Crystal structure and Coordination number Questions in English

(N/A) The crystal structure of $NaCl$ consists of a face-centered cubic $(fcc)$ lattice where $Cl^{-}$ ions occupy the corners and face centers,while $Na^{+}$ ions occupy the octahedral voids (edge centers and body center). The formation of $NaCl$ from its gaseous ions is described by the Born-Haber cycle:
$(i)$ Ionization enthalpy $(\Delta_{i} H)$ :
$Na_{(g)} \rightarrow Na_{(g)}^{+} + e^{-} \ldots \Delta_{i} H = 495.8 \ kJ \ mol^{-1}$
$(ii)$ Electron gain enthalpy $(\Delta_{eg} H)$ :
$Cl_{(g)} + e^{-} \rightarrow Cl_{(g)}^{-} \ldots \Delta_{eg} H = -348.7 \ kJ \ mol^{-1}$
Sum of $(i) + (ii) = 147.1 \ kJ \ mol^{-1}$ (Endothermic step).
$(iii)$ Lattice enthalpy $(\Delta_{L} H)$ :
$Na_{(g)}^{+} + Cl_{(g)}^{-} \rightarrow NaCl_{(s)} \ldots \Delta_{L} H = -788 \ kJ \ mol^{-1}$
Total enthalpy change $= (i) + (ii) + (iii) = -640.9 \ kJ \ mol^{-1}$.
Since the total enthalpy change is negative,the formation of $NaCl$ is an exothermic process,and the crystal lattice is stable.

(N/A) - $A$ primitive cubic unit cell has atoms only at its corners. Each atom at the corner is shared between $8$ adjacent unit cells as shown in the figure. Four unit cells are in the same layer and four unit cells are in the layer above or below.
- Only $\frac{1}{8}$ of an atom,molecule,or ion is included in a particular unit cell.
- In the following figure,primitive cubic unit cells are shown: $(a)$ open structure,$(b)$ space-filling structure,$(c)$ actual portion of atoms included in each unit cell. In figure $(a)$,each small sphere represents the center of the particle at its position. It does not represent the actual size. Such structures are called open structures. Overall,there are $8$ atoms at the corners of each cubic unit cell.
- $\therefore$ Total number of atoms in one unit cell $= 8 \times \frac{1}{8} = 1$.

(B) In a body-centered cubic $(bcc)$ unit cell:
$1$. There is one atom at each of the $8$ corners of the cube. Each corner atom contributes $\frac{1}{8}$ to the unit cell.
Contribution from corners $= 8 \times \frac{1}{8} = 1$ atom.
$2$. There is one atom at the body center of the cube,which is entirely contained within the unit cell.
Contribution from the body center $= 1 \times 1 = 1$ atom.
$3$. Total number of atoms per unit cell $= 1 + 1 = 2$ atoms.

(C) In a face-centered cubic $(fcc)$ unit cell,atoms are present at all the corners and at the center of all the faces of the cube.
$(i)$ Each atom at the corner is shared by $8$ adjacent unit cells,so the contribution of each corner atom to the unit cell is $\frac{1}{8}$. Since there are $8$ corners,the total contribution from corner atoms is $8 \times \frac{1}{8} = 1$ atom.
$(ii)$ Each atom at the face center is shared by $2$ adjacent unit cells,so the contribution of each face-centered atom to the unit cell is $\frac{1}{2}$. Since there are $6$ faces,the total contribution from face-centered atoms is $6 \times \frac{1}{2} = 3$ atoms.
Therefore,the total number of atoms per unit cell in an $(fcc)$ structure is $1 + 3 = 4$ atoms.

(B) In a body-centered cubic $(BCC)$ unit cell,atoms are present at all the $8$ corners and one atom is present at the center of the body.
Each corner atom contributes $1/8$ to the unit cell.
Contribution from $8$ corners = $8 \times (1/8) = 1$ atom.
The atom at the center of the body belongs entirely to the unit cell,contributing $1$ atom.
Total number of atoms per unit cell = $1 + 1 = 2$ atoms.
Therefore,the correct option is $B$.

