Question

In diamond, carbon atoms occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is $356pm$, then the radius of carbon atom is:
(A) $77.07pm$
(B) $154.14pm$
(C) $251.7pm$
(D) $89pm$

Answer 1

Hint: Face-centred cubic lattice has lattice points at the eight corners of the unit cell and also at the centres of each face of the unit cell. A unit cell is the smallest possible representation of a crystal. The coordination number of the face centred cubic lattice is twelve.
Formula used: $\dfrac{{\sqrt 3 }}{4}a = 2r$

Complete answer:
 Diamond is an allotrope of carbon where each carbon is covalently bonded with four other carbon atoms. They are arranged in a face centered cubic crystal structure which is called diamond lattice. Four atoms are bonded to four other atoms within the volume of the cell. For the six faces of the cube, six atoms are present on the middle of the face. Out of the eight cube corners, four atoms bond to an atom inside the cube. The other four will bond to the adjacent cubes of the crystal.
For a diamond unit cell,
$\dfrac{{\sqrt 3 }}{4}a = 2r$
Where $a = $edge length
$r = $Radius of the centre atom
Given that $a = 356pm$
So, $\dfrac{{\sqrt 3 }}{4} \times 356 = 2r$
$r = 77.07pm$
Therefore, option A is the correct answer.

Note:
 Unit cells can be defined as the smallest portion of a crystal lattice which shows the three-dimensional pattern of the complete crystal. A crystal can be considered as the repetition of the same unit cell in three dimensions. A unit cell is the least volume consuming repeating structure that can be possible in any solid. It repeats itself and forms a network called lattice. The crystal lattice is the three-dimensional structural arrangement of atoms or ions or molecules in a crystalline solid as points.