Crystallography and Lattice Questions in English

(A) The seven crystal systems are Cubic,Orthorhombic,Tetragonal,Monoclinic,Triclinic,Rhombohedral,and Hexagonal.
By observing the Bravais lattices for these systems,we find that the 'Simple' (or primitive) unit cell is present in all seven crystal systems.
| Sr. No. | Type of System | Bravais lattices present |
| :--- | :--- | :--- |
| $1$ | Cubic | Simple,Face-centered,Body-centered |
| $2$ | Orthorhombic | Simple,Base-centered,Face-centered,Body-centered |
| $3$ | Tetragonal | Simple,Body-centered |
| $4$ | Monoclinic | Simple,Base-centered |
| $5$ | Triclinic | Simple |
| $6$ | Rhombohedral | Simple |
| $7$ | Hexagonal | Simple or primitive |

(B) For a $BCC$ (Body-Centered Cubic) unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by the formula: $\sqrt{3} a = 4 r$.
Given that $a = 351 \ pm$, we can calculate $r$ as follows:
$r = \frac{\sqrt{3} \times 351}{4} \approx \frac{1.732 \times 351}{4} \approx 151.98 \ pm$.
Rounding this value gives $r \approx 152 \ pm$.

(D) In a $Simple \ cubic$ unit cell,particles are present only at the corners. Each corner particle is shared by $8$ unit cells. Therefore,the number of particles per unit cell is $8 \times (1/8) = 1$.

(C) In a face-centred cubic $(FCC)$ unit cell,atoms are present at each corner and at each face centre.
Number of lattice points $=$ Number of corners $+$ Number of face centres
$= 8 + 6$
$= 14$

(A) In a simple cubic unit cell,the atoms are present only at the corners of the cube.
These atoms touch each other along the edge of the cube.
Let '$a$' be the edge length of the unit cell and '$r$' be the radius of the atom.
Since the atoms touch along the edge,the edge length '$a$' is equal to the sum of the radii of the two touching atoms.
Therefore,$a = r + r = 2r$.

(A) For a $BCC$ (Body-Centered Cubic) structure, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by the formula: $\sqrt{3} a = 4 r$.
Given that the edge length $a = 600 \ pm$.
Substituting the value of $a$ into the formula: $r = \frac{\sqrt{3} a}{4} = \frac{\sqrt{3} \times 600}{4}$.
Calculating the value: $r = \sqrt{3} \times 150 \ pm$.

(A) The condition $a \neq b \neq c$ and $\alpha \neq \beta \neq \gamma \neq 90^{\circ}$ represents the triclinic crystal system.
Among the given options,$K_2Cr_2O_7$ crystallizes in the triclinic system.
Therefore,the correct option is $A$.

(B) The given unit cell parameters $a=b \neq c$ and $\alpha=\beta=90^{\circ}$,$\gamma=120^{\circ}$ correspond to the hexagonal crystal system.
Among the given options,graphite crystallizes in the hexagonal system.

(B) The crystal system with dimensions $a \neq b \neq c$,$\alpha = \gamma = 90^{\circ}$ and $\beta \neq 90^{\circ}$ is known as the Monoclinic system.

(A) In a crystal lattice,a face is common to two adjacent unit cells.
Therefore,each face of a unit cell is shared equally by $2$ unit cells.

(C) The crystal system defined by the parameters $a \neq b \neq c$ and $\alpha = \beta = \gamma = 90^{\circ}$ is the Orthorhombic system.
The Orthorhombic crystal system has $4$ types of Bravais lattices: Primitive,Body-centered,Face-centered,and End-centered.
Therefore,$x = \text{Orthorhombic}$ and $y = 4$.

(B) There are $7$ crystal systems in total.
Among these,the face-centered unit cell $(FCC)$ is observed only in the cubic and orthorhombic crystal systems.
Therefore,the total number of crystal systems that have face-centered unit cells is $2$.

(D) There are $3$ body-centred lattices possible among the $14$ Bravais lattices.
These are:
$(I)$ Body-centred cubic $(BCC)$
$(II)$ Body-centred tetragonal
$(III)$ Body-centred orthorhombic

$(A)$ For a crystal system, the parameters $a = b \neq c$ and $\alpha = \beta = \gamma = 90^{\circ}$ correspond to the tetragonal crystal system.
In this case, $a = b = 200 \text{ pm}$ and $c = 300 \text{ pm}$, which satisfies the condition $a = b \neq c$.
Therefore, the correct crystal system is tetragonal.

(C) For the hexagonal crystal system,the axial distances are $a = b \neq c$.
The axial angles are $\alpha = \beta = 90^{\circ}$ and $\gamma = 120^{\circ}$.
Thus,the correct option is $a = b \neq c, \alpha = \beta = 90^{\circ}, \gamma = 120^{\circ}$.

(C) White tin,an allotropic form of tin,is an example of a tetragonal system.
For a tetragonal system,the unit cell parameters are defined as $a=b \neq c$ and $\alpha=\beta=\gamma=90^{\circ}$.
The given reasoning states $a=b=c$ and $\alpha=\beta=\gamma \neq 90^{\circ}$,which describes a rhombohedral system,not a tetragonal one.
Therefore,$Assertion (A)$ is true but $Reasoning (R)$ is false.

(B) In a sodium chloride $(NaCl)$ crystal,the structure is $fcc$ (face-centered cubic).
$Cl^{-}$ ions are present at the corners and face centers,while $Na^{+}$ ions occupy all octahedral voids.
The distance between the nearest $Na^{+}$ and $Cl^{-}$ ions is equal to half of the edge length $(a)$ of the unit cell.
Given,distance between $Na^{+}$ and $Cl^{-} = Y \ pm$.
Therefore,$Y = \frac{a}{2}$.
This implies that the edge length $a = 2 Y \ pm$.
Thus,option $(b)$ is the correct answer.

(C) The angle $\phi$ between two planes $(h_1 k_1 l_1)$ and $(h_2 k_2 l_2)$ in a cubic system is given by the formula:
$\cos \phi = \frac{h_1 h_2 + k_1 k_2 + l_1 l_2}{\sqrt{h_1^2 + k_1^2 + l_1^2} \sqrt{h_2^2 + k_2^2 + l_2^2}}$
For the planes $(100)$ and $(110)$:
$h_1=1, k_1=0, l_1=0$ and $h_2=1, k_2=1, l_2=0$
Substituting these values:
$\cos \phi = \frac{(1 \times 1) + (0 \times 1) + (0 \times 0)}{\sqrt{1^2 + 0^2 + 0^2} \sqrt{1^2 + 1^2 + 0^2}}$
$\cos \phi = \frac{1}{\sqrt{1} \times \sqrt{2}} = \frac{1}{\sqrt{2}}$
$\phi = \cos^{-1}(\frac{1}{\sqrt{2}}) = 45^{\circ}$

(D) The Miller indices $(h, k, l)$ are the reciprocals of the fractional intercepts of the plane along the crystallographic axes $(a, b, c)$.
Given intercepts are $1a$,$1/2b$,and $3c$.
The fractional intercepts are $1, 1/2, 3$.
Taking the reciprocals: $h = 1/1 = 1$,$k = 1/(1/2) = 2$,$l = 1/3$.
To express these as the smallest set of integers,multiply by the least common multiple (which is $3$):
$h = 1 \times 3 = 3$,$k = 2 \times 3 = 6$,$l = (1/3) \times 3 = 1$.
Thus,the Miller indices are $(3$ $6$ $1)$.