Q10.62CP - Chemistry: Molecular Nature Of Matter And Change

Question

Draw a Lewis structure for each species:

$\begin{array}{l}\left(a\right){\mathbf{PF}}_5;\left(b\right){\mathbf{CCl}}_4;\left(c\right){\mathbf{H}}_3{\mathbf{O}}^{+};\left(d\right){\mathbf{ICl}}_3;\left(e\right){\mathbf{BeH}}_2;\left(f\right){\mathbf{PH}}_2^{-};\\ \left(g\right){\mathbf{GeBr}}_4;\left(h\right){\mathbf{CH}}_3^{-};\left(i\right){\mathbf{BCl}}_3;\left(j\right){\mathbf{BrF}}_4^{-};\left(k\right){\mathbf{XeO}}_3;\left(l\right){\mathbf{TeF}}_4.\end{array}$ 

Step-by-Step Solution

Answer

The Lewis structure for the species are:

Step 1: Definition of Lewis Structure

Lewis structure is the two dimensional structure which contains electron dot symbols of every atom present and the bonding pairs that hold them together. 

The steps involved while writing Lewis structures are as follows:  

1. Consider the molecular formula of the molecule and place the atom which has the lowest electronegativity as center.
2. Place other atoms relative to each other. 
3. Find the total number of valence electrons. 
4. Draw single bonds, subtract 2 electrons for each bond.
5. From the remaining valence electrons give each atom 8 electrons. Doing this we get the Lewis structure for the molecule. 

Step 2: Lewis Structure of Species

Let’s draw Lewis structure for all the species given one by one.

${\mathbf{PF}}_5$, Phosphorus pentafluoride

Consider molecule ${\mathbf{PF}}_5$, the central atom is P (less electronegative atom), which is surrounded by five fluorine atoms with total number of valence electrons 40.

$$\begin{gathered} \mathbf{Central}\mathbf{ }\mathbf{atom}\mathbf{=}\mathbf{P}\mathbf{;}\mathbf{ }\mathbf{Surrounding}\mathbf{ }\mathbf{atoms}\mathbf{=}\mathbf{5}\mathbf{ }\mathbf{F}\mathbf{ }\mathbf{atoms} \\ \mathbf{Valence}\mathbf{ }\mathbf{electron}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{P}\mathbf{=}\mathbf{5}\mathbf{ }\mathbf{electrons}\mathbf{;} \\ \mathbf{Valence}\mathbf{ }\mathbf{electrons}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{F}\mathbf{=}\mathbf{7}\mathbf{ }\mathbf{electrons}\mathbf{;} \\ \mathbf{Total}\mathbf{ }\mathbf{no}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{valence}\mathbf{ }\mathbf{electrons}\mathbf{=}\mathbf{5}\mathbf{+}\mathbf{5}\mathbf{x}\mathbf{7}\mathbf{=}\mathbf{40}\mathbf{ }\mathbf{electrons} \end{gathered}$$

P bonds with five F atoms resulting in the formation of bonds. 

The formula for formal charge is:   

$\begin{array}{c}\mathbf{Formal}\mathbf{ }\mathbf{charge}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{an}\mathbf{ }\mathbf{atom}\mathbf{=}\mathbf{no}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{valence}\mathbf{ }{\mathbf{e}}^{-}\mathbf{–}\mathbf{[}\mathbf{no}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{unshared}\mathbf{ }\text{\hspace{0.17em}}\mathbf{valence}\mathbf{ }{\mathbf{e}}^{-}\\ \mathbf{+}\frac{1}{2}\mathbf{no}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{shared}\mathbf{ }\mathbf{valence}\mathbf{ }{\mathbf{e}}^{-}\mathbf{]}\end{array}$

For the molecule ${\mathbf{PF}}_5$ the formal charge on each atom would be,

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{P}=5{\mathrm{e}}^{-} -\left(0{\mathrm{e}}^{- }+\frac{1}{2}\mathrm{x}10{\mathrm{e}}^{-}\right) \\ =5{\mathrm{e}}^{-} -5{\mathrm{e}}^{-}\\ =0\end{array}$

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{F}=7{\mathrm{e}}^{-} -\left(6{\mathrm{e}}^{-} +\frac{1}{2}\mathrm{x}2{\mathrm{e}}^{-}\right)\\ =7{\mathrm{e}}^{-} -7{\mathrm{e}}^{-}\\ =0\end{array}$

Therefore, formal charge on P and F is 0.

