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CHXII01:THE SOLID STATE

318748 In a crystalline solid, having formula $X Y_{2} O_{4}$ , oxide ions are arranged in cubic close packed lattice while, cations $X$ are present in tetrahedral voids and cations $Y$ are present in octahedral voids. The percentage of tetrahedral voids occupied by $X$ is

1

12.5 %

2

25 %

3

50 %

4

75 %

Explanation:

In a cubic close packed lattice of oxide ions, there would be two tetrahedral and one octahedral void per oxide ion.Since, the formula shows the presence of 4 oxide ions, the number of tetrahedral voids is eight and that of octahedral voids is four. Out of the eight tetrahedral voids, one is occupied by $X$ .
$∴$ Percentage of tetrahedral voids
occupied $= \frac{1}{8} \times 100 = 12.5 %$

CHXII01:THE SOLID STATE

318749 Total number of voids in 0.5 mole of a compound forming hexagonal closed packed structure are

1

6.022 \times 10^{23}

2

3.011 \times 10^{23}

3

9.033 \times 10^{23}

4

4.516 \times 10^{23}

Explanation:

In hexagonal close packed structure, there are 6 atoms per unit cell.
$∴$ Number of octahedral voids $= 6$
$∴$ Number of tetrahedral voids
$= 2 \times 6 = 12$
Total number of voids $= 6 + 12 = 18$
$∴$ Total number of voids per atom
$= \frac{18}{6} = 3$
$∵$ In 1 mole of a compound, the total number of voids $= 3 \times 6.023 \times 10^{23}$
$∴$ In 0.5 mole compound, the total number of voids
$= 3 \times 0.5 \times 6.022 \times 10^{23} = 9.033 \times 10^{23}$

CHXII01:THE SOLID STATE

318750 In chrysoberyl, a compound containing Beryllium, Aluminium and Oxygen, oxide ions form cubic close packed structure. Aluminium ions occupy $\frac{1}{4}$ th of octahedral voids. The formula of the compound is:

1

{BeAlO}_{4}

2

{BeAl}_{2} O_{4}

3

{Be}_{2} {AlO}_{2}

4

BeAlO

Explanation:

$Be Al O N =$ No. of oxide ions involved in $CCP$
$\frac{N}{4} : \frac{2 N}{4} : N ∴ octahedral voids = N$
1:2:4 Tetrahedral voids $= 2 N$

CHXII01:THE SOLID STATE

318751 Titanium crystallises in a face centred cubic lattice. It reacts with carbon or hydrogen interstitially by allowing atoms of these elements to occupy holes in the host lattice. Hydrogen occupies tetrahedral holes but carbon occupies octahedral holes the formula of titanium carbide and hydride are

1

{TiC}_{2}, TiH

2

{TiC}_{2}, {TiH}_{4}

3

TiC, {TiH}_{2}

4

{TiC}_{3}, {TiH}_{2}

Explanation:

Number of Ti atoms $= 4$
Number of $H$ atoms $= 2 \times 4 = 8$
Number of $C$ atoms $= 4$
Formula: $TiC, {TiH}_{2}$

CHXII01:THE SOLID STATE

318748 In a crystalline solid, having formula $X Y_{2} O_{4}$ , oxide ions are arranged in cubic close packed lattice while, cations $X$ are present in tetrahedral voids and cations $Y$ are present in octahedral voids. The percentage of tetrahedral voids occupied by $X$ is

1

12.5 %

2

25 %

3

50 %

4

75 %

Explanation:

In a cubic close packed lattice of oxide ions, there would be two tetrahedral and one octahedral void per oxide ion.Since, the formula shows the presence of 4 oxide ions, the number of tetrahedral voids is eight and that of octahedral voids is four. Out of the eight tetrahedral voids, one is occupied by $X$ .
$∴$ Percentage of tetrahedral voids
occupied $= \frac{1}{8} \times 100 = 12.5 %$

CHXII01:THE SOLID STATE

318749 Total number of voids in 0.5 mole of a compound forming hexagonal closed packed structure are

1

6.022 \times 10^{23}

2

3.011 \times 10^{23}

3

9.033 \times 10^{23}

4

4.516 \times 10^{23}

Explanation:

