Crystal structure and Coordination number Questions in English

(B) The $CsCl$ crystal structure is a body-centered cubic $(bcc)$ type lattice.
In this structure,each $Cs^+$ ion is located at the center of the cube and is surrounded by $8$ $Cl^-$ ions located at the corners of the cube.
Similarly,each $Cl^-$ ion is surrounded by $8$ $Cs^+$ ions.
Therefore,the coordination number of both $Cs^+$ and $Cl^-$ is $8$.

(D) The number of $X$ atoms at the corners of the cube is $8 \times \frac{1}{8} = 1$.
The number of $Y$ atoms at the face centers of the cube is $6 \times \frac{1}{2} = 3$.
Therefore,the ratio of $X:Y$ is $1:3$.
The molecular formula of the compound is $XY_3$.

(B) In the $NaCl$ crystal structure,both $Na^{+}$ and $Cl^{-}$ ions occupy octahedral sites,resulting in a coordination number of $6:6$.
Therefore,the statement that $Na^{+}$ ions have a coordination number of $4$ is incorrect.

(B) In the $CsCl$ crystal structure,$Cl^{-}$ ions occupy the corners of a simple cubic unit cell,and the $Cs^{+}$ ion occupies the body center.
Each $Cs^{+}$ ion is surrounded by $8$ $Cl^{-}$ ions,and each $Cl^{-}$ ion is surrounded by $8$ $Cs^{+}$ ions.
Therefore,the coordination number of both $Cs^{+}$ and $Cl^{-}$ is $8$.

(D) In an $NaCl$ structure,$A$ atoms (cations) occupy the corners and face centers,while $B$ atoms (anions) occupy the edge centers and body center.
Initial number of $A$ atoms per unit cell $= 8 \times \frac{1}{8} (\text{corners}) + 6 \times \frac{1}{2} (\text{face-centers}) = 1 + 3 = 4$.
Initial number of $B$ atoms per unit cell $= 12 \times \frac{1}{4} (\text{edge-centers}) + 1 (\text{body-center}) = 3 + 1 = 4$.
Removing face-centered atoms along one of the axes means removing $2$ face-centered $A$ atoms.
New number of $A$ atoms $= 4 - 2 = 2$.
Wait,the question states $A$ atoms occupy corners. In $NaCl$,$A$ is typically the cation. If $A$ is at corners and face centers,and $B$ is at edge centers and body center,removing $2$ face-centered $A$ atoms leaves $A = 8 \times \frac{1}{8} + 4 \times \frac{1}{2} = 1 + 2 = 3$.
Number of $B$ atoms remains $4$.
Therefore,the stoichiometry is $A_3B_4$.

(A) In the $CsCl$ crystal structure,the $Cs^{+}$ ion occupies the body-centered position of the cubic unit cell.
Each $Cs^{+}$ ion is surrounded by $8$ $Cl^{-}$ ions at the corners of the cube.
Conversely,each $Cl^{-}$ ion is surrounded by $8$ $Cs^{+}$ ions.
Therefore,the coordination number for both $Cs^{+}$ and $Cl^{-}$ is $8:8$.

(B) The radius ratio is calculated as: $\frac{r_{+}}{r_{-}} = \frac{180 \ pm}{187 \ pm} = 0.962$.
Since the radius ratio $0.962$ lies in the range of $0.732 - 1.000$, the coordination number of the ions is $8$.
Therefore, the crystal structure of the compound is $CsCl$ type.

(B) In a hexagonal close packed $(hcp)$ structure,each atom is in contact with $12$ other atoms.
Therefore,the coordination number of a metal crystallizing in an $hcp$ structure is $12$.

(C) The structure of $MgO$ is similar to $NaCl$,which crystallizes in a rock salt structure.
In the $NaCl$ crystal lattice,both the cation $(Na^+)$ and the anion $(Cl^-)$ occupy octahedral voids.
Since $Mg^{2+}$ ions occupy the octahedral sites in the $MgO$ lattice,each $Mg^{2+}$ ion is surrounded by $6$ $O^{2-}$ ions.
Therefore,the coordination number of magnesium is $6$.

(D) The coordination number of an ionic crystal is determined by the radius ratio $(r_+ / r_-)$.
For a radius ratio in the range of $0.225 - 0.414$,the coordination number is $4$ (tetrahedral).
For a radius ratio in the range of $0.414 - 0.732$,the coordination number is $6$ (octahedral).
For a radius ratio in the range of $0.732 - 1.000$,the coordination number is $8$ (cubic).