The Lewis structure of ${\mathbf{PF}}_5$ would be as below:

${\mathbf{CCl}}_4$, Carbon tetrachloride

Consider molecule ${\mathbf{CCl}}_4$, the central atom is C (less electronegative atom), which is surrounded by four chlorine atoms with total number of valence electrons 32. .

$$\begin{gathered} \mathbf{Central}\mathbf{ }\mathbf{atom}\mathbf{=}\mathbf{P}\mathbf{;}\mathbf{ }\mathbf{Surrounding}\mathbf{ }\mathbf{atoms}\mathbf{=}\mathbf{4}\mathbf{ }\mathbf{F}\mathbf{ }\mathbf{atoms} \\ \mathbf{Valence}\mathbf{ }\mathbf{electron}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{=}\mathbf{4}\mathbf{ }\mathbf{electrons}\mathbf{;} \\ \mathbf{Valence}\mathbf{ }\mathbf{electrons}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Cl}\mathbf{=}\mathbf{7}\mathbf{ }\mathbf{electrons}\mathbf{;} \\ \mathbf{Total}\mathbf{ }\mathbf{no}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{valence}\mathbf{ }\mathbf{electrons}\mathbf{=}\mathbf{4}\mathbf{+}\mathbf{4}\mathbf{x}\mathbf{7}\mathbf{=}\mathbf{32}\mathbf{ }\mathbf{electrons} \end{gathered}$$

C bonds with five Cl atoms resulting in the formation of bonds. 

The formula for formal charge is:   

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{of} \mathrm{an} \mathrm{atom}=\mathrm{no}. \mathrm{of} \mathrm{valence} {\mathrm{e}}^{-}–[\mathrm{no}. \mathrm{of} \mathrm{unshared} \text{\hspace{0.17em}}\mathrm{valence} {\mathrm{e}}^{-}\\ +\frac{1}{2}\mathrm{no}. \mathrm{of} \mathrm{shared} \mathrm{valence} {\mathrm{e}}^{-}]\end{array}$

For the molecule ${\mathbf{CCl}}_4$ the formal charge on each atom would be,

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{C}=4{\mathrm{e}}^{-} -\left(0{\mathrm{e}}^{- }+\frac{1}{2}\mathrm{x}8{\mathrm{e}}^{-}\right)\\ =4{\mathrm{e}}^{-} -4{\mathrm{e}}^{-}\\ =0\end{array}$

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{Cl}=7{\mathrm{e}}^{-} -\left(6{\mathrm{e}}^{-} +\frac{1}{2}\mathrm{x}2{\mathrm{e}}^{-}\right)\\ =7{\mathrm{e}}^{-} -7{\mathrm{e}}^{-}\\ =0\end{array}$

Therefore, formal charge on C and Cl is 0. 

The Lewis structure of ${\mathbf{CCl}}_4$ would be as below:

${\mathbf{H}}_3{\mathbf{O}}^{+}$, Hydronium ion

Consider molecule ${\mathbf{H}}_3{\mathbf{O}}^{+}$, the central atom is O. Even though H has least electronegative, it does not have more bonding sites. Hence, O having more bonding sites than H is taken as the central atom. Here O is surrounded by three H atoms with total number of valence electrons 8.

Central atom = O; 

Surrounding atoms = 3H atoms

Valence electron of O = 6 electrons; 

Valence electrons of H = 1 electron; 

Total no. of valence electrons$=6+3\times 1=9 \mathrm{electrons}$.

O bonds with three H atoms resulting in the formation of bonds. But as there is + charge, we have to remove one electron, then the total no. of valence electrons would be = 9-1 = 8 electrons. 

