Predict the shapes of the following molecules on the basis of hybridisation.

$B C l_{3}, C H_{4}, C O_{2}, N H_{3}$ .

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Solution

Predicting shapes of given molecules using hybridisation

Hybridization is the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies which results in the formation of new set of orbitals of equivalent energies and shape.

Hybridization helps in deciding the structure of the molecules.

Electron pair types

Lone Pair (lp)

Bond Pair (bp)

Total number of hybrid orbitals = No. of lone pairs on central atom + bond pair (i.e. Number of sigma bonds formed by central atom)

Hybridisation can be predicted by using the number of total hybridised orbitals

Molecules Hybrid orbitals=lone pairs +sigma bonds Hybridisation Shape

$B C l_{3}$ $0 + 3 = 3$ $s p^{2}$
Trigonal planar

$C H_{4}$ $0 + 4 = 4$ $s p^{3}$
Tetrahedral

$C O_{2}$ $0 + 2 = 2$ $s p$
Linear

$N H_{3}$ $1 + 3 = 4$ $s p^{3}$
Pyramidal

There, $N H_{3}$ is having pyramidal shape despite having $s p^{3}$ hybridisation because central atom $(N)$ contains one lone pair of electron due to which when this lone pair and $3$ sigma bonds will correspond to $4$ vertices of tetrahedron, it will give rise to pyramidal shape.

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Standard XII Chemistry

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Molecules	Hybrid orbitals=lone pairs +sigma bonds	Hybridisation	Shape
$B C l_{3}$	$0 + 3 = 3$	$s p^{2}$	Trigonal planar
$C H_{4}$	$0 + 4 = 4$	$s p^{3}$	Tetrahedral
$C O_{2}$	$0 + 2 = 2$	$s p$	Linear
$N H_{3}$	$1 + 3 = 4$	$s p^{3}$	Pyramidal

Predict the shapes of the following molecules on the basis of hybridisation. BCl3,CH4,CO2,NH3.

Predict the shapes of the following molecules on the basis of hybridisation.

$B C l_{3}, C H_{4}, C O_{2}, N H_{3}$ .