Crystallography and Lattice Questions in English

(C) In a triclinic crystal system,the unit cell parameters are defined by the absence of any symmetry in both edge lengths and axial angles.
Specifically,the edge lengths are unequal $(a \ne b \ne c)$ and the axial angles are unequal and not equal to $90^o$ $(\alpha \ne \beta \ne \gamma \ne 90^o)$.
Therefore,the correct option is $C$.

(A) In a $NaCl$ structure,the edge length $a$ is related to the ionic radii by the formula $a = 2(r_{Sr^{2+}} + r_{O^{2-}})$.
Given $a = 0.514 \ nm$ and $r_{Sr^{2+}} = 0.113 \ nm$.
Substituting the values: $0.514 = 2(0.113 + r_{O^{2-}})$.
$0.257 = 0.113 + r_{O^{2-}}$.
$r_{O^{2-}} = 0.257 - 0.113 = 0.144 \ nm$.
Converting to meters: $0.144 \ nm = 0.144 \times 10^{-9} \ m = 1.44 \times 10^{-10} \ m$.

(B) The crystal system defined by the parameters $a \neq b \neq c$ and $\alpha \neq \beta \neq \gamma \neq 90^{\circ}$ is the Triclinic system.
This is the most unsymmetrical crystal system among the seven primitive crystal systems.

(B) The crystal system defined by the axial distances $a \neq b \neq c$ and the axial angles $\alpha = \gamma = 90^{\circ}, \beta \neq 90^{\circ}$ is known as the monoclinic system.
In this system,none of the edge lengths are equal.
Two of the interfacial angles are equal to $90^{\circ}$,while the third angle $\beta$ is not equal to $90^{\circ}$.
An example of a monoclinic crystal system is $Na_2SO_4 \cdot 10H_2O$.
Therefore,the correct option is $B$.

(C) In a face-centered cubic $(FCC)$ crystal system,the axis passing through the centers of opposite faces is a $C_4$ axis of symmetry.
This means that rotating the unit cell by $90^{\circ}$ $(360^{\circ}/4)$ about this axis results in an identical orientation.
Therefore,it is known as a four-fold axis of symmetry.

(B) For a cubic crystal system,the axial lengths are equal $(a = b = c)$ and the axial angles are all equal to $90^o$ $(\alpha = \beta = \gamma = 90^o)$.
Thus,the correct option is $B$.

(C) $SnO_2$ (Tin dioxide) crystallizes in the tetragonal crystal system.
For the tetragonal crystal system,the axial lengths are $a = b \neq c$ and the axial angles are $\alpha = \beta = \gamma = 90^o$.
Therefore,the correct option is $C$.

(D) The condition $a \ne b \ne c$ and $\alpha = \beta = \gamma = 90^o$ represents the orthorhombic crystal system.
$BaSO_4$ (barium sulfate) and $KNO_3$ (potassium nitrate) crystallize in the orthorhombic system.
$TiO_2$ (titanium dioxide) exists in different forms,but the rutile form is tetragonal $(a = b \ne c, \alpha = \beta = \gamma = 90^o)$.
$NaCl$ (sodium chloride) crystallizes in the cubic system $(a = b = c, \alpha = \beta = \gamma = 90^o)$.
Since the question asks for which substance the orthorhombic condition does $NOT$ hold,both $TiO_2$ and $NaCl$ do not fit the criteria. However,in standard textbook contexts for this specific question,$NaCl$ is the classic example of a cubic system,making it the most distinct answer.

(B) $CdS$ (Cadmium sulfide) typically crystallizes in the Wurtzite structure,which belongs to the Hexagonal crystal system.

(D) The crystal system with dimensions $a \neq b \neq c$ and angles $\alpha = \gamma = 90^{\circ}, \beta \neq 90^{\circ}$ is known as the monoclinic crystal system.
This system is often described as having a structure similar to a matchbox,where the sides are rectangular but the end faces are tilted.
Among the given options,monoclinic sulphur crystallizes in the monoclinic system.

