Given: Molar mass of $\mathrm{Fe}$ and $\mathrm{O}$ is 56 and $16 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively. $\mathrm{N}_{\mathrm{A}}=6.0 \times 10^{23} \mathrm{~mol}^{-1}$
Explanation:
$\mathrm{d=\frac{z\times m}{a^3}}$
$\mathrm{4=\frac{z\times72}{6\times10^{23}(5\times10^{-8})^3}}$
$\mathrm{4=\frac{z\times72}{6\times125\times10^{-1}}}$
$=\mathrm{z \approx4}$
A metal M forms hexagonal close-packed structure. The total number of voids in 0.02 mol of it is _________ $\times$ $10^{21}$ (Nearest integer).
(Given $\mathrm{N_A=6.02\times10^{23}}$)
Explanation:
Number of voids
$=0.02\times3\times6.02\times10^{23}$
$\simeq36\times10^{21}$
Ionic radii of cation $\mathrm{A}^{+}$ and anion $\mathrm{B}^{-}$ are 102 and $181 \,\mathrm{pm}$ respectively. These ions are allowed to crystallize into an ionic solid. This crystal has cubic close packing for $\mathrm{B}^{-}$. $\mathrm{A}^{+}$ is present in all octahedral voids. The edge length of the unit cell of the crystal AB is __________ pm. (Nearest Integer)
Explanation:
a = 2(rA+ + rB–)
a = 2 (102 + 181)
a = 566 pm
Metal $\mathrm{M}$ crystallizes into a fcc lattice with the edge length of $4.0 \times 10^{-8} \mathrm{~cm}$. The atomic mass of the metal is ____________ $\mathrm{g} / \mathrm{mol}$. (Nearest integer)
$\left(\right.$ Use : $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}$, density of metal, $\mathrm{M}=9.03 \mathrm{~g} \mathrm{~cm}^{-3}$ )
Explanation:
$ \begin{aligned} &=\frac{9.03 \times 6.02 \times 64 \times 10^{-1}}{4} \\\\ &=86.9 \mathrm{~g} \mathrm{~mol}^{-1} \\\\ &\approx 87 \mathrm{~g} \mathrm{~mol}^{-1} \end{aligned} $
An element M crystallises in a body centred cubic unit cell with a cell edge of $300 \,\mathrm{pm}$. The density of the element is $6.0 \mathrm{~g} \mathrm{~cm}^{-3}$. The number of atoms present in $180 \mathrm{~g}$ of the element is ____________ $\times 10^{23}$. (Nearest integer)
Explanation:
Let mass of 1 atom of $M$ is $A$
Edge length $=300 \,\mathrm{pm}$
Density $=6 \mathrm{~g} / \mathrm{cm}^3$
$ \therefore 6 \mathrm{~g} / \mathrm{cm}^3=\frac{\mathrm{Z} \times \mathrm{A}}{\left(300 \times 10^{-10}\right)^3}=\frac{2 \times \mathrm{A}}{27 \times 10^{-24}} $
$\mathrm{A}=81 \times 10^{-24} \mathrm{~g}$
$\therefore$ Atomic mass $=48.6 \mathrm{~g}$
$\therefore$ Mole in $180 \mathrm{~g}=\frac{180}{48.6}=3.7$ moles
Atoms of $\mathrm{M}=3.7 \times 6 \times 10^{23}$ $=22.22 \times 10^{23}$ atoms
Metal deficiency defect is shown by Fe0.93O. In the crystal, some Fe2+ cations are missing and loss of positive charge is compensated by the presence of Fe3+ ions. The percentage of Fe2+ ions in the Fe0.93O crystals is __________. (Nearest integer)
Explanation:
Let the number of $\mathrm{O}^{-2}$ ions be 100 and the number of $\mathrm{Fe}^{+2}$ ions be $\mathrm{X}$ The number of $\mathrm{Fe}^{+3}$ ions be $(93-\mathrm{X})$
$ \begin{aligned} &\therefore \,X(2)+(93-X) 3=200 \\\\ &279-X=200 \\\\ &X=79 \\\\ &\therefore \quad \% \text { of } F^{+2} \text { ions } =\frac{79}{93} \times 100 \\\\ &\simeq 85 \% \end{aligned} $
