$\mathrm{CsCl}$ has body centered type structure in which $\mathrm{Cs}^{+}$ occupies at corner of a cube and $\mathrm{Cl}^{-}$occupies the centre of the cube.
$2 \mathrm{r}_{\mathrm{Cs}^{+}}+2 \mathrm{r}_{\mathrm{Cl}^{-}}=\sqrt{3} \mathrm{a}$ (where a is the edge length of the cube)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2023 (Online) 10th April Evening Shift
The correct relationships between unit cell edge length '$a$' and radius of sphere '$r$' for face-centred and body-centred cubic structures respectively are :
A.
$r=2 \sqrt{2} a$ and $\sqrt{3} r=4 a$
B.
$r=2 \sqrt{2} a$ and $4 r=\sqrt{3} a$
C.
$2 \sqrt{2} r=a$ and $\sqrt{3} r=4 a$
D.
$2 \sqrt{2} r=a$ and $4 r=\sqrt{3} a$
Correct Answer: D
Explanation:
In a face-centered cubic (FCC) unit cell, atoms are present at the corners as well as at the centers of the faces. Hence, the diagonal of the face of the unit cell is equal to 4 times the radius of an atom.
This gives us the equation:
$\sqrt{2} a = 4r$
Which simplifies to:
$a = 2\sqrt{2}r$
In a body-centered cubic (BCC) unit cell, atoms are present at the corners and at the center of the unit cell. The body diagonal of the unit cell is equal to 4 times the radius of an atom.
This gives us the equation:
$\sqrt{3} a = 4r$
Which simplifies to:
$a = \frac{4}{\sqrt{3}}r$
Therefore, Option D is correct:
$2\sqrt{2}r = a$ and $4r = \sqrt{3}a$
2023
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2023 (Online) 6th April Morning Shift
A compound is formed by two elements $\mathrm{X}$ and $\mathrm{Y}$. The element $\mathrm{Y}$ forms cubic close packed arrangement and those of element $\mathrm{X}$ occupy one third of the tetrahedral voids. What is the formula of the compound?
A.
$\mathrm{XY}_{3}$
B.
$\mathrm{X_3Y}_{2}$
C.
$\mathrm{X_3Y}$
D.
$\mathrm{X_2Y}_{3}$
Correct Answer: D
Explanation:
A compound is formed by two elements $\mathrm{X}$ and $\mathrm{Y}$. The element $\mathrm{Y}$ forms a cubic close-packed (CCP) arrangement, and element $\mathrm{X}$ occupies one third of the tetrahedral voids.
In a CCP structure, there are 4 atoms of $\mathrm{Y}$ in a unit cell. This means there are 8 tetrahedral voids in the CCP structure.
Since $\mathrm{X}$ occupies one third of the tetrahedral voids, there are $\frac{1}{3} \times 8 = \frac{8}{3}$ atoms of $\mathrm{X}$ in the formula unit.
The ratio of $\mathrm{X}$ to $\mathrm{Y}$ in the formula unit is $\frac{8}{3} : 4 = \frac{2}{3} : 1$. Multiplying both parts of the ratio by 3, we get $2 : 3$.
So, the formula of the compound is $\mathrm{X_2Y_{3}}$
2023
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2023 (Online) 1st February Morning Shift
Which of the following represents the lattice structure of $\mathrm{A_{0.95}O}$ containing $\mathrm{A^{2+},A^{3+}}$ and $\mathrm{O^{2-}}$ ions?
A.
B only
B.
A and B only
C.
A only
D.
B and C only
Correct Answer: C
Explanation:
Applying electrical neutrality principle in metal
defficiency defect.
3 A2+ are replaced by 2A3+, thus one vacant site per
pair of A3+ is created.
2023
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2023 (Online) 25th January Morning Shift
A cubic solid is made up of two elements X and Y. Atoms of X are present on every alternate corner and one at the center of cube. Y is at $\frac{1}{3}^{\mathrm{rd}}$ of the total faces. The empirical formula of the compound is :
A.
$\mathrm{{X_2}{Y_{1.5}}}$
B.
$\mathrm{{X_{3}}{Y_2}}$
C.
$\mathrm{X{Y_{2.5}}}$
D.