(D) The crystal systems are classified based on their unit cell parameters.
$CuSO_4 \cdot 5H_2O$ (Copper sulfate pentahydrate) has a triclinic structure.
$H_3BO_3$ (Boric acid) has a triclinic structure.
$K_2Cr_2O_7$ (Potassium dichromate) has a triclinic structure.
$Na_2SO_4 \cdot 10H_2O$ (Glauber's salt) has a monoclinic structure.
Therefore,$Na_2SO_4 \cdot 10H_2O$ does not have a triclinic structure.

(A) In crystal systems,axial distances are equal when $a = b = c$. This condition is satisfied by the cubic crystal system. Among the given options,$BaSO_4$ (Barium sulfate) crystallizes in the orthorhombic system,while Calcite and Graphite have different axial ratios. However,in the context of standard textbook questions regarding axial distances,the question often refers to the cubic system. Since none of the options are cubic,we re-evaluate: $BaSO_4$ is orthorhombic $(a \neq b \neq c)$,Calcite is rhombohedral $(a = b = c)$,and Graphite is hexagonal. Thus,Calcite $(CaCO_3)$ is the correct answer as it belongs to the rhombohedral (trigonal) system where $a = b = c$.

(B) trigonal (or trigonal prismatic) structure possesses a specific set of symmetry elements.
In a trigonal system,there are $3$ vertical planes of symmetry passing through the principal axis and $1$ horizontal plane of symmetry perpendicular to it,along with other symmetry operations.
However,for a standard trigonal prism,the total number of planes of symmetry is $4$ (vertical) + $1$ (horizontal) = $5$.
Thus,the correct option is $B$.

(B) According to the classification of crystal systems,there are $7$ primitive crystal systems. Among these,the orthorhombic system is unique because it exhibits all four types of unit cells: primitive,body-centered,face-centered,and end-centered. Therefore,the body-centered unit cell is found in the orthorhombic crystal system.

(D) The Miller indices are determined by taking the reciprocals of the Weiss indices.
$1$. Weiss indices are: $-1, 2, \infty$.
$2$. Taking the reciprocals: $\frac{1}{-1}, \frac{1}{2}, \frac{1}{\infty} = -1, 0.5, 0$.
$3$. To convert to the smallest set of integers,multiply by $2$: $-2, 1, 0$.
$4$. Thus,the Miller indices are $(\bar{2}10)$.

(C) The Miller indices are given as $(h, k, l) = (2, 3, 1)$.
To find the Weiss indices,we take the reciprocals of the Miller indices: $1/h, 1/k, 1/l$.
Reciprocals: $1/2, 1/3, 1/1$.
To convert these into the smallest set of integers,we multiply by the least common multiple $(LCM)$ of the denominators,which is $6$.
Weiss indices: $(6 \times 1/2, 6 \times 1/3, 6 \times 1/1) = (3, 2, 6)$.

(B) For a $bcc$ structure, the relationship between the edge length $a$ and the atomic radius $r$ is given by:
$4r = \sqrt{3}a$
Given $a = 351 \, pm$ and $\sqrt{3} \approx 1.732$:
$r = \frac{\sqrt{3} \times 351}{4} = \frac{1.732 \times 351}{4} = 151.98 \, pm \approx 151.9 \, pm$.

(D) In graphite,the carbon atoms are arranged in regular hexagons in flat parallel layers. This structure corresponds to a $Hexagonal$ crystal system.

(D) Graphite crystallizes in a hexagonal crystal system,not a tetragonal one. For a hexagonal system,the parameters are $a = b \neq c$,$\alpha = \beta = 90^o$,and $\gamma = 120^o$.
For a tetragonal system,the parameters are $a = b \neq c$ and $\alpha = \beta = \gamma = 90^o$.
Since the Assertion incorrectly identifies the crystal system of graphite and the Reason provides incorrect parameters for a tetragonal system,both the Assertion and the Reason are incorrect.

