(a). Of the lanthanides, cerium `(Z=58)` forms a tetrapositive ion, `Ce^(4+)` in aqueous solution. Why?
(b), The `+3` oxidation states of lanthanum `(Z=57)`, gadolinium `(Z=64)` and lutetium `(Z=71)` are especially stable. Why?

Explanation

(a). `Ce^(3+)` having the configuration `4f^15d^06s^0` can easily lose an electron to acquire the configuration `4f^0` and form `Ce^(4+)`. In fact this is the only `+4` state lanthanide which exists in solution.
(b). This is because they have empthy half filled and completely filled 4f sub-shell respectively.
(c). Due to the consequency of lanthanoid contaction, `Hf(Z=72)` has size similar to that of `Zr(Z=40)`. Hence, their properties are similar. For the same reason, Nb and Ta have similar size and hence similar properties.
(d). `La(OH)_3` is more basic than `Lu(OH)_3`. As the size of the lanthanoid ions decreases from `La^(3+)` to `Lu^(3+)`, the covalent character of the hydroxides increases (Fajan's rules). Hence, the basic strength decreases from `La(OH)_3` to `Lu(OH)_3`