Crystal structure and Coordination number Questions in English

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

(A) In the $CaF_2$ (fluorite) structure,the $Ca^{2+}$ ions form a face-centred cubic $(FCC)$ lattice,while the $F^-$ ions occupy all the tetrahedral voids.

(A) For a body-centered cubic $(bcc)$ arrangement,the coordination number is $8$.
The limiting radius ratio $({r_+}/{r_-})$ for a coordination number of $8$ is in the range of $0.732 - 1.000$.

(D) In a face-centered cubic $(F.C.C)$ lattice,each face of the unit cell is shared by $2$ adjacent unit cells.
Since there are $6$ faces in a cubic unit cell,each face is shared equally by $6$ other unit cells.

(C) The $NaCl$ crystal structure is a face-centered cubic $(fcc)$ lattice.
In a unit cell of $NaCl$,the number of $Na^{+}$ ions is $4$ ($12$ edges $\times$ $1/4$ + $1$ center = $4$) and the number of $Cl^{-}$ ions is $4$ ($8$ corners $\times$ $1/8$ + $6$ faces $\times$ $1/2$ = $4$).
Therefore,each unit cell contains $4$ $NaCl$ units.

(C) For a tetrahedral void,the coordination number is $4$.
The limiting radius ratio $\frac{r_{c^+}}{r_{a^-}}$ for a tetrahedral arrangement lies in the range of $0.225 - 0.414$.

(B) In the $NaCl$ (rock salt) structure:
Number of $Na^{+}$ ions $= 12$ (at edge centers) $\times \frac{1}{4} + 1$ (at body center) $\times 1 = 3 + 1 = 4$.
Number of $Cl^{-}$ ions $= 8$ (at corners) $\times \frac{1}{8} + 6$ (at face centers) $\times \frac{1}{2} = 1 + 3 = 4$.
Thus,there are $4$ formula units of $NaCl$ per unit cell.

(D) The structure of $TlCl$ is similar to $CsCl$,which crystallizes in a body-centered cubic $(BCC)$ lattice.
In a $BCC$ lattice,the coordination number of the cation is $8$.
According to the radius ratio rule,for a coordination number of $8$,the radius ratio $(r_+/r_-)$ lies in the range of $0.732 - 1.000$.
Therefore,the correct option is $(D)$.

(C) The zinc blende $(ZnS)$ structure is a face-centered cubic $(fcc)$ lattice where the $S^{2-}$ ions occupy the lattice points and $Zn^{2+}$ ions occupy half of the tetrahedral voids.
This results in a $4:4$ coordination number.
Among the given options,$CuCl$ exhibits a structure similar to zinc blende,where $Cu^+$ ions occupy tetrahedral sites in a cubic close-packed arrangement of $Cl^-$ ions.

(D) The $Na_2O$ crystal exhibits an antifluorite structure.
In this structure,the oxide ions $(O^{2-})$ form a face-centered cubic $(fcc)$ lattice,and the sodium ions $(Na^+)$ occupy all the tetrahedral voids.
This is the inverse of the fluorite $(CaF_2)$ structure,hence it is called antifluorite.

(B) The $ZnS$ (Zinc blende) crystal structure is based on a face-centered cubic $(fcc)$ lattice of $S^{2-}$ ions,with $Zn^{2+}$ ions occupying half of the tetrahedral voids.
It exhibits a $4:4$ coordination number.

(D) In an $fcc$ (face-centered cubic) unit cell:
$1.$ Atoms at the corners: There are $8$ corners,and each corner atom contributes $\frac{1}{8}$ to the unit cell. So,$\frac{1}{8} \times 8 = 1$ atom.
$2.$ Atoms at the face centers: There are $6$ faces,and each face-centered atom contributes $\frac{1}{2}$ to the unit cell. So,$\frac{1}{2} \times 6 = 3$ atoms.
$3.$ Total number of atoms $(Z)$ = $1 + 3 = 4$ atoms.

