Chapter 6, Problem 68A

(a)

Interpretation Introduction

Interpretation: To write the electronic configuration of liquid in Group 7A with a $-1$ charge.

Concept introduction:

The electron configuration of an element defines how its electrons are arranged throughout its atomic orbitals. Standard notation is used to represent atomic electron configurations, placing all electron-containing atomic subshells in sequential order. The number of electrons held by the subshell is indicated by superscript.

Answer to Problem 68A

  ${\text{Br}}^{-}:\left[\text{Ar}\right]4s²3d¹4p^6$

Explanation of Solution

Group 7A elements are fluorine, chlorine, bromine, iodine, and astatine. Out of all halogen elements, bromine is liquid. The atomic number of Bromine is 35 and after gaining 1 electron, its electronic configuration is as shown:

  ${\text{Br}}^{-}:\left[\text{Ar}\right]4s²3d¹4p^6$

(b)

Interpretation Introduction

Interpretation: To write the electronic configuration of metalloids in period 3 with a $+4$ charge.

Concept introduction:

The electron configuration of an element defines how its electrons are arranged throughout its atomic orbitals. Standard notation is used to represent atomic electron configurations, placing all electron-containing atomic subshells in sequential order. The number of electrons held by the subshell is indicated by superscript.

Answer to Problem 68A

  ${\text{Si}}^{\text{+4}}:\left[\text{Ne}\right]$

Explanation of Solution

The only metalloid in period 3 is Si. Its atomic number is 14 and its electronic configuration is as follows:

  $\text{Si}:\left[\text{Ne}\right]\text{ 3}s²3p^2$

After losing 4 electrons, its electronic configuration is as shown:

  ${\text{Si}}^{\text{+4}}:\left[\text{Ne}\right]$

(c)

Interpretation Introduction

Interpretation: To write the electronic configuration of group 6A gaseous element having $-2$ charge.

Concept introduction:

The electron configuration of an element defines how its electrons are arranged throughout its atomic orbitals. Standard notation is used to represent atomic electron configurations, placing all electron-containing atomic subshells in sequential order. The number of electrons held by the subshell is indicated by superscript.

Answer to Problem 68A

  ${\text{O}}^{2-}:\left[\text{He}\right]2s^{2}2p^6$

Explanation of Solution

Group 6A elements are oxygen, sulfur, selenium, tellurium, and polonium. Out of all group 6A elements, oxygen is gas. The atomic number of Oxygen is 8 and after gaining $-2$ electron, its electronic configuration is as shown:

  ${\text{O}}^{2-}:\left[\text{He}\right]2s^{2}2p^6$

(d)

Interpretation Introduction

Interpretation: To write the electronic configuration of alkali earth metal in period 3 with +2 charge.

Concept introduction:

The electron configuration of an element defines how its electrons are arranged throughout its atomic orbitals. Standard notation is used to represent atomic electron configurations, placing all electron-containing atomic subshells in sequential order. The number of electrons held by the subshell is indicated by superscript.

Answer to Problem 68A

  ${\text{Mg}}^{\text{2+}}:1s^{2}2s^{2}2p^6$

Explanation of Solution

The only alkali earth metal in period 3 is magnesium. Its atomic number is 12 and its electronic configuration is as shown:

  $\text{Mg}:1s^{2}2s^{2}2p^{6}3s^2$

After losing 2 electrons, its electronic configuration will be:

  ${\text{Mg}}^{\text{2+}}:1s^{2}2s^{2}2p^6$