In a solid AB, A atoms are in ccp arrangement and B atoms occupy all the octahedral sites. If two atoms from the opposite faces are removed, then the resultant stoichiometry of the compound is AxBy. The value of x is ____________. [nearest integer]
Explanation:
$A=4$
If atoms from opposite faces are removed then
$ \begin{aligned} & A=4-x \times \frac{1}{x} \\\\ & A=3 \end{aligned} $
Value of $x=3$
The distance between Na+ and Cl$-$ ions in solid NaCl of density 43.1 g cm$-$3 is _______________ $\times$ 10$-$10 m. (Nearest Integer)
(Given : NA = 6.02 $\times$ 1023 mol$-$1)
Explanation:
$ 43.1=\frac{4 \times 58.5}{a^{3} \times 6.02 \times 10^{23}} $
$ \begin{aligned} & a^{3}=0.9 \times 10^{-23} \\\\ & =9 \times 10^{-24} \end{aligned} $
$a=2.08 \times 10^{-8} \mathrm{~cm}$
$=2.08 \times 10^{-10} \mathrm{~m}$
for $\mathrm{NaCl}$, distance between $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}=\frac{a}{2}$
$=1.04 \times 10^{-10} \mathrm{~m}$
Atoms of element X form hcp lattice and those of element Y occupy ${2 \over 3}$ of its tetrahedral voids. The percentage of element X in the lattice is ____________. (Nearest integer)
Explanation:
Number of particles of type $Y=\frac{2}{3} \times 12=8$
$\therefore$ Percentage of element $X=\frac{6}{14} \times 100$
$ \begin{aligned} &=\frac{300}{7} \\\\ &=42.85 \\\\ &\simeq 43 \% \end{aligned} $
Explanation:
cell formula $\to$ A4B4
Empirical formula $\to$ AB $\to$ (x = 1)
[Given : NA = 6.022 $\times$ 1023 mol$-$1]
Explanation:
$7.62 = {{4 \times M/6.022 \times {{10}^{23}}} \over {{{(0.4518 \times {{10}^{ - 7}}cm)}^3}}} $
$\Rightarrow $ M = 105.8 g/mol
Explanation:
Therefore number of carbon atoms per unit cell = 8
[Atomic Mass : K : 39.1 u, Br : 79.9 u NA = 6.023 $\times$ 1023]
Explanation:
Since mole percentage of SrBr2 dropped is 10$-$5 to that of total moles of KBr.
Hence,
No. of cationic vacancy $ = {{{{10}^{-5}}} \over {100}} \times {1 \over {119}} \times {N_A}$
$ = {1 \over {119}} \times {10^{ - 7}} \times 6.022 \times {10^{23}}$
$ = 5 \times {10^{ - 2}} \times {10^{ - 7}} \times {10^{23}} = 5 \times {10^{14}}$
Explanation:
Tetrahedral Void = (Z $\times$ 2) = 12
Octahedral Void = [Z $\times$ 1] = 6
No. of void per unit cell = 18
No. of unit cell = $\left( {{{0.581 \times {N_A}} \over {70 \times 6}}} \right)$
No. of void = $\left( {{{0.581 \times 6.023 \times {{10}^{23}}} \over {70 \times 6}}} \right) \times 18$
$ = 14.99 \times {10^{21}}$
[Assume each lattice point has a single atom]
[Assume $\sqrt 3 $ = 1.73, $\sqrt 2 $ = 1.41]
Explanation:
$a = {{4R} \over {\sqrt 3 }} = 27$
$R = {{27\sqrt 3 } \over 4}$
For FCC unit cell
$\sqrt 2 a = 4R$
$ \Rightarrow $ $a = {4 \over {\sqrt 2 }}\left( {{{27\sqrt 3 } \over 4}} \right)$
$ \Rightarrow $ $a = 27\sqrt {{3 \over 2}} $
$ \Rightarrow $ $a = 33.12 \approx 33$
Explanation:
Let us assume, the crystal has fcc or ccp lattice which has octahedral voids.