(C) The radius ratio determines the coordination number of an ionic crystal.
For a radius ratio in the range $[0.225 - 0.414]$,the coordination number is $4$.
For a radius ratio in the range $[0.414 - 0.732]$,the coordination number is $6$.
For a radius ratio in the range $[0.732 - 1.0]$,the coordination number is $8$.
Therefore,for the range $0.414 - 0.732$,the coordination number is $6$.

(B) In $Na_2O$,the structure is an antifluorite structure.
Each oxide ion $(O^{2-})$ is surrounded by $8$ $Na^{+}$ ions.
Each sodium ion $(Na^{+})$ is surrounded by $4$ oxide ions $(O^{2-})$.
Therefore,the coordination number of sodium is $4$ and the coordination number of oxygen is $8$.

(B) The radius ratio rule states that for an ionic crystal,the coordination number depends on the ratio of the radius of the cation $(r^+)$ to the radius of the anion $(r^-)$.
For a radius ratio in the range $0.732 - 1.000$,the coordination number is $8$.
This corresponds to a body-centred cubic $(BCC)$ type structure.

(C) In the $CsCl$ crystal lattice,the structure is body-centered cubic $(bcc)$.
Each $Cs^{+}$ ion is located at the center of the cube and is surrounded by $8$ $Cl^{-}$ ions located at the corners of the cube.
Therefore,the coordination number of $Cs^{+}$ is $8$,and the coordination number of $Cl^{-}$ is also $8$,resulting in a $8:8$ coordination ratio.

(A) $NaCl$ crystallizes in a face-centered cubic $(fcc)$ lattice structure.
In this structure,$Cl^-$ ions occupy the corners and face centers of the cube,while $Na^+$ ions occupy all the octahedral voids.

(C) In the $NaCl$ crystal lattice,the structure is a face-centered cubic $(fcc)$ arrangement.
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 for both $Na^{+}$ and $Cl^{-}$ ions is $6:6$.

(B) The zinc blende $(ZnS)$ structure consists of a face-centered cubic $(fcc)$ lattice of $S^{2-}$ ions,where $Zn^{2+}$ ions occupy half of the tetrahedral voids.
In this structure,each $Zn^{2+}$ ion is surrounded by $4$ $S^{2-}$ ions,and each $S^{2-}$ ion is surrounded by $4$ $Zn^{2+}$ ions.
Therefore,the coordination number of both $Zn^{2+}$ and $S^{2-}$ ions is $4:4$.

(D) Rock salt $(NaCl)$ is an ionic crystalline compound consisting of $Na^{+}$ and $Cl^{-}$ ions.
In the $NaCl$ crystal lattice,each $Na^{+}$ ion is octahedrally surrounded by $6 \, Cl^{-}$ ions,and each $Cl^{-}$ ion is surrounded by $6 \, Na^{+}$ ions.
Therefore,the coordination number of the $Na^{+}$ ion is $6$.

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

(A) In a hexagonal close-packed $(hcp)$ structure,each atom is in contact with $12$ other atoms (six in its own layer,three in the layer above,and three in the layer below).
Therefore,the coordination number of a metal crystallizing in an $hcp$ structure is $12$.

(A) The correct option is $A$.
In the $NaCl$ crystal structure,which is a face-centered cubic $(fcc)$ lattice,each $Na^{+}$ ion is octahedrally surrounded by $6$ $Cl^{-}$ ions.
Similarly,each $Cl^{-}$ ion is surrounded by $6$ $Na^{+}$ ions.
Therefore,the coordination number for both $Na^{+}$ and $Cl^{-}$ ions is $6$.

(B) In the $CaF_2$ (fluorite) structure,the $Ca^{2+}$ ions form a face-centered cubic $(fcc)$ lattice.
Each $Ca^{2+}$ ion is surrounded by $8$ $F^{-}$ ions,and each $F^{-}$ ion is surrounded by $4$ $Ca^{2+}$ ions.
Therefore,the coordination number of the cation $(Ca^{2+})$ is $8$ and the anion $(F^{-})$ is $4$.
Thus,the correct arrangement is $8:4$.

(A) The antifluorite structure is characteristic of compounds with the formula $A_2B$,where the anions $(B^{2-})$ form a face-centered cubic $(fcc)$ lattice and the cations $(A^+)$ occupy all the tetrahedral voids.
Among the given options,$Na_2O$ follows the $A_2B$ stoichiometry,where $Na^+$ ions occupy all tetrahedral voids and $O^{2-}$ ions form an $fcc$ lattice.
Therefore,the correct option is $(A)$.