The formula for formal charge is:   

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{of} \mathrm{an} \mathrm{atom}=\mathrm{no}.{\mathrm{ofvalencee}}^{-}–[\mathrm{no}. \mathrm{of} \mathrm{unshared} \text{\hspace{0.17em}}\mathrm{valence} {\mathrm{e}}^{-}\\ +\frac{1}{2}\mathrm{no}. \mathrm{of} \mathrm{shared} \mathrm{valence} {\mathrm{e}}^{-}]\end{array}$

For the molecule ${\mathbf{H}}_3{\mathbf{O}}^{+}$ the formal charge on each atom would be,

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{O}=6{\mathrm{e}}^{-} -\left(2{\mathrm{e}}^{- }+\frac{1}{2}\mathrm{x}6{\mathrm{e}}^{-}\right)\\ =6{\mathrm{e}}^{-} -5{\mathrm{e}}^{-}\\ =+1\end{array}$

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{H}=1{\mathrm{e}}^{-} -\left(0{\mathrm{e}}^{-} +\frac{1}{2}\mathrm{x}2{\mathrm{e}}^{-}\right)\\ =1{\mathrm{e}}^{-} -1{\mathrm{e}}^{-}\\ =0\end{array}$

Therefore, formal charge on O is +1 and H is 0. 

The Lewis structure of ${\mathbf{H}}_3{\mathbf{O}}^{+}$ would be as below:

${\mathbf{ICl}}_3$, Iodine trichloride

Consider molecule ${\mathbf{ICl}}_3$, the central atom is iodine (less electronegative atom), which is surrounded by three chlorine atoms with total number of valence electrons 28.

$$\begin{gathered} \mathbf{Central}\mathbf{ }\mathbf{atom}\mathbf{=}\mathbf{I}\mathbf{;}\mathbf{ }\mathbf{Surrounding}\mathbf{ }\mathbf{atoms}\mathbf{=}\mathbf{3}\mathbf{ }\mathbf{Cl}\mathbf{ }\mathbf{atoms} \\ \mathbf{Valence}\mathbf{ }\mathbf{electron}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{=}\mathbf{7}\mathbf{ }\mathbf{electrons}\mathbf{;} \\ \mathbf{Valence}\mathbf{ }\mathbf{electrons}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Cl}\mathbf{=}\mathbf{7}\mathbf{ }\mathbf{electrons}\mathbf{;} \\ \mathbf{Total}\mathbf{ }\mathbf{no}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{valence}\mathbf{ }\mathbf{electrons}\mathbf{=}\mathbf{7}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{7}\mathbf{=}\mathbf{28}\mathbf{ }\mathbf{electrons} \end{gathered}$$

Iodine bonds with three Cl atoms resulting in the formation of bonds.

The formula for formal charge is:   

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{of} \mathrm{an} \mathrm{atom}=\mathrm{no}. \mathrm{of} \mathrm{valence} {\mathrm{e}}^{-}–[\mathrm{no}. \mathrm{of} \mathrm{unshared} \text{\hspace{0.17em}}\mathrm{valence} {\mathrm{e}}^{-}\\ +\frac{1}{2}\mathrm{no}. \mathrm{of} \mathrm{shared} \mathrm{valence} {\mathrm{e}}^{-}]\end{array}$

For the molecule ${\mathbf{ICl}}_3$ the formal charge on each atom would be,

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{I}=7{\mathrm{e}}^{-} -\left(4{\mathrm{e}}^{- }+\frac{}{}\mathrm{x}6{\mathrm{e}}^{-}\right)\\ =7{\mathrm{e}}^{-} -7{\mathrm{e}}^{-}\\ =0\end{array}$

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{Cl}=7{\mathrm{e}}^{-} -\left(6{\mathrm{e}}^{-} +\frac{}{}\mathrm{x}2{\mathrm{e}}^{-}\right)\\ =7{\mathrm{e}}^{-} -7{\mathrm{e}}^{-}\\ =0\end{array}$

Therefore, formal charge on I and Cl is 0. 

The Lewis structure of ${\mathbf{ICl}}_3$ would be as below:

${\mathbf{BeH}}_2$

Consider molecule ${\mathbf{BeH}}_2$, the central atom is Be.   Even though H has least electronegative, it does not have more bonding sites. Hence, Be having more bonding sites than H is taken as the central atom. Here Be is surrounded by two H atoms with total number of valence electrons 4.

