Question

What is the hybridization of the central atom of each of the following molecules? Match the...

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

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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

sp{\rm{sp}}

3

0

Trigonal planar

Sp2{\rm{S}}{{\rm{p}}^{\rm{2}}}

2

1

Trigonal planar

Sp2{\rm{S}}{{\rm{p}}^{\rm{2}}}

4

0

Tetrahedral

Sp3{\rm{S}}{{\rm{p}}^3}

3

1

Tetrahedral

Sp3{\rm{S}}{{\rm{p}}^3}

2

2

Tetrahedral

Sp3{\rm{S}}{{\rm{p}}^3}

5

0

Trigonal bipyramidal

Sp3d{\rm{S}}{{\rm{p}}^{\rm{3}}}{\rm{d}}

4

1

Trigonal bipyramidal

Sp3d{\rm{S}}{{\rm{p}}^{\rm{3}}}{\rm{d}}

3

2

Trigonal bipyramidal

Sp3d{\rm{S}}{{\rm{p}}^{\rm{3}}}{\rm{d}}

2

3

Trigonal bipyramidal

Sp3d{\rm{S}}{{\rm{p}}^{\rm{3}}}{\rm{d}}

6

0

Octahedral

Sp3d2{\rm{S}}{{\rm{p}}^{\rm{3}}}{{\rm{d}}^{\rm{2}}}

5

1

Octahedral

Sp3d2{\rm{S}}{{\rm{p}}^{\rm{3}}}{{\rm{d}}^{\rm{2}}}

4

2

Octahedral

Sp3d2{\rm{S}}{{\rm{p}}^{\rm{3}}}{{\rm{d}}^{\rm{2}}}

COH
HSH
Electronic geometry = trigonal planar
Hybridization = sp?

H2S
H
Electronic geometry = tetrahedral
Hybridization = sp3

PBr5
—Br:
Br:
Electronic geometry = trigonal bipyramidal
Hybridization = spºd

Ans:

Thehybridizationofthecentralatomofthemoleculesis:COH2=sp2H2S=sp3PBr5=sp3d\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}

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