$\mathrm{{X_{2.5}}Y}$
Correct Answer: B
Explanation:
$\begin{aligned} & \text { Number of } X \text { particles }=4 \times \frac{1}{8}+1=1.5 \\\\ & \text { Number of } Y \text { particles }=6 \times \frac{1}{3} \times \frac{1}{2}=1 \\\\ & \therefore \text { Empirical formula }=X_{1.5} Y_1=X_3 Y_2\end{aligned}$
2023
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2023 (Online) 13th April Evening Shift
Sodium metal crystallizes in a body centred cubic lattice with unit cell edge length of $4~\mathop A\limits^o $. The radius of sodium atom is __________ $\times ~10^{-1}$ $\mathop A\limits^o $ (Nearest integer)
Correct Answer: 17
Explanation:
In a body-centered cubic (BCC) lattice, the relationship between the edge length (a) and the atomic radius (r) is given by :
$\sqrt{3}a = 4r$
Given the unit cell edge length (a) of sodium metal as 4 Å :
$a = 4 ~\mathop A\limits^o$
We can now solve for the radius (r) of the sodium atom :
So, the radius of the sodium atom is 17.32 × 10⁻¹ $\mathop A\limits^o$.
2023
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2023 (Online) 11th April Morning Shift
An atomic substance A of molar mass $12 \mathrm{~g} \mathrm{~mol}^{-1}$ has a cubic crystal structure with edge length of $300 ~\mathrm{pm}$. The no. of atoms present in one unit cell of $\mathrm{A}$ is ____________. (Nearest integer)
Given the density of $\mathrm{A}$ is $3.0 \mathrm{~g} \mathrm{~mL}^{-1}$ and $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}$
Correct Answer: 4
Explanation:
Given:
Atomic substance A with molar mass $M = 12 \, \text{g mol}^{-1}$
Body-centered unit cells can be found in the following crystal systems among those listed:
Cubic: Body-centered cubic (BCC) is one of the lattice structures that cubic systems can have.
Tetragonal: A body-centered tetragonal system is also a possibility.
Orthorhombic: The orthorhombic system can have a body-centered orthorhombic lattice.
Therefore, the number of crystal systems from the list where a body-centered unit cell can be found is 3.
2023
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2023 (Online) 1st February Evening Shift
A metal M crystallizes into two lattices :- face centred cubic (fcc) and body centred cubic (bcc) with unit cell edge length of $2.0$ and $2.5\,\mathop A\limits^o $ respectively. The ratio of densities of lattices fcc to bcc for the metal M is ____________. (Nearest integer)
Correct Answer: 4
Explanation:
$\begin{aligned} & d_1 \text {, Density of fcc lattice of metal } M=\frac{4 \times M}{N_0\left(a_{\mathrm{fcc}}\right)^3} \\\\ & d_2 \text {, Density of bcc lattice of metal } M=\frac{2 \times M}{N_0\left(a_{\mathrm{bcc}}\right)^3} \\\\ & \frac{d_1}{d_2}=\frac{4}{2}\left(\frac{a_{\mathrm{bcc}}}{a_{\mathrm{fcc}}}\right)^3=2\left(\frac{2.5}{2}\right)^3=3.90 \simeq 4\end{aligned}$
2023
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2023 (Online) 30th January Evening Shift
Iron oxide FeO, crystallises in a cubic lattice with a unit cell edge length of 5.0 Å. If density of the $\mathrm{FeO}$ in the crystal is $4.0 \mathrm{~g} \mathrm{~cm}^{-3}$, then the number of $\mathrm{FeO}$ units present per unit cell is __________. (Nearest integer)
Given: Molar mass of $\mathrm{Fe}$ and $\mathrm{O}$ is 56 and $16 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively. $\mathrm{N}_{\mathrm{A}}=6.0 \times 10^{23} \mathrm{~mol}^{-1}$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2023 (Online) 29th January Evening Shift
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 $\mathrm{N_A=6.02\times10^{23}}$)
Correct Answer: 36
Explanation:
Number of voids
$=0.02\times3\times6.02\times10^{23}$
$\simeq36\times10^{21}$
2022
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 30th June Morning Shift
An element X has a body centred cubic (bcc) structure with a cell edge of 200 pm. The density of the element is 5 g cm$-$3. The number of atoms present in 300 g of the element X is _______________.
