Write the formula ammonium sulphate . Also name the elements presents in them.

Explanation

(i) The ions `La^(3+), Lu^(2+), Ce^(4+)` and `Yb^(2+)` have `4 f^0` and `4f^(14)` electronic configuration.
`La^(3+) = [Xe] 4f^0, Yb^(2+) = [Xe] 4f^(14)`
`Ce^(4+) = [Xe] 4f^0 , Lu^(3+) = [Xe] 4f^(14)`
Since these ions do not have unpaired electrons, therefore, these ions are diamagnetic in nature. All other ions except `La^(3+), Lu^(3+), Ce^(4+) and Yb^(2+)` are Paramagnetic because they have unpaired electrons.
(ii) The common oxidation state of all the Lanthanoids is `+3,` Some lanthanoids show +2 and +4 oxidation state also.
For example : Cerium (Ce) and terbium (Tb) attain `f^0` and `f^7` configuration respectively when they get +4 oxidation state.
`Ce^(4+) : [Xe] 4f^0`
`Th^(4+) :[Xe] 4f^7`
Clearly europium and Ytterbium get `f^7` and `f^(14)` configuration in +2 oxidation state as shown:
`Eu^(2+) : [Xe] 4f^7`
`Ye^(2+) : [Xe] 4f^(14)`