Chapter 7: Problem 117
Apply Praseodymium is a lanthanide element that reacts with hydrochloric acid, forming praseodymium(III) chloride. It also reacts with nitric acid, forming praseodymium(III) nitrate. Praseodymium has the electron configuration \([\mathrm{Xe}] 4 \mathrm{f}^{3} 6 \mathrm{s}^{2}\) a. Examine the electron configuration, and explain how praseodymium forms a \(3+\) ion. b. Write the correct formulas for both compounds formed by praseodymium.
Short Answer
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.