Chapter 7.3, Problem 39SSC

Interpretation Introduction

(a)

Interpretation:

The subscripts most likely to be used if ionic compound is formed between alkali metal and halogen.

Concept introduction:

The naming of ionic compound is such that the cation comes first and anion comes next. The subscripts are numeric indicating number of atoms that are involved in bonding. To get electrically neutral compound, there is balancing of the polyatomic ions that forms ionic bonds.

Answer to Problem 39SSC

The subscript to be used if ionic compound is formed between alkali metal and halogen is 1, 1.

Explanation of Solution

The binary ionic compound is written with the metal ion first and then the nonmetal next. The charge of the metal ion and non metal ion is balanced by cancelling out each other depending on the number of ions of each.

The alkali metal belongs to Group 1. It has electronic configuration as ${\text{1s}}^2{\text{ 2s}}^1$. The electron in the outermost shell can be easily removed and hence, it is said to be reactive. It loses electrons to become a monatomic ion. If X is the element it will become ${\text{X}}^{+}$

The halogens belong of Group 7. It has electronic configuration as ${\text{ns}}^2{\text{np}}^5$ having 7 valence electrons. As it is in need of one more electron, it will easily take up an electron to become a negatively charged ion. If Y is the element, it will become ${\text{Y}}^{-}$

Combining alkali metal ion and halogen ion, it will be written as

Therefore, the subscripts for this ionic compound is 1, 1

(b)

Interpretation Introduction

Interpretation:

The subscripts most likely to be used if ionic compound is formed between alkali metal and non metal from group 16.

Concept introduction:

The naming of ionic compound is such that the cation comes first and anion comes next. The subscripts are numeric indicating number of atoms that are involved in bonding. To get electrically neutral compound, there is balancing of the polyatomic ions that forms ionic bonds.

Answer to Problem 39SSC

The subscript to be used if ionic compound is formed between alkali metal and non metal from group 16 is 2, 1

Explanation of Solution

The binary ionic compound is written with the metal ion first and then the nonmetal next. The charge of the metal ion and non metal ion is balanced by cancelling out each other depending on the number of ions of each.

The alkali metal belongs to Group 1. It has electronic configuration as ${\text{1s}}^2{\text{ 2s}}^1$. The electron in the outermost shell can be easily removed and hence, it is said to be reactive. It loses electrons to become a monatomic ion. If X is the element it will become ${\text{X}}^{+}$

The elements of Group 16 have electronic configuration as ${\text{ns}}^2{\text{np}}^4$ having 6 valence electrons. As it is in need of two more electron, it will take up two electrons to become a negatively charged ion. If Y is the element, it will become ${\text{Y}}^{2-}$

Combining alkali metal ion and non metal ion of group 16, it will be written as

Therefore, the subscripts for this ionic compound is 2, 1

(c)

Interpretation Introduction

Interpretation:

The subscripts most likely to be used if ionic compound is formed between alkaline earth metal and halogen

Concept introduction:

The naming of ionic compound is such that the cation comes first and anion comes next. The subscripts are numeric indicating number of atoms that are involved in bonding. To get electrically neutral compound, there is balancing of the polyatomic ions that forms ionic bonds.

Answer to Problem 39SSC

The subscript to be used if ionic compound is formed between alkaline earth metal and halogen

is 1, 2.

Explanation of Solution

The binary ionic compound is written with the metal ion first and then the nonmetal next. The charge of the metal ion and non metal ion is balanced by cancelling out each other depending on the number of ions of each.

The alkaline earth metal belongs to Group 2. It has electronic configuration as ${\text{ns}}^2$. It has 2 valence electrons in the outermost shell. Hence, it loses two electrons to become a diatomic ion. If X is the element it will become ${\text{X}}^{2+}$

The halogens belong of Group 7. It has electronic configuration as ${\text{ns}}^2{\text{np}}^5$ having 7 valence electrons. As it is in need of one more electron, it will easily take up an electron to become a negatively charged ion. If Y is the element, it will become ${\text{Y}}^{-}$

Combining alkali metal ion and non metal ion of group 16, it will be written as

Therefore, the subscripts for this ionic compound is 1, 2

(d)

Interpretation Introduction

Interpretation:

The subscripts most likely to be used if ionic compound is formed between alkaline earth metal and non metal from group 16.

Concept introduction:

The naming of ionic compound is such that the cation comes first and anion comes next. The subscripts are numeric indicating number of atoms that are involved in bonding. To get electrically neutral compound, there is balancing of the polyatomic ions that forms ionic bonds.

Answer to Problem 39SSC

The subscript to be used if ionic compound is formed between alkaline earth metal and non metal form group 16 is 1, 1.

Explanation of Solution

The binary ionic compound is written with the metal ion first and then the nonmetal next. The charge of the metal ion and non metal ion is balanced by cancelling out each other depending on the number of ions of each.

The alkaline earth metal belongs to Group 2. It has electronic configuration as ${\text{ns}}^2$. It has 2 valence electrons in the outermost shell. Hence, it loses two electrons to become a diatomic ion. If X is the element it will become ${\text{X}}^{2+}$

The elements of Group 16 have electronic configuration as ${\text{ns}}^2{\text{np}}^4$ having 6 valence electrons. As it is in need of two more electron, it will take up two electrons to become a negatively charged ion. If Y is the element, it will become ${\text{Y}}^{2-}$

Combining alkaline earth metal ion and non metal ion of group 16, it will be written as

Therefore, the subscripts for this ionic compound is 1, 1