WBJEE 2024 Physics & Chemistry Paper

To identify the ion having the $4 f^{6}$ electronic configuration, we need to look at the electron configurations of the given ions. The f-block elements, also known as the lanthanides, typically fill their 4 f orbitals as we move across the series.The atomic numbers and electron configurations of some relevant lanthanides are as follows:Samarium(Sm) : Atomic number $=62$ , Configuration: $\mathrm{[Xe]} 4 f^{6} 6 s^{2}$Gadolinium ( Gd) : Atomic number $=64$ , Configuration: $\mathrm{[Xe]} 4 f^{7} 5 d^{1} 6 s^{2}$Terbium (Tb) : Atomic number $=65$ , Configuration: $\mathrm{[Xe]} 4 f^{9} 6 s^{2}$ Now let's examine the electronic configurations of the ions given in the options: $\square$${G d}^{3+}$ : Gadolinium loses three electrons to form the ion. This typically means losing two 6 s electrons and one 5d electron:$[\mathrm{X e}] 4 f^{7}$$[\mathrm{X e}] 4 f^{7}$$\mathrm{Sm}^{3+}$ : Samarium loses three electrons to form the ion, usually removing the two 6 s electrons and one 4 f electron:$[\mathrm{X e}] 4 f^{5}$$\mathrm{Sm}^{2+}$ : Samarium loses two electrons to form the ion, typically losing only the two 6 s electrons:$[\mathrm{X e}] 4 f^{6}$$\mathrm{Tb}^{3+}$ : Terbium loses three electrons to form the ion, generally removing the two 6 s electrons and one 4 f electron:$[\mathrm{X e}] 4 f^{8}$ Among the options, the ion with the $4 f^{6}$ electronic configuration is $\mathrm{Sm}^{2+}$. This configuration is obtained after Samarium loses its two 6 s electrons.Therefore, the correct answer is: Option C: $\mathrm{Sm}^{2+}$