Let's consider the elements: - Pr (Praseodymium) has an atomic number of 59. Its ground state electron configuration is [Xe]4f36s2. - Pr3+ (Praseodymium ion) will lose 3 electrons. It loses the two 6 s electrons and one from the 4f orbital. So, the configuration is [Xe]4f2. −Nd3+ (Neodymium ion) with an atomic number of 60 , in its ground state has the configuration [Xe]4f46s2. When it loses 3 electrons (two from 6s and one from 4f ), its configuration is [Xe] 4f3. - Pm3+ (Promethium ion) with an atomic number of 61 , in its ground state has the configuration [Xe]4f56s2. When it loses 3 electrons (two from 6s and one from 4f ), its configuration is [Xe]4f4. So, the 4f2 electronic configuration is found in Pr3+. Hence, Option B is correct.