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

Question #160785

For each of the following complexes:

(NH4)[FeBrx], µeff = 5.92 BM

Nax[Nb(CO)], µeff = 0 BM

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

Expert's answer

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.

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