$ (1) La^{1+}, Pr, \quad (2) Ce^{4+}, La^{3+} \quad (3) Ce^{3}, Pr^{4+} \quad (4) La^{2+}, Nd $

$\begin {array} {1 1} (1)\;and\;(2) \\ (2) \;and \;(3) \\ (1)\;and \;(3) \\ (2)\;and\;(4) \end {array}$

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Answer: (2) and (3)

The Lanthanoids have an electronic configuartion of $6s^2$ common but variable occupancy of $4f$ and $5d$ subshells.

The first element in the series is $La$ with the configuration $[Xe] 4f^05d^16s^2$, the next two elements are $Ce$ and $Pr$ with the configurations $[Xe]4f5d616s^2$ and $[Xe]4f^36s^2$ respectively.

From the choices given, to arrive at the $X^{+n}$ we need to lose the correct number of electrons. It follows that the electron configuration of the given Lanthanoids are as follows:

$ \begin{array} {lllll} \text{Element} & Ln & Ln^{2+} & Ln^{3+} & Ln^{4}\\ 1.\;La & 4f^05d^16s^2 & 5d^1 & 4f^0 & \\ 2. \; Ce& 4f5fd^16s^2&4f^2&4f^1&4f^0\\ 3.\;Pr & 4f^36s^2&4f^3&4f^2&4f^1 \end{array}$

Therefore, we can see that $Ce^{4+}, La^{3+}$ and $Ce^{3}, Pr^{4+}$ have the same electron configuration.

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