(A) In an $fcc$ unit cell,there are $8$ corners and $6$ face centers.
Number of atoms $A$ at corners = $8 \times \frac{1}{8} = 1$.
If one atom $A$ is missing,the number of atoms $A = 1 - \frac{1}{8} = \frac{7}{8}$.
Number of atoms $B$ at face centers = $6 \times \frac{1}{2} = 3$.
The ratio of $A:B = \frac{7}{8} : 3 = 7 : 24$.
Therefore,the molecular formula is $A_7B_{24}$.

(C) In an $fcc$ unit cell,octahedral voids $(O.V.)$ are located at the centre of each edge and at the body centre.
The distance between two nearest octahedral voids located at the centres of two adjacent edges is calculated using the Pythagorean theorem.
The coordinates of the octahedral voids at the centres of two adjacent edges (e.g.,at $(a/2, 0, 0)$ and $(0, a/2, 0)$) result in a distance of:
$d = \sqrt{(\frac{a}{2} - 0)^2 + (0 - \frac{a}{2})^2 + (0 - 0)^2}$
$d = \sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2}$
$d = \sqrt{\frac{a^2}{4} + \frac{a^2}{4}} = \sqrt{\frac{2a^2}{4}} = \sqrt{\frac{a^2}{2}} = \frac{a}{\sqrt{2}}$

(A) Let the number of oxide ions $(O^{2-})$ in the $ccp$ lattice be $4$.
Number of octahedral voids $(O.V.)$ = $4$.
Number of tetrahedral voids $(T.V.)$ = $2 \times 4 = 8$.
$M_1$ occupies $50 \%$ of $O.V. = 0.50 \times 4 = 2$.
$M_2$ occupies $12.5 \%$ of $T.V. = 0.125 \times 8 = 1$.
The formula of the crystal is $(M_1)_2(M_2)_1O_4$.
For the crystal to be electrically neutral,the sum of oxidation states must be zero:
$2 \times (\text{O.N. of } M_1) + 1 \times (\text{O.N. of } M_2) + 4 \times (-2) = 0$
$2 \times (\text{O.N. of } M_1) + (\text{O.N. of } M_2) = 8$.
Checking option $A$: $2(+2) + (+4) = 4 + 4 = 8$. This satisfies the condition.

(B) In a unit cell:
Number of $W$ atoms at the corners $= 8 \times \frac{1}{8} = 1$
Number of $O$ atoms at the centre of edges $= 12 \times \frac{1}{4} = 3$
Number of $Na$ atoms at the centre of the cube $= 1 \times 1 = 1$
Therefore,the ratio of atoms is $Na:W:O = 1:1:3$.
Hence,the formula of the compound is $NaWO_{3}$.

(A) In a body-centered cubic $(BCC)$ unit cell,an atom is present at the center of the cube.
This central atom is surrounded by $8$ corner atoms,each of which is at an equal distance from the center.
Therefore,the coordination number of an atom in a $BCC$ structure is $8$.

(D) In a crystal lattice,if the number of lattice points (atoms) is $N$,then the number of octahedral voids is equal to $N$.
Therefore,the number of octahedral voids per lattice site is given by the ratio $\frac{N}{N} = 1$.

(B) In a cubic close packed $(CCP)$ arrangement,the number of anions $(A^-)$ per unit cell is $4$.
The number of octahedral voids in a $CCP$ lattice is equal to the number of atoms forming the lattice,which is $4$.
Since cations $(B^+)$ occupy all octahedral sites,the number of cations per unit cell is $4$.
The formula of the unit cell is $A_4B_4$,which simplifies to the empirical formula $AB$.
Comparing $AB$ with $A_xB$,we get $x = 1$.

(D) In a face-centered cubic $(FCC)$ lattice,the number of atoms at the corners is $8 \times (1/8) = 1$ and the number of atoms at the face centers is $6 \times (1/2) = 3$,totaling $4$ atoms.
In the diamond structure,carbon atoms occupy all the $FCC$ lattice points ($4$ atoms) and $50\,\%$ of the tetrahedral voids.
The total number of tetrahedral voids in an $FCC$ unit cell is $8$.
Number of carbon atoms in tetrahedral voids $= 50\,\% \text{ of } 8 = 4$.
Total number of carbon atoms per unit cell $= 4 \text{ (lattice points)} + 4 \text{ (tetrahedral voids)} = 8$.