(N/A) The significance of a 'lattice point' is that each point represents one constituent particle of a solid,which may be an atom,a molecule (group of atoms),or an ion.

(N/A) unit cell is characterised by six parameters:
$(i)$ The dimensions along the three edges,$a, b,$ and $c$. These edges may or may not be mutually perpendicular.
$(ii)$ The angles between the edges,which are $\alpha$ (between edges $b$ and $c$),$\beta$ (between edges $a$ and $c$),and $\gamma$ (between edges $a$ and $b$).

(N/A) $(i)$ Hexagonal unit cell:
For a hexagonal unit cell,$a = b \neq c$ and $\alpha = \beta = 90^{\circ}, \gamma = 120^{\circ}$.
Monoclinic unit cell:
For a monoclinic cell,$a \neq b \neq c$ and $\alpha = \gamma = 90^{\circ}, \beta \neq 90^{\circ}$.
$(ii)$ Face-centred unit cell:
In a face-centred unit cell,the constituent particles are present at all the corners and at the centre of each of the six faces.
End-centred unit cell:
An end-centred unit cell contains particles at all the corners and one at the centre of any two opposite faces.

(N/A) $(i)$ Crystal lattice: The regular three-dimensional arrangement of constituent particles (atoms,ions,or molecules) in space is called a crystal lattice.
$(ii)$ Unit cell: $A$ unit cell is the smallest three-dimensional portion of a crystal lattice that,when repeated in different directions,generates the entire crystal lattice.

(N/A) $(i)$ There are $14$ lattice points in a face-centred cubic unit cell ($8$ from the corners $+ 6$ from the faces).
$(ii)$ There are $14$ lattice points in a face-centred tetragonal unit cell ($8$ from the corners $+ 6$ from the faces).
$(iii)$ There are $9$ lattice points in a body-centred unit cell ($1$ from the centre $+ 8$ from the corners).

(D) crystal lattice is a regular three-dimensional arrangement of points in space.
The characteristics are:
$1$. Each point in a lattice is called a lattice point or lattice site.
$2$. Each point in a crystal lattice represents one constituent particle (atom,molecule,or ion).
$3$. Lattice points are joined by straight lines to bring out the geometry of the lattice.
Therefore,all the given statements are correct.

(N/A) $1$. When tiles are laid on a floor,a repeated pattern is formed. If we mark a specific point on each tile (e.g.,the center) and ignore the tiles themselves,we obtain a set of points. This set of points acts as a scaffolding upon which the pattern is built. This scaffolding is known as a $Space \ Lattice$.
$2$. The structural unit (in this case,the tile) placed on each point of the lattice to develop the pattern is called the $Motif$ or $Basis$.
$3$. $A$ crystal structure is generated when these motifs are placed on the points of a space lattice. The lattice represents the pattern of points indicating the positions of these motifs.
$4$. $A$ $Space \ Lattice$ is a hypothetical framework for a crystal structure. When motifs are placed identically on each point of the space lattice,the actual crystal structure is obtained.
$5$. In summary,the combination of a $Motif$ and a $Space \ Lattice$ results in a crystal structure.

(N/A) two-dimensional lattice is a regular arrangement of points in a $2D$ plane,where each point is called a lattice point.
In a $2D$ lattice,the environment of every point is identical.
$A$ hypothetical two-dimensional crystal is formed by placing a structural unit (an atom,molecule,or ion) at each lattice point of a $2D$ lattice.
This model helps in understanding the basic principles of crystal structures,such as unit cells and symmetry,in a simplified $2D$ space before extending them to $3D$ structures.