(C) In a face-centred cubic $(FCC)$ unit cell, the ions are in contact along the edge of the unit cell if we consider the rock salt structure (like $NaCl$).
For an $FCC$ unit cell, the edge length $a = 2(r_c + r_a)$.
Given, $a = 508 \ pm$ and $r_c = 110 \ pm$.
Substituting the values: $508 = 2(110 + r_a)$.
$254 = 110 + r_a$.
$r_a = 254 - 110 = 144 \ pm$.

(A) If the pressure on a $NaCl$ structure is increased,its coordination number increases.
$NaCl$ crystal has a $6:6$ coordination number.
When pressure is applied,the ions are forced closer to each other,resulting in a more compact structure.
The coordination number increases from $6:6$ to $8:8$,and the structure changes from $NaCl$ type to $CsCl$ type structure.

(A) The correct answer is $(A)$.
In the $CsCl$ crystal lattice,each $Cs^{+}$ ion is surrounded by $8$ $Cl^{-}$ ions,and each $Cl^{-}$ ion is surrounded by $8$ $Cs^{+}$ ions.
Therefore,the coordination number for $CsCl$ is $8:8$.

(B) In the $CaF_2$ crystal structure (fluorite structure),each $Ca^{2+}$ ion is coordinated by $8$ $F^{-}$ ions,forming a cubic geometry.
Since the stoichiometry of the compound is $1:2$ $(Ca:F)$,the ratio of the coordination numbers must be the inverse of the ratio of the number of ions.
Let the coordination number of $Ca^{2+}$ be $CN(Ca^{2+}) = 8$ and the coordination number of $F^{-}$ be $CN(F^{-})$.
According to the relation $N_{Ca} \times CN(Ca^{2+}) = N_{F} \times CN(F^{-})$,where $N$ is the number of ions per unit cell.
For $CaF_2$,$1 \times 8 = 2 \times CN(F^{-})$.
Therefore,$CN(F^{-}) = 4$.

(C) The coordination number is determined by the radius ratio $(r_+ / r_-)$ of the cation and anion.
For a radius ratio in the range $[0.414 - 0.732]$,the coordination number is $6$.
Since the given radius ratio is $0.525$,which falls within the range $[0.414 - 0.732]$,the coordination number is $6$.

(B) In the $ZnS$ (zinc blende) crystal structure,each $Zn^{2+}$ ion is tetrahedrally surrounded by $4$ $S^{2-}$ ions.
Therefore,the coordination number of $Zn^{2+}$ is $4$.

(B) The structure of $MgO$ is similar to $NaCl$ (rock salt structure).
In the $NaCl$ crystal lattice,both the cation $(Na^+)$ and the anion $(Cl^-)$ occupy octahedral voids.
Since $MgO$ follows the same structure,the $Mg^{2+}$ ions occupy octahedral voids,and the $O^{2-}$ ions form a face-centered cubic $(fcc)$ lattice.
The coordination number of an ion in an octahedral void is $6$.
Therefore,the coordination number of $Mg^{2+}$ is $6$.

(C) $NaCl$ crystallizes in a face-centered cubic $(FCC)$ lattice structure.
In an $FCC$ unit cell,the number of $Na^+$ ions is $4$ (from $12$ edges $\times \frac{1}{4} + 1$ center $= 4$) and the number of $Cl^-$ ions is $4$ (from $8$ corners $\times \frac{1}{8} + 6$ faces $\times \frac{1}{2} = 4$).
Therefore,the total number of $NaCl$ units per unit cell is $4$.

(C) $CsCl$ crystallizes in a body-centered cubic $(BCC)$ type structure.
In this structure,each $Cs^+$ ion is surrounded by $8$ $Cl^-$ ions,and each $Cl^-$ ion is surrounded by $8$ $Cs^+$ ions.
Therefore,the coordination number of both $Cs^+$ and $Cl^-$ is $8$.

(B) For a radius ratio in the range of $0.732 - 1.000$,the coordination number is $8$. Since $CsCl$ has a coordination number of $8$,it corresponds to this radius ratio.

(C) For an octahedral arrangement,the coordination number is $6$. The minimum limiting radius ratio $(r_+/r_-)$ for an octahedral void is $0.414$.

(B) For an ionic crystal with a coordination number of $6$,the limiting radius ratio range is $0.414 - 0.732$.

(D) In a calcium fluoride $(CaF_2)$ crystal structure,the $Ca^{2+}$ ions form a face-centered cubic $(fcc)$ lattice.
Each $Ca^{2+}$ ion is surrounded by $8$ $F^-$ ions,and each $F^-$ ion is surrounded by $4$ $Ca^{2+}$ ions.
Therefore,the coordination number ratio of $Ca^{2+} : F^-$ is $8 : 4$.

