Chapter 7, Problem 117A

Interpretation Introduction

(a)

Interpretation:

The formation of +3 ions in praseodymium needs to be explained.

Concept introduction:

Electronic configurations of the atoms of the elements in the ground state are determined with the help of Aufbau principle, hund’s rule and Paulie’s Exclusion principle.

Answer to Problem 117A

Praseodymium loses 3 electrons in 4f orbital because 4f orbital has higher energy than 6s orbital.

Explanation of Solution

Given the electronic configuration of praseodymium is $\left[\text{Xe}\right]{\text{4f}}^{\text{3}}{\text{6s}}^{\text{2}}$. Praseodymium has 2 electrons in 6s orbital. Praseodymium loses 3 electrons in 4f orbital because 4f orbital has higher energy than 6s orbital.

(b)

Interpretation Introduction

Interpretation:

The correct formulas for both compounds that formed by praseodymium needs to be determined.

Concept introduction:

Electronic configurations of the atoms of the elements in the ground state are determined with the help of Aufbau principle, hund’s rule and Paulie’s Exclusion principle.

Answer to Problem 117A

The correct formulas for both compounds that formed by praseodymium PrN and ${\text{PrCl}}_{\text{3}}$.

Explanation of Solution

From the reaction of hydrochloric and Praseodymium ionic compounds are formed. With praseodymium (III) cation, ${\text{PrCl}}_{\text{3}}$ chloride anion carries -1 charge and form 1:3 compound. From the reaction of nitric acid and Praseodymium ionic compounds are formed. praseodymium (III) nitride, PrN carries -3 charge and form 1:1 compound.