$$\begin{gathered} \mathbf{Central}\mathbf{ }\mathbf{atom}\mathbf{=}\mathbf{Be}\mathbf{;}\mathbf{ }\mathbf{Surrounding}\mathbf{ }\mathbf{atoms}\mathbf{=}\mathbf{2}\mathbf{H}\mathbf{ }\mathbf{atoms} \\ \mathbf{Valence}\mathbf{ }\mathbf{electron}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Be}\mathbf{=}\mathbf{2}\mathbf{ }\mathbf{electrons}\mathbf{;} \\ \mathbf{Valence}\mathbf{ }\mathbf{electrons}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{ }\mathbf{electron}\mathbf{;} \\ \mathbf{Total}\mathbf{ }\mathbf{no}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{valence}\mathbf{ }\mathbf{electrons}\mathbf{=}\mathbf{2}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{1}\mathbf{=}\mathbf{4}\mathbf{ }\mathbf{electrons} \end{gathered}$$

Be bonds with two H atoms resulting in the formation of bonds.  

The formula for formal charge is:   

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{of} \mathrm{an} \mathrm{atom}=\mathrm{no}. \mathrm{of} \mathrm{valence} e^{-}–[\mathrm{no}. \mathrm{of} \mathrm{unshared} \text{\hspace{0.17em}}\mathrm{valence} e^{-}\\ +\frac{1}{2}\mathrm{no}. \mathrm{of} \mathrm{shared} \mathrm{valence} e^{-}]\end{array}$

For the molecule ${\mathbf{BeH}}_2$ the formal charge on each atom would be,

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{Be}=2{\mathrm{e}}^{-} -\left(0{\mathrm{e}}^{- }+\frac{1}{2}\mathrm{x}4{\mathrm{e}}^{-}\right)\\ =2{\mathrm{e}}^{-} -2{\mathrm{e}}^{-}\\ =0\end{array}$

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{H}=1{\mathrm{e}}^{-} -\left(0{\mathrm{e}}^{-} +\frac{1}{2}\mathrm{x}2{\mathrm{e}}^{-}\right)\\ =1{\mathrm{e}}^{-} -1{\mathrm{e}}^{-}\\ =0\end{array}$

Therefore, formal charge on Be and H is 0. 

The Lewis structure of ${\mathbf{BeH}}_2$ would be as below:

${\mathbf{PH}}_2^{-}$, Phosphanide

Consider molecule ${\mathbf{PH}}_2^{-}$, the central atom is P.   Even though H has least electronegative, it does not have more bonding sites. Hence, P having more bonding sites than H is taken as the central atom. Here P is surrounded by two H atoms with total number of valence electrons 8.

$$\begin{gathered} \mathbf{Central}\mathbf{ }\mathbf{atom}\mathbf{=}\mathbf{P}\mathbf{;}\mathbf{ }\mathbf{Surrounding}\mathbf{ }\mathbf{atoms}\mathbf{=}\mathbf{2}\mathbf{ }\mathbf{H}\mathbf{ }\mathbf{atoms} \\ \mathbf{Valence}\mathbf{ }\mathbf{electron}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{P}\mathbf{=}\mathbf{5}\mathbf{ }\mathbf{electrons}\mathbf{;} \\ \mathbf{Valence}\mathbf{ }\mathbf{electrons}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{electron}\mathbf{;} \\ \mathbf{Total}\mathbf{ }\mathbf{no}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{valence}\mathbf{ }\mathbf{electrons}\mathbf{=}\mathbf{5}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{1}\mathbf{=}\mathbf{7}\mathbf{ }\mathbf{electrons} \end{gathered}$$

Due to negative charge, one electron is added. 

$\mathbf{Total}\mathbf{ }\mathbf{no}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{valence}\mathbf{ }\mathbf{electrons}\mathbf{=}\mathbf{7}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{8}\mathbf{ }\mathbf{electrons}$

P bonds with two H atoms resulting in the formation of bonds. 

The formula for formal charge is: 

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{of} \mathrm{an} \mathrm{atom}=\mathrm{no}. \mathrm{of} \mathrm{valence} {\mathrm{e}}^{-}–[\mathrm{no}.\mathrm{of} \mathrm{unshared} \text{\hspace{0.17em}}{\mathrm{valencee}}^{-}\\ +\frac{1}{2}\mathrm{no}. \mathrm{of} \mathrm{shared} \mathrm{valence} {\mathrm{e}}^{-}]\end{array}$

For the molecule ${\mathbf{PH}}_2^{-}$ the formal charge on each atom would be,

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{P}=5{\mathrm{e}}^{-} -\left(4{\mathrm{e}}^{- }+\frac{1}{2}\mathrm{x}4{\mathrm{e}}^{-}\right)\\ =5{\mathrm{e}}^{-} -6{\mathrm{e}}^{-}\\ =-1\end{array}$

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{H}=1{\mathrm{e}}^{-} -\left(0{\mathrm{e}}^{-} +\frac{1}{2}\mathrm{x}2{\mathrm{e}}^{-}\right)\\ =1{\mathrm{e}}^{-} -1{\mathrm{e}}^{-}\\ =0\end{array}$

Therefore, formal charge on Pis -1and H is 0. 

