Solid State — NEET UG practice

1. An edge-centred octahedral void sits at the midpoint of an edge of the unit cell. To find its contribution, we need to know how many unit cells share that edge.

2. Every edge of a cube is shared among 4 neighbouring unit cells, since 4 unit cells meet along any given edge in a crystal lattice.

3. Therefore, only a fraction of this void actually 'belongs' to one unit cell, and that fraction is found by dividing the whole void equally among the 4 sharing cells.

$\text{Contribution per unit cell} = \frac{1}{4}$

Hence, the answer is B.

1. We are given the amount of the compound in moles, and we know that 1 mole of any substance contains Avogadro's number ($N_A$) of particles.

Given $n = 5\ \text{mol}$

Therefore $\text{Number of particles} = 5N_A$

2. For a cubic close packed (ccp) structure, each close-packed particle generates 2 tetrahedral voids per particle, so we apply this relation.

$\text{Number of tetrahedral voids} = 2 \times \text{Number of particles}$

3. Substituting the number of particles found in step 1:

$= 2 \times 5N_A = 10N_A$

4. Now we substitute $N_A = 6.023 \times 10^{23}$ to get the numerical value.

$10 \times 6.023 \times 10^{23} = 6.023 \times 10^{24}$

Hence, the number of tetrahedral voids is $6.022 \times 10^{24}$, so the answer is D.

1. In a ccp (fcc) structure, the atoms touch each other along the face diagonal of the cube, not along the edge. This geometric fact gives us the relation between $r$ and $a$.

Given $4r = \sqrt{2}\,a$

2. We now rearrange this equation to express $a$ in terms of $r$, since that is what the question asks for.

Therefore $a = \frac{4r}{\sqrt{2}}$

3. Simplifying the fraction by rationalising:

$a = 2\sqrt{2}\,r$

Hence, the correct relation is $a = 2\sqrt{2}r$, so the answer is D.