Given : Avogadro constant, NA = 6.0 $\times$ 1023 mol$-$1.
$12 \mathrm{~g}$ of element contain $=N_{\mathrm{A}}$ atoms
$300 \mathrm{~g}$ of element contains $=N_{\mathrm{A}} \times \frac{300}{12}=25 N_{\mathrm{A}}$
2022
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 28th June Morning Shift
The incorrect statement about the imperfections in solids is :
A.
Schottky defect decreases the density of the substance.
B.
Interstitial defect increases the density of the substance.
C.
Frenkel defect does not alter the density of the substance.
D.
Vacancy defect increases the density of the substance.
Correct Answer: D
Explanation:
The vacancy defect increases the density of substance.
It does not change the density of the crystal. It only creates cationic vacancies. Frenkel
Defect causes vacancy defect at its original site and an interstitial defect at its new location. Therefore, it
does not change the density of the solid.
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 29th July Morning Shift
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)
Correct Answer: 566
Explanation:
In cubic close packing, octahedral voids form at
edge centers and body center of the cube
a = 2(rA+ + rB–)
a = 2 (102 + 181)
a = 566 pm
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 28th July Evening Shift
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, $\mathrm{M}=9.03 \mathrm{~g} \mathrm{~cm}^{-3}$ )
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 28th July Morning Shift
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)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 27th June Morning Shift
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)
Correct Answer: 85
Explanation:
$\mathrm{Fe}_{0.93} \mathrm{O}$
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 $(93-\mathrm{X})$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 26th June Evening Shift
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]
for $\mathrm{NaCl}$, distance between $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}=\frac{a}{2}$
$=1.04 \times 10^{-10} \mathrm{~m}$
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 24th June Morning Shift
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)
Correct Answer: 43
Explanation:
Since $X$ occupies hop lattice,
Number of particles of type $X$ in a unit cell $=6$
Number of particles of type $Y=\frac{2}{3} \times 12=8$
$\therefore$ Percentage of element $X=\frac{6}{14} \times 100$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 26th August Morning Shift
Given below are two statements.
Statement I : Frenkel defects are vacancy as well as interstitial defects.
Statement II : Frenkel defect leads to colour in ionic solids due to presence of F-centres.
Choose the most appropriate answer for the statements from the options given below :
A.
Statement I is false but Statement II is true
B.
Both Statement I and Statement II are true
C.
Statement I is true but Statement II is false
D.
Both Statement I and Statement II are false
Correct Answer: C
Explanation:
To answer this question, let's analyze each statement separately.
Statement I: Frenkel defects are vacancy as well as interstitial defects.
This statement is true. A Frenkel defect, also known as a dislocation defect, occurs in a crystalline solid when an atom or ion leaves its normal site and moves to an interstitial site, creating a vacancy defect at its original position and an interstitial defect at the new position. However, it's crucial to note that the vacancy and interstitial are created by the same atom or ion moving from one site to another within the crystal. So, the Frenkel defect involves both a vacancy and an interstitial defect, but they are correlated because they involve the displacement of the same atom or ion.
Statement II: Frenkel defect leads to color in ionic solids due to presence of F-centres.
This statement is false. The color in ionic solids is usually caused by F-centers (color centers), which are defects formed when an anion vacancy is occupied by one or more electrons. These electrons can absorb visible light, giving the crystal a characteristic color. However, the Frenkel defect, involving an atom or ion moving from its lattice site to an interstitial site, does not directly lead to the creation of F-centers. Instead, the coloration associated with F-centers is most commonly linked with Schottky defects (which involve paired vacancies of cations and anions but do not include displaced atoms to interstitial sites) or directly with anion vacancies that are not necessarily related to Frenkel defects.
Conclusion: Given the explanations above, the correct answer is Option C: Statement I is true but Statement II is false.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th July Evening Shift
Select the correct statements
(A) Crystalline solids have long range order.
(B) Crystalline solids are isotropic
(C) Amorphous solid are sometimes called pseudo solids.
(D) Amorphous solids soften over a range of temperatures.
(E) Amorphous solids have a definite heat of fusion.
Choose the most appropriate answer from the options given below.
A.
