Chapter 14, Problem 113A

(a)

Interpretation Introduction

Interpretation: To write the names of the given compounds.

Concept Introduction: The given compound contains both metal and non-metal. The cations in the compound contain positive charge and the anion in the compound contain the negative charge. The cation which is a metal will be named first and the anion which is a non-metal will be named after the metal.

Answer to Problem 113A

Tin (II) Bromide.

Explanation of Solution

In the given compound, ${\text{SnBr}}_{\text{2}}$ , the tin will be a metal and the bromine will be a non-metal. The tin forms a cation and the oxidation state of tin will be +2. The bromine atom will form and anion and the oxidation state of bromine will be -1. Therefore, tin is named first along with its oxidation state and then the bromine anion with suffix -ide.

(b)

Interpretation Introduction

Interpretation: To write the names of the given compounds.

Concept Introduction: The given compound contains both metal and non-metal. The cations in the compound contain positive charge and the anion in the compound contain the negative charge. The cation which is a metal will be named first and the anion which is a non-metal will be named after the metal.

Answer to Problem 113A

Barium Sulphate.

Explanation of Solution

In the given compound, ${\text{BaSO}}_{\text{4}}$ , the Barium will be a metal and the sulfate ion will be the anion. The barium forms a cation and the oxidation state of tin will be +2. The Sulfate ion will form and anion and the oxidation state will be -2. Therefore, Barium is named first and then the sulfate anion.

(c)

Interpretation Introduction

Interpretation: To write the names of the given compounds.

Concept Introduction: The given compound contains both metal and non-metal. The cations in the compound contain positive charge and the anion in the compound contain the negative charge. The cation which is a metal will be named first and the anion which is a non-metal will be named after the metal.

Answer to Problem 113A

Magnesium Hydroxide

Explanation of Solution

In the given compound, ${\text{Mg(OH)}}_{\text{2}}$ , the magnesium will be a metal and the hydroxide ion will be the anion. The Magnesium forms a cation and the oxidation state of tin will be +2. The hydroxide ion will form and anion and the oxidation state will be -1. Therefore, Magnesium is named first and then the Hydroxide anion with suffix -ide.

(d)

Interpretation Introduction

Interpretation: To write the names of the given compounds.

Concept Introduction: In case of a compound which contain two halogens, the halogen with positive charge is written first and the halogen with the negative charge is written next. These types of compounds are called as inter-halogen compounds and they are named as Halogen halide.

Answer to Problem 113A

Iodine penta-fluoride.

Explanation of Solution

In the given compound, ${\text{IF}}_{\text{5}}$ , the iodine will have a positive charge and the fluorine will have a negative charge. Therefore, according to the rule, the iodine with positive charge is named first and then the fluorine with negative charge. Since, there are 5 fluorine atoms it is named as penta-fluoride.