The, Lewis structure of ${\mathbf{PH}}_2^{-}$ would be as below:

${\mathbf{GeBr}}_4$, Germanium tetrabromide

Consider molecule ${\mathbf{GeBr}}_4$, the central atom is Ge (less electronegative atom), which is surrounded by four Br atoms with total number of valence electrons 32.

$$\begin{gathered} \mathbf{Central}\mathbf{ }\mathbf{atom}\mathbf{=}\mathbf{Ge}\mathbf{;}\mathbf{ }\mathbf{Surrounding}\mathbf{ }\mathbf{atoms}\mathbf{=}\mathbf{4}\mathbf{ }\mathbf{Br}\mathbf{ }\mathbf{atoms} \\ \mathbf{Valence}\mathbf{ }\mathbf{electron}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Ge}\mathbf{=}\mathbf{4}\mathbf{ }\mathbf{electrons}\mathbf{;} \\ \mathbf{Valence}\mathbf{ }\mathbf{electrons}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Br}\mathbf{=}\mathbf{7}\mathbf{ }\mathbf{electrons}\mathbf{;} \\ \mathbf{Total}\mathbf{ }\mathbf{no}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{valence}\mathbf{ }\mathbf{electrons}\mathbf{=}\mathbf{4}\mathbf{+}\mathbf{4}\mathbf{x}\mathbf{7}\mathbf{=}\mathbf{32}\mathbf{ }\mathbf{electrons} \end{gathered}$$

Germanium bonds with four Br atoms resulting in the formation of bonds.

The formula for formal charge is: 

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{of} \mathrm{an} \mathrm{atom}=\mathrm{no}. \mathrm{of} \mathrm{valence} {\mathrm{e}}^{-}–[\mathrm{no}. \mathrm{of} \mathrm{unshared} \text{\hspace{0.17em}}{\mathrm{valencee}}^{-}\\ +\frac{1}{2}\mathrm{no}.\mathrm{of} \mathrm{shared} \mathrm{valence} {\mathrm{e}}^{-}]\end{array}$

For the molecule ${\mathbf{GeBr}}_4$ the formal charge on each atom would be,

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{Ge}=4{\mathrm{e}}^{-} -\left(0{\mathrm{e}}^{- }+\frac{1}{2}\mathrm{x}8{\mathrm{e}}^{-}\right)\\ =4{\mathrm{e}}^{-} -4{\mathrm{e}}^{-}\\ =0\end{array}$

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{Br}=7{\mathrm{e}}^{-} -\left(6{\mathrm{e}}^{-} +\frac{1}{2}\mathrm{x}2{\mathrm{e}}^{-}\right)\\ =7{\mathrm{e}}^{-} -7{\mathrm{e}}^{-}\\ =0\end{array}$

 Therefore, formal charge on Ge and Br is 0. 

The Lewis structure of ${\mathbf{GeBr}}_4$ would be as below:

${\mathbf{\text{CH}}}_{\mathbf{3}}^{\mathbf{-}}$, Methyl anion

Consider molecule ${\mathbf{\text{CH}}}_{\mathbf{3}}^{\mathbf{-}}$, the central atom is C (less electronegative atom), which is surrounded by three Hydrogen atoms with total number of valence electrons 8.

