Following are the characteristic features of a solid state:

    I.            Definite shape, volume and mass

    II.           Intermolecular distances are short and intermolecular forces are strong

    III.          They are incompressible and rigid

    IV.          The constituent particles have fixed positions and can oscillate only about their mean positions

Amorphous solids consist of particles of irregular shape, so it displays short range order.
Amorphous solids soften over a range of temperature, and hence doesn’t show sharp melting point.
The value of a physical property will be same from any directions in case of amorphous solids. So, they are said to be isotropic in nature.
Amorphous solids do not have a definite heat of fusion.

Polar and non-polar molecular solids are soft and non-conductors of electricity.
Ionic solids are non-conductors of electricity in solid state, as the constituent ions are not free to move (but in molten state or when dissolved in water the ions are free to move and hence they conduct electricity)

Choose the correct option:
$\begin{array}{cccc}& \text{Crystal system}& \text{Axial distance}& \text{Axial angles}\\ \text{(a)}& \text{Tetragonal}& a=b\ne c& \alpha =\beta =\gamma ={90}^{\circ }\\ \text{(b)}& \text{Monoclinic}& a\ne b\ne c& \alpha =\gamma ={90}^{\circ }\beta \ne {90}^{\circ }\\ \text{(c)}& \text{Rhombohedral}& \text{a = b = c}& \alpha =\beta =\gamma \ne {90}^{\circ }\\ \text{(d)}& \text{Triclinic}& a\ne b\ne c& \alpha \ne \beta \ne \gamma \ne {90}^{\circ }\end{array}$

The bcc cell consistes of 8 atoms at the corners and one atom at centre.
$\therefore n=(8\times \frac{1}{8})+1=2$. 
The fcc cell consists of 8 atoms at the eight corners and one atom at each of the six faces.This atom at the faces is shared by two unit cells.
$\therefore n=8\times \frac{1}{8}+(6\times \frac{1}{2})=4$

In face-centered cubic arrangement, 74% of the crystal space is filled

•  Vacant space = 100 – 74 = 26%

Tetrahedral sites are double compared to octahedral sites; and X is present in two-third of them

a)  So, ratio of X and Z equals $2\times \left(\frac{2}{3}\right):1=4:3$
$\Rightarrow$ formula of the compound = $X_4{Z}_3$ 

In $△EFD$, 
b = a + a = 2a
Now in $△AFD$
$c^2=a^2+b^2=a^2+2a^2=3a^2$
$\to c=\sqrt{3a}$
The length of the body diagonal 'c' is equal to 4r, where r is the radius of the sphere (atom), as the three spheres along the diagonal touch each other.
Therefore, $\sqrt{3a}=4r$
$a=\frac{4r}{\sqrt{3}}$
$\to r=\frac{\sqrt{3}}{4}a$

Volume of unit cell $=(120pm{)}^3=(120\times {10}^{-12}{)}^3{m}^3=({12}^3\times {10}^{-33})m^3$
Volume of 408 g of the element $=\frac{mass}{density}=\frac{408}{6.8}=60c{m}^3=6\times {10}^{-5}{m}^3$
So, number of unit cells present in 408 g of the elements $=\frac{6\times {10}^{-5}}{{12}^3\times {10}^{-33}}=3.472\times {10}^{25}$ unit cells
Since each jcc unit cell consist of 7 atoms,
therefore the total number of atoms presents in 408 g of the given element
$=7\times 3.472\times {10}^{25}=2.43\times {10}^{26}atoms$

    I.      8 corner atom $\times \left(\frac{1}{8}\right)$ atom per unit cell = 1 atom

    II.     4 face-centered atoms $\times \left(\frac{1}{2}\right)$ atom per unit cell = 2 atom

    III.    12 edge-centered atoms $\times \left(\frac{1}{4}\right)$ atom per unit cell = 3 atom

    IV.     1 body centered atom $\times$ 1 atom per unit cell = 1 atom

So, Total number of atoms per unit cell $=1+2+3+1=7$ atoms