Q14.145CP - Chemistry: Molecular Nature Of Matter And Change

Question

Thionyl chloride (SOCl₂) is a sulfur oxohalide used industrially to dehydrate metal halide hydrates. 

(a) Write a balanced equation for its reaction with magnesium chloride hexahydrate, in which SO₂  and HCl form along with the metal halide. 

(b) Draw a Lewis structure of (SOCl₂)  with minimal formal charges.

Step-by-Step Solution

Answer

1. The balanced equation for its reaction with magnesium chloride hexahydrate, including HCl, is: ${\mathrm{MgCl}}_2.6{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)}+6{\mathrm{SOCl}}_{2\left(\mathrm{l}\right)}\to {\mathrm{MgCl}}_{2\left(\mathrm{s}\right)}+6\mathrm{S}{{\mathrm{O}}_2}_{\left(\mathrm{g}\right)}+12{\mathrm{HCl}}_{\left(\mathrm{g}\right)}$

2. The Lewis structure of SOCl₂, with minimal formal charges, is as follows:

Step 1: (a) Write a balanced equation for its reaction with magnesium chloride hexahydrate, including and HCl.

When thionyl chloride SOCl₂ is introduced to magnesium chloride hexahydrate $\left({\mathrm{MgCl}}_2+6{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)}\right)$ , the thionyl chloride absorbs the water from the magnesium chloride hexahydrate to create magnesium chloride MgCl₂, sulphur dioxide SO₂, and hydrogen chloride (HCl).

This reaction's balanced equation is as follows:

${\mathrm{MgCl}}_2.6{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)}+6{\mathrm{SOCl}}_{2\left(\mathrm{l}\right)}\to {\mathrm{MgCl}}_{2\left(\mathrm{s}\right)}+6\mathrm{S}{{\mathrm{O}}_2}_{\left(\mathrm{g}\right)}+12{\mathrm{HCl}}_{\left(\mathrm{g}\right)}$

Step 2: (b) Draw a Lewis structure of SOCl 2 with minimal formal charges

Make the SOCl₂ Lewis structure:

In a Lewis structure, the center atom is the least electronegative. Sulfur is the least electronegative atom among Sulphur, Oxygen, and Chlorine; hence it is the center atom in this case.

There are 26 valence electrons in all (6 electrons from S, 6 electrons from O, and 7 electrons from each Cl).

Draw the following skeletal structure:

Subtract six electrons from total valence electrons to account for the three bonds in the skeletal structure. Distribute the remaining 20 electrons on terminal atoms, then the center atom, to complete the octet of each atom.

Calculate each atom's formal charge as follows:

$$\begin{gathered} \mathrm{Formal} \mathrm{charge}=\mathrm{valance}\mathrm{electrons}-\left(\mathrm{unshared}\mathrm{electrons}-\frac{1}{2}\mathrm{bonding}\mathrm{electrons}\right) \\ \mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{S}=6-\left(2+\frac{1}{2}\times 6\right)=+1 \\ \mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{O}=6-\left(6+\frac{1}{2}\times 2\right)=-1 \\ \mathrm{Formal} \mathrm{charge} \mathrm{on} \mathrm{Cl}=7-\left(6+\frac{1}{2}\times 2\right)=0 \end{gathered}$$

Make a double bond between oxygen and sulphur atoms utilising one of the lone pairs on the oxygen atom to reduce formal charges.