Answer

In square close-packed layer, a molecule is in contact with four of its neighbours. Therefore, the two-dimensional coordination number of a molecule in square close-packed layer is 4.

Answer

Number of close-packed particles $=0.5 \times 6.022 \times 10^{23}=3.011 \times 10^{23}$

Therefore, number of octahedral voids $=3.011 \times 10^{23}$

And, number of tetrahedral voids $=2 \times 3.011 \times 10^{23}=6.022 \times 10^{23}$

Therefore, total number of voids $=3.011 \times 10^{23}+6.022 \times 10^{23}=9.033 \times 10^{23}$

Answer

The ccplattice is formed by the atoms of the element $\mathrm{N}$.

Here, the number of tetrahedral voids generated is equal to twice the number of atoms of the element $\mathrm{N}$.

According to the question, the atoms of element M occupy $\frac{1}{3}^{\text {rd }}$ of the tetrahedral voids.

Therefore, the number of atoms of $M$ is equal to $2 \times \frac{1}{3}=\frac{2}{3}$ rd of the number of atoms of $\mathrm{N}$.

Therefore, ratio of the number of atoms of $\mathrm{M}$ to that of $\mathrm{N}$ is $\mathrm{M}: \mathrm{N}=\frac{2}{3}: 1$ $=2: 3$

Thus, the formula of the compound is $\mathrm{M_2} \mathrm{~N_3}$.

Answer

Hexagonal close-packed lattice has the highest packing efficiency of $74 \%$. The packing efficiencies of simple cubic and body-centred cubic lattices are $52.4 \%$ and $68 \%$ respectively.

Answer

It is given that density of the element, $d=2.7 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$

Molar mass, $\mathrm{M}=2.7 \times 10^{-2} \mathrm{~kg} \mathrm{~mol}^{-1}$

Edge length, $a=405 \mathrm{pm}=405 \times 10^{-12} \mathrm{~m}$

$=4.05 \times 10^{-10} \mathrm{~m}$

It is known that, Avogadro’s number, $N_{A}=6.022 \times 1023 \mathrm{~mol}^{-1}$

Applying the relation,

$$ \begin{aligned} d & =\frac{z, M}{a^{3} \cdot \mathrm{N_A}} \\ \end{aligned} $$

$$ \begin{aligned} z & =\frac{d \cdot a^{3} \mathrm{~N_A}}{M} \\ & =\frac{2.7 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3} \times\left(4.05 \times 10^{-10} \mathrm{~m}\right)^{3} \times 6.022 \times 10^{23} \mathrm{~mol}^{-1}}{2.7 \times 10^{-2} \mathrm{~kg} \mathrm{~mol}^{-1}} \\ & =4.004 \\ & =4 \end{aligned} $$

This implies that four atoms of the element are present per unit cell. Hence, the unit cell is face-centred cubic (fcc) or cubic close-packed ( $c c p$ ).