(A), (B), (E) only
B.
(B), (D) only
C.
(C), (D) only
D.
(A), (C), (D) only
Correct Answer: D
Explanation:
(A) Crystalline solids have definite arrangement of constituent particles and have long range order.
(C), (D) Different constituent particles of an amorphous solid have different bond strengths and soften over a range of temperatures.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th July Morning Shift
The parameters of the unit cell of a substance are a = 2.5, b = 3.0, c = 4.0, $\alpha$ = 90$^\circ$, $\beta$ = 120$^\circ$, $\gamma$ = 90$^\circ$. The crystal system of the substance is :
A.
Hexagonal
B.
Orthorhombic
C.
Monoclinic
D.
Triclinic
Correct Answer: C
Explanation:
a $\ne$ b $\ne$ c and $\alpha$ = $\gamma$ = 90$^\circ$ $\ne$ $\beta$ are parameters of monoclinic unit cell.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 20th July Morning Shift
Given below are two statements. One is labelled as Assertion A nd the other is labelled as Reason R.
Assertion A : Sharp glass edge becomes smooth on heating it upto its melting point.
Reason R : The viscosity of glass decreases on melting.
Choose the most appropriate answer from the options given below.
A.
A is true but R is false
B.
Both A and R are true but R is NOT the correct explanation of A.
C.
A is false but R is true.
D.
Both A and R are true and R is the correct explanation of A.
Correct Answer: B
Explanation:
On heating the glass, it melts and takes up rounded shape at the
edges, which has minimum surface area. This is due to the property
of surface tension of liquids and not due to decrease in viscosity.
Viscosity generally decreases as the temperature increases.
Hence, both A and R are true but R is not the correct explanation
of A.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 18th March Evening Shift
A hard substance melts at high temperature and is an insulator in both solid and in molten state. This solid is most likely to be a/an :
A.
Covalent solid
B.
Molecular solid
C.
Ionic solid
D.
Metallic solid
Correct Answer: A
Explanation:
Covalent or network solid are insulator (except graphite) and have very high melting point.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 18th March Morning Shift
In a binary compound, atoms of element A form a hcp structure and those of element M occupy 2/3 of the tetrahedral voids of the hcp structure. The formula of the binary compound is :
A.
M2A3
B.
M4A3
C.
M4A
D.
MA3
Correct Answer: B
Explanation:
In HCP unit cell,
Z = 6 so A = 6
Also we know, in HCP
Tetrahedral voids = 2Z = 12
$ \therefore $ No. of M = ${2 \over 3}$ [TV] = ${2 \over 3}$ $\times$ 12 = 8
$ \therefore $ Formula = M8A6 = M4A3
2021
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 31st August Evening Shift
The empirical formula for a compound with a cubic close packed arrangement of anions and with cations occupying all the octahedral sites in AxB. The value of x is _____________. (Integer answer)
Correct Answer: 1
Explanation:
Anions forms CCP or FCC(A$-$) = 4A$-$ per unit cell Cations occupy all octahedral voids (B+) = 4B+ per unit cell
cell formula $\to$ A4B4
Empirical formula $\to$ AB $\to$ (x = 1)
2021
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 22th July Evening Shift
A copper complex crystallising in a CCP lattice with a cell edge of 0.4518 nm has been revealed by employing X-ray diffraction studies. The density of a copper complex is found to be 7.62 g cm$-$3. The molar mass of copper complex is ____________ g mol$-$1. (Nearest integer)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 20th July Evening Shift
Diamond has a three dimensional structure of C atoms formed by covalent bonds. The structure of diamond has face centred cubic lattice where 50% of the tetrahedral voids are also occupied by carbon atoms. The number of carbon atoms present per unit cell of diamond is ____________.
Correct Answer: 8
Explanation:
Carbon atoms occupy FCC lattice points as well as half of the tetrahedral voids.
Therefore number of carbon atoms per unit cell = 8
2021
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 17th March Evening Shift
KBr is doped with 10$-$5 mole percent of SrBr2. The number of cationic vacancies in 1g of KBr crystal is ____________ 1014. (Round off to the Nearest Integer).