(B) The $ABCABC$ packing sequence corresponds to the face-centered cubic $(FCC)$ or cubic close-packed $(CCP)$ structure.
In any close-packed structure,if the number of atoms per unit cell is $Z$,the number of octahedral voids is $Z$ and the number of tetrahedral voids is $2Z$.
Therefore,the number of tetrahedral voids is $2Z$.

(A) In a body-centred cubic $(BCC)$ lattice,each atom at the corner is surrounded by $8$ nearest neighbours located at the body centres of the adjacent unit cells.
Alternatively,the atom at the body centre is surrounded by $8$ atoms located at the corners of the same unit cell.
Therefore,the coordination number of a $BCC$ lattice is $8$.

(B) In a sodium chloride $(NaCl)$ crystal,which adopts a rock salt structure,each $Na^{+}$ ion is surrounded by $6$ $Cl^{-}$ ions,and each $Cl^{-}$ ion is surrounded by $6$ $Na^{+}$ ions.
Therefore,the coordination number of $Na^{+}:Cl^{-}$ is $6:6$,meaning there are $6$ equidistant oppositely charged ions.

(D) The $BCC$ (Body-Centered Cubic) unit cell consists of $8$ atoms at the corners and $1$ atom at the body center.
Number of atoms in $BCC$ $(n)$ = $(8 \times \frac{1}{8}) + 1 = 1 + 1 = 2$.
Thus,$Na$ has $2$ atoms per unit cell.
The $FCC$ (Face-Centered Cubic) unit cell consists of $8$ atoms at the corners and $1$ atom at each of the $6$ faces.
Each face-centered atom is shared by $2$ unit cells.
Number of atoms in $FCC$ $(n)$ = $(8 \times \frac{1}{8}) + (6 \times \frac{1}{2}) = 1 + 3 = 4$.
Thus,$Mg$ has $4$ atoms per unit cell.
Therefore,the number of atoms for $Na$ and $Mg$ are $2$ and $4$ respectively.

(C) The $AB_2$ type of structure is characteristic of the fluorite structure,which is exhibited by $CaF_2$.
In this structure,the cation $Ca^{2+}$ occupies the face-centered cubic lattice sites,while the anions $F^-$ occupy all the tetrahedral voids.
The dissociation can be represented as: $CaF_2 \rightleftharpoons Ca^{2+} + 2F^-$.

(B) Potassium $(K)$ crystallizes in a body-centred cubic $(bcc)$ lattice structure.

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

(B) The coordination number of $8$ is characteristic of the body-centred cubic $(BCC)$ lattice structure.
In a $BCC$ unit cell,each atom at the center is surrounded by $8$ nearest neighbors,and each corner atom is also surrounded by $8$ nearest neighbors.
Therefore,the crystal class is body-centred cube.

(D) In a close-packed structure,the number of octahedral voids is equal to the number of spheres present in the lattice.
For a $fcc$ (face-centered cubic) unit cell,the number of spheres per unit cell is $4$.
The number of octahedral voids per unit cell is also $4$.
Therefore,the number of octahedral sites per sphere is $\frac{4}{4} = 1$.

(A) The body-centered cubic $(BCC)$ structure is characterized by atoms at the corners and one atom at the center of the unit cell.
Among the given options,$Sodium$ $(Na)$ crystallizes in a $BCC$ structure at room temperature.
$Magnesium$ $(Mg)$ and $Zinc$ $(Zn)$ typically exhibit hexagonal close-packed $(HCP)$ structures,while $Copper$ $(Cu)$ exhibits a face-centered cubic $(FCC)$ structure.
Therefore,the correct option is $A$.

(B) The fluorite structure is named after the mineral fluorite $(CaF_2)$.
In this structure,the cations $(Ca^{2+})$ form a face-centered cubic $(fcc)$ lattice,and the anions $(F^-)$ occupy all the tetrahedral voids.
Among the given options,$SrF_2$ crystallizes in the fluorite structure,similar to $CaF_2$.