Number of lattice sites occupied = 8 corner + 6 face centres = 14
Number of octahedral voids = 12 edge centres + 1 body centre = 13
Number of octahedral void(s) per lattice site
$ = {{13} \over {14}} = 0.928 \simeq 1$
Explanation:
Density of copper, $d = {{Z \times M} \over {{a^3} \times {N_A}}}$
Given, Z = 4, for fcc lattice,
M = 63.54 g mol$-$1
= 63.54 $\times$ 10$-$3 kg mol$-$1,
a = 3.596 $\mathop A\limits^o $ = 3.596 $\times$ 10$-$10 m,
NA = 6.022 $\times$ 1023 mol$-$1
On putting given values, we get
$ \Rightarrow d = {{4 \times (63.54 \times {{10}^{ - 3}})} \over {{{(3.596 \times {{10}^{ - 10}})}^3} \times (6.022 \times {{10}^{23}})}}$ kg/m3
$ = 9076.26 \simeq 9077$ kg/m3
[Assume that the lattice is made up of atoms.]
Explanation:
bcc has a coordination number of 8 and contains 2 atoms per unit cell.
This is because each atom touches four atoms in the layer above it, four in the layer below it and none in its own layer.
Explanation:
Edge length of a cubic unit cell (a) = 405 pm = 4.05 × 10–8 cm
density of the element (d) = 2.7 gm/cc
d = ${{Z \times M} \over {{N_A} \times {{\left( a \right)}^3}}}$
$ \Rightarrow $ 2.7 = ${{Z \times 27} \over {6 \times {{10}^{23}} \times {{\left( {4.05 \times {{10}^{ - 8}}} \right)}^3}}}$
$ \Rightarrow $ Z = 4
The element has fcc unit cell
$ \therefore $ $\sqrt 2 $a = 4r
$ \Rightarrow $ r = ${{1.414 \times 405} \over 4}$
= 143 pm
= 143 × 10–12 m
The density (in $\mathrm{g} \mathrm{cm}^{-3}$ ) of the metal which forms a cubic close packed (ccp) lattice with an axial distance (edge length) equal to 400 pm is ___________.
Use: Atomic mass of metal $=105.6 \mathrm{amu}$ and Avogadro's constant $=6 \times 10^{23} \mathrm{~mol}^{-1}$
Explanation:
• In a ccp (fcc) unit cell, number of atoms per cell, n = 4
• Edge length
$a=400\text{ pm}=400\times10^{-12}\text{ m}=4.0\times10^{-8}\text{ cm}$
• Volume of cell
$V=a^3=(4.0\times10^{-8}\text{ cm})^3=6.4\times10^{-23}\text{ cm}^3$
• Mass of one atom
$m_{\rm atom}=\frac{M}{N_A}=\frac{105.6\text{ g/mol}}{6.0\times10^{23}\text{ mol}^{-1}} =1.76\times10^{-22}\text{ g}$
• Mass of unit cell
$m_{\rm cell}=n\;m_{\rm atom}=4\times1.76\times10^{-22}=7.04\times10^{-22}\text{ g}$
• Density
$\rho=\frac{m_{\rm cell}}{V} =\frac{7.04\times10^{-22}}{6.4\times10^{-23}} \approx11\;\text{g/cm}^3$
Answer: 11 g·cm⁻³.