$$\begin{gathered} \mathbf{Central}\mathbf{ }\mathbf{atom}\mathbf{=}\mathbf{C}\mathbf{;}\mathbf{ }\mathbf{Surrounding}\mathbf{ }\mathbf{atoms}\mathbf{=}\mathbf{3}\mathbf{ }\mathbf{H}\mathbf{ }\mathbf{atoms} \\ \mathbf{Valence}\mathbf{ }\mathbf{electron}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{=}\mathbf{4}\mathbf{ }\mathbf{electrons}\mathbf{;} \\ \mathbf{Valence}\mathbf{ }\mathbf{electrons}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{ }\mathbf{electron}\mathbf{;} \\ \mathbf{Negative}\mathbf{ }\mathbf{charge}\mathbf{=}\mathbf{1} \\ \mathbf{Total}\mathbf{ }\mathbf{no}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{valence}\mathbf{ }\mathbf{electrons}\mathbf{=}\mathbf{4}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{1}\mathbf{=}\mathbf{7}\mathbf{ }\mathbf{electrons} \end{gathered}$$

But as there is negative charge, we have to add one electron, then the total no. of valence electrons would be,$\mathbf{7}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{8}$ electrons.

Carbon bonds with three H atoms resulting in the formation of bonds.

The formula for formal charge is:  

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{of} \mathrm{an} \mathrm{atom}=\mathrm{no}. \mathrm{of} \mathrm{valence} {\mathrm{e}}^{-}–[\mathrm{no}. \mathrm{of} \mathrm{unshared} \text{\hspace{0.17em}}\mathrm{valence} {\mathrm{e}}^{-}\\ +\frac{1}{2}\mathrm{no}.\mathrm{of} \mathrm{shared} \mathrm{valence} {\mathrm{e}}^{-}]\end{array}$

For the molecule ${\mathbf{\text{CH}}}_{\mathbf{3}}^{\mathbf{-}}$ the formal charge on each atom would be,

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{C}=4{\mathrm{e}}^{-} -\left(2{\mathrm{e}}^{- }+\frac{1}{2}\mathrm{x}6{\mathrm{e}}^{-}\right)\\ =4{\mathrm{e}}^{-} -5{\mathrm{e}}^{-}\\ =-1\end{array}$

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{H}=1{\mathrm{e}}^{-} -\left(0 +\frac{1}{2}\mathrm{x}2{\mathrm{e}}^{-}\right)\\ =1{\mathrm{e}}^{-} -1{\mathrm{e}}^{-}\\ =0\end{array}$

Therefore, formal charge on Cis -1 and H is 0. 

The, the Lewis structure of ${\mathbf{\text{CH}}}_{\mathbf{3}}^{\mathbf{-}}$ would be as below:  

${\mathbf{\text{BCl}}}_{\mathbf{3}}$, Boron trichloride

Consider molecule ${\mathbf{\text{BCl}}}_{\mathbf{3}}$, the central atom is boron (less electronegative atom), which is surrounded by three chlorine atoms with total number of valence electrons 24.

$$\begin{gathered} \mathbf{Central}\mathbf{ }\mathbf{atom}\mathbf{=}\mathbf{B}\mathbf{;}\mathbf{ }\mathbf{Surrounding}\mathbf{ }\mathbf{atoms}\mathbf{=}\mathbf{3}\mathbf{ }\mathbf{Cl}\mathbf{ }\mathbf{atoms} \\ \mathbf{Valence}\mathbf{ }\mathbf{electrons}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{B}\mathbf{=}\mathbf{3}\mathbf{ }\mathbf{electrons}\mathbf{;} \\ \mathbf{Valence}\mathbf{ }\mathbf{electrons}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Cl}\mathbf{=}\mathbf{7}\mathbf{ }\mathbf{electrons}\mathbf{;} \\ \mathbf{Total}\mathbf{ }\mathbf{no}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{valence}\mathbf{ }\mathbf{electrons}\mathbf{=}\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{7}\mathbf{=}\mathbf{24}\mathbf{ }\mathbf{electrons} \end{gathered}$$

Boron bonds with three Cl atoms resulting in the formation of bonds.

