D and F Block Elements

Step 1. Recall that $4 f^7$ means half-filled 4 f shell (7 electrons).Step 2. Check each ion(A) $\mathrm{Pr}^{3+}$Praseodymium (Pr) = atomic number 59Neutral configuration: $[X e] 4 f^3 6 s^2$When it loses 3 electrons $\rightarrow$ remove $6 s^2$ and one $4 f$ :$\mathrm{Pr}^{3+}:[X e] 4 f^2$Not $4 f^7$.(B) $\mathrm{Lu}^{3+}$Lutetium (Lu) = atomic number 71Neutral configuration: $[X e] 4 f^{14} 5 d^1 6 s^2$Removing three electrons $\rightarrow 6 s^2$ and $5 d^1$ :$\mathrm{Lu}^{3+}:[X e] 4 f^{14}$Not $4 f^7$.(C) $\mathrm{Eu}^{2+}$Europium $(\mathrm{Eu}) =$ atomic number 63 Neutral configuration: $[X e] 4 f^7 6 s^2$Losing two electrons $\rightarrow$ remove both $6 s$ electrons:$\mathrm{Eu}^{2+}:[X e] 4 f^7$、 Matches $4 f^7$.(D) $\mathrm{Ce}^{4+}$Cerium (Ce) = atomic number 58Neutral: $[X e] 4 f^1 5 d^1 6 s^2$Losing 4 electrons $\rightarrow 4 f^1 5 d^1 6 s^2 \rightarrow$ remove all $\rightarrow$$\mathrm{Ce}^{4+}:[X e]$$\text{Not} 4 f^7$. Correct Answer: Option C - Eu ${ }^{2+}$