Q25 E - Chemistry (OpenStax)

Question

 Write the Lewis symbols of the ions in each of the following ionic compounds and the Lewis symbols of the atom from which they are formed: (a) \({\rm{MgS}}\) (b) \({\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\) (c) \({\rm{GaC}}{{\rm{l}}_{\rm{3}}}\) (d) \({{\rm{K}}_{\rm{2}}}{\rm{O}}\) (e) \({\rm{L}}{{\rm{i}}_{\rm{3}}}{\rm{N}}\) (f) \({\rm{KF}}\) .

Step-by-Step Solution

Answer

The Lewis symbol for each ion and atom are:

a) 

b) 

c) 

d) 

e) 

f) 

Define Chemical Bonding

Lewis structures show each atom along with its position in the structure of the molecule by using its chemical informations.

Explanation

Certain procedures must be performed while drawing the Lewis structure: -

(1) Count the number of valence electrons in total.

(2) If it's a cation, deduct the charges from the valence electrons; if it's an anion, add the charges to the valence electrons.

(3) Finally, we create a skeleto that depicts the atom's total electrons as well as charges (If there is any).

Writing the Lewis symbol

(a)

In the case of: \({\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}\).

The valence electrons total is: \({\rm{2}}\).

The charge is: \({\rm{(2 + )}}\).

As a result, total electrons will be shown in the Lewis dot structure as: \({\rm{0}}\).

Then, In the case of: \({{\rm{S}}^{{\rm{2 - }}}}\).

The valence electrons total is: \({\rm{6}}\).

The charge is: \({\rm{(2 - )}}\).

As a result, total electrons will be shown in the Lewis dot structure as: \({\rm{8}}\).

Writing the Lewis symbol

(b)

In the case of: \({\rm{A}}{{\rm{l}}^{{\rm{3 + }}}}\).

The valence electrons total is: \({\rm{3}}\).

The charge is: \({\rm{(3 + )}}\).

As a result, total electrons will be shown in the Lewis dot structure as: \({\rm{0}}\).

Then, In the case of: \({{\rm{O}}^{{\rm{2 - }}}}\).

The valence electrons total is: \({\rm{6}}\).

The charge is: \({\rm{(2 - )}}\).

As a result, total electrons will be shown in the Lewis dot structure as: \({\rm{8}}\).

Writing the Lewis symbol

(c)

In the case of: \({\rm{G}}{{\rm{a}}^{{\rm{3 + }}}}\).

The valence electrons total is: \({\rm{3}}\).

The charge is: \({\rm{(3 + )}}\).

As a result, total electrons will be shown in the Lewis dot structure as: \({\rm{0}}\).

Then, In the case of: \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\).

The valence electrons total is: \({\rm{7}}\).

The charge is: \({\rm{( - )}}\).

As a result, total electrons will be shown in the Lewis dot structure as: \({\rm{8}}\).

Writing the Lewis symbol

(d)

In the case of: \({{\rm{K}}^{\rm{ + }}}\).

The valence electrons total is: \({\rm{1}}\).

The charge is: \({\rm{( + )}}\).

As a result, total electrons will be shown in the Lewis dot structure as: \({\rm{0}}\).

Then, In the case of: \({{\rm{O}}^{{\rm{2 - }}}}\).

The valence electrons total is: \({\rm{6}}\).

The charge is: \({\rm{(2 - )}}\).

As a result, total electrons will be shown in the Lewis dot structure as: \({\rm{8}}\).

Writing the Lewis symbol

(e)

In the case of: \({\rm{L}}{{\rm{i}}^{\rm{ + }}}\).

The valence electrons total is: \({\rm{1}}\).

The charge is: \({\rm{( + )}}\).

As a result, total electrons will be shown in the Lewis dot structure as: \({\rm{0}}\).

Then, In the case of: \({{\rm{N}}^{{\rm{3 - }}}}\).

The valence electrons total is: \({\rm{5}}\).

The charge is: \({\rm{(3 - )}}\).

As a result, total electrons will be shown in the Lewis dot structure as: \({\rm{8}}\).

Writing the Lewis symbol

(f)

In the case of: \({{\rm{K}}^{\rm{ + }}}\).

The valence electrons total is: \({\rm{1}}\).

The charge is: \({\rm{( + )}}\).

As a result, total electrons will be shown in the Lewis dot structure as: \({\rm{0}}\).

Then, In the case of: \({{\rm{F}}^{\rm{ - }}}\).

The valence electrons total is: \({\rm{7}}\).

The charge is: \({\rm{( - )}}\).

As a result, total electrons will be shown in the Lewis dot structure as: \({\rm{8}}\).