Explain oxidation states of lanthanoids.

Explanation

(1) The common oxidation state of the Lanthanoids is 3+ due to the loss of 2 electrons from outermost 6s orbital amd one electron from the penultimate 5d-sub-shell.
(2) `Gd^(3+)` and `Lu^(3+)`e show extra stability due to their half filled and completely filled f-sub -shells.
`Gd^(3+) = [Xe]4f^(7)`
`Lu^(3+) = [Xe]4f^(14)`
(3) Ce and Tb attain the `4f^(0)` and `4f^(7)` configurations in the 4+ oxidation states. Eu and Yb attain the `4f^(7)` and `4f^(14)` configurations in the 2+ oxidation states. Sm and Tm also show the 2+ oxidation state although their stability can be explained based on thermodynamic factors.
(4) Some lanthnoids show 2+ and 4+ oxidation states even though they do not have stable electronic configuration of `4f^(0)` , `4f^(7)` or `4f^(14)`
E.g. `Pr^(4+) (4f^(1)),Nd^(2+),Sm^(2+)(4f^(6)),Dy^(4+)(4f^(8)),` etc