When forming an ion, what is the expected charge on palladium and what is the electron configuration of that ion? +8, [kr] 552 +2, [Kr] 4d +10, [kr] +2, [Kr] 552 4d +2, [kr] 552 406
The Correct Answer and Explanation is:
Correct Answer:
+2, [Kr] 4d⁸
Explanation:
Palladium (symbol Pd, atomic number 46) is a transition metal in the d-block of the periodic table. To determine the expected charge and electron configuration of the ion, we must first look at the ground-state electron configuration of a neutral palladium atom.
1. Ground-State Electron Configuration of Neutral Pd:
Neutral Pd has 46 electrons. Its actual ground-state electron configuration is:
[Kr] 4d¹⁰
This is unusual. Based on the Aufbau principle, one might expect [Kr] 5s² 4d⁸, but Pd is an exception due to stability gained from a fully filled 4d subshell (4d¹⁰) and an empty 5s orbital. Hence:
Pd: [Kr] 4d¹⁰
2. Common Ion Formed by Palladium:
Palladium most commonly forms a +2 oxidation state in compounds, though it can also form +4. However, +2 is far more stable and prevalent in nature and industrial chemistry (e.g., PdCl₂, Pd(NO₃)₂).
When forming the +2 ion (Pd²⁺), palladium loses two electrons. Since the 5s orbital is higher in energy and typically ionized first, and it is already empty, the two electrons are removed from the 4d orbital.
So:
Pd²⁺: [Kr] 4d⁸
This is the most stable and common ionic state for Pd.
3. Incorrect Options Explained:
- +8 charge is not possible for palladium; it exceeds its typical valence capabilities.
- [Kr] 552 is a non-existent or miswritten configuration.
- [Kr] 552 406 is incorrect due to invalid orbital notation and ordering.
- [Kr] 4d is incomplete.
Final Answer:
Pd²⁺ has a charge of +2 and an electron configuration of [Kr] 4d⁸.
