Why is +3 oxidation state most stable in lanthamoids?
Asked by aaa | 29th Nov, 2014, 07:53: PM
- The electronic configuration of lanthanoids is [Xe]4fn 6s2. The most stable electronic configuration is +3. It is achieved by loosing 2 electrons from 6s orbital and one electron from 4f orbital.
- The 4 f electrons are inner core electrons. There has been a further contraction of the 4f distribution. Its maximum is close to breaking out to the left from under the umbrella provided by the outer maxima of the 4s, 4p and 4d distributions. A significantly larger proportion of the 4f electron density now lies between the umbrella and the nucleus where it is exposed to a considerably greater nuclear charge.
- This contraction in the 4f electron distribution with increasing effective nuclear charge might explain why it is hard to obtain lanthanide oxidation states greater than +3.
- Because there are 18 electrons in the 4s, 4p and 4d sub-shells, the contraction occurs through a region of especially high electron density. With increase in oxidation state, the nuclear charge experienced by the remaining 4f electrons therefore increases especially sharply and at an early stage they become too tightly bound to engage in further bonding.
- Hence it is difficult to remove further electrons from 4f orbital as they are very closely bound by the nuclear charge.
Answered by Prachi Sawant | 1st Dec, 2014, 11:43: AM
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