Chapter 8, Problem 5QP

Write Lewis dot symbols for the following atoms and ions: $\left(\text{a}\right)\text{ I, }\left(\text{b}\right){\text{ I}}^{\text{-}}\text{, }\left(\text{c}\right)\text{ S, }\left(\text{d}\right){\text{ S}}^{\text{2-}}\text{, }\left(\text{e}\right)\text{ P, }\left(\text{f}\right){\text{ P}}^{\text{3-}}\text{, }\left(\text{g}\right)\text{ Na, }\left(\text{h}\right){\text{ Na}}^{\text{+}}\text{, }\left(\text{i}\right)\text{ Mg, }\left(\text{j}\right){\text{ Mg}}^{\text{2+}}\text{, }\left(\text{k}\right)\text{ Al, }\left(\text{l}\right){\text{ As}}^{\text{3+}}\text{, }\left(\text{m}\right)\text{ Pb, }\left(\text{n}\right){\text{ Pb}}^{\text{2+}}\text{.}$

Interpretation Introduction

Interpretation:

The Lewis dot symbols for the given atoms and ions are to be written.

Concept introduction:

Lewis dot symbols contain dots that give information about the valence electrons.

Lewis dot symbols show both bond and lone pairs as dots.

Lewis structures show bonds as lines and lone pairs as dots.

Answer to Problem 5QP

Solution:

a)

b)

c)

d)

e)

f)

g)

$\stackrel{·}{\text{Na}}$

h)

i)

$\stackrel{··}{\text{Mg}}$

j)

${\left[\text{Mg}\right]}^{2+}$

k)

$\underset{·}{\overset{·}{\text{Al}·}}$

l)

${\left[·\text{As}·\right]}^{3+}$

m)

$\underset{·}{\overset{·}{·\text{Pb}·}}$

n)

${\left[·\text{Pb}·\right]}^{2+}$

Explanation of Solution

a) Atom—$\text{I}$

The number of valence electrons in $\text{I}$

atom is

$=7$

So, there are $3$ lone pairs present.

So, the Lewis structure for $\text{I}$ is as follows:

b) Ion—${\text{I}}^{-}$

The number of valence electrons in ${\text{I}}^{-}$ ion is

$\begin{array}{l}=7+1\\ =8\end{array}$

So, there are $4$ lone pairs present.

So, the Lewis structure for ${\text{I}}^{-}$

is as follows:

c) Atom—$\text{S}$

The number of valence electrons in $\text{S}$

atom is

$=6$

So, there are 3 lone pairs present.

So, the Lewis structure for $\text{S}$

is as follows:

$\underset{··}{\overset{··}{\text{:S}}}$

d) Ion—${\text{S}}^{2-}$

The number of valence electrons in ${\text{S}}^{2-}$ ion is

$\begin{array}{l}=6+2\\ =8\end{array}$

So, there are $4$ lone pairs present.

So, the Lewis structure for ${\text{S}}^{2-}$

is as follows:

${\left[\underset{··}{\overset{··}{\text{:S:}}}\right]}^{2-}$

e) Atom—$\text{P}$

The number of valence electrons in $\text{P}$

atom is

$\begin{array}{l}=5+0\\ =5\end{array}$

So, there are $2$ lone pairs present.

So, the Lewis structure for $\text{P}$

is as follows:

f) Ion—${\text{P}}^{3-}$

The number of valence electrons in ${\text{P}}^{3-}$ ion is

$\begin{array}{l}=5+3\\ =8\end{array}$

So, there are $4$ lone pairs present.

So, the Lewis structure for ${\text{P}}^{3-}$

is as follows:

g) Atom—$\text{Na}$

The number of valence electrons in $\text{Na}$

atom is

$=1$

So, there are no lone pairs present.

So, the Lewis structure for $\text{Na}$

is as follows:

$\stackrel{·}{\text{Na}}$

h) Ion—${\text{Na}}^{+}$

The number of valence electrons in ${\text{Na}}^{+}$ ion is

$\begin{array}{l}=1-1\\ =0\end{array}$

So, the Lewis structure for ${\text{Na}}^{+}$

is as follows:

${\left[\text{Na}\right]}^{+}$

i) Atom—$\text{Mg}$

The number of valence electrons in $\text{Mg}$

atom is

$=2$

So, the Lewis structure for $\text{Mg}$

is as follows:

$\stackrel{··}{\text{Mg}}$

j) Ion—${\text{Mg}}^{2+}$

The number of valence electrons in ${\text{Mg}}^{2+}$ ion is

$\begin{array}{l}=2-2\\ =0\end{array}$

So, the Lewis structure for ${\text{Mg}}^{2+}$

is as follows:

${\left[\text{Mg}\right]}^{2+}$

k) Atom—$\text{Al}$

The number of valence electrons in $\text{Al}$

atom is

$=3$

So, the Lewis structure for $\text{Al}$

is as follows:

$\underset{·}{\overset{·}{\text{Al}·}}$

l) Ion—${\text{As}}^{3+}$

The number of valence electrons in ${\text{As}}^{3+}$ ion is

$=2$

So, the Lewis structure for ${\text{Al}}^{3+}$

is as follows:

${\left[·\text{As}·\right]}^{3+}$

m) Atom—$\text{Pb}$

The number of valence electrons in $\text{Pb}$

atom is

$=4$

So, the Lewis structure for $\text{Pb}$

is as follows:

$\underset{·}{\overset{·}{·\text{Pb}·}}$

n) Ion—${\text{Pb}}^{2+}$

The number of valence electrons in ${\text{Pb}}^{2+}$ ion is

$\begin{array}{l}=4-2\\ =2\end{array}$

So, the Lewis structure for ${\text{Pb}}^{2+}$

is as follows:

${\left[·\text{Pb}·\right]}^{2+}$