(A) Number of atoms of $A$ per unit cell = $8 \times \frac{1}{8} = 1$.
Number of atoms of $B$ per unit cell = $1$ (at the body center).
Ratio of $A:B = 1:1$.
Therefore,the formula of the compound is $AB$.

(D) In $Na_2O$,the $O^{2-}$ ions form a face-centered cubic $(fcc)$ lattice,and the $Na^+$ ions occupy all the tetrahedral voids. This arrangement is known as the antifluorite structure.

(D) In a cubic unit cell,each corner atom is shared by $8$ adjacent unit cells. Therefore,the contribution of an atom at the corner to a single unit cell is $\frac{1}{8}$.

(A) The structure of $NaCl$ (rock salt type) has a coordination number of $6:6$.
When high pressure is applied to $NaCl$,the ions are forced closer together,leading to an increase in the coordination number from $6$ to $8$.
This results in a phase transition to the $CsCl$ (cesium chloride) type structure,which has a coordination number of $8:8$.

(C) In a body-centered cubic $(BCC)$ unit cell,the atoms touch along the body diagonal.
The length of the body diagonal is given by $\sqrt{3} \, a$,where $a$ is the edge length of the unit cell.
Since the body diagonal consists of two radii of the corner atoms and one full diameter of the body-centered atom,the total length is $4r = \sqrt{3} \, a$.
The distance between the nearest neighbors (the center atom and a corner atom) is the distance from the center of the unit cell to a corner,which is half of the body diagonal.
Therefore,the distance $d = \frac{\sqrt{3} \, a}{2}$.

(B) In an $FCC$ unit cell,the number of $A$ atoms at the corners is $8 \times \frac{1}{8} = 1$.
Initially,there are $6$ face centers,so the number of $B$ atoms is $6 \times \frac{1}{2} = 3$.
After removing one $B$ atom from a face center,the number of $B$ atoms becomes $6 - 1 = 5$ face centers remaining,so the number of $B$ atoms is $5 \times \frac{1}{2} = \frac{5}{2}$.
The ratio of $A:B$ is $1 : \frac{5}{2}$,which simplifies to $2 : 5$.
Therefore,the molecular formula is $A_2B_5$.

(D) In the solid state,$Zn$ has a hexagonal close-packed $(HCP)$ structure with a coordination number of $12$.
Upon melting,the ordered lattice structure breaks down,and the atoms become more randomly arranged.
Experimental studies on liquid metals like $Zn$ show that the average coordination number decreases significantly in the molten state,typically falling to a value of approximately $4$ to $6$ due to the loss of long-range order.
Among the given options,$4$ is the most accepted value for the coordination number of $Zn$ in the molten state.

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

(B) In a face-centered cubic $(FCC)$ unit cell,each face is shared between two adjacent unit cells.
Therefore,an atom located at the face center contributes only $1/2$ or $0.5$ of its volume to a single unit cell.
Thus,the correct option is $B$.

(A) In a $NaCl$ crystal structure, the cation occupies the octahedral void.
For an octahedral void, the limiting radius ratio is $\frac{r}{R} = 0.414$.
Given the radius of the anion $(R) = 250 \ pm$.
The ideal radius of the cation $(r) = 0.414 \times 250 \ pm = 103.5 \ pm$.

(B) Number of $X$ atoms per unit cell $= 8 \times \frac{1}{8} = 1$
Number of $Y$ atoms per unit cell $= 1$ (body center)
Number of $Z$ atoms per unit cell $= 6 \times \frac{1}{2} = 3$ (face centers)
Therefore,the formula is $XYZ_3$.

(C) In the $NaCl$ crystal structure,the $Cl^-$ ions form a face-centered cubic $(fcc)$ lattice.
This means the $Cl^-$ ions are located at the corners of the cube and at the centers of each face of the cube.

(B) In a $bcc$ (body-centered cubic) unit cell,there is $1$ atom at the center and $8$ corner atoms,each contributing $1/8$ to the unit cell.
Total number of atoms = $(8 \times 1/8) + 1 = 1 + 1 = 2$.

(A) Number of $A$ atoms per unit cell $= 8 \text{ (corners)} \times \frac{1}{8} = 1.$
Number of $B$ atoms per unit cell $= 6 \text{ (faces)} \times \frac{1}{2} = 3.$
Therefore,the ratio of $A:B = 1:3$.
The formula of the substance is $AB_3$.