WBJEE 2023 Physics & Chemistry Solved Paper

Let's consider the elements:- $\mathrm{Pr}$ (Praseodymium) has an atomic number of 59. Its ground state electron configuration is $[\mathrm{Xe}] 4 f^{3} 6 s^{2}$.- $\mathrm{Pr}^{3+}$ (Praseodymium ion) will lose 3 electrons. It loses the two 6 s electrons and one from the $4 f$ orbital. So, the configuration is $[\mathrm{Xe}] 4 f^{2}$.$-\mathrm{Nd}^{3+}$ (Neodymium ion) with an atomic number of 60 , in its ground state has the configuration $[X e] 4 f^{4} 6 s^{2}$. When it loses 3 electrons (two from $6 s$ and one from $4 f$ ), its configuration is [Xe] $4 f3$.- $\mathrm{Pm}^{3+}$ (Promethium ion) with an atomic number of 61 , in its ground state has the configuration $[X e] 4 f^{5} 6 s^{2}$. When it loses 3 electrons (two from $6 s$ and one from $4 f$ ), its configuration is $[\mathrm{Xe}] 4 f^{4}$.So, the $4 f^{2}$ electronic configuration is found in $\mathrm{Pr}^{3+}$. Hence, Option B is correct.