Question

Assertion: Nearest neighbour distance in bcc unit cell is greater than that of fcc having same edge length.
Reason: Bcc has greater packing efficiency than fcc.
(a) Both assertion and reason are correct and reason is the correct explanation for assertion
(b) Both assertion and reason are correct but reason is not the correct explanation for the assertion
(c) Assertion is correct but reason is incorrect
(d) Assertion is incorrect but reason is correct

Answer 1

Bcc or body centred cubic system consists of each atom in its corners as well as its body centre. Fcc or face centered cubic system consists of an atom in its corners as well as in the faces.

Complete step by step solution:
In the Bcc packing system, an atom is present in the body centre other than atoms occupying the corners. In the case of the fcc packing system, atoms are percent in each face as well as the corners. From this we can say that the number of atoms in fcc is more than bcc. The neighbouring atoms in fcc are more close to each other than in bcc.
Therefore, the assertion is correct.
For body centered cubic lattice, he relationship between edge length a and the radius r of the unit cell is $a=\dfrac{4r}{\sqrt{3}}$
The volume occupied by 1 atom is $\dfrac{4}{3}\pi {{r}^{3}}$
There are 2 atoms in a Bcc unit cell (atoms in the corner constitute 1 atom and other atom is at body center so 2 atoms).
So, the total volume becomes =$2\times \dfrac{4}{3}\pi {{r}^{3}}=\dfrac{8}{3}\pi {{r}^{3}}$
Packing efficiency =Volume occupied by atoms in a unit cell divided by the total volume of the unit cell
Volume of a bcc unit cell is ${{a}^{3}}$ ,where a is the edge length. For a bcc unit cell $a=\dfrac{4}{\sqrt{3}}r$
So, total volume of bcc unit cell will be equal to =${{a}^{3}}={{(\dfrac{4}{\sqrt{3}}r)}^{3}}=\dfrac{64{{r}^{3}}}{3\sqrt{3}}$
Packing efficiency=$\dfrac{\dfrac{8}{3}\pi {{r}^{3}}}{\dfrac{64{{r}^{3}}}{3\sqrt{3}}}\times 100$=68.04%
For fcc unit cell,
The volume of a fcc unit cell =${{a}^{3}}$, here a the edge length for fcc unit cell =$\dfrac{4}{\sqrt{2}}r$
SO, total volume of a fcc unit cell will be=${{(\dfrac{4}{\sqrt{2}}r)}^{3}}=\dfrac{64{{r}^{3}}}{2\sqrt{2}}$
Volume for 1 atom=$\dfrac{4}{3}\pi {{r}^{3}}$
We know that in a fcc unit cell there are 4 atoms (one atom contributed by the corners and 3 atoms contributed by the faces).
Packing efficiency=$\dfrac{4(\dfrac{4}{3}\pi {{r}^{3}})}{\dfrac{64{{r}^{3}}}{2\sqrt{2}}}\times 100$= 74%.
Comparing the two packing efficiencies of bcc and fcc, we can conclude that packing efficiency of fcc is greater than bcc. Hence the reason is incorrect.

The answer to the question is option (c). Assertion is correct and reason is wrong.

Note: The volume depends on the edge length of the unit cell. The edge length of bcc and fcc is not the same. Because the number of atoms present in bcc is 2 and bcc is 4, that is why for bcc $a=\dfrac{4}{\sqrt{3}}r$ and for fcc a=$\dfrac{4}{\sqrt{2}}r$.