Q19 E - Chemistry (OpenStax)

Question

Strike-anywhere matches contain a layer of \({\rm{KCl}}{{\rm{O}}_{\rm{3}}}\) and a layer of \({{\rm{P}}_{\rm{4}}}{{\rm{S}}_{\rm{3}}}\). The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets fire to the wooden stem of the match. \({\rm{KCl}}{{\rm{O}}_{\rm{3}}}\) contains the \({\rm{Cl}}{{\rm{O}}_{\rm{3}}}^{\rm{ - }}\) ion. \({{\rm{P}}_{\rm{4}}}{{\rm{S}}_{\rm{3}}}\) is an unusual molecule with the skeletal structure.

 

1. Write Lewis structures for \({{\rm{P}}_{\rm{4}}}{{\rm{S}}_{\rm{3}}}\) and the \({\rm{Cl}}{{\rm{O}}_{\rm{3}}}^{\rm{ - }}\) ion. 
2. Describe the geometry about the \({\rm{P}}\) atoms, the \({\rm{S}}\) atom, and the \({\rm{Cl}}\) atom in these species. 
3. Assign a hybridization to the \({\rm{P}}\) atoms, the  \({\rm{S}}\)atom, and the \({\rm{Cl}}\) atom in these species. 
4. Determine the oxidation states and formal charge of the atoms in \({{\rm{P}}_{\rm{4}}}{{\rm{S}}_{\rm{3}}}\) and the \({\rm{Cl}}{{\rm{O}}_{\rm{3}}}^{\rm{ - }}\) ion.

Step-by-Step Solution

Answer

(a)

 

(b) P's geometry is trigonal pyramidal, S's geometry is bent, and Cl's geometry is trigonal pyramidal.

(c) P, S, and Cl have \({\rm{s}}{{\rm{p}}^{\rm{3}}}\) hybridization due to their tetrahedral electron-pair geometry.

(d) \({\rm{P}}\) has an oxidation state of \({\rm{ + 1}}\), \({\rm{S}}\) has \({\rm{ - }}\frac{{\rm{4}}}{{\rm{3}}}\), while \({\rm{Cl}}\) has \({\rm{ + 5}}\).\({\rm{P}}\) has a formal charge of \({\rm{0}}\), \({\rm{S}}\) has \({\rm{0}}\), \({\rm{Cl}}\) has \({\rm{ + 2}}\), and \({\rm{O}}\) has  \({\rm{ - 1}}\)

Step 1: Define Lewis structures

A Lewis Structure gives the simplified representation of the valence shell electrons within a molecule.

The total number of electrons that an atom gains or loses in order to form a chemical bond with another atom is called as oxidation number.

Step 2: Explanation

(a) \({{\rm{P}}_{\rm{4}}}{{\rm{S}}_{\rm{3}}}\) and the \({\rm{Cl}}{{\rm{O}}_{\rm{3}}}^{\rm{ - }}\) ion have the following Lewis structures,

Step 3: Explanation

(b) The phosphorus atoms in \({{\rm{P}}_{\rm{4}}}{{\rm{S}}_{\rm{3}}}\) are surrounded by four zones of electron density that form a tetrahedral electron-pair geometry, with a trigonal pyramidal molecular geometry around the \({\rm{P}}\) atoms.

In \({{\rm{P}}_{\rm{4}}}{{\rm{S}}_{\rm{3}}}\), the sulphur atoms are surrounded by four areas of electron density that form a tetrahedral electron-pair geometry, and the molecular geometry around the \({\rm{S}}\) atoms is twisted.

The chlorine atom is surrounded by four areas of electron density that form a tetrahedral electron-pair geometry and a trigonal pyramidal molecular geometry around it.

Step 4: Explanation

(c) Tetrahedral electron-pair geometry in \({\rm{P}}\), \({\rm{S}}\), and \({\rm{Cl}}\) results in \({\rm{s}}{{\rm{p}}^{\rm{3}}}\) hybridization.

Step 5: Explanation

(d) \({\rm{P}}\) has an oxidation state of \({\rm{ + 1}}\), \({\rm{S}}\) has a state of \({\rm{ - }}\frac{{\rm{4}}}{{\rm{3}}}\), while \({\rm{Cl}}\) has a state of \({\rm{ + 5}}\).\({\rm{P}}\) has a formal charge of \({\rm{0}}\), \({\rm{S}}\) has a formal charge of \({\rm{0}}\), \({\rm{Cl}}\) has a formal charge of \({\rm{ + 2}}\), and \({\rm{O}}\) has a formal charge of \({\rm{ - 1}}\).