Question

Give the geometry and type of hybridisation of water molecule, ethylene and acetylene molecule.

Answer 1

To answer this question you must recall the VSEPR theory. The Valence shell electron pair repulsion theory proposes that the hybridized orbitals in an atom arrange themselves in such a way so as to minimize the repulsion between them, hence determining the geometry of a molecule on the basis of its hybridization.

Complete Answer:
First we consider a water molecule. The central atom is oxygen. We know that the atomic number of oxygen is 8 and it has 6 valence electrons. In water molecules, It shares one electron each with two hydrogen atoms and has 4 extra electrons which are present as lone pairs. The hybridization of oxygen in the molecule is $s{p^3}$ and geometry must be tetrahedral.
In ethylene or ethene, there is a double bond between two carbon atoms. Carbon has 4 valence electrons, one of which is used for the formation of a pi bond. Thus, the hybridization of carbon is $s{p^2}$ and the shape of the molecule in trigonal planar.
In acetylene or ethyne, there is a triple bond between two carbon atoms. Carbon has 4 valence electrons, two of which are used for the formation of 2 $\pi$ bonds. Thus, the hybridization of carbon is $sp$ and the shape of the molecule is linear.

Note: The concept of mixing of atomic orbitals in order to form new hybrid orbitals that possess different shapes and energies as compared to the original parent atomic orbitals is known as hybridisation. Hybrid orbitals are suitable to form chemical bonds of equal energies. Also hybridization of orbitals leads to the formation of more stable compounds because hybrid orbitals have lower energy than the unhybrid orbitals.