$(i)\,\,\,\,\,$ Remove all the anions $(X)$ except the central one
$(ii)\,\,\,\,$ Replace all the face centered cations $(M)$ by anions $(X)$
$(iii)\,\,$ Remove all the corner cations $(M)$
$(iv)\,\,\,\,$ Replace the central anion $(X)$ with cation $(M)$
The value of $\,\,\left( {{{number\,\,of\,\,anions} \over {number\,\,of\,\,cations}}} \right)\,\,$ in $Z$ is ___________.
Explanation:
The unit cell of initial structure of ionic solid MX looks like
In NaCl type of solids cations (Na+) occupy the octahedral voids while anions (Cl$-$) occupy the face centre positions.
However, as per the demand of problem the position of cations and anions are swapped.
We also know that (for 1 unit cell)
(A) Total number of atoms at FCC = 4
(B) Total number of octahedral voids = 4 (as no. of atoms at FCC = No. of octahedral voids)
Now taking the conditions one by one
(i) If we remove all the anions except the central one than number of left anions.
= 4 $-$ 3 = 1
(ii) If we replace all the face centred cations by anions than effective number of cations will be = 4 $-$ 3 = 1 Likewise effective number of anions will be = 1 + 3 = 4
(iii) If we remove all the corner cations then effective number of cations will be 1 $-$ 1 = 0
(iv) If we replace central anion with cation then effective number of cations will be 0 + 1 = 1 Likewise effective number of anions will be 4 $-$ 1 = 3
Thus, as the final outcome, total number of cations present in Z after fulfilling all the four sequential instructions = 1 Likewise, total number of anions = 3
Hence, the value of ${{Number\,of\,anions} \over {Number\,of\,cations}} = {3 \over 1} = 3$
Explanation:
(i) Edge length of face centred cubic (FCC) unit cell (a) $=400 \mathrm{pm}$
(ii) Density of the substance $(\rho)=8 \mathrm{~g} \mathrm{~cm}^{-3}$
(iii) Mass of the crystal $(m)=256 \mathrm{~g}$
To Find: The value of $\mathrm{N}$ in $\mathrm{N} \times 10^{24}$
Formula: Density of the cell $(\rho)$
$ \begin{aligned} & =\frac{\text { Mass of crystal }(\mathrm{M}) \times \begin{array}{l} \text { No. of atoms in unit cell } \\ \end{array}}{\text { Volume of unit cell } \times \text { No. of atoms of pure substance in mass m }} \\ \end{aligned} $
Calculations: A face-centred cubic (F.C.C.) lattice has atoms of pure substance at the corner of the unit cell and/or atom each at the centre of the face of the unit cell.
Since there are eight atoms at the corners and one atom at each of the six faces (i.e., six atoms in total):
Total number of atoms present per unit cell $(z)$
$ =\frac{1}{8} \times 8+\frac{1}{2} \times 6=4 $
Density of the cell $(\rho)$
$ \begin{aligned} & =\frac{256 \mathrm{~g} \times 4}{\left(400 \times 10^{-10}\right)^3 \times \mathrm{N}^1} \\\\ & 8 \mathrm{~g} \mathrm{~cm}^{-3}=\frac{256 \mathrm{~g} \times 4}{\left(400 \times 10^{-10}\right) \times \mathrm{N}^1} \\\\ & \mathrm{~N}^{\prime}=\frac{256 \mathrm{~g} \times 4}{8 \mathrm{~g} \mathrm{~cm}^{-3} \times(400)^3 \times 10^{-30}} \\\\ & \mathbf{N}^{\prime}=2 \times 10^{24} \text { atoms } \end{aligned} $
Hence, the crystal of pure substance of mass $256 \mathrm{~g}$ contains $2 \times 10^{24}$ atoms.
Now, $\mathrm{N}^{\prime}=\mathrm{N} \times 10^{24}$
Therefore, value of $\mathrm{N}=2$.
Explanation:
(i) The shape of a regular octahedron :
When the octahedron is truncated from all edges.
(ii) Each face of truncated octahedron is hexagonal.
Since, there are 8 faces, there are eight hexagons.
