Q10.83CP - Chemistry: Molecular Nature Of Matter And Change

Question

Use Lewis structures to determine which two of the following are unstable: (a) ${\mathbf{SF}}_2^{}$ ; (b) ${\mathbf{SF}}_3^{}$  (c)   ${\mathbf{SF}}_4^{}$(d)  ${\mathbf{SF}}_5^{}$; (e).${\mathbf{SF}}_6^{}$

 

Step-by-Step Solution

Answer

a)  ${\mathbf{SF}}_2^{}$ has a zero formal charge on the central sulfur atom hence, it is a stable Lewis structure.

b) ${\mathbf{SF}}_3^{}$  exists as ${\mathbf{SF}}_3^{}$  in which S atom has -1 formal charge,${\mathbf{SF}}_3^{}$ because of this -1 formal charge    is quite unstable.

c)  ${\mathbf{SF}}_4^{}$ has a zero formal charge on the central S-atom hence, it is a stable Lewis structure.

d) ${\mathbf{SF}}_5^{}$ exists as  ${\mathbf{SF}}_5^{}$ in which the central S-atom has a -1 formal charge and this makes the${\mathbf{SF}}_5^{}$  unstable.

e)  ${\mathbf{SF}}_6^{}$has a zero formal charge on the central S-atom hence, it is a stable Lewis structure.

 

Step 1: Subpart (a) The formal charge of . SF 2

In${\mathbf{SF}}_2$ one sulfuratom forms two single bonds with two fluorine atoms. As sulfur has two lone pairs of electrons and fluorine has three lone pair pairs of electrons,thereforeit forms two sigma bonds and the hybridization is spand the geometry islinearas shown below.

 

The formal charge on sulfur,

$\begin{array}{c}\text{formal\hspace{0.17em}charge = }\left(\begin{array}{l}\text{valene}\\ \text{electrons}\end{array}\right)\text{ - }\left(\begin{array}{l}\text{nonbonding\hspace{0.17em}valence}\\ \text{electrons}\end{array}\right)-\frac{\text{bonding\hspace{0.17em}electrons}}{\text{2}}\\ \text{= 6 - 4 - }\frac{\text{4}}{\text{2}}\\ \text{= 0}\end{array}$ 

 Hence, it is a stable Lewis structure.

Step 2:Subpart (b) The formal charge on SF 3 .

${\mathbf{SF}}_3$ exists as ${\mathbf{SF}}_3$  ion. As sulphur has six valence electrons and fluorine has seven valence electrons, therefore sulphur forms three single covalent bonds with three fluorine atoms.

Three lone pair of electrons remains on fluorine and sulphur atom consists a negative charge. Therefore, the formal charge in ${\mathbf{SF}}_3$  is -1.

Hybridization of ${\mathbf{SF}}_3$  is ${\mathbf{sp}}^2$ and the geometry will be trigonal planar.

The formal charge on sulfur,

$\begin{array}{c}\text{formal\hspace{0.17em}charge = }\left(\begin{array}{l}\text{valene}\\ \text{electrons}\end{array}\right)\text{ - }\left(\begin{array}{l}\text{nonbonding\hspace{0.17em}valence}\\ \text{electrons}\end{array}\right)-\frac{\text{bonding\hspace{0.17em}electrons}}{\text{2}}\\ {{\text{SF}}_{\text{3}}}^{-}\text{ = 6 - 4- }\frac{\text{6}}{\text{2}}\\ \text{= -1}\end{array}$ 

 

Hence, it is anunstable Lewis structure.

Step 3: Subpart (c) The formal charge on. SF 4

In Lewis structure of,${\mathbf{SF}}_4$ sulphur is formed four sigma bonds with four fluorine atoms. Thus, the remaining valence electrons on Sulphur is two and one electron on Sulphur is due to the negative charge.

The formal charge on sulfur,

 

 $\begin{array}{c}\text{formal\hspace{0.17em}charge = }\left(\begin{array}{l}\text{valene}\\ \text{electrons}\end{array}\right)\text{ - }\left(\begin{array}{l}\text{nonbonding\hspace{0.17em}valence}\\ \text{electrons}\end{array}\right)-\frac{\text{bonding\hspace{0.17em}electrons}}{\text{2}}\\ \text{= 6 - 2 - }\frac{\text{8}}{\text{2}}\\ \text{= 0}\end{array}$

Hence, it is a stable Lewis structure.

Step 4: Subpart (d) formal cha rge on . SF 5

${\mathbf{SF}}_5$ exists as   ${\mathbf{SF}}_5$ ion. In the Lewis structure of  ${\mathbf{SF}}_5$Sulphur forms five sigma bonds with five fluorine. Therefore, there is remained only one non-bonded valence electron on Sulphur atom. The hybridization of${\mathbf{SF}}_5$  is  ${\mathbf{sp}}^3\mathbf{d}$ and the geometry will be trigonal bipyramidal.

 

The formal charge on sulfur,

 

 $\begin{array}{c}\text{formal\hspace{0.17em}charge = }\left(\begin{array}{l}\text{valene}\\ \text{electrons}\end{array}\right)\text{ - }\left(\begin{array}{l}\text{nonbonding\hspace{0.17em}valence}\\ \text{electrons}\end{array}\right)-\frac{\text{bonding\hspace{0.17em}electrons}}{\text{2}}\\ \text{= 6 - 2 - }\frac{\text{10}}{\text{2}}\\ \text{= -1}\end{array}$

Hence, it is an unstable Lewis structure.

Step 5: Subpart (e) The formal charge on . SF 6

In the Lewis structure of,${\mathbf{SF}}_6^{}$ Sulphur is formed six covalent sigma bonds with six fluorine atoms. Thus, no non-bonding electron remains on Sulphur atom. The hybridization of  ${\mathbf{SF}}_6^{}$  is${\mathbf{sp}}^3{\mathbf{d}}^2$ . The geometry will be octahedral.

The formal charge on sulfur,

 

$\begin{array}{c}\text{formal\hspace{0.17em}charge = }\left(\begin{array}{l}\text{valene}\\ \text{electrons}\end{array}\right)\text{ - }\left(\begin{array}{l}\text{nonbonding\hspace{0.17em}valence}\\ \text{electrons}\end{array}\right)-\frac{\text{bonding\hspace{0.17em}electrons}}{\text{2}}\\ \text{= 6 - 0 - }\frac{\text{12}}{\text{6}}\\ \text{= 0}\end{array}$ 

Hence, it is a stable Lewis structure.