Question

What is the hybridization of the central atom of each of the following molecules? Match the atoms below with their number---sp, sp2, sp3, sp3d, sp3d2

COH2

H2S

PBr5

Answer 1

Concepts and reason

The hybridization of an organic compound can be explained using the electronic geometry of the compound. Electronic geometry gives information about both bond pair and lone pair of electrons. The geometry of a molecule can be described by Valence Shell Electron Pair Repulsion (VSEPR) theory.

Fundamentals

A Lewis structure is used to determine the electronic geometry as well as molecular geometry of a molecule.

A Lewis structure is a skeletal structure with all the elements and non-bonding electrons shown. Two electrons indicate one electron pair.

The table for hybridization with respect to electronic geometry using different combinations of bonds and lone pairs of electrons present on central atom is below:

Number of atom/group attached to central atom	Number of lone pair of electrons on central atom	Electron geometry	Hybridization of the central atom
2	0	Linear	${\rm{sp}}$
3	0	Trigonal planar	${\rm{S}}{{\rm{p}}^{\rm{2}}}$
2	1	Trigonal planar	${\rm{S}}{{\rm{p}}^{\rm{2}}}$
4	0	Tetrahedral	${\rm{S}}{{\rm{p}}^3}$
3	1	Tetrahedral	${\rm{S}}{{\rm{p}}^3}$
2	2	Tetrahedral	${\rm{S}}{{\rm{p}}^3}$
5	0	Trigonal bipyramidal	${\rm{S}}{{\rm{p}}^{\rm{3}}}{\rm{d}}$
4	1	Trigonal bipyramidal	${\rm{S}}{{\rm{p}}^{\rm{3}}}{\rm{d}}$
3	2	Trigonal bipyramidal	${\rm{S}}{{\rm{p}}^{\rm{3}}}{\rm{d}}$
2	3	Trigonal bipyramidal	${\rm{S}}{{\rm{p}}^{\rm{3}}}{\rm{d}}$
6	0	Octahedral	${\rm{S}}{{\rm{p}}^{\rm{3}}}{{\rm{d}}^{\rm{2}}}$
5	1	Octahedral	${\rm{S}}{{\rm{p}}^{\rm{3}}}{{\rm{d}}^{\rm{2}}}$
4	2	Octahedral	${\rm{S}}{{\rm{p}}^{\rm{3}}}{{\rm{d}}^{\rm{2}}}$

Ans:

$\begin{array}{l}\\{\rm{The}}\,{\rm{hybridization}}\,{\rm{of}}\,{\rm{the}}\,{\rm{central}}\,{\rm{atom}}\,{\rm{of}}\,{\rm{the}}\,{\rm{molecules}}\,{\rm{is:}}\\\\{\rm{CO}}{{\rm{H}}_{\rm{2}}}\,{\rm{ = }}\,{\rm{s}}{{\rm{p}}^{\rm{2}}}\\\\{{\rm{H}}_{\rm{2}}}{\rm{S = s}}{{\rm{p}}^{\rm{3}}}\\\\{\rm{PB}}{{\rm{r}}_{\rm{5}}}{\rm{ = s}}{{\rm{p}}^{\rm{3}}}{\rm{d}}\\\end{array}$