Question

The crystal system of a compound with unit cell dimensions a=0.387, b=0.387, c=0.504 nm and $\alpha =\beta ={{90}^{\text{o}}}$ and $\gamma ={{120}^{\text{o}}}$ is:
A. cubic
B. hexagonal
C. orthorhombic
D. rhombohedral

Answer 1

Hint: Unit cell is the smallest portion of a crystal lattice which when repeated generates the entire crystal lattice. Unit cells are 3-dimensional structures, there are three different edge length and axial angles present whose magnitude can be same or different.

Complete answer:
Let us first understand what is crystal lattice; the three dimensional arrangement of constituent particles in a crystal, the arrangement of points in space is called crystal lattice. There are 14 possible lattices. Such lattices are Bravais Lattices. The characteristics of a crystal lattice are:
(a) Each point in a lattice is known as a lattice point. Each point in a crystal lattice represents a particle which may be an atom or an ion.
(c) Lattice points are joined together by straight lines to bring out the geometry of the lattice.
The characteristics of unit cell are which defines such crystals:
(a) Have dimensions along the three edges a, b and c. These edges may or may not be mutually perpendicular.
(b) axial angles between the edges, $\alpha $ (between b and c), $\beta $(between a and c) and $\gamma $ (between a and b). So, a unit cell is featured by six parameters $\text{a,b,c,}\alpha \text{,}\beta ,\gamma $.
The edge length a=0.387 nm and b=0.387 nm which means that both are the same. But c is not equal to 0.387 nm, so, a=b$\ne $c. Let us find the crystal which satisfies all these conditions.

S. No.	CRYSTAL SYSTEM	VARIATIONS	EDGE LENGTHS	ANGLES	EXAMPLES
1	Cubic	Primitive, body-centred, face-centred	a=b=c	$\alpha =\beta =\gamma =9{{0}^{\text{o}}}$	Zinc blende, copper
2	Hexagonal	Primitive	a=b$\ne $c	$\gamma =12{{0}^{\text{o}}}, \beta =\alpha =9{{0}^{\text{o}}}$	Graphite, ZnO
3	Orthorhombic	Primitive, body-centred, face-centred, end-centred	a$\ne $b$\ne $c	$\alpha =\beta =\gamma =9{{0}^{\text{o}}}$	Rhombic sulphur
4	Rhombohedral	Primitive	a=b=c	$\alpha =\beta =\gamma \ne 9{{0}^{\text{o}}}$	Calcite $\left( \text{CaC}{{\text{O}}_{3}} \right)$ and cinnabar $\text{HgS}$

The correct answer of this question is option ‘b’ Hexagonal crystal lattice as it satisfies unit cell dimensions a=0.387, b=0.387, c=0.504 nm and $\alpha =\beta ={{90}^{\text{o}}}$ and $\gamma ={{120}^{\text{o}}}$.

Note: There can be confusion between rhombohedral and hexagonal because the axial angles values are almost the same. As, in rhombohedral $\alpha =\beta =\gamma \ne 9{{0}^{\text{o}}}$ and in hexagonal also $\alpha =\beta =\gamma \ne 9{{0}^{\text{o}}}$, but the value of third angle is ${{120}^{\text{o}}}$, which is variable is rhombohedral. But the factor in edge lengths that make these lattices different.