The coordination number of Al in the crystalline state of AlCl$_3$ is ___________.
Explanation:
AlCl3 exists as a close packed lattice of chloride ions Cl– with Al3+ occupying octahedral holes. Hence, coordination number of Al3+ is = 6.
An element crystallises in fcc lattice having edge length 400 pm. Calculate the maximum diameter of the atom which can be placed in interstitial site such that the structure is not distorted.
Explanation:
Types of Voids in FCC:
In a face-centered cubic (FCC) structure, there are two types of empty spaces, called voids: tetrahedral voids and octahedral voids.
Ratio for Void Sizes:
For a tetrahedral void: $\frac{r}{R}=0.225$
For an octahedral void: $\frac{r}{R}=0.414$
Here, $r$ is the radius of the smaller atom that goes into the empty site (the interstitial atom), and $R$ is the radius of the atom making up the structure.
Choosing the Correct Void:
The largest atom that can fit into the FCC structure without changing its shape fits in the octahedral void, because this void is bigger than the tetrahedral one. So, we use the octahedral void ratio: $\frac{r}{R}=0.414$
Finding the Radius ($R$) of the Main Atom:
In FCC, the relation between the radius $R$ and the edge length $a$ is: $R=\frac{a}{2\sqrt{2}}$
The question says $a = 400$ pm.
Calculate $R$:
$R=\frac{400}{2\sqrt{2}}$
Find $r$ (radius of the maximum atom that fits in an octahedral site):
Substitute the value of $R$ into the octahedral void ratio:
$r=0.414\times\frac{400}{2\sqrt{2}}=58.55~\text{pm}$
Calculate Diameter:
The diameter of the atom is twice the radius:
Diameter = $2r = 2 \times 58.55 = 117.1~\text{pm}$
Final Answer:
The largest atom that can fit into the empty site in an FCC lattice (without changing the structure) has a diameter of 117.1 pm.
Note:
Only the octahedral void is considered for calculating the maximum possible diameter, because it is the largest void in FCC structures.
The arrangement of X$-$ ions around A+ ion in solid AX is given in the figure (not drawn to scale). If the radius of X$-$ is 250 pm, the radius of A+ is
A compound MpXq has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below. The empirical formula of the compound is
The number of atoms in this HCP unit cell is :
The volume of this HCP unit cell is :
The empty space in this HCP unit cell is :
Match the crystal system/unit cells mentioned in Column I with their characteristic features mentioned in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 $\times$ 4 matrix given in the ORS.
| Column I | Column II | ||
|---|---|---|---|
| (A) | Simple cubic and face-centred cubic | (P) | have these cell parameters $a=b=c$ and $\alpha=\beta=\gamma$ |
| (B) | cubic and rhombohedral | (Q) | are two crystal systems |
| (C) | cubic and tetragonal | (R) | have only two crystallographic of 90$^\circ$ |
| (D) | hexagonal and monoclinic | (S) | belong to same crystal system |
If $\mathrm{r}_{\mathrm{z}}=\frac{\sqrt{3}}{2} r_{\mathrm{y}} ; \mathrm{r}_{\mathrm{y}}=\frac{8}{\sqrt{3}} \mathrm{r}_{\mathrm{x}} ; M_{\mathrm{z}}=\frac{3}{2} M_{\mathrm{y}}$ and $M_{\mathrm{z}}=3 M_{\mathrm{x}}$, then the correct statement(s) is(are) :
[Given: $M_x, M_y$, and $M_z$ are molar masses of metals $x, y$, and $z$, respectively.
$\mathrm{r}_{\mathrm{x}}, \mathrm{r}_{\mathrm{y}}$, and $\mathrm{r}_{\mathrm{z}}$ are atomic radii of metals $\mathrm{x}, \mathrm{y}$, and $\mathrm{z}$, respectively.]