The formula for formal charge is:  

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{of} \mathrm{an} \mathrm{atom}=\mathrm{no}. \mathrm{of} \mathrm{valence} {\mathrm{e}}^{-}–[\mathrm{no}. \mathrm{of} \mathrm{unshared} \text{\hspace{0.17em}}\mathrm{valence} {\mathrm{e}}^{-}\\ +\frac{1}{2}\mathrm{no}.\mathrm{of} \mathrm{shared} \mathrm{valence} {\mathrm{e}}^{-}]\end{array}$

For the molecule ${\mathbf{BCl}}_{\mathbf{3}}$ the formal charge on each atom would be,

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{B}=3{\mathrm{e}}^{-} -\left(0{\mathrm{e}}^{- }+\frac{1}{2}\mathrm{x}6{\mathrm{e}}^{-}\right)\\ =3{\mathrm{e}}^{-} -3{\mathrm{e}}^{-}\\ =0\end{array}$

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{Cl}=7{\mathrm{e}}^{-} -\left(6{\mathrm{e}}^{-} +\frac{1}{2}\mathrm{x}2{\mathrm{e}}^{-}\right)\\ =7{\mathrm{e}}^{-} -7{\mathrm{e}}^{-}\\ =0\end{array}$

Therefore, formal charge on B and Cl is 0. 

The Lewis structure of ${\mathbf{BCl}}_{\mathbf{3}}$ would be as below:  

${\mathbf{BrF}}_{\mathbf{4}}^{\mathbf{-}}$, Bromine tetrafluoride anion

Consider molecule ${\mathbf{BrF}}_{\mathbf{4}}^{\mathbf{-}}$, the central atom is bromine (less electronegative atom), which is surrounded by four fluorine atoms with total number of valence electrons 36.

$$\begin{gathered} \mathbf{Central}\mathbf{ }\mathbf{atom}\mathbf{=}\mathbf{Br}\mathbf{;}\mathbf{ }\mathbf{Surrounding}\mathbf{ }\mathbf{atoms}\mathbf{=}\mathbf{4}\mathbf{ }\mathbf{F}\mathbf{ }\mathbf{atoms} \\ \mathbf{Valence}\mathbf{ }\mathbf{electron}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Br}\mathbf{=}\mathbf{7}\mathbf{electrons}\mathbf{;} \\ \mathbf{Valence}\mathbf{ }\mathbf{electrons}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{F}\mathbf{=}\mathbf{7}\mathbf{ }\mathbf{electrons}\mathbf{;} \\ \mathbf{Total}\mathbf{ }\mathbf{no}\mathbf{.}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{valence}\mathbf{ }\mathbf{electrons}\mathbf{=}\mathbf{7}\mathbf{+}\mathbf{4}\mathbf{x}\mathbf{7}\mathbf{=}\mathbf{35}\mathbf{ }\mathbf{electrons} \end{gathered}$$

But as there is negative charge, we have to add one electron, then the total no. of valence electrons would be $\mathbf{=}\mathbf{36}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{36}$ electrons.

Bromine bonds with four F atoms resulting in the formation of bonds.

The formula for formal charge is:  

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{of} \mathrm{an} \mathrm{atom}=\mathrm{no}. \mathrm{of} \mathrm{valence} {\mathrm{e}}^{-}–[\mathrm{no}. \mathrm{of} \mathrm{unshared} \text{\hspace{0.17em}}\mathrm{valence} {\mathrm{e}}^{-}\\ +\frac{1}{2}\mathrm{no}.\mathrm{of} \mathrm{shared} \mathrm{valence} {\mathrm{e}}^{-}]\end{array}$

For the molecule ${\mathbf{BrF}}_{\mathbf{4}}^{\mathbf{-}}$ the formal charge on each atom would be,

$\begin{array}{c}\mathbf{Formal}\mathbf{ }\mathbf{charge}\mathbf{ }\mathbf{on}\mathbf{ }\mathbf{Br}\mathbf{=}\mathbf{7}{\mathbf{e}}^{-}\mathbf{ }\mathbf{-}\left(\mathbf{4}{\mathbf{e}}^{- }\mathbf{+}\frac{1}{2}\mathbf{x}\mathbf{8}{\mathbf{e}}^{-}\right)\\ \mathbf{=}\mathbf{7}{\mathbf{e}}^{-}\mathbf{ }\mathbf{-}\mathbf{8}{\mathbf{e}}^{-}\\ =-1\end{array}$

$\begin{array}{c}\mathbf{Formal}\mathbf{ }\mathbf{charge}\mathbf{ }\mathbf{on}\mathbf{ }\mathbf{F}\mathbf{=}\mathbf{7}{\mathbf{e}}^{-}\mathbf{ }\mathbf{-}\left(\mathbf{6}{\mathbf{e}}^{-}\mathbf{ }\mathbf{+}\frac{1}{2}\mathbf{x}\mathbf{2}{\mathbf{e}}^{-}\right)\\ \mathbf{=}\mathbf{7}{\mathbf{e}}^{-}\mathbf{ }\mathbf{-}\mathbf{7}{\mathbf{e}}^{-}\\ =0\end{array}$

Therefore, formal charge on Br is -1 and F is 0. 