(N/A) Based on the study of crystals,it can be concluded that the shape of any crystal belongs to one of the seven regular shapes. These basic regular shapes are known as the seven crystal systems.
To determine which system a given crystal belongs to,one must measure the angles between the faces and the number of axes required to define the unit cell.
The seven crystal systems are:
$1$. Cubic: $a = b = c$ and $\alpha = \beta = \gamma = 90^{\circ}$
$2$. Tetragonal: $a = b \neq c$ and $\alpha = \beta = \gamma = 90^{\circ}$
$3$. Orthorhombic: $a \neq b \neq c$ and $\alpha = \beta = \gamma = 90^{\circ}$
$4$. Monoclinic: $a \neq b \neq c$ and $\alpha = \gamma = 90^{\circ}, \beta \neq 90^{\circ}$
$5$. Hexagonal: $a = b \neq c$ and $\alpha = \beta = 90^{\circ}, \gamma = 120^{\circ}$
$6$. Rhombohedral (Trigonal): $a = b = c$ and $\alpha = \beta = \gamma \neq 90^{\circ}$
$7$. Triclinic: $a \neq b \neq c$ and $\alpha \neq \beta \neq \gamma \neq 90^{\circ}$

(N/A) The French mathematician Bravais showed that there are only $14$ possible three-dimensional lattices. These are called Bravais lattices.
The characteristics of primitive and centered unit cells are listed below:

$S.No.$ Crystal System	Possible Variations	Edge Lengths	Axial Angles	Examples
$(1)$ Cubic	Primitive,Body-centered,Face-centered	$a=b=c$	$\alpha=\beta=\gamma=90^{\circ}$	$NaCl, Cu, ZnS$
$(2)$ Tetragonal	Primitive,Body-centered	$a=b \neq c$	$\alpha=\beta=\gamma=90^{\circ}$	White $Sn, SnO_2, TiO_2, CaSO_4$
$(3)$ Orthorhombic	Primitive,Body-centered,Face-centered,End-centered	$a \neq b \neq c$	$\alpha=\beta=\gamma=90^{\circ}$	Rhombic sulphur,$KNO_3, BaSO_4$
$(4)$ Hexagonal	Primitive	$a=b \neq c$	$\alpha=\beta=90^{\circ}, \gamma=120^{\circ}$	Graphite,$ZnO, CdS$
$(5)$ Rhombohedral	Primitive	$a=b=c$	$\alpha=\beta=\gamma \neq 90^{\circ}$	Calcite $(CaCO_3), HgS$
$(6)$ Monoclinic	Primitive,End-centered	$a \neq b \neq c$	$\alpha=\gamma=90^{\circ}, \beta \neq 90^{\circ}$	Monoclinic sulphur,$Na_2SO_4 \cdot 10H_2O$
$(7)$ Triclinic	Primitive	$a \neq b \neq c$	$\alpha \neq \beta \neq \gamma \neq 90^{\circ}$	$K_2Cr_2O_7, CuSO_4 \cdot 5H_2O, H_3BO_3$

(B) The $14$ Bravais lattices are the distinct spatial arrangements of lattice points in a crystal system. These are derived from the $7$ crystal systems based on the variations in unit cell parameters and centering (primitive,body-centered,face-centered,and end-centered). Therefore,the total number of Bravais lattices is $14$.

The $14$ Bravais lattices are the distinct spatial arrangements of points in a crystal lattice that maintain the symmetry of the unit cell.
They are classified into $7$ crystal systems based on the parameters of the unit cell ($a, b, c$ and $\alpha, \beta, \gamma$).
$1$. Cubic: $a=b=c, \alpha=\beta=\gamma=90^{\circ}$ (e.g.,$NaCl$).
$2$. Tetragonal: $a=b\neq c, \alpha=\beta=\gamma=90^{\circ}$ (e.g.,$SnO_2$).
$3$. Orthorhombic: $a\neq b\neq c, \alpha=\beta=\gamma=90^{\circ}$ (e.g.,$KNO_3$).
$4$. Hexagonal: $a=b\neq c, \alpha=\beta=90^{\circ}, \gamma=120^{\circ}$ (e.g.,$ZnO$).
$5$. Rhombohedral: $a=b=c, \alpha=\beta=\gamma\neq 90^{\circ}$ (e.g.,$CaCO_3$).
$6$. Monoclinic: $a\neq b\neq c, \alpha=\gamma=90^{\circ}, \beta\neq 90^{\circ}$ (e.g.,$Monoclinic \ S$).
$7$. Triclinic: $a\neq b\neq c, \alpha\neq \beta\neq \gamma\neq 90^{\circ}$ (e.g.,$K_2Cr_2O_7$).