The correct statement(s) regarding defects in solids is (are)
The edge length of unit cell of a metal having molecular weight $75 \mathrm{~g} \mathrm{~mol}^{-1}$ is 5 $\mathop A\limits^o $ which crystallizes in cubic lattice. If the density is $2 \mathrm{~g} / \mathrm{cc}$, find the radius of metal atom. $\left(\mathrm{N}_{\mathrm{A}}=6 \times 10^{23}\right)$. Give the answer in $\mathrm{pm}$.
Explanation:
Given,
$\mathrm{M}_{w}=75 \mathrm{~g} \mathrm{~mol}^{-1}$
$a=5$ $\mathop A\limits^o $ = 5 $\times$ 10$^{-8}$ cm
$\rho=2$ g cm$^{-3}$
$\mathrm{N_A=6\times10^{23}}$
To Find: Radius of metal atom ($r$)
To find the value of $r$, we need to find out the type of unit cell. For that, the value of Z is required. On the basis of the value of Z the type of unit cell can be identified and using suitable formula, radius of metal atom will be determined.
Step 1 : Determination of $Z_{\text {eff }}$
The formula for calculating $\mathrm{Z}_{\text {eff }}$ will be obtained from the formula for density.
The formula for calculating density is,
$\rho=\frac{\mathrm{Z} \times \mathrm{M}_{w}}{a^{3} \times \mathrm{N}_{\mathrm{A}}}$
Here, $\rho$ is the density of unit cell, Z is the number of atoms present in a unit cell, $\mathrm{M}_{w}$ is the molecular weight, $a$ is the edge length of a unit cell and $\mathrm{N}_{\mathrm{A}}$ is the Avogadro's number.
On rearranging this formula,
$\mathrm{Z}=\frac{\rho \times a^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{M}_{w}}$ ...... (i)
Substituting the respective values in equation (i),
$Z=\frac{2 \times\left(5 \times 10^{-8}\right)^{3} \times\left(6 \times 10^{23}\right)}{75}$
$\therefore \quad \mathrm{Z}=1.98 \approx 2$
The value of Z is 2 for BCC (body centered cubic) unit cell.
Step 1: Determination of radius of metal atom.
For $\mathrm{BCC}$, the radius of atom is given by,
$4 r=\sqrt{3} a$
Here, $r$ denotes the radius of the atom.
$\begin{array}{ll} \therefore & r=\frac{\sqrt{3}}{4} a \\ \therefore & r=\frac{\sqrt{3}}{4} \times 5=2.165 ~\mathop A\limits^o \\ \therefore & r=216.5 ~\mathrm{pm} \end{array}$
The radius of metal atom is $216.5 ~\mathrm{pm}$.
(a) Find the density of the lattice
(b) If the density of lattice is found to be 20 kg m-3, then predict the type of defect
Explanation:
AB has a rock salt (NaCl) structure. This type of crystal structure possesses fcc unit cell and contains four formula units per unit cell, i.e., Z = 4.
In case of a rock salt structure, the edge-length (a) of the unit cell = 2 $\times$ (radius of cation + radius of anion)
Therefore, the edge-length (a) of the unit cell of AB crystal = 2 $\times$ Y1/3 nm = 2Y1/3 $\times$ 10$-$9 m.
We know, $\rho = {{Z \times M} \over {N \times {a^3}}}$
Given : M = 6.022 Y g mol$-$1 = 6.022 $\times$ 10$-$3 Y kg mol$-$1
$\therefore$ $\rho = {{4 \times 6.022 \times {{10}^{ - 3}}Y} \over {6.022 \times {{10}^{23}} \times {{(2{Y^{1/3}} \times {{10}^{ - 9}})}^3}}} = 5.0$ kg m$-$3
(1) Density of the crystal = 5.0 kg m$-$3
(2) The observed density (= 20 kg m$-$3) is higher than that of the calculated density. This indicates that the crystal structure of AB is likely to have non-stoichiometric defect in the form of metal excess or metal deficiency defect or to have impurity defect in the form of substitutional impurity defect or interstitial impurity defect.