Chapter 6, Problem 6.25P

(a)

Interpretation Introduction

Interpretation:

The equation of calcium fluoride from calcium and fluorine has to be described in terms of electronic configurations.

Answer to Problem 6.25P

The equation is,

  $\text{Ca}\left(\left[\text{Ar}\right]{\text{4s}}^{\text{2}}\right)\text{+2F}\left(\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{5}}\right)\stackrel{}{\to }{\text{Ca}}^{\text{2+}}\left(\left[\text{Ar}\right]\right){\text{+2F}}^{\text{-}}\left(\left[\text{Ne}\right]\right)$

Explanation of Solution

Calcium has a dipositive charge $\left({\text{Ca}}^{\text{2+}}\right)$ and fluorine belongs to the group 17 and forms a mononegative anion $\left({\text{F}}^{\text{-}}\right)$ known as fluoride.  The chemical formula is,

  ${\text{Ca}}^{\text{2+}}{\text{+F}}^{\text{-}}\stackrel{}{\to }{\text{CaF}}_{\text{2}}$

The electronic configuration of the species is,

Calcium: $\left[\text{Ar}\right]{\text{4s}}^{\text{2}}$

Calcium loses two electrons to attain the electronic configuration of argon.

${\text{Ca}}^{\text{2+}}\text{:}\left[\text{Ar}\right]$

Fluorine: $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{5}}$

Fluorine gains one electron to attain the electronic configuration of neon.

${\text{F}}^{\text{-}}\text{:}\left[\text{Ne}\right]$

The chemical equation is,

  $\text{Ca}\left(\left[\text{Ar}\right]{\text{4s}}^{\text{2}}\right)\text{+2F}\left(\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{5}}\right)\stackrel{}{\to }{\text{Ca}}^{\text{2+}}\left(\left[\text{Ar}\right]\right){\text{+2F}}^{\text{-}}\left(\left[\text{Ne}\right]\right)$

(b)

Interpretation Introduction

Interpretation:

The equation of strontium bromide from strontium and bromine has to be described in terms of electronic configurations.

Answer to Problem 6.25P

The equation is,

  $\text{Sr}\left(\left[\text{Kr}\right]{\text{5s}}^{\text{2}}\right)\text{+2Br}\left(\left[\text{Ar}\right]{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{5}}\right)\stackrel{}{\to }{\text{Sr}}^{\text{2+}}\left(\left[\text{Kr}\right]\right){\text{+2Br}}^{\text{-}}\left(\left[\text{Kr}\right]\right)$

Explanation of Solution

Strontium has a dipositive charge $\left({\text{Sr}}^{\text{2+}}\right)$ and bromine belongs to the group 17 and forms a mononegative anion $\left({\text{Br}}^{\text{-}}\right)$ known as bromide.  The chemical formula is,

  ${\text{Sr}}^{\text{2+}}{\text{+Br}}^{\text{-}}\stackrel{}{\to }{\text{SrBr}}_{\text{2}}$

The electronic configuration of the species is,

Strontium: $\left[\text{Kr}\right]{\text{5s}}^{\text{2}}$

Strontium loses two electrons to attain the electronic configuration of krypton.

${\text{Sr}}^{\text{2+}}\text{:}\left[\text{Kr}\right]$

Bromine: $\left[\text{Ar}\right]{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{5}}$

Bromine gains one electron to attain the electronic configuration of krypton.

${\text{Br}}^{\text{-}}\text{:}\left[\text{Kr}\right]$

The chemical equation is,

  $\text{Sr}\left(\left[\text{Kr}\right]{\text{5s}}^{\text{2}}\right)\text{+2Br}\left(\left[\text{Ar}\right]{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{5}}\right)\stackrel{}{\to }{\text{Sr}}^{\text{2+}}\left(\left[\text{Kr}\right]\right){\text{+2Br}}^{\text{-}}\left(\left[\text{Kr}\right]\right)$

(c)

Interpretation Introduction

Interpretation:

The equation of aluminium oxide from aluminium and oxygen has to be described in terms of electronic configurations.

Answer to Problem 6.25P

The equation is,

  $\text{2Al}\left(\left[\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{1}}\right)\text{+2O}\left(\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{4}}\right)\stackrel{}{\to }{\text{3Al}}^{\text{3+}}\left(\left[\text{Ne}\right]\right){\text{+3O}}^{\text{-}}\left(\left[\text{Ne}\right]\right)$

Explanation of Solution

Aluminium has a tripositive charge $\left({\text{Al}}^{\text{3+}}\right)$ and oxygen belongs to the group 16 and forms a dinegative anion $\left({\text{O}}^{\text{2-}}\right)$ known as oxide.  The chemical formula is,

  ${\text{Al}}^{\text{3+}}{\text{+O}}^{\text{2-}}\stackrel{}{\to }{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$

The electronic configuration of the species is,

Aluminium: $\left[\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{1}}$

Aluminium loses three electrons to attain the electronic configuration of neon.

${\text{Al}}^{\text{3+}}\text{:}\left[\text{Ne}\right]$

Oxygen: $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{4}}$

Oxygen gains two electrons to attain the electronic configuration of neon.

${\text{O}}^{\text{2-}}\text{:}\left[\text{Ne}\right]$

The chemical equation is,

  $\text{2Al}\left(\left[\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{1}}\right)\text{+2O}\left(\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{4}}\right)\stackrel{}{\to }{\text{3Al}}^{\text{3+}}\left(\left[\text{Ne}\right]\right){\text{+3O}}^{\text{-}}\left(\left[\text{Ne}\right]\right)$