Question

Which of the following elements shows a +4 oxidation state with the given configuration?

• A. Ce
• B. Tb
• C. Eu
• D. Lu

Hint:

To determine the oxidation state of an element, look at its electron configuration and consider how many electrons can be removed to achieve a stable configuration. Transition metals and lanthanides often show multiple oxidation states.

Answer 1

The Correct Option is A

We are tasked with identifying which of the given elements shows a +4 oxidation state with the corresponding electron configuration.
Step 1: Analyze the electron configurations Let’s first look at the elements listed: - Cerium (Ce): Ce has an atomic number of 58, with an electron configuration of \( [Xe] 4f^1 5d^1 6s^2 \). In the +4 oxidation state, cerium loses all its 4f and 5d electrons, resulting in a configuration of \( [Xe] \), which corresponds to a +4 oxidation state. - Terbium (Tb): Tb has an atomic number of 65, and its electron configuration is \( [Xe] 4f^9 6s^2 \). In the +4 oxidation state, it would lose electrons from the 4f and 6s orbitals, but Tb commonly exhibits a +3 oxidation state, not +4. - Europium (Eu): Eu has an atomic number of 63, with the electron configuration \( [Xe] 4f^7 6s^2 \). Europium commonly exhibits +2 and +3 oxidation states, but it does not typically form a +4 state. - Lutetium (Lu): Lu has an atomic number of 71, with the electron configuration \( [Xe] 4f^{14} 5d^1 6s^2 \). In the +3 oxidation state, Lu typically loses its 5d and 6s electrons, and it does not form a +4 state.
Step 2: Conclusion From the analysis of the electron configurations, we see that Cerium (Ce) is the only element among the options that shows a +4 oxidation state, as it can lose all its 4f and 5d electrons. Thus, the correct answer is \( \boxed{\text{Ce}} \).