Chapter 9.1, Problem 6LC

(a)

Interpretation Introduction

Interpretation: The type of ion and symbol for the ion formed by the given element is to be identified.

Concept Introduction : There are two types of charged species, anion (negatively charged species) and cation (positively charged species). Anion is formed by the gaining of electron/s whereas a cation is formed by losing electron/s.

Answer to Problem 6LC

The ion formed by the given element is ${\text{K}}^{+}$ (cation).

Explanation of Solution

The given element is potassium. It belongs to group 1 which is alkali metals. It has 1 valence electron. In order to complete the octet, it can lose 1 electron to form a cation with a +1 charge. The symbol of the ion is ${\text{K}}^{+}$ .

(b)

Interpretation Introduction

Interpretation: The type of ion and symbol for the ion formed by the given element is to be identified.

Concept Introduction : There are two types of charged species, anion (negatively charged species) and cation (positively charged species). Anion is formed by the gaining of electron/s whereas a cation is formed by losing electron/s.

Answer to Problem 6LC

The ion formed by the given element is ${\text{O}}^{2-}$ (anion).

Explanation of Solution

The given element is oxygen. It belongs to group 16 and has 6 valence electrons. In order to complete the octet, it can gain 2 electrons to form an anion with a -2 charge. The symbol of the ion is ${\text{O}}^{2-}$ .

(c)

Interpretation Introduction

Interpretation: The type of ion and symbol for the ion formed by the given element is to be identified.

Concept Introduction : There are two types of charged species, anion (negatively charged species) and cation (positively charged species). Anion is formed by the gaining of electron/s whereas a cation is formed by losing electron/s.

Answer to Problem 6LC

The ion formed by the given element is ${\text{Br}}^{-}$ (anion).

Explanation of Solution

The given element is bromine. It belongs to group 17 in the periodic table. It has 7 valence electrons. Thus, in order to complete the octet, it can gain 1 electron to form an anion with a -1 charge. The symbol is ${\text{Br}}^{-}$ .

(d)

Interpretation Introduction

Interpretation: The type of ion and symbol for the ion formed by the given element is to be identified.

Concept Introduction : There are two types of charged species, anion (negatively charged species) and cation (positively charged species). Anion is formed by the gaining of electron/s whereas a cation is formed by losing electron/s.

Answer to Problem 6LC

The ion formed by the given element is ${\text{Sn}}^{2+}$ (cation).

Explanation of Solution

The given element is tin. It can lose 2 electrons to form a cation with a +2 charge. The symbol is ${\text{Sn}}^{2+}$ .

(e)

Interpretation Introduction

Interpretation: The type of ion and symbol for the ion formed by the given element is to be identified.

Concept Introduction : There are two types of charged species, anion (negatively charged species) and cation (positively charged species). Anion is formed by the gaining of electron/s whereas a cation is formed by losing electron/s.

Answer to Problem 6LC

The ion formed by the given element is ${\text{Be}}^{2+}$ (cation).

Explanation of Solution

The given element is beryllium. It belongs to group 2 which is alkaline earth metal. It has 2 valence electrons. In order to complete the octet, it can lose 2 electrons to form a cation with a +2 charge. The symbol is ${\text{Be}}^{2+}$ .

(f)

Interpretation Introduction

Interpretation: The type of ion and symbol for the ion formed by the given element is to be identified.

Concept Introduction : There are two types of charged species, anion (negatively charged species) and cation (positively charged species). Anion is formed by the gaining of electron/s whereas a cation is formed by losing electron/s.

Answer to Problem 6LC

The ion formed by the given element is ${\text{Co}}^{3+}$ (cation).

Explanation of Solution

The given element is cobalt. It can lose 3 electrons to form a cation with a +3 charge. The symbol is ${\text{Co}}^{3+}$ .