(B) In the $ZnS$ (zinc blende) structure,the $S^{2-}$ ions form a face-centered cubic $(FCC)$ lattice.
There are $8$ tetrahedral voids per unit cell in an $FCC$ lattice.
In $ZnS$,the $Zn^{2+}$ ions occupy only alternate tetrahedral voids,meaning $4$ out of $8$ tetrahedral voids are filled.
Therefore,the correct option is $(B)$.

(B) The rock salt structure is the crystal structure of $NaCl$.
Among the given options,$MgO$ crystallizes in the rock salt structure because the ratio of the ionic radii $r_{Mg^{2+}} / r_{O^{2-}}$ falls within the range required for octahedral coordination,similar to $NaCl$.

(C) The fluorite structure $(CaF_2)$ consists of a face-centered cubic $(fcc)$ arrangement of $Ca^{2+}$ ions.
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 $Ca^{2+}$ is $8$ and the coordination number of $F^-$ is $4$,resulting in an $8:4$ coordination ratio.

(C) The number of $A$ ions at the eight corners of the cube is $8 \times \frac{1}{8} = 1$.
The number of $B$ ions at the six face centres of the cube is $6 \times \frac{1}{2} = 3$.
Thus,the ratio of $A$ to $B$ is $1:3$.
Therefore,the empirical formula of the compound is $AB_3$.

(A) In a $bcc$ (body-centered cubic) unit cell,the packing efficiency is $68\%$.
This means that $68\%$ of the total volume of the unit cell is occupied by the atoms.
Therefore,the vacant space (void space) is calculated as: $100\% - 68\% = 32\%$.
Thus,the correct option is $A$.

(C) In a cubic close packed $(CCP)$ or face-centered cubic $(FCC)$ unit cell,the total number of atoms per unit cell is $Z = 4$.
The number of octahedral voids is equal to the number of atoms present in the unit cell.
Therefore,the number of octahedral voids $= 4$.

(A) The coordination number in both $HCP$ (Hexagonal Close Packing) and $CCP$ (Cubic Close Packing) arrangements is $12$.
In a $bcc$ (Body-Centered Cubic) arrangement,the coordination number is $8$.

(D) In $NaCl$ (rock salt) structure:
Number of $Na^{+}$ ions $= 12$ (at edge centers) $\times \frac{1}{4} + 1$ (at body center) $\times 1 = 3 + 1 = 4$.
Number of $Cl^{-}$ ions $= 8$ (at corners) $\times \frac{1}{8} + 6$ (at face centers) $\times \frac{1}{2} = 1 + 3 = 4$.
Since there are $4$ $Na^{+}$ ions and $4$ $Cl^{-}$ ions per unit cell,the number of formula units per unit cell is $4$.

(B) In an $NaCl$ type structure,the ions are located at the corners and face centers of the unit cell.
For an $NaCl$ type crystal,the distance between the cation $(K^{+})$ and the anion $(F^{-})$ along the edge of the unit cell is given by $d = \frac{a}{2}$,where $a$ is the edge length of the unit cell.
Therefore,the distance between $K^{+}$ and $F^{-}$ ions is $\frac{a}{2} \ cm$.

(A) In a $bcc$ (body-centered cubic) unit cell,the number of atoms per unit cell is $2$.
Given the number of unit cells $= 12.08 \times 10^{23}$.
Total number of atoms $= (\text{Number of atoms per unit cell}) \times (\text{Total number of unit cells})$.
Total number of atoms $= 2 \times 12.08 \times 10^{23} = 24.16 \times 10^{23}$ atoms.

(B) An atom located at the body center of a cube is entirely contained within that specific unit cell.
It is not shared with any other adjacent unit cells.
Therefore,the contribution of an atom at the center of the cube to the unit cell is $1$.

(B) The coordination number $6$ corresponds to an octahedral geometry in ionic crystals.
For an octahedral arrangement,the limiting radius ratio $(r_+/r_-)$ lies in the range of $0.414$ to $0.732$.
Therefore,the correct range is between $0.41$ and $0.73$.

