$\mathrm{Ce}, \mathrm{Pr}$ and Nd are lanthanoids and have incomplete $4 f$ shell while $\mathrm{Th}, \mathrm{Pa}$ and U are actinoids and have $5 f$ shell incomplete. In the beginning, when $5 f$-orbitals begin to be occupied, they will penetrate less into the inner core of electrons. The 5 -electrons will therefore, be more effectively shielded from the nuclear charge than $4 f$ electrons of the corresponding lanthanoids.

Therefore, outer electrons are less firmly held and they are available for bonding in the actinoids.

Separation of Zr and Hf are quite difficult because of lanthanoid contraction. Due to lanthanoid contraction, they have almost same size ( $\mathrm{Zr}=160 \mathrm{pm}$ and $\mathrm{Hf}=159 \mathrm{~pm}$ ) and thus, similar chemical properties. That's why it is very difficult to separate them by chemical methods.

It is due to the fact that after losing one more electron Ce acquires stable $4 f^{\circ}$ electronic configuration. So, Ce shows +4 oxidation state also alongwith +3 oxidation state.

When oxalic acid is added to acidic solution of $\mathrm{KMnO}_4$, its colour disappear due to reduction of $\mathrm{MnO}_4^{-}$ion to $\mathrm{Mn}^{2+}$. Chemical reaction occurring during this neutralisation reaction is as follows

$5 \mathrm{C}_2 \mathrm{O}_4^{2-}+\underset{\text { (Coloured) }}{2 \mathrm{MnO}_4^{-}}+16 \mathrm{H}^{+} \longrightarrow \underset{\text { (Colourless) }}{2 \mathrm{Mn}^{2+}}+8 \mathrm{H}_2 \mathrm{O}+10 \mathrm{CO}_2$

When orange solution containing $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ ion is treated with an alkali, a yellow solution of

$\mathrm{CrO}_4^{2-}$ is obtained. On the same way, $\underset{\substack{\text { Dischromate } \\ \text { (orange) }}}{\mathrm{Cr}_2 \mathrm{O}_7^{2-}} \xrightarrow[\mathrm{H}^{+}]{\mathrm{OH}^{-}} \underset{\begin{array}{c}\text { Chromate } \\ \text { (yellow) }\end{array}}{\mathrm{CrO}_4{ }^{2-}}$

when $\mathrm{H}^{+}$ions are added to yellow solution, an orange solution is obtained due to interconversion.