The discussion focuses on the hybridization of the chromate ion (CrO4^2-), clarifying that it is best described using d3s hybridization rather than sp3 hybridization. Participants emphasize that in the highly charged chromium VI state, the d orbitals are energetically more favorable for bonding than the p orbitals. The d3s hybrids align along the diagonals of a cube, leading to a tetrahedral shape when accounting for slight substitutions of d for p orbitals. Additionally, the inner 3s orbital does not participate in bonding due to its low energy and full occupancy.
PREREQUISITESChemistry students, educators, and professionals interested in transition metal chemistry and the structural analysis of complex ions.
Guys please help me.Karan Punjabi said:Can some one explain the structure of chromate ion on the basis of hybridisation cause I'm a bit confused I googled about it and there are posts like explaining it as sp3 hybridised ion. Explain please.
The inner s orbital ie 3s would be affected by hybridisation or not? And how we are considering hybridisation between 2 orbitals of different shells?DrDu said:If you really want to describe transition metal compounds in terms of hybrids, I would rather go for ##d^3s## than ##sp^3##, as the p orbitals in a highly charged ion like chromium VI are energetically much higher than the d orbitals. The d3s hybrids are directed along the diagonals of a cube. If you allow for a small substitution of d for p, the two directions along the diagonal become unequal and you will get a more tetrahedral shape.
Got it. One more thing chromate ion is a complex right?DrDu said:The inner 3s orbital is way too low in energy and completely filled. Hence it won't participate in bonding. There is no prinicpal difficulty with hybridizing orbitals of different sub-shells as long as they are of comparable energy and size.