Question

Describe the molecular geometry and hybridization of the N, P, or S atoms in each of the following compounds. (a) \({{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}\), phosphoric acid, used in cola soft drinks (b) \({\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{N}}{{\rm{O}}_{\rm{3}}}\) , ammonium nitrate, a fertilizer and explosive (c) \({{\rm{S}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}\), disulfur dichloride, used in vulcanizing rubber (d) \({{\rm{K}}_{\rm{4}}}{\rm{[}}{{\rm{O}}_{\rm{3}}}{\rm{POP}}{{\rm{O}}_{\rm{3}}}{\rm{]}}\), potassium pyrophosphate, an ingredient in some toothpastes

Short answer

(a)\({{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}\)is tetrahedral with\({\rm{s}}{{\rm{p}}^{\rm{3}}}\)hybridization.

(b)\({\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{N}}{{\rm{O}}_{\rm{3}}}\)→\({\rm{N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}\)is tetrahedral with\({\rm{s}}{{\rm{p}}^{\rm{3}}}\)hybridization and\({\rm{N}}{{\rm{O}}_{\rm{3}}}^{\rm{ - }}\)is trigonal planar with\({\rm{s}}{{\rm{p}}^{\rm{2}}}\)hybridization.

(c)\({{\rm{S}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}\)is bent with\({\rm{s}}{{\rm{p}}^{\rm{3}}}\)hybridization.

(d) \({{\rm{K}}_{\rm{4}}}{\rm{[}}{{\rm{O}}_{\rm{3}}}{\rm{POP}}{{\rm{O}}_{\rm{3}}}{\rm{]}}\) is tetrahedral with \({\rm{s}}{{\rm{p}}^{\rm{3}}}\) hybridization

Step-by-step solution

Define hybridisation

Atomic orbitals that fuse to form newly hybridized orbitals influences molecular geometry and bonding properties.This process is called as hybridisation.

Explanation

(a) \({{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}\): The phosphorus atom is surrounded by four electron-density zones that form a tetrahedral electron-pair shape. In a tetrahedral arrangement, the hybridization is \({\rm{s}}{{\rm{p}}^{\rm{3}}}\) and the molecular geometry is tetrahedral.

Explanation

(b)

\({\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{N}}{{\rm{O}}_{\rm{3}}}\): In \({\rm{N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}\), the nitrogen atom is surrounded by four electron density areas that form a tetrahedral electron-pair geometry. In a tetrahedral arrangement, the hybridization is \({\rm{s}}{{\rm{p}}^{\rm{3}}}\) and the molecular geometry is tetrahedral.

Three zones of electron density surround the nitrogen atom in \({\rm{N}}{{\rm{O}}_{\rm{3}}}^{\rm{ - }}\), arranging themselves in a trigonal planar electron-pair shape. In a trigonal planar configuration, the hybridization is \({\rm{s}}{{\rm{p}}^{\rm{2}}}\), and the molecular geometry is trigonal planar.

Explanation

(c)

\({{\rm{S}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}\): The sulphur atoms are encircled by four zones of electron density that form a tetrahedral electron-pair geometry. In a tetrahedral structure, the hybridization is \({\rm{s}}{{\rm{p}}^{\rm{3}}}\) and the molecular geometry is bent.

Explanation

(d)

\({{\rm{K}}_{\rm{4}}}{\rm{[}}{{\rm{O}}_{\rm{3}}}{\rm{POP}}{{\rm{O}}_{\rm{3}}}{\rm{]}}\): The phosphorus atoms are encircled by four zones of electron density that form a tetrahedral electron-pair geometry. In a tetrahedral arrangement, the hybridization is \({\rm{s}}{{\rm{p}}^{\rm{3}}}\) and the molecular geometry is tetrahedral