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Correct answer is: (B) 6

Correct answer is: (C) 1.5 \(\times\) 10²³

58.5 g of NaCl = 1 mole = 6.023 \(\times\) 10²³ NaCl

units. One unit cell contains 4 NaCl units.

Hence, the number of unit cells present

= \(\cfrac{6.023\times10^{23}}{4}=1.5\times10^{23}\)

Correct answer: (C) 1.5×10²³

(i) The defect in which equal number of cations and anions are missing from the lattice.

(ii) Due to dislocation of smaller ion from its normal site to an interstitial site. 

(iii) Anionic vacancies are occupied by unpaired electron. 

(i) Stoichiometric defect.

(ii) Schottky defect NaCl

(iii) Density of crystal decreases.

Correct answer is: (C) 4

Correct answer is: (B) body centered cubic

Total number of spheres in body centered cubic unit cell = 1/8 \(\times\) 8 + 1 = 2 spheres (atoms, ions or molecules).

Ferromagnetism : 

(i) Substance which are attracted most easily in magnetic field is called Ferromagnetic substance. 

(ii) Examples: Fe, Co, Ni, CrO₂ and Alnico (alloy of Co, Fe and Cu). 

(iii) It occurs due to alignment of all magnetic moments (due to unpaired e^-) in the same directions.

Ferrimagnetism : 

(i) A substance which is weakly attracted by the magnetic field is called ferrimagnetic substances. 

(ii) Examples: Fe₃O₄, ferrites having the formula M²⁺Fe₂O₄ (where M²⁺ = Cu²⁺ or Zn²⁺ ).

(iii) In these substances the alignment of magnetic moment in opposite direction are in unequal numbers.

The type of structure is formed by ccp/fcc arrangement.

The type of structure is formed by hcp arrangement.

(c) Glucose It is a molecular solid where as others are amorphous solids.

Correct answer: (A) \(a=b=c,\,\alpha=\beta=\gamma\neq90^\circ\)

Correct answer is: (A) 1

(i) Frenkel defect: The defect in which the smaller ion/cation is dislocated to interstitial site.

Example: Silver halides, ZnS.

(ii) F centres: The anion vacancy occupied by an electron.

Example: NaCl, KCl, LiCl 

(c) Frozen elements of group 18 It is atomic solid where as others are metallic solids.

Correct answer is: (A) a = b = c, \(\alpha\) = \(\beta\) = \(\gamma\) \(\neq\) 90°

(d) M₃N₂

If the total number of M atoms is n, then the number of tetrahedral voids = 2 n. Given that \(\Big(\frac{1}{3}\Big)^{rd}\)of tetrahedral voids are occupied i.e.,\(\Big(\frac{1}{3}\Big)\) x 2 n are occupied by N atoms.

∴ M : N ⇒ n :\(\Big(\frac{2}{3}\Big)\)

1 :\(\Big(\frac{1}{3}\Big)\)3 : 2 ⇒ M₃N₂

(a) Glass, plastic (Amorphous solids)

(i) Tetrahedral void is surrounded by 4 constituent particles (atoms/molecules/ions).

Octahedral voids is surrounded by 6 constituent particles (atoms/molecules/ions).

OR

Radius ratio (r⁺ /r^-) fo, tetrahedral void is 0.225 & radius ratio for Octahedral voids is 0.414.

(ii) A regular three dimensional arrangement of point in space is called a crystal lattice.

Unit cell is the smallest portion of a crystal lattice which, when repeated in three directions, generates an entire lattice/ unit cell is the miniature of crystal lattice/microscopic edition of the crystal lattice.

(a) Solid CO₂ It is a polar molecular solid where as others are hydrogen bonded molecular solids.

Tetrahedral void 

1. A single triangular void in a crystal is surrounded by four (4) spheres and is called a tetrahedral void 

2. A sphere of second layer is above the void of the first layer, a tetrahedral void is formed 

3. This constitutes four spheres, three in the lower and one in upper layer. When the centres of these four spheres are joined a tetrahedron is formed

4. The radius of the sphere which can be accommodated in an octahedral hole without disturbing the structure should not exceed 0.414 times that of the structure forming sphere 

5. Radius of an tetrahedral void \(\frac{r}{R}\) = 0.225

Octahedral void 

1. A double triangular void like c is surrounded by six(6) spheres and is called an octahedral void 

2. The voids in the first layer are partially covered by the spheres of layer now such a void is called a octahedral void 

3. This constitutes six spheres, three in the lower layer and three in the upper layer. When the centers of these six spheres are joined an octahedron is formed 

4. The sphere which can be placed in a tetrahedral hole without disturbing the close packed structure should not have a radius larger than 0.225 times the radius of the sphere forming the structure

5. Radius of a octahedral void   \(\frac{r}{R}\) = 0.225

(d) Cyclic It is a ring structure where as other primitive crystal systems.

Correct answer is: (D) X₄Y₃

Let the number of atoms of element Y in hcp unit cell be n.