(C) The radius ratio is calculated as:
$Radius \ ratio = \frac{r_{+}}{r_{-}} = \frac{148 \ pm}{195 \ pm} \approx 0.759$.
Since the radius ratio lies in the range $0.732 - 1.000$, the coordination number is $8$, which corresponds to the $Cesium \ chloride \ (CsCl)$ structure.

(A) In a $ccp$ arrangement,the number of $A$ atoms per unit cell is $4$ (corners + face centers).
The number of octahedral sites in a $ccp$ unit cell is equal to the number of atoms,which is $4$.
When two atoms from opposite faces are removed,we remove two $A$ atoms from the face centers.
New number of $A$ atoms = $4 - (2 \times \frac{1}{2}) = 4 - 1 = 3$.
The $B$ atoms occupy all octahedral sites. There are $4$ octahedral sites (one at the body center and $12$ at the edges,each shared by $4$ unit cells: $1 + 12 \times \frac{1}{4} = 4$).
Since the removal of face-centered $A$ atoms does not affect the octahedral sites,the number of $B$ atoms remains $4$.
Thus,the stoichiometry is $A_{3} B_{4}$.
The value of $x$ is $3$.

(D) The chemical formula of calcium fluoride is $CaF_2$.
In the $CaF_2$ crystal structure (fluorite structure),$Ca^{2+}$ ions form a face-centered cubic $(fcc)$ lattice.
The number of $Ca^{2+}$ ions per unit cell is $4$ (since $8$ corners $\times 1/8 + 6$ faces $\times 1/2 = 4$).
Since the stoichiometry is $1:2$,for every $Ca^{2+}$ ion,there are $2$ $F^{-}$ ions.
Therefore,the number of $F^{-}$ ions per unit cell is $4 \times 2 = 8$.

(C) The correct option is $C$.
$A$ face-centered cubic $(fcc)$ unit cell contains atoms at all the corners and at the center of all faces of the cube.
Thus,the number of atoms in $fcc = (\frac{1}{8} \times 8) + (\frac{1}{2} \times 6) = 1 + 3 = 4$.
$A$ body-centered cubic $(bcc)$ unit cell has an atom at each of its corners and also one atom at its body center.
Thus,the number of atoms in $bcc = (\frac{1}{8} \times 8) + 1 = 1 + 1 = 2$.

(D) Total corners in a cube are $8$. Atoms of $X$ are at alternate corners,so number of $X$ atoms at corners = $8 \times \frac{1}{2} \times \frac{1}{8} = 0.5$.
One $X$ atom is at the center of the cube,so total $X = 0.5 + 1 = 1.5$.
Total faces in a cube are $6$. Atoms of $Y$ are at $\frac{1}{3}$ of the total faces,so number of $Y$ atoms = $6 \times \frac{1}{3} \times \frac{1}{2} = 1$.
Wait,let us re-evaluate: $Y$ is at $\frac{1}{3}$ of the total faces,meaning $6 \times \frac{1}{3} = 2$ faces. Each face contributes $\frac{1}{2}$ to the unit cell.
So,$Y = 2 \times \frac{1}{2} = 1$.
The ratio $X:Y = 1.5:1 = 3:2$.
The formula is $X_3 Y_2$.

(C) In a $CsCl$ crystal structure,the $Cs^{+}$ ion is located at the body center,and $Cl^{-}$ ions are located at the corners of the cube.
The body diagonal of the cube is given by $\sqrt{3} a$.
Since the $Cs^{+}$ ion at the center touches the $Cl^{-}$ ions at the corners along the body diagonal,the length of the body diagonal is equal to $2(r_{Cs^{+}} + r_{Cl^{-}})$.
Therefore,$2(r_{Cs^{+}} + r_{Cl^{-}}) = \sqrt{3} a$.
This simplifies to $r_{Cs^{+}} + r_{Cl^{-}} = \frac{\sqrt{3}}{2} a$.

(B) truncated octahedron is an Archimedean solid.
It is formed by cutting off the corners of a regular octahedron.
It consists of $6$ square faces and $8$ hexagonal faces.
Therefore,the number of hexagonal faces present in a truncated octahedron is $8$.

(B) In the given unit cell,$X$ atoms are at the corners and face centers (typical $ccp$ arrangement).
Number of $X$ atoms $= (8 \text{ corners} \times \frac{1}{8}) + (6 \text{ face centers} \times \frac{1}{2}) = 1 + 3 = 4$.
$M$ atoms are located at the edges and the body center.
Number of $M$ atoms $= (4 \text{ edges} \times \frac{1}{4}) + (1 \text{ body center} \times 1) = 1 + 1 = 2$.
Ratio of $M:X = 2:4 = 1:2$.
Therefore,the empirical formula is $MX_2$.

(C) In $NaCl$ structure,cations $(M)$ occupy $FCC$ positions and anions $(X)$ occupy octahedral voids:
Initial positions:
$M$: $8$ corners + $6$ face-centers
$X$: $12$ edge-centers + $1$ body-center (central)
Following the instructions:
$(i)$ Remove all $X$ except the central one: Remaining $X = 1$ (body-center).
$(ii)$ Replace face-centered $M$ with $X$: $X = 1$ (body-center) + $6$ (face-centers); $M = 8$ (corners).
$(iii)$ Remove $M$ at corners: $M = 0$.
$(iv)$ Replace central $X$ with $M$: $M = 1$ (body-center); $X = 6$ (face-centers).
In the resulting structure $Z$:
Number of anions $(X) = 6 \times \frac{1}{2} = 3$
Number of cations $(M) = 1 \times 1 = 1$
Ratio $\left(\frac{\text{number of anions}}{\text{number of cations}}\right) = \frac{3}{1} = 3$.

(A) The figure shows an $A^{+}$ ion surrounded by three $X^{-}$ ions in a triangular planar arrangement, which corresponds to a trigonal planar void.
For a trigonal planar void, the limiting radius ratio is given by:
$\frac{r_{A^{+}}}{r_{X^{-}}} = 0.155$
Given $r_{X^{-}} = 250 \ pm$, we calculate:
$r_{A^{+}} = 0.155 \times 250 \ pm = 38.75 \ pm$.
However, looking at the options provided and the geometry of the figure, it represents a coordination number of $3$. If we assume the question implies the minimum radius for stability in this geometry, the value is $38.75 \ pm$. Given the options, there might be a discrepancy in the provided figure or options. Re-evaluating the figure, it shows a central ion in a triangular hole. Based on standard chemistry problems of this type, if the intended geometry was octahedral, the ratio is $0.414$, giving $103.5 \ pm$, which is closest to $104 \ pm$.

(C) For an $fcc$ unit cell, the relation between edge length $(a)$ and atomic radius $(r)$ is $a = 2\sqrt{2}r$.
Given $r = 141.4 \ pm = 1.414 \times 10^{-8} \ cm$.
Substituting the value of $r$: $a = 2 \times 1.414 \times 1.414 \times 10^{-8} \ cm = 4.0 \times 10^{-8} \ cm$.
The volume of the unit cell is $V = a^3 = (4.0 \times 10^{-8} \ cm)^3 = 64 \times 10^{-24} \ cm^3 = 6.4 \times 10^{-23} \ cm^3$.

(C) In a $bcc$ (body-centered cubic) unit cell,the number of particles $(Z)$ is $2$.
The packing efficiency of a $bcc$ unit cell is $68\%$.
The volume occupied by the particles is given by: $\text{Volume} = \text{Packing Efficiency} \times \text{Volume of unit cell}$.
$\text{Volume} = 0.68 \times 8.0 \times 10^{-23} \ cm^3$.
$\text{Volume} = 5.44 \times 10^{-23} \ cm^3$.

(C) In a crystal lattice,a $bcc$ (body-centered cubic) unit cell is a type of cubic system.
In any cubic unit cell,each corner particle is shared by $8$ adjacent unit cells.
This is because each corner is a point where $8$ cubes meet in a three-dimensional space.

(C) In a simple cubic $(SC)$ unit cell,each particle is at the corner of the cube.
Each particle is in contact with $6$ nearest neighbors (one along each axis: $x, -x, y, -y, z, -z$).
Therefore,the coordination number of a particle in a simple cubic structure is $6$.

(B) In a $bcc$ unit cell,there are $8$ corner atoms,each contributing $\frac{1}{8}$ to the unit cell,and $1$ body-centered atom,which contributes $1$ to the unit cell.
Total number of particles = $(\frac{1}{8} \times 8) + 1 = 1 + 1 = 2$.

(D) In $hcp$ structures,each sphere is surrounded by $12$ neighbouring spheres: $6$ in its own layer,$3$ in the layer above,and $3$ in the layer below.
Therefore,the coordination number of any sphere in an $hcp$ structure is $12$.

(B) In a face-centered cubic $(fcc)$ unit cell,atoms are present at the corners and at the centers of each face.
Number of atoms at corners $= 8 \times \frac{1}{8} = 1$.
Number of atoms at face centers $= 6 \times \frac{1}{2} = 3$.
Total number of atoms $= 1 + 3 = 4$.

(A) In a face-centered cubic $(fcc)$ lattice,each atom is in contact with $4$ atoms in its own layer,$4$ atoms in the layer above,and $4$ atoms in the layer below.
Therefore,the total coordination number of an atom in an $fcc$ structure is $4 + 4 + 4 = 12$.

(A) For a simple cubic unit cell,the number of atoms per unit cell $(n)$ is $1$.
Total volume occupied by atoms in the unit cell $= n \times \frac{4}{3} \pi r^3$.
Given $r = 3 \times 10^{-8} \ cm$.
Volume $= 1 \times \frac{4}{3} \times 3.14159 \times (3 \times 10^{-8} \ cm)^3$.
Volume $= \frac{4}{3} \times 3.14159 \times 27 \times 10^{-24} \ cm^3$.
Volume $= 4 \times 3.14159 \times 9 \times 10^{-24} \ cm^3$.
Volume $= 113.097 \times 10^{-24} \ cm^3 = 1.13 \times 10^{-22} \ cm^3$.

(C) In a base-centred unit cell,particles are present at the $8$ corners and at the centers of $2$ opposite faces.
Contribution from $8$ corners $= 8 \times \frac{1}{8} = 1$.
Contribution from $2$ face centres $= 2 \times \frac{1}{2} = 1$.
Total number of particles $= 1 + 1 = 2$.

(A) The number of atoms in one mole of a metal is $6.022 \times 10^{23}$.
In a simple cubic unit cell,the number of atoms per unit cell is $1$.
Therefore,the number of unit cells $= \frac{\text{Total number of atoms}}{\text{Number of atoms per unit cell}} = \frac{6.022 \times 10^{23}}{1} = 6.022 \times 10^{23}$.

(C) The number of atoms of $A$ per unit cell (at corners) = $8 \times (1/8) = 1$.
The number of atoms of $B$ per unit cell (at face centres) = $6 \times (1/2) = 3$.
Therefore,the ratio of atoms $A:B = 1:3$.
Hence,the formula of the compound is $AB_3$.

(A) The $ccp$ (cubic close-packed) structure is also known as $fcc$ (face-centered cubic) structure.
Among the given options,$Cu$ (Copper) crystallizes in the $fcc$ $(ccp)$ structure.
$Zn$ (Zinc) and $Mg$ (Magnesium) crystallize in $hcp$ (hexagonal close-packed) structure.
$Po$ (Polonium) crystallizes in a simple cubic structure.

(D) For $fcc$ crystal structure, the relation between edge length $a$ and atomic radius $r$ is given by $4r = \sqrt{2}a$.
Therefore, $r = \frac{\sqrt{2}a}{4}$.
Substituting the given value $a = 393 \ pm$:
$r = \frac{1.414 \times 393}{4} = 138.93 \ pm$.

(B) In a hexagonal close-packed $(hcp)$ crystal lattice,each atom is in contact with $12$ other atoms.
Therefore,the coordination number of the $hcp$ crystal lattice is $12$.

(B) The effective number of particles in a unit cell is calculated as follows:
For a simple cubic unit cell,$Z = 1$.
For a body-centred cubic $(BCC)$ unit cell,$Z = 2$.
For a face-centred cubic $(FCC)$ unit cell,$Z = 8 \times (1/8) + 6 \times (1/2) = 1 + 3 = 4$.
Therefore,the unit cell with four particles is the face-centred cubic $(FCC)$ unit cell.

(A) In a simple cubic lattice,each sphere is in contact with $6$ nearest neighbors. Therefore,the coordination number is $6$.

(C) In a face-centered cubic $(fcc)$ crystal lattice,each atom is in contact with $12$ nearest neighbors.
Therefore,the coordination number of a particle in an $fcc$ lattice is $12$.

(C) In one $BCC$ crystal unit cell,the number of atoms $(Z) = 2$.
Total number of atoms $= Z \times \text{Number of unit cells}$.
Total number of atoms $= 2 \times 1.8 \times 10^{20} = 3.6 \times 10^{20}$.

(A) In a face-centred cubic $(FCC)$ unit cell,particles are present at each of the $8$ corners and at the centre of each of the $6$ faces of the cube.

(C) In an $FCC$ unit cell,the number of atoms at the corners is $8 \times \frac{1}{8} = 1$.
Since ions of $A$ are at the corners,the number of $A$ atoms per unit cell is $1$.
In an $FCC$ unit cell,the number of atoms at the face centres is $6 \times \frac{1}{2} = 3$.
Since ions of $B$ are at the face centres,the number of $B$ atoms per unit cell is $3$.
Therefore,the ratio of $A:B$ is $1:3$.
The formula of the compound is $AB_3$.

(A) In a $BCC$ unit cell,the number of particles $(Z) = 2$.
The relation between edge length '$a$' and radius '$r$' is $\sqrt{3} a = 4 r$,which gives $r = \frac{\sqrt{3} a}{4}$.
The volume occupied by particles is calculated as $Z \times \text{Volume of one sphere}$.
$\text{Volume} = 2 \times \frac{4}{3} \pi r^3$.
Substituting $r$: $\text{Volume} = 2 \times \frac{4}{3} \pi (\frac{\sqrt{3} a}{4})^3$.
$\text{Volume} = \frac{8}{3} \pi \times \frac{3 \sqrt{3} a^3}{64}$.
$\text{Volume} = \frac{\sqrt{3} \pi a^3}{8}$.

(D) In $BCC$ structure,atoms are present at each corner and at the body center of the unit cell.
$No. \text{ of atoms per unit cell } (Z) = 8 \times \frac{1}{8} + 1 \times 1$
$= 1 + 1 = 2$
Therefore,there are $2$ particles per unit cell in a $BCC$ structure.

(B) The correct answer is in between $0.414$ and $0.732$.
For an ionic crystal with a coordination number of $6$,the structure is octahedral.
The radius ratio $(r_+ / r_-)$ for an octahedral geometry is mathematically derived to be in the range of $0.414$ to $0.732$.
An example of such a crystal is $NaCl$.

(D) The packing efficiency of a $bcc$ (body-centered cubic) unit cell is $68 \%$.
The void space is calculated as $100 \% - \text{Packing Efficiency}$.
Therefore,the void space $= 100 \% - 68 \% = 32 \%$.

(C) In an $fcc$ unit cell,the number of atoms per unit cell is $Z = 4$.
The mass of the unit cell is equal to the mass of $4$ atoms.
Mass of unit cell $= 4 \times (6 \times 10^{-23} \ g) = 24 \times 10^{-23} \ g$.
Converting to scientific notation,we get $2.4 \times 10^{-22} \ g$.

(C) In a cubic close packing $(ccp)$ or face-centered cubic $(fcc)$ structure,each atom is surrounded by $12$ nearest neighbors.
Specifically,an atom at a face center is in contact with $4$ atoms in its own plane,$4$ atoms in the plane above,and $4$ atoms in the plane below.
Therefore,the coordination number is $12$.