[Atomic Mass : K : 39.1 u, Br : 79.9 u NA = 6.023 $\times$ 1023]
Correct Answer: 5
Explanation:
For every Sr+2 ion, 1 cationic vacancy is created. Hence, no. of Sr+2 ion = Number of cationic vacancies
Since mole percentage of SrBr2 dropped is 10$-$5 to that of total moles of KBr.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 16th March Evening Shift
Ga (atomic mass 70 u) crystallizes in a hexagonal close packed structure. The total number of voids in 0.581 g of Ga is __________ $\times$ 1021. (Round off to the Nearest Integer). [Given : NA = 6.023 $\times$ 1023]
Correct Answer: 15
Explanation:
For HCP structure Z = 6
Tetrahedral Void = (Z $\times$ 2) = 12
Octahedral Void = [Z $\times$ 1] = 6
No. of void per unit cell = 18
No. of unit cell = $\left( {{{0.581 \times {N_A}} \over {70 \times 6}}} \right)$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 16th March Morning Shift
A certain element crystallises in a bcc lattice of unit cell edge length 27$\mathop A\limits^o $. If the same element under the same conditions crystallises in the fcc lattice, the edge length of the unit cel in $\mathop A\limits^o $ will be ____________. (Round off to the Nearest Integer).
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 26th February Evening Shift
The number of octahedral voids per lattice site in a lattice is _________. (Rounded off to the nearest integer)
Correct Answer: 1
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$
2021
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 25th February Evening Shift
The unit cell of copper corresponds to a face centered cube of edge length 3.596 $\mathop A\limits^o $ with one copper atom at each lattice point. The calculated density of copper in kg/m3 is ___________. [Molar mass of Cu : 63.54 g; Avogadro Number = 6.022 $\times$ 1023]
Correct Answer: 9077
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,
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 24th February Morning Shift
The coordination number of an atom in a body-centered cubic structure is _______.
[Assume that the lattice is made up of atoms.]
Correct Answer: 8
Explanation:
Coordination number is the number of nearest neighbours of a central atom in the structure.
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.
2020
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 6th September Evening Slot
A crystal is made up of metal ions 'M1' and 'M2'
and oxide ions. Oxide ions form a ccp lattice
structure. The cation 'M1' occupies 50% of
octahedral voids and the cation 'M2' occupies
12.5% of tetrahedral voids of oxide lattice. The
oxidation numbers of 'M1' and 'M2' are,
respectively :
Let charge on M1 and M2 are +x and +y respectively. And O4 has -8 charge.
As crystal is neutral. So metals must have +8
charge in total.
$ \therefore $ +2x + y = 8 ....(1)
By checking options we found eq (1) satisfy when
x = +2
y = +4
2020
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 5th September Evening Slot
An element crystallises in a face-centred cubic (fcc) unit cell with cell edge $a$. The distance
between the centres of two nearest octahedral voids in the crystal lattice is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 5th September Morning Slot
A diatomic molecule X2
has a body-centred cubic
(bcc) structure with a cell edge of 300 pm. The
density of the molecule is 6.17 g cm–3. The number
of molecules present in 200 g of X2
is :
(Avogadro constant (N
A) = 6 $ \times $ 1023 mol–1
)
Number of moles in 200 gm = ${{{200} \over {50}}}$ = 4
$ \therefore $ Number of molecules = 4$ \times $NA
2020
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 8th January Evening Slot
Which of the following compounds is likely to
show both Frenkel and Schottky defects in its
crystalline form?
A.
CsCl
B.
AgBr
C.
ZnS
D.
KBr
Correct Answer: B
Explanation:
Since AgBr has intermediate radius ratio. AgBr shows both, Frenkel as well as Schottky
defects.
ZnS only shows Frenkel defects.
KBr, CsCl only shows Schottky defects.
2020
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 3rd September Morning Slot
An element with molar mass 2.7 $ \times $ 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its
density is 2.7 $ \times $ 103
kg m-3, the radius of the element is approximately ______ $ \times $ 10-12 m (to the
nearest integer).
Correct Answer: 143
Explanation:
Molar mass of an element (M) = 27 gm mol–1
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 = ${{Z \times M} \over {{N_A} \times {{\left( a \right)}^3}}}$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 12th April Morning Slot
An element has a face-centred cubic (fcc) structure with a cell edge of $a$. The distance between the centres of
two nearest tetrahedral voids in the lattice is :
A.
