Chapter 6, Problem 6.26P

(a)

Interpretation Introduction

Interpretation:

The equation of gallium (III) fluoride from gallium and fluorine has to be described in terms of electronic configurations.

Answer to Problem 6.26P

The equation is,

  $\text{Ga}\left(\left[\text{Ar}\right]{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{1}}\right)\text{+3F}\left(\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{5}}\right)\stackrel{}{\to }{\text{Ga}}^{\text{2+}}\left(\left[\text{Ar}\right]{\text{3d}}^{\text{10}}\right){\text{+3F}}^{\text{-}}\left(\left[\text{Ne}\right]\right)$

Explanation of Solution

Gallium has a tripositive charge $\left({\text{Ga}}^{\text{3+}}\right)$ and fluorine belongs to the group 17 and forms a mononegative anion $\left({\text{F}}^{\text{-}}\right)$ known as fluoride.  The chemical formula is,

  ${\text{Ga}}^{\text{3+}}{\text{+F}}^{\text{-}}\stackrel{}{\to }{\text{GaF}}_{\text{3}}$

The electronic configuration of the species is,

Gallium: $\left[\text{Ga}\right]{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{1}}$

Gallium loses three electrons to attain the electronic configuration of argon.

${\text{Ga}}^{\text{3+}}\text{:}\left[\text{Ar}\right]{\text{3d}}^{\text{10}}$

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,

  $\text{Ga}\left(\left[\text{Ar}\right]{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{1}}\right)\text{+3F}\left(\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{5}}\right)\stackrel{}{\to }{\text{Ga}}^{\text{2+}}\left(\left[\text{Ar}\right]{\text{3d}}^{\text{10}}\right){\text{+3F}}^{\text{-}}\left(\left[\text{Ne}\right]\right)$

(b)

Interpretation Introduction

Interpretation:

The equation of silver chloride from silver and chloride has to be described in terms of electronic configurations.

Answer to Problem 6.26P

The equation is,

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

Explanation of Solution

Silver has a monopositive charge $\left({\text{Ag}}^{\text{+}}\right)$ and chlorine belongs to the group 17 and forms a mononegative anion $\left({\text{Cl}}^{\text{-}}\right)$ known as chloride.  The chemical formula is,

  ${\text{Ag}}^{\text{+}}{\text{+Cl}}^{\text{-}}\stackrel{}{\to }\text{AgCl}$

The electronic configuration of the species is,

Silver: $\left[\text{Kr}\right]{\text{5s}}^{\text{1}}{\text{4d}}^{\text{10}}$

Strontium loses one electron to attain the electronic configuration of krypton.

${\text{Ag}}^{\text{+}}\text{:}\left[\text{Kr}\right]{\text{4d}}^{\text{10}}$

Chlorine: $\left[\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{5}}$

Chlorine gains one electron to attain the electronic configuration of argon.

${\text{Cl}}^{\text{-}}\text{:}\left[\text{Ar}\right]$

The chemical equation is,

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

(c)

Interpretation Introduction

Interpretation:

The equation of lithium nitride from lithium and nitrogen has to be described in terms of electronic configurations.

Answer to Problem 6.26P

The equation is,

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

Explanation of Solution

Lithium has a monopositive charge $\left({\text{Li}}^{\text{+}}\right)$ and nitrogen belongs to the group 15 and forms a trinegative anion $\left({\text{N}}^{\text{3-}}\right)$ known as nitride.  The chemical formula is,

  ${\text{Li}}^{\text{+}}{\text{+N}}^{\text{3-}}\stackrel{}{\to }{\text{Li}}_{\text{3}}\text{N}$

The electronic configuration of the species is,

Lithium: $\left[\text{He}\right]{\text{2s}}^1$

Lithium loses one electron to attain the electronic configuration of helium.

${\text{Li}}^{\text{+}}\text{:}\left[\text{He}\right]$

Nitrogen: $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{3}}$

Nitrogen gains three electrons to attain the electronic configuration of neon.

${\text{N}}^{\text{3-}}\text{:}\left[\text{Ne}\right]$

The chemical equation is,

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