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Correct answer is: (C) oxygen

Correct answer is: (C) both (A) and (B)

Paramagnetic: Sodium ₁₁Na,

Diamagnetic: ₁₇Cl^- anion, Magnesium ₁₂Mg, ₂₀Ca²⁺ cation

Ferromagnetic: Iron ₂₆Fe, ₂₇Co atom, ₂₈Ni atom

Option : (d) ferromagnetic

Correct answer is: (B) in between 0.73 and 0.41

Correct answer is: (A) diamagnetic materials

According to the formula, MgAl₂O₄ . If there are 4 oxide ions, there will be 1 Mg²⁺ ions and 2 Al³⁺. But if the 4 O^2– ions are ccp in arrangement, there will be 4 octahedral and 8 tetrahedral voids.

(i) Percentage of tetrahedral voids occupied by Mg²⁺ = (1/8) x 100 = 12.5%

(ii) Percentage of octahedral voids occupied by Al³⁺ = (2/4) x 100 = 50%

The temperature at which they are changed into paramagnetic is called curie temperature.

This is because the realignment of electrons spin or their magnetic moments which are now oriented in one particular direction. 

Consequences of Schottky defect :

• Since the number of ions (cations and anions) decreases but volume remains unchanged, the density of a substance decreases.
• As the number of missing cations and anions is equal, the electrical neutrality of the compound remains same.
• This defect arises in ionic crystals like NaCl, AgBr, KCl, etc.

(1) The magnetic properties of a substance arise due to the presence of electrons.

(2) An electron while revolving around the nucleus, also spins around its own axis and generates a magnetic moment and magnetic properties.

(3) If an atom or a molecule contains one or more unpaired electrons spinning in same direction, clockwise or anticlockwise, then the substance is associated with net magnetic moment and magnetic properties. They experience a net force of attraction when placed in the magnetic field.

This phenomenon is called paramagnetism and the substance is said to be paramagnetic.

For example, O₂, Cu²⁺, Fe³⁺, Cr³⁺, NO, etc.

Correct answer is: (A) Paramagnetic substances are attracted by the magnetic field.

Option : (c) ferromagnetic

Option : (b) \(\frac{8}{3}\pi r^3\)

When constituent particles are present only at the corner positions of a unit cell, it is called as primitive unit cell.

These are also called simple unit cells. In all, there are seven types of primitive unit cells. They are cubic, tetragonal, orthorhombic, monoclinic, triclinic, hexagonal and rhombohedral.

Option : (d) 14

Correct answer is: (C) metallic

Option : (a) Frenkel defect

Correct answer is: (C) KBr

Option : (a) 500 pm

Correct answer is: (B) AgBr

(b) 6.023 x 10²²

In bcc unit cell, 2 atoms = 1 unit cell

Number of atoms in 8g of element is, Number of moles = \(\Big(\frac{8g}{40g,mol^{-1}}\Big)\) = 0.2 mol

1 mole contains 6.023 x 10²³ atoms 0.2 mole contains 0.2 x 6.023 x 10²³ atoms

\(\Big(\frac{1\,unitcell}{2\,atoms}\Big)\)0.2 x 6.023 x 10²³ = 6.023 x 10²² unit cells

The solid where both Frankel and schottky defect occur is AgBr.

Due to F-Centre KCl appears violet in colour.

Because Silicon is a semiconductor and in semiconductors electrons and move from valance band to conduction band only when the gap between the is filled by energy. So, rise in temperature in semiconductors like silicon increases their conductivity.

Frenkel defect is not found in pure alkali metal halides. This is because of the larger size of the cations of the alkali metals, due to the large size the cations cannot fit into the interstitial sites.

The conductivity is increased by adding appropriate amount of suitable impurity. This process is called Doping.

Both crystal structures have same density because the percentageof occupied space is same.

Correct answer is: (C) AB₃C

An atom at the corner of a cube is shared among 8 unit cells. As there are 8 corners in a cube, number of corner atom [A] per unit cell

= 8 \(\times\cfrac{1}{8}\) = 1

A face-centered atom in a cube is shared by two unit cells. As there are 6 faces in a cube, number of face-centered atoms [B] per 

unit cell = 6 \(\times\cfrac{1}{2}\) = 3

An atom in the body of the cube is not shared by other cells.

Thus, number of atoms [C] at the body centre per unit cell = 1

Hence, the formula of the solid is AB₃C.

Correct answer: (C) AuCu₃

Vacancy and Schottky defect.

fcc< bcc <sc

Correct answer is: (C) AuCu₃

One-eighth of each corner atom (Au) and one half of each face centered atom (Cu) are contained within the unit cell of the compound. Thus, the number of Au atoms per unit cell

= 8 \(\times\cfrac{1}{8}\) = 1 and the number of Cu atoms per

unit cell = 6\(\times\cfrac{1}{2}=3.\)

The formula of the compound is AuCu₃.