$a$
B.
${3 \over 2}a$
C.
${a \over 2}$
D.
$\sqrt 2 a$
Correct Answer: C
Explanation:
In FCC, tetrahedral voids are located on the
body diagonal at a
distance of ${{\sqrt 3 a} \over 4}$ from the
corner. Together they form a smaller cube of
edge length ${a \over 2}$.
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 9th April Evening Slot
10 mL of 1mM surfactant solution forms a
monolayer covering 0.24 cm2 on a polar
substrate. If the polar head is approximated as
cube, what is its edge length?
A.
1.0 pm
B.
2.0 nm
C.
0.1 nm
D.
2.0 pm
Correct Answer: D
Explanation:
No of moles formed = 10-3 $ \times $ ${{10} \over {1000}}$ = 10-5
$ \therefore $ No of molecules formed = 10-5 $ \times $ NA
In unimolecular layer formation each cube occupy an area = a2
$ \therefore $ Total area occupied = 10-5 $ \times $ NA $ \times $ a2
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 8th April Evening Slot
Consider the bcc unit cells of the solids 1 and
2 with the position of atoms as shown below.
The radius of atom B is twice that of atom A.
The unit cell edge length is 50% more in solid
2 than in 1. What is the approximate packing
efficiency in solid 2?
A.
45%
B.
90%
C.
75%
D.
65%
Correct Answer: B
Explanation:
Given that, The radius of atom B is twice that of atom A.
Let r = radius of atom A then radius of atom B = 2r
Let edge length of the cube = $a$
We know length of body diagonal = $a\sqrt 3 $
Here length of body diagonal in solid 2 = r + r + 2(2r) = 6r
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 8th April Evening Slot
0.27 g of a long chain fatty acid was dissolved
in 100 cm3 of hexane. 10 mL of this solution
was added dropwise to the surface of water in
a round watch glass. Hexane evaporates and a
monolayer is formed. The distance from edge
to centre of the watch glass is 10 cm. What is
the height of the monolayer?
[Density of fatty acid = 0.9 g cm–3, $\pi $ = 3]
A.
10–8 m
B.
10–2 m
C.
10–4 m
D.
10–6 m
Correct Answer: D
Explanation:
In 100 ml of hexane solution contains 0.27 g of fatty acid.
$ \therefore $ In 10 ml of hexane solution contains 0.027 g of fatty acid.
Volume of fatty acid present on the round glass = ${{0.027} \over {0.9}}$
As here Area of fatty acid layer = Area of round plate = $\pi {r^2}$
$ \therefore $ Volume of fatty acid layer = $\pi {r^2}$ $ \times $ h
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 8th April Morning Slot
Element 'B' forms ccp structure and 'A' occupies half of the octahedral voids, while oxygen atoms
occupy all the tetrahedral voids, The structure of bimetallic oxide is :
A.
AB2O4
B.
A2BO4
C.
A4B2O
D.
A2B2O
Correct Answer: A
Explanation:
We know, for cubic unit cell, only FCC has octahedral and tetrahedral voids.
B forms ccp structure means B forms FCC structure.
For FCC, z = 4
We know for octahedral voids z = 4. In this lattice, A present in half of octahedral voids.
$ \therefore $ For A, z = 2
For tetrahedral voids, 2z = 8. In this lattice, oxygen present in all the tetrahedral voids.
$ \therefore $ Simplest formula :
${A_{4 \times {1 \over 2}}}{B_4}{O_8}$
= ${A_2}{B_4}{O_8}$
= $A{B_2}{O_4}$
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 11th January Evening Slot
The radius of the largest sphere which fits properly at the centre of the edge of a body centred cubic unit cell is : (Edge length is represented by 'a')
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 11th January Morning Slot
A solid having density of 9$ \times $103 kg m–3 forms face centred cubic crystals of edge length $200\sqrt 2 $ pm. What is the molar mass of the solid?
[Avogadro constant $ \cong $ 6 $ \times $ 1023 mol–1
, $\pi $ $ \cong $ 3]
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 10th January Evening Slot
A compound of formula A2B3 has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms :
A.
hcp lattice - A, ${1 \over 3}$ Tetrahedral voids-B
