P ELONGKOVAN

Question

For each of the following complexes:

(NH4)[FeBr_x], µeff = 5.92 BM

Na_x[Nb(CO)], µeff = 0 BM

i. Explain briefly the hybrid orbitals involved with respect to their structure.

Answer 1

This µeff value corresponds to iron in the +3 oxidation state. Iron has a 3d⁶ 4s² configuration, without three electrons it is 3d⁵. Bromine ions are ligands of a weak field, therefore, electrons in the d-orbital will not pair, electron pairs of bromine will be located in the s and p orbitals. Hybridization state sp³.

Since µeff = 0, then niobium has no unpaired electrons, and its usual configuration is 4d⁴ 5c¹. This means the oxidation state of niobium is -1. This is niobium monocarbonyl, which means the molecule is linear and only one orbital is involved in the formation of a bond. Hence, hybridization does not occur.