(C) In an end-centered unit cell (also known as base-centered),the constituent particles are located at all the $8$ corners of the unit cell and at the centers of any two opposite faces.

(A) In an orthorhombic crystal system,the axial lengths are unequal $(a \neq b \neq c)$.
However,the axial angles are all equal to $90^{\circ}$.
Therefore,the relationship is $\alpha = \beta = \gamma = 90^{\circ}$.

(A) In a primitive cubic unit cell,atoms are present only at the corners of the cube.
Each corner atom is shared by $8$ adjacent unit cells.
Therefore,the contribution of each corner atom to a single unit cell is $\frac{1}{8}$.
$A$ cube has $8$ corners.
Total number of atoms per unit cell = $8 \times \frac{1}{8} = 1$.

(A) primitive cubic unit cell is a type of crystal lattice where the constituent particles are present only at the corners of the unit cell.
According to the Bravais lattice system,there is only $1$ type of primitive cubic unit cell,which is the simple cubic unit cell.
Therefore,the correct answer is $1$.

(D) In the $14$ Bravais lattices,the body-centred $(BCC)$ unit cells are found in the following crystal systems:
$1$. Cubic system: $1$ $(BCC)$
$2$. Tetragonal system: $1$ $(BCC)$
$3$. Orthorhombic system: $1$ $(BCC)$
Therefore,the total number of body-centred unit cells is $1 + 1 + 1 = 3$.

(D) The given parameters are $a \neq b \neq c$ and $\alpha = \gamma = 90^{\circ}, \beta \neq 90^{\circ}$.
These are the characteristic parameters of a monoclinic crystal system.

(N/A) crystal lattice is a regular,repeating three-dimensional arrangement of points in space. Each point represents a constituent particle (atom,molecule,or ion).
It is formed by the regular,periodic,and translational repetition of a structural unit called the unit cell in three dimensions.
The unit cell is the smallest repeating structural unit of a crystal lattice that,when repeated in all directions,generates the entire crystal structure. The figure illustrates how a unit cell,by translational displacement,builds up a two-dimensional lattice.

(N/A) unit cell is characterized by six parameters:
$1$. Its dimensions along three edges $a, b$,and $c$. These edges may or may not be mutually perpendicular.
$2$. Angles between the edges: $\alpha$ (between $b$ and $c$),$\beta$ (between $a$ and $c$),and $\gamma$ (between $a$ and $b$).
Unit cells are broadly classified into two categories:
$(a)$ Primitive Unit Cells: When constituent particles are present only on the corner positions of a unit cell,it is called a primitive unit cell.
$(b)$ Centred Unit Cells: When a unit cell contains one or more constituent particles present at positions other than corners in addition to those at corners,it is called a centred unit cell. These are of three types:
$(i)$ Body-Centred Unit Cells: Such a unit cell contains one constituent particle at its body-centre besides the ones at its corners.
$(ii)$ Face-Centred Unit Cells: Such a unit cell contains one constituent particle present at the centre of each face,besides the ones at its corners.
$(iii)$ End-Centred Unit Cells: Such a unit cell contains one constituent particle at the centre of each of two opposite faces,besides the ones at its corners.

(C) The end-centered unit cell is found in the Orthorhombic and Monoclinic crystal systems.
For the Orthorhombic system,the parameters are $a \neq b \neq c$ and $\alpha=\beta=\gamma=90^{\circ}$.
Comparing this with the given options,option $C$ represents the Orthorhombic system.