Option : (c) vacancy defect

(c) 6

\(\frac{r_{c1}}{rA^-}\) = \(\Big(\frac{0.98\times10^{-10}}{1.81\times10^{-10}}\Big)\)

= 0.54. It is in the range of 0.414, 0.732,hence the coordination number of each ion is 6.

Correct answer is: (B) simple cubic

(c) Frenkel defect

(d)  g\(\Big(\frac{\sqrt{3}}{2}\Big)\times 400\)pm

g\(\sqrt{3a}\) = r_Cs⁺ + 2rCs^- + r_Cs⁺

\(\Big(\frac{\sqrt{3}}{2}\Big)\)a  =  (r_Cs⁺ + rCs^- )

\(\Big(\frac{\sqrt3}{2}\Big)\)400 = inter ionic distance

(a)\(\Big(\frac{100}{0.414}\Big)\)

For an fcc structure

(c) 32%

Packing efficiency = 68%. Therefore empty space percentage = (100-68) = 32%

Correct answer is: (B) XY

The number of ions of ‘Y’ in ccp unit cell = 4. Number of octahedral voids in ccp unit cell = Number of ions of ‘Y’ in ccp unit cell = 4 Since, ions of ‘X’ occupy all the octahedral voids, the number of ions of ‘X’ in ccp unit cell = 4

\(\therefore\) The ratio of number of ions of ‘X’ to the number of ions of ‘Y’ = 4 : 4 i.e., 1 : 1. Hence, the molecular formula of the given compound is XY.

Correct answer is: (B) 32%

Correct answer is: (B) a / 2 cm

Correct answer is: (A) 26 %

1. In metals a large number of outermost electrons of atoms occupy conduction bands.

2. Band formation in metals results in delocalisation of outermost electrons forming metal ions or cations.

3. The metallic cations occupying crystal lattice sites vibrate about mean positions.

4. As temperature increases the vibrational motion increases which interrupts flow of electrons decreasing electrical conductivity.

Rhenium oxide (ReO₃)

 N = (4 x 98.5)/[5.22 x (500 x 10^-10) ³] = 6.023 x 10²³.

(a)Glass objects from sncient civilizations are found to become milky because glass gets converted into crystalline solid due to heating and cooling for a long period of time. 

(b)Glass is supercooled liquid as it has tendency to flow. so, glass panes are thicker at the bottom than the top. 

Given: Density (d) = 10 g cm^-3

Edge length (a) = 100 pm = 100×10^–12 m = 100×10¹⁰ cm

Mass of crystal = 2 g

To find: Number of atoms

Formula: Density =\(\frac{Mass}{Volume}\)

Density=\(\frac{Mass}{Volume}\)

∴ Volume =\(\frac{Mass}{Density}\)

∴  Volume = \(\frac{2\,g}{10\,g\,cm^{-3}}=0.2\,cm^3\)

Volume of unit cell = a³ = (100×10^–10 cm)3 = 1×10^-24 cm³

Number of unit cells in 2 g of crystal =\(\frac{total\,volume}{volume\,of\,unit\,cell}\)

=\(\frac{0.2\,cm^3}{1\times10^{-24}cm^3}\)

= 0.2×10²⁴ unit cells

∵ The given unit cell is of fcc type, therefore, it contains 4 atoms.

∴ 0.2×10²⁴ unit cells will contain 4×0.2×10²⁴ = 0.8×10²⁴ atoms

= 0.8×10²⁴ atoms = 8×10²³ atoms

Ans: Number of atoms present in 2 g of given crystal is 8×10²³ atoms.

Given: Unit cell is cubic close-packed structure.

Metallic radius (r) of aluminium = 125 pm.

To find:

i. The edge length (a) .

ii. Number of unit cells in 1.00 cm3 of aluminium.

Formulae:

i. a =\(\sqrt{8}\)r = 2 \(\sqrt{2}\)r 

ii. Volume of the unit cell = a³

Calculation: 

i. For cubic close packing, a = 2\(\sqrt{2}\) r 

= 2\(\sqrt{2}\)×125 = 353.5 pm = 353.5×10^-10 cm 

ii. Volume of the unit cell = a³ = (353.5×10^-10)³ = 4.4174×10^-23 

Number of unit cells in 1.00 cm3 of aluminium =\(\frac{1.00}{4.4174\times10^{-23}}=2.2638\times10^{22}\)

Ans: i. The length of the side of the unit cell = 353.5×10^-10 cm. 

ii. Number of unit cells in 1.00 cm3 of aluminium = 2.2638×10²².

The answer is (b) A₂B.