\(\therefore\) Number of tetrahedral voids = 2n

As 2/3^rd of the tetrahedral voids are occupied by atoms of element X,

Number of atoms of element X = 2n\(\times \cfrac{2}{3}=\cfrac{4n}{3}\)

\(\therefore\) Ratio of atoms of element X : atoms of element Y =\(\cfrac{4n}{3}\):n = 4 : 3

The formula of the compound is X₄Y₃.

Hexagonal close packing 

1. aba arrangement 

2. In this case, the spheres of the third layer are exactly aligned with those of the first layer 

3. In HCP, tetrahedral voids of the second layer may be covered by the spheres of the third layer

Cubic close packing: 

1. abc arrangement 

2. In this case, the spheres of the third layer are not aligned with those of the first layer or second layer. Only when fourth layer is placed, its spheres are aligned with the first layer 

3. In cep third layer may be placed above the second layer in a manner such that its sphere cover the octahedral voids

Correct answer is: (D) 8

Correct answer is: (B) 7

Correct answer is: (C) hexagons and pentagons

Correct answer is: (C) AB₂

Correct answer: (A) X₂Y₄Z

Correct answer is: (B) XYZ₃

Atoms of X per unit cell = 8 \(\times \cfrac{1}{8}\) = 1

Atoms of Y per unit cell = 1

Atoms of Z per unit cell = 6 \(\times \cfrac{1}{2}\) = 3

Hence, the formula is XYZ₃

Given: Oxide (O^2-) ions form hcp lattice and ferric (Fe³⁺) ions occupy 2/3rd of the octahedral voids.

To find: Formula of the compound.

Calculation: Let the number of oxide ions (O^2- ) in hcp unit cell = n

 \(\therefore\) Number of octahedral voids = n

As\(\cfrac{2}{3}\) rd of the octahedral voids are occupied by ferric ions, the number of ferric ions present

=n\(\times\frac{2}{3}=\frac{2n}{3}\)

\(\therefore\) Ratio of Fe³⁺ : O^2- =\(\frac{2n}{3}:n=2:3\)

Ans: The formula of ferric oxide is Fe₂O₃.

Correct answer is: (B) NaWO₃

W at corner;\(\cfrac{1}{8}\times 8=1\)

O at centres of edges; \(\cfrac{1}{2}\times 6=3\)

Na at centre of cube = 1

Na : W : O

1 : 1 : 3

Correct answer is: (B) XY₃, 4.38 g/cm³

Correct answer is: (D) 367.64 pm, 8.54 g cm^-3

Correct answer is: (B) Schottky defect

Correct answer is: (A) solid

Correct answer: (A) solid

Diamond is three dimensional net work covalent solid with strong inter atomic forces where as sulphur consists of packered ring structure (S₈) in which atoms are held together by weak vander waal forces. 

Correct answer is: (C) does not change

1. Defect in crystalline structure : Any deviation from orderly and stoichiometrically perfect arrangement of atoms, ions or molecules in the crystal lattice is called a defect in the crystalline structure.

2. Defects are created during the crystallisation process. If the process occurs at faster rate, the defects are more.

3. The properties of solids are affected due to imperfactions.

The defects in crystalline solids are of two types viz., 

(1) Point defect and 

(2) Line defect. 

(1) Point defects are further classified as : 

(a) Vacancy defect or Schottky defect 

(b) Interstitial defect or Frenkel defect 

(c) Impurity defect

This is further classified as :

• Substitution impurity defect 
• Interstitial impurity defect.

(2) Line defects are further classified as :

• Edge dislocation 
• Screw dislocation. 

When Fe₃O₄ is heated to 850 K it loses ferrimagnetism and becomes paramagnetic.

There are two tetrahedral holes.

Correct answer is: (C) 181pm

2r⁺ + 2r^– = 552; r⁺ + r^– = \(\cfrac{552}{2}\)= 276

r^– = 276 – 95 = 181 pm.

Correct answer is: (A) 200 pm

Correct answer is: (C) 6 × 10²³ mol^-1

Correct answer is: (B) face centered cubic lattice

1. Amorphous solids do not have perfectly ordered crystalline structure.

2. They have short range order of regular pattern hence periodically repeating regular pattern is over a short distance. 

3. The thermal energy required to break the structure and separate constituent particles is not uniform. 

4. Hence the temperature needed to melt the solid is not same, therefore amorphous solids do not have sharp melting points but melt over a range of temperature.

No. of Y atoms per unit cell in ccp lattice = 4 

No. of tetrahedral voids= 2*4 = 8 

No. of tetrahedral voids occupied by X = 2/3*8 = 16/3 

Therefore formula of the compound = X_16/3 Y₄ 

= X₁₆ Y₁₂ 

= X₄ Y₃

Number of element Y = n, Number of element X = 2 n x\(\frac{2}{3}\) .

As number of tetrahedral void = 2n

X:Y = \(\frac{4n}{3}\):n

Formula = X₄Y₃