The Lewis structure of ${\mathbf{BrF}}_{\mathbf{4}}^{\mathbf{-}}$ would be as below:

XeO₃ , Xenon trioxide 

Consider molecule XeO₃, the central atom is xenon (less electronegative atom), which is surrounded by three oxygen atoms with total number of valence electrons 26.; .

Central atom = Xe; Surrounding atoms = 3 O atoms

Valence electron of Xe = 8 electrons; 

Valence electron of O = 6 electrons;

Total no. of electrons = 8+3 x 6 = 26 electrons 

Xenon bonds with three O atoms resulting in the formation of bonds.

The formula for formal charge is:  

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{of} \mathrm{an} \mathrm{atom}=\mathrm{no}. \mathrm{of} \mathrm{valence} {\mathrm{e}}^{-}–[\mathrm{no}. \mathrm{of} \mathrm{unshared} \text{\hspace{0.17em}}\mathrm{valence} {\mathrm{e}}^{-}\\ +\frac{1}{2}\mathrm{no}.\mathrm{of} \mathrm{shared} \mathrm{valence} {\mathrm{e}}^{-}]\end{array}$

For the molecule XeO₃ the formal charge on each atom would be, 

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{Xe}=8{\mathrm{e}}^{-} -\left(2{\mathrm{e}}^{- }+\frac{1}{2}\mathrm{x}6{\mathrm{e}}^{-}\right)\\ =8{\mathrm{e}}^{-} -5{\mathrm{e}}^{-}\\ =+3\end{array}$

$\begin{array}{c}\text{Formal charge on O }={\text{ 6e}}^{\text{-}}\text{ - }\left({\text{6e}}^{\text{-}}\text{ + }\frac{\text{1}}{\text{2}}{\text{ x 2e}}^{\text{-}}\right)\\ {\text{= 6e}}^{\text{-}}{\text{ - 7e}}^{\text{-}}\\ \text{= -1}\end{array}$

Therefore, formal charge on Xe is +3 and O is -1. 

The Lewis structure of XeO₃ would be as below: 

 

TeF₄, Tellurium tetrafluoride

Consider molecule TeF₄ , the central atom is tellurium (less electronegative atom), which is surrounded by four fluorine atoms with total number of valence electrons 34.

Central atom = Te; Surrounding atoms = 4 F atoms

Valence electron of Te = 6 electrons; 

Valence electron of F = 7 electrons;

Total no. of electrons = 6+4 x 7 = 34 electrons

Tellurium bonds with four F atoms resulting in the formation of bonds.

The formula for formal charge is:  

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{of} \mathrm{an} \mathrm{atom}=\mathrm{no}. \mathrm{of} \mathrm{valence} {\mathrm{e}}^{-}–[\mathrm{no}. \mathrm{of} \mathrm{unshared} \text{\hspace{0.17em}}\mathrm{valence} {\mathrm{e}}^{-}\\ +\frac{1}{2}\mathrm{no}.\mathrm{of} \mathrm{shared} \mathrm{valence} {\mathrm{e}}^{-}]\end{array}$

For the molecule TeF₄ the formal charge on each atom would be, 

$\begin{array}{c}\mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{Te}=6{\mathrm{e}}^{-} -\left(2{\mathrm{e}}^{- }+\frac{1}{2}\mathrm{x}8{\mathrm{e}}^{-}\right)\\ =6{\mathrm{e}}^{-} -6{\mathrm{e}}^{-}\\ =0\\ \mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{F}=7{\mathrm{e}}^{-} -\left(6{\mathrm{e}}^{-} +\frac{1}{2}\mathrm{x}2{\mathrm{e}}^{-}\right) \\ =7{\mathrm{e}}^{-} -7{\mathrm{e}}^{-}\\ =0\end{array}$

Therefore, formal charge on Te and F is 0. 

The Lewis structure of TeF₄ would be as below: