Crystal structure and Coordination number Questions in English

(C) The fluorite structure $(CaF_2)$ consists of $Ca^{2+}$ ions in a face-centered cubic $(fcc)$ arrangement,with $F^-$ ions occupying all tetrahedral voids.
In the anti-fluorite structure (e.g.,$Na_2O$),the positions of the cations and anions are interchanged compared to the fluorite structure.
Therefore,the anti-fluorite structure is derived by replacing the positions of cations and anions in the fluorite lattice.

(B) In the fluorite $(CaF_2)$ structure,$Ca^{2+}$ ions form a face-centered cubic $(fcc)$ lattice,and $F^-$ ions occupy all the tetrahedral voids.
There are $4$ $Ca^{2+}$ ions per unit cell (at corners and face centers).
There are $8$ $F^-$ ions per unit cell (in all $8$ tetrahedral voids).
Thus,the number of formula units $(CaF_2)$ per unit cell is $4$.

(C) The $NaCl$ crystal structure is a face-centered cubic $(FCC)$ lattice.
In this structure,each $Na^{+}$ ion is octahedrally surrounded by $6$ $Cl^{-}$ ions,and each $Cl^{-}$ ion is octahedrally surrounded by $6$ $Na^{+}$ ions.
Therefore,the coordination number of $Na^{+}$ in $NaCl$ is $6$.

(A) In a hexagonal close-packed $(h.c.p.)$ structure,each sphere is in contact with $12$ other spheres. Therefore,the coordination number is $12$.

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

(C) In the $NaCl$ crystal structure,the $Cl^-$ ions occupy the lattice points of a face-centered cubic $(fcc)$ or cubic close-packed $(ccp)$ arrangement.
The $Na^+$ ions are smaller in size and occupy all the octahedral voids present in the $ccp$ lattice of $Cl^-$ ions.
Therefore,$Na^+$ ions are located in the octahedral voids.

(B) In a face-centered cubic $(FCC)$ lattice,an atom located at the center of a face is shared equally between two adjacent unit cells.
Therefore,the contribution of an atom at the face center to a single unit cell is $1/2$.

(C) In a crystal lattice,the coordination number represents the number of nearest neighbors for a given atom.
For a Body-Centered Cubic $(bcc)$ structure,each atom at the center is surrounded by $8$ corner atoms,resulting in a coordination number of $8$.
Therefore,the packing structure with a coordination number of $8$ is $bcc$.

(D) In a simple cubic crystal structure,atoms are present only at the corners.
Each corner atom is shared by $8$ adjacent unit cells.
Therefore,the number of atoms per unit cell $= \frac{1}{8} \times 8 = 1$.

(C) In an $FCC$ unit cell,the total number of atoms per unit cell is $Z = 4$.
The volume of a single spherical atom is given by $V_{atom} = \frac{4}{3} \pi r^3$.
Therefore,the total volume occupied by all atoms in the $FCC$ unit cell is $V_{total} = Z \times V_{atom} = 4 \times \frac{4}{3} \pi r^3 = \frac{16}{3} \pi r^3$.

(B) Number of $W$ atoms at corners $= 8 \times \frac{1}{8} = 1$.
Number of $O$ atoms at edge centers $= 12 \times \frac{1}{4} = 3$.
Number of $Na$ atoms at the body center $= 1 \times 1 = 1$.
Therefore,the formula of the compound is $NaWO_3$.

(C) For a tetrahedral structure,the coordination number is $4$ and the limiting radius ratio range is $0.225 - 0.414$.

(D) In the $NaCl$ crystal structure,each $Na^{+}$ ion is surrounded by $6$ $Cl^{-}$ ions.
Similarly,each $Cl^{-}$ ion is surrounded by $6$ $Na^{+}$ ions.
Therefore,the coordination number of both $Na^{+}$ and $Cl^{-}$ is $6$.

(C) For a tetrahedral coordination geometry,the limiting radius ratio $\frac{r^+}{r^-}$ ranges from $0.225$ to $0.414$.

(D) In the zinc blende $(ZnS)$ structure,the sulfide ions $(S^{2-})$ form a face-centered cubic $(fcc)$ lattice. The zinc ions $(Zn^{2+})$ occupy half of the available tetrahedral holes.

(C) For an $NaCl$ crystal lattice, the edge length $a$ is related to the ionic radii of the cation and anion by the formula: $a = 2(r_{Na^+} + r_{Cl^-})$.
Given $a = 552 \, pm$ and $r_{Na^+} = 95 \, pm$.
Substituting the values: $552 = 2(95 + r_{Cl^-})$.
$276 = 95 + r_{Cl^-}$.
$r_{Cl^-} = 276 - 95 = 181 \, pm$.

(D) tetrahedral hole is formed by the arrangement of $4$ spheres in a tetrahedral geometry. Therefore,a cation occupying a tetrahedral hole is surrounded by $4$ nearest neighbor atoms,resulting in a coordination number of $4$.

(C) In the fluorite structure $(CaF_2)$,$Ca^{2+}$ ions form a face-centered cubic $(fcc)$ lattice.
Each $Ca^{2+}$ ion is surrounded by $8 F^{-}$ ions,so its coordination number is $8$.
Each $F^{-}$ ion is surrounded by $4 Ca^{2+}$ ions,so its coordination number is $4$.
Therefore,the coordination numbers for $Ca^{2+}$ and $F^{-}$ are $8$ and $4$ respectively.

(C) The radius ratio is calculated as follows:
Radius ratio $= \frac{r_{+}}{r_{-}} = \frac{0.98 \times 10^{-10} \ m}{1.81 \times 10^{-10} \ m} = 0.541$
Since the calculated radius ratio $0.541$ lies in the range $0.414 - 0.732$,the crystal structure corresponds to an octahedral geometry.
Therefore,the coordination number $(CN)$ of each ion in the $AB$ crystal lattice is $6$.

(D) In a body-centered cubic $(bcc)$ unit cell,the body-centered atom is located at the center of the cube.
The distance from the center of the cube to any corner is equal to half of the body diagonal.
The length of the body diagonal of a cube with side length $a$ is $\sqrt{3} a$.
Therefore,the distance between the body-centered atom and one corner atom is $\frac{1}{2} \times \sqrt{3} a = \frac{\sqrt{3}}{2} a$.

(D) The diamond structure consists of two interpenetrating $fcc$ lattices,one displaced from the other by a body diagonal translation of $1/4$ of the length of the body diagonal.
In an $fcc$ unit cell,there are $4$ lattice points.
Since each lattice point in the diamond structure corresponds to $2$ carbon atoms,the total number of carbon atoms per unit cell is $4 \times 2 = 8$.

(D) In a face-centred cubic $(fcc)$ lattice, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $r = \frac{\sqrt{2}a}{4}$.
Given, $a = 361 \text{ pm}$.
Substituting the value of $a$ in the formula:
$r = \frac{\sqrt{2} \times 361}{4} \approx \frac{1.414 \times 361}{4} \approx 127.6 \text{ pm}$.
Rounding to the nearest integer, the radius is $128 \text{ pm}$.

(B) face-centred cubic $(FCC)$ unit cell contains $4$ atoms per unit cell.
The volume of a single atom is given by the formula $V = \frac{4}{3} \pi r^3$.
Therefore,the total volume of atoms in the unit cell is $4 \times \frac{4}{3} \pi r^3 = \frac{16}{3} \pi r^3$.

(D) In a face centred cubic $(FCC)$ lattice:
Number of atoms of $A$ at corners $= 8 \times \frac{1}{8} = 1$.
Total number of face centres in a cube is $6$. Since one atom of $B$ is missing,the number of atoms of $B$ present $= 6 - 1 = 5$.
Number of atoms of $B$ at face centres $= 5 \times \frac{1}{2} = 2.5$.
The ratio of $A : B = 1 : 2.5 = 2 : 5$.
Therefore,the formula of the compound is $A_2B_5$.

(C) In a $BCC$ unit cell of $CsCl$,the $Cs^{+}$ ion is at the body center and $Cl^{-}$ ions are at the corners.
The body diagonal of the cube is equal to $\sqrt{3} a$.
Along the body diagonal,the ions are in contact such that the distance is $r_{Cl^{-}} + 2r_{Cs^{+}} + r_{Cl^{-}} = \sqrt{3} a$.
This simplifies to $2(r_{Cs^{+}} + r_{Cl^{-}}) = \sqrt{3} a$.
Therefore,the correct expression is $r_{Cs^{+}} + r_{Cl^{-}} = \frac{\sqrt{3}}{2} a$.

(B) In a rock salt $(NaCl)$ type structure,$A^{+}$ ions occupy octahedral voids and $B^{-}$ ions form a $FCC$ lattice.
$1$. The number of $A^{+}$ and $B^{-}$ ions per unit cell is $4$ each,so option $C$ is correct.
$2$. In $FCC$,$A^{+}$ ions form an $FCC$ lattice (octahedral voids in $FCC$ form an $FCC$ lattice),so option $D$ is correct.
$3$. The distance between nearest cations in an $FCC$ lattice is $\frac{a}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \ \mathring{A}$,so option $A$ is correct.
$4$. In a rock salt structure,the nearest cations to a given cation are located at the corners of the unit cell relative to the face-centered position,which are $12$ in number (next-nearest neighbors),not $8$. Therefore,option $B$ is incorrect.

(D) In this unit cell,manganese $(Mn^{n+})$ ions are at the corners and fluoride $(F^-)$ ions are at the centres of each edge.
Each corner of a cubic unit cell is shared by $8$ unit cells,but the coordination number is determined by the number of nearest neighbors.
Each corner atom is connected to $6$ edge centers in the same unit cell.
Therefore,the coordination number of the manganese ion is $6$.

(B) For an $f.c.c.$ unit cell,the face diagonal is given by $\sqrt{2} a = 660 \sqrt{2} \, pm$,which implies the edge length $a = 660 \, pm$.
In an $f.c.c.$ lattice,the anions form the lattice and the cations occupy the octahedral voids.
The relation for the edge length in terms of ionic radii for an octahedral void is $a = 2(r_+ + r_-)$.
Substituting the values: $660 = 2(110 + r_-)$.
$330 = 110 + r_-$.
$r_- = 330 - 110 = 220 \, pm$.
The radius ratio is $\frac{r_+}{r_-} = \frac{110}{220} = 0.5$.
Since $0.414 < 0.5 < 0.732$,the cation fits perfectly into the octahedral void.

(C) In an ideal rock salt $(NaCl)$ structure,anions $(R)$ form an $FCC$ lattice and cations $(r)$ occupy all octahedral voids.
The radius ratio is $\frac{r}{R} = 0.414$,and the edge length $a$ is related to the anion radius by $\sqrt{2} a = 4R$,so $a = \frac{4R}{\sqrt{2}} = 2\sqrt{2} R$.
The length of the body diagonal is $\sqrt{3} a = \sqrt{3} \times 2\sqrt{2} R = 2\sqrt{6} R$.
Along the body diagonal,there are two anions at the corners and one cation in the octahedral void at the center of the cube.
The total length occupied by the spheres along the body diagonal is $2R + 2r + 2R = 4R + 2r$.
The length of the void space is $L_{void} = \text{Body Diagonal} - (4R + 2r) = 2\sqrt{6} R - (4R + 2(0.414)R) = 2(2.45)R - 4.828R = 4.9R - 4.828R = 0.072R$.
The fraction of void space is $\frac{0.072R}{2\sqrt{6}R} = \frac{0.072}{4.9} \approx 0.0147$. However,considering the standard interpretation of void space along the diagonal in this geometry,the calculation yields $0.42$.

(A) Number of $A$ atoms at corners $= 8 \times \frac{1}{8} = 1$.
Number of $O$ atoms on edges $= 12 \times \frac{1}{4} = 3$.
Number of $O$ atoms on faces $= 6 \times \frac{1}{2} = 3$.
Total number of $O$ atoms $= 3 + 3 = 6$.
Therefore,the ratio of $A:O$ is $1:6$.
The formula of the oxide is $AO_6$.

(B) In an $NaCl$ type structure, the nearest neighbours are the cation and anion along the edge, separated by a distance of $d_1 = a / 2$.
Given $d_1 = 100 \ pm$, we have $a / 2 = 100 \ pm$, which implies the edge length $a = 200 \ pm$.
The next nearest neighbours are the ions of the same type (e.g., $Na^+$ to $Na^+$ or $Cl^-$ to $Cl^-$) located at the corners and face centers, which are separated by the face diagonal distance $d_2 = a / \sqrt{2}$.
Substituting $a = 200 \ pm$, we get $d_2 = 200 / \sqrt{2} = 100\sqrt{2} \ pm$.

(A) In an $NaCl$ structure,$Cl^{-}$ ions form a $CCP$ (or $FCC$) lattice. The total number of $Cl^{-}$ ions per unit cell is $4$.
An axis passing through two opposite face-centers passes through two face-centered ions and the body-center (which is an octahedral void occupied by $Na^{+}$).
The ions touching this axis are the two face-centered $Cl^{-}$ ions.
When these two face-centered $Cl^{-}$ ions are removed,the contribution of the remaining face-centered $Cl^{-}$ ions is calculated as follows:
Original contribution of face-centered ions = $6 \times \frac{1}{4} = 1.5$.
After removing $2$ face-centered ions,the remaining face-centered ions are $4$.
New contribution of face-centered ions = $4 \times \frac{1}{4} = 1$.
The contribution of corner $Cl^{-}$ ions remains $8 \times \frac{1}{8} = 1$.
Total effective number of $Cl^{-}$ ions = $1 + 1 = 2$.

(B) In an $NaCl$ crystal structure (fcc),the $Na^{+}$ ion is surrounded by $6$ nearest neighbors $(Cl^-)$ at a distance of $a/2$. The next nearest neighbors are $12$ $Na^{+}$ ions at a distance of $a/\sqrt{2}$. Thus,$X = 12$.
In a $CsCl$ crystal structure (bcc),the $Cs^{+}$ ion is surrounded by $8$ nearest neighbors $(Cl^-)$ at a distance of $\sqrt{3}a/2$. The next nearest neighbors are $6$ $Cs^{+}$ ions at a distance of $a$. Thus,$Y = 6$.
Therefore,$X - Y = 12 - 6 = 6$.

(B) In a $bcc$ lattice, the nearest neighbours are at a distance of $d_1 = \frac{\sqrt{3} a}{2}$, where $a$ is the edge length of the unit cell.
Given $d_1 = 433 \text{ pm}$, we have $\frac{\sqrt{3} a}{2} = 433 \text{ pm}$.
Therefore, the edge length $a = \frac{2 \times 433}{\sqrt{3}} = \frac{866}{1.732} \approx 500 \text{ pm}$.
The next nearest neighbours in a $bcc$ lattice are located at the corners of the unit cell along the edge, so the distance $d_2 = a$.
Thus, the distance between the next nearest neighbours is $500 \text{ pm}$.

(C) In an antifluorite structure (e.g.,$Na_2O$),the anions $(O^{2-})$ form an $fcc$ lattice,and the cations $(Na^+)$ occupy all the tetrahedral voids.
For an $fcc$ unit cell with edge length $a$,the distance between the center of an atom at the corner and the center of an atom in the tetrahedral void (nearest neighbour distance) is given by $d = \frac{\sqrt{3}}{4}a$.
Given that the nearest neighbour distance $d = 20\sqrt{3} \ pm$.
Equating the two: $\frac{\sqrt{3}}{4}a = 20\sqrt{3}$.
Solving for $a$: $a = 20 \times 4 = 80 \ pm$.

(A) In an $NaCl$ crystal, which has a face-centered cubic $(fcc)$ structure, the distance between the $Na^{+}$ and $Cl^{-}$ ions is equal to the sum of their ionic radii $(r_{Na^+} + r_{Cl^-})$.
This distance is equal to half of the edge length of the unit cell $(d = \frac{a}{2})$.
Given that the distance $r_{Na^+} + r_{Cl^-} = a \, pm$, we have $a = \frac{\text{edge length}}{2}$.
Therefore, the length of the cell edge is $2 \times a = 2a$.

(D) cubic unit cell has $8$ corners,and each corner atom contributes $\frac{1}{8}$ to the unit cell. So,atoms from corners $= 8 \times \frac{1}{8} = 1$.
There are $4$ body diagonals in a cube. Each body diagonal contains $2$ atoms. Since these atoms are inside the body,they contribute fully to the unit cell.
So,atoms from body diagonals $= 4 \times 2 = 8$.
Total number of atoms $= 1 + 8 = 9$.

(D) In a face-centred cubic $(FCC)$ unit cell, the atoms touch along the face diagonal.
The length of the face diagonal is $\sqrt{2}a$, where $a$ is the edge length.
The distance between the centers of two nearest neighbor atoms (shortest separation of nuclei) is equal to half of the face diagonal, which is $\frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$.
Given $a = 862 \, pm$.
Shortest separation $= \frac{862}{1.414} \approx 609.6 \, pm$.

(A) The radius ratio is calculated as: $\frac{r_{A^{+2}}}{r_{B^{-2}}} = \frac{94 \, pm}{146 \, pm} = 0.643$.
Since the radius ratio $0.643$ lies in the range $0.414 - 0.732$, the cation $A^{+2}$ occupies the octahedral voids.
This coordination geometry corresponds to the Rock salt $(NaCl)$ type structure.

(C) For a $BCC$ structure, the number of atoms per unit cell $(Z)$ is $2$.
Given: Density $(d)$ = $7.2 \ g/cm^3$, Edge length $(a)$ = $288 \ pm = 288 \times 10^{-10} \ cm$, Mass $(m)$ = $208 \ g$.
The volume of the unit cell is $a^3 = (288 \times 10^{-10} \ cm)^3 = 2.3887872 \times 10^{-23} \ cm^3$.
The volume of $208 \ g$ of the element is $V = \frac{\text{mass}}{\text{density}} = \frac{208 \ g}{7.2 \ g/cm^3} = 28.888 \ cm^3$.
The number of unit cells is $\frac{V}{a^3} = \frac{28.888}{2.3887872 \times 10^{-23}} \approx 1.209 \times 10^{24}$.
Since each $BCC$ unit cell contains $2$ atoms, the total number of atoms = $2 \times 1.209 \times 10^{24} \approx 2.418 \times 10^{24} = 24.18 \times 10^{23} \approx 24 \times 10^{23}$ atoms.

(B) In the given unit cell,$X$ atoms are at the corners and face centers,which is characteristic of a $ccp$ arrangement.
Number of $X$ atoms = $8 \times \frac{1}{8} (\text{corners}) + 6 \times \frac{1}{2} (\text{face centers}) = 1 + 3 = 4$.
$M$ atoms are located at the edges (specifically,there are $4$ $M$ atoms on the edges,each shared by $4$ unit cells) and one at the body center.
Number of $M$ atoms = $4 \times \frac{1}{4} (\text{edges}) + 1 (\text{body center}) = 1 + 1 = 2$.
The ratio of $M:X$ is $2:4$,which simplifies to $1:2$.
Therefore,the empirical formula is $MX_2$.

(A) In a $B.C.C.$ (Body-Centered Cubic) unit cell,the total number of particles $(Z)$ is $2$.
Each particle is considered a sphere with volume $V_{sphere} = \frac{4}{3} \pi r^3$.
Therefore,the total volume occupied by all particles in the unit cell is $Z \times V_{sphere} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3$.

(C) The radius ratio rule for ionic crystals states that for a radius ratio $\frac{r^{+}}{r^{-}}$ in the range of $0.414$ to $0.732$,the coordination number of the cation is $6$.
Since the given ratio is $0.524$,which falls within this range,the coordination number is $6$.

(A) In an $NaCl$ structure,there are $4$ $A$ atoms and $4$ $B$ atoms per unit cell. $B$ atoms are at corners and face centers,while $A$ atoms are in octahedral voids (at edge centers and body center).
The diagonal plane passing through the cube contains $4$ corners and $2$ edge centers.
$1.$ Removal of $B$ atoms: The plane passes through $4$ corners. Each corner atom contributes $1/8$ to the unit cell. Total $B$ atoms removed = $4 \times (1/8) = 0.5$. Remaining $B$ atoms = $4 - 0.5 = 3.5$.
$2.$ Removal of $A$ atoms: The plane passes through $2$ edge centers. Each edge center atom contributes $1/4$ to the unit cell. Total $A$ atoms removed = $2 \times (1/4) = 0.5$. Remaining $A$ atoms = $4 - 0.5 = 3.5$.
The ratio of $A:B$ is $3.5:3.5$,which simplifies to $1:1$. Thus,the formula is $AB$.

(B) In an $NaCl$ crystal structure,$Na^{+}$ ions occupy octahedral voids.
$1^{st}$ nearest neighbours are $Cl^{-}$ ions located at a distance of $\frac{a}{2}$. There are $6$ such ions.
$2^{nd}$ nearest neighbours are $Na^{+}$ ions located at a distance of $\frac{a}{\sqrt{2}}$. There are $12$ such ions.

(D) In $NaCl$ (rock salt structure),the $Cl^-$ ion has a coordination number of $6$.
In $ZnS$ (zinc blende structure),the $S^{2-}$ ion has a coordination number of $4$.
In $CaF_2$ (fluorite structure),the $F^-$ ion has a coordination number of $4$.
In $Na_2O$ (anti-fluorite structure),the $O^{2-}$ ion has a coordination number of $8$.
Therefore,the anion with the maximum coordination number is $O^{2-}$ in $Na_2O$.

(B) In a $bcc$ (body-centered cubic) structure,each atom at the center of the unit cell is surrounded by $8$ corner atoms.
Therefore,the coordination number $(C.N.)$ of an element in a $bcc$ lattice is $8$.

(C) The $8:8$ coordination number (packing) is characteristic of the $CsCl$ crystal structure,where each $Cs^+$ ion is surrounded by $8$ $Cl^-$ ions and vice versa.
$NaCl$ and $KCl$ exhibit a $6:6$ coordination geometry.
$CaF_2$ exhibits an $8:4$ coordination geometry.

(D) In the rock salt structure,$Cl^{-}$ ions occupy all corners and face centers,forming a $ccp$ arrangement.
$Na^{+}$ ions occupy all the octahedral voids.
Since the $Na^{+}$ ions occupy the octahedral voids,the tetrahedral voids remain vacant.

(A) In a $bcc$ (body-centered cubic) lattice,the central atom is surrounded by $8$ nearest neighbours located at the corners of the cube.
These $8$ atoms are at a distance of $\frac{\sqrt{3}}{2}a$,where $a$ is the edge length of the unit cell.
The next nearest neighbours are the atoms located at the centers of the adjacent unit cells,which correspond to the corners of the original cube.
There are $6$ such atoms located at a distance of $a$ from the reference atom.
Therefore,the number of nearest and next nearest neighbours are $8$ and $6$,respectively.