(C) There are $7$ crystal systems in total.
Among these,the body-centered unit cell is found in $3$ crystal systems:
$1$. Cubic
$2$. Tetragonal
$3$. Orthorhombic
Therefore,the total number is $3$.

(A) Analysis of the options:
$(A)$ Simple cubic and face-centred cubic belong to the same crystal system (Cubic),so $(A-s)$. They also have parameters $a=b=c$ and $\alpha=\beta=\gamma=90^{\circ}$,so $(A-p)$.
$(B)$ Cubic and rhombohedral are two distinct crystal systems,so $(B-q)$. Cubic has $a=b=c$ and $\alpha=\beta=\gamma=90^{\circ}$,so $(B-p)$.
$(C)$ Cubic and tetragonal are two distinct crystal systems,so $(C-q)$.
$(D)$ Hexagonal and monoclinic are two distinct crystal systems,so $(D-q)$. Hexagonal has $\alpha=\beta=90^{\circ}, \gamma=120^{\circ}$ and Monoclinic has $\alpha=\gamma=90^{\circ}, \beta \neq 90^{\circ}$. Both have exactly two angles equal to $90^{\circ}$,so $(D-r)$.
Thus,the correct matching is $A-p, s; B-p, q; C-q; D-q, r$.

(C) Substances that possess the same crystal structure and similar chemical composition are called isomorphous.
$NaNO_3$ and $CaCO_3$ are isomorphous (both have calcite structure).
$K_2SO_4$ and $K_2SeO_4$ are isomorphous.
$NaF$ and $MgO$ are isomorphous (both have rock salt structure).
$NaCl$ and $KCl$ are not isomorphous because the ionic radii of $Na^+$ and $K^+$ are significantly different,leading to different crystal lattice parameters despite both having a face-centered cubic structure.

(D) There are $14$ distinct Bravais lattices that can be formed within the $7$ crystal systems.

(B) The tetragonal crystal system consists of two types of unit cells: $1$. Primitive (simple) and $2$. Body-centered. Therefore,the total number of different types of unit cells is $2$.

(A) There are $7$ primitive crystal systems in nature.
These $7$ crystal systems can be further classified into $14$ types of Bravais lattices based on the arrangement of lattice points within the unit cell.

(A) There are $7$ distinct crystal systems in crystallography. These $7$ crystal systems give rise to $14$ possible Bravais lattices (also known as space lattices) based on the arrangement of lattice points within the unit cells. Therefore,the total number of crystal systems is $7$.

(D) Two or more substances having the same crystal structure are called isomorphous.
They show the same atomic ratio (Iso- Same,Morphous- Form).
Examples include $NaF$ and $MgO$ ($1:1$ ratio),and $NaNO_3$ and $CaCO_3$ ($1:1:3$ ratio).
$NaCl$ and $KCl$ have similar properties but different crystal structures.

(C) The number of unit cells is calculated by dividing the total volume by the volume of a single unit cell.
Number of unit cells = $\frac{\text{Total Volume}}{\text{Volume of one unit cell}}$
Number of unit cells = $\frac{1 \ cm^3}{3.448 \times 10^{-23} \ cm^3}$
Number of unit cells = $0.29002 \times 10^{23} = 2.9 \times 10^{22}$
Therefore,the correct option is $C$.

(A) The triclinic crystal system is one of the seven crystal systems in crystallography. In this system,the unit cell is characterized by the least symmetry: all three sides have different lengths,and the angles between them are all different and not $90^{\circ}$.
Characteristics of the Triclinic System:
- The unit cell has $a \neq b \neq c$ (all sides are of different lengths).
- The angles between the axes are $\alpha \neq \beta \neq \gamma$,and all are not $90^{\circ}$.
There is only one type of unit cell in the triclinic system,which is the primitive (simple) unit cell. Due to the lack of symmetry,no other types of unit cells (like body-centered or face-centered) are possible.
Therefore,the total number of different types of unit cells present in the triclinic crystal system is $1$.