Question

What is the hybrid state of carbon in ethyne, graphite and diamond?
(A) $s{{p}^{2}},\text{ }sp,\text{ }s{{p}^{3}}$
(B) $sp,\ s{{p}^{2}},\ s{{p}^{3}}$
(C) $s{{p}^{3}},\ s{{p}^{2}},\ sp$
(D) $s{{p}^{2}},\ s{{p}^{3}},\ sp$

Answer 1

By knowing the structure of ethyne, graphite and diamond we can predict its hybridization easily. As ethyne molecules have linear structure, it is in sp hybridized state. Graphite is in hexagonal form and hence it is in$s{{p}^{2}}$ hybridized state. Diamonds have tetrahedron structure and it is in $s{{p}^{3}}$ hybridized state.

Complete Solution :
First of all what is hybridization? Hybridization is the process in which two atomic orbital with comparable energy of the same atom combine to form equal numbers of the hybrid orbitals of the same energy.
Carbon’s atomic number is 6. Its electronic configuration is $1{{s}^{2}}2{{s}^{2}}2{{p}^{2}}$.
First let us consider ethyne molecules. The chemical formula is $CH\equiv CH$. In ethyne molecules one 2s electron and one of the three 2p orbitals combine to form two sp hybridized orbitals. Hence ethyne molecules have linear structure and bond angle of $180{}^\circ$.
The graphite is $s{{p}^{2}}$ hybridized because it forms a stable chemical bond with three other carbon atoms. The carbon atoms in graphite are $s{{p}^{2}}$ hybridized and thus form a hexagonal ring.
Diamond has a perfect tetrahedron structure as the carbon in diamond forms a chemical bond with four other carbon atoms. Because of its tetrahedron structure, diamond is in $s{{p}^{3}}$ hybridized state.
So, the correct answer is “Option B”.

Hybridisation	Shape
sp	Linear
sp2	Trigonal planar
sp3	Tetrahedral
sp3d	Trigonal bipyramidal
sp3d2	Octahedral

Note: The hybrid orbital and molecular orbitals are not the same. When two atomic orbitals of the same atoms interact, it is called hybrid orbitals. Whereas when two atomic orbitals of different atoms combine, it is called molecular orbitals.