(A) For a body centred cubic $(BCC)$ unit cell,the number of spheres is $1 + (8 \times \frac{1}{8}) = 2$.
For a face centred cubic $(FCC)$ unit cell,the number of spheres is $(8 \times \frac{1}{8}) + (6 \times \frac{1}{2}) = 1 + 3 = 4$.
Thus,the number of spheres in $(i)$ $BCC$ is $2$ and in $(ii)$ $FCC$ is $4$.

(B) In a $bcc$ structure,the ions are in contact along the body diagonal.
The body diagonal of a cube with edge length $a$ is $\sqrt{3} a$.
The shortest interionic distance $(d)$ between the center ion and the corner ion is half of the body diagonal.
$d = \frac{\sqrt{3}}{2} a$
Given $a = 4.3 \ \mathring{A}$,
$d = \frac{1.732}{2} \times 4.3 = 0.866 \times 4.3 = 3.7238 \ \mathring{A} \approx 3.72 \ \mathring{A}$.

(C) tetrahedral void is a simple triangular void in a crystal lattice and is surrounded by $4$ spheres arranged tetrahedrally around it.
An octahedral void is a bi-triangular void (formed by two triangles,one pointing upwards and the other downwards) and is surrounded by $6$ spheres.

(D) In a face-centred cubic $(FCC)$ unit cell,atoms are present at the corners and at the centre of each face.
An atom present at the corner is shared by $8$ adjacent unit cells,so its contribution is $1/8$.
An atom present at the centre of a face is shared by $2$ adjacent unit cells,so its contribution to each unit cell is $1/2$.
Therefore,the correct option is $(D)$.

(C) In a $CsCl$ crystal,the structure is body-centered cubic $(BCC)$.
In this structure,the $Cs^+$ ion is at the body center and $Cl^-$ ions are at the corners.
The body diagonal of the cube is equal to $\sqrt{3} a$.
The distance between the body center and a corner is half of the body diagonal,which is $\frac{\sqrt{3} a}{2}$.
Thus,the interionic distance $(d)$ is $\frac{\sqrt{3} a}{2}$.

(B) The radius ratio rule for ionic crystals states that for a radius ratio in the range $0.225 - 0.414$,the coordination number is $4$ and the crystal structure is tetrahedral.
Since the given value $0.40$ falls within this range,the structure is tetrahedral.

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

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

(D) In a $NaCl$ crystal (fcc structure), the distance between the cation $(Na^{+})$ and the nearest anion $(Cl^{-})$ along the edge of the unit cell is equal to half of the edge length $(a)$.
Given that the distance between $Na^{+}$ and $Cl^{-}$ is $X \ pm$, we have:
$X = a/2$
Therefore, the edge length of the unit cell is:
$a = 2X \ pm$
Thus, the correct option is $(D)$.

(D) In a $bcc$ (body-centered cubic) lattice,each atom is in contact with $8$ nearest neighbors.
Therefore,the coordination number of potassium in a $bcc$ lattice is $8$.

(B) In a body-centered cubic $(BCC)$ lattice,each atom at the center of the unit cell is surrounded by $8$ atoms located at the corners of the unit cell.
Therefore,the coordination number of a body-centered cubic lattice is $8$.

(C) Atoms $A$ are present at the $8$ corners of the cube. The contribution of each corner atom to the unit cell is $\frac{1}{8}$.
Number of $A$ atoms per unit cell $= 8 \times \frac{1}{8} = 1$.
Atoms $B$ are present at the body center of the cube. The contribution of a body-centered atom is $1$.
Number of $B$ atoms per unit cell $= 1 \times 1 = 1$.
Therefore,the ratio of $A:B$ is $1:1$,and the simplest formula of the compound is $AB$.

(D) . $Cu$ (Copper) crystallizes in a face-centered cubic $(FCC)$ or cubic close-packed $(CCP)$ structure.
In an $FCC$ lattice,each atom is surrounded by $12$ nearest neighbors.
Therefore,the coordination number for $Cu$ is $12$.