In hexagonal close packed structure, there are 6 atoms per unit cell.
$∴$ Number of octahedral voids $= 6$
$∴$ Number of tetrahedral voids
$= 2 \times 6 = 12$
Total number of voids $= 6 + 12 = 18$
$∴$ Total number of voids per atom
$= \frac{18}{6} = 3$
$∵$ In 1 mole of a compound, the total number of voids $= 3 \times 6.023 \times 10^{23}$
$∴$ In 0.5 mole compound, the total number of voids
$= 3 \times 0.5 \times 6.022 \times 10^{23} = 9.033 \times 10^{23}$

CHXII01:THE SOLID STATE

318750 In chrysoberyl, a compound containing Beryllium, Aluminium and Oxygen, oxide ions form cubic close packed structure. Aluminium ions occupy $\frac{1}{4}$ th of octahedral voids. The formula of the compound is:

1

{BeAlO}_{4}

2

{BeAl}_{2} O_{4}

3

{Be}_{2} {AlO}_{2}

4

BeAlO

Explanation:

$Be Al O N =$ No. of oxide ions involved in $CCP$
$\frac{N}{4} : \frac{2 N}{4} : N ∴ octahedral voids = N$
1:2:4 Tetrahedral voids $= 2 N$

CHXII01:THE SOLID STATE

318751 Titanium crystallises in a face centred cubic lattice. It reacts with carbon or hydrogen interstitially by allowing atoms of these elements to occupy holes in the host lattice. Hydrogen occupies tetrahedral holes but carbon occupies octahedral holes the formula of titanium carbide and hydride are

1

{TiC}_{2}, TiH

2

{TiC}_{2}, {TiH}_{4}

3

TiC, {TiH}_{2}

4

{TiC}_{3}, {TiH}_{2}

Explanation:

Number of Ti atoms $= 4$
Number of $H$ atoms $= 2 \times 4 = 8$
Number of $C$ atoms $= 4$
Formula: $TiC, {TiH}_{2}$

CHXII01:THE SOLID STATE

318748 In a crystalline solid, having formula $X Y_{2} O_{4}$ , oxide ions are arranged in cubic close packed lattice while, cations $X$ are present in tetrahedral voids and cations $Y$ are present in octahedral voids. The percentage of tetrahedral voids occupied by $X$ is

1

12.5 %

2

25 %

3

50 %

4

75 %

Explanation:

In a cubic close packed lattice of oxide ions, there would be two tetrahedral and one octahedral void per oxide ion.Since, the formula shows the presence of 4 oxide ions, the number of tetrahedral voids is eight and that of octahedral voids is four. Out of the eight tetrahedral voids, one is occupied by $X$ .
$∴$ Percentage of tetrahedral voids
occupied $= \frac{1}{8} \times 100 = 12.5 %$

CHXII01:THE SOLID STATE

318749 Total number of voids in 0.5 mole of a compound forming hexagonal closed packed structure are

1

6.022 \times 10^{23}

2

3.011 \times 10^{23}

3

9.033 \times 10^{23}

4

4.516 \times 10^{23}

Explanation:

In hexagonal close packed structure, there are 6 atoms per unit cell.
$∴$ Number of octahedral voids $= 6$
$∴$ Number of tetrahedral voids
$= 2 \times 6 = 12$
Total number of voids $= 6 + 12 = 18$
$∴$ Total number of voids per atom
$= \frac{18}{6} = 3$
$∵$ In 1 mole of a compound, the total number of voids $= 3 \times 6.023 \times 10^{23}$
$∴$ In 0.5 mole compound, the total number of voids
$= 3 \times 0.5 \times 6.022 \times 10^{23} = 9.033 \times 10^{23}$

CHXII01:THE SOLID STATE

318750 In chrysoberyl, a compound containing Beryllium, Aluminium and Oxygen, oxide ions form cubic close packed structure. Aluminium ions occupy $\frac{1}{4}$ th of octahedral voids. The formula of the compound is:

1

{BeAlO}_{4}

2

{BeAl}_{2} O_{4}

3

{Be}_{2} {AlO}_{2}

4

BeAlO

Explanation:

$Be Al O N =$ No. of oxide ions involved in $CCP$
$\frac{N}{4} : \frac{2 N}{4} : N ∴ octahedral voids = N$
1:2:4 Tetrahedral voids $= 2 N$

CHXII01:THE SOLID STATE

318751 Titanium crystallises in a face centred cubic lattice. It reacts with carbon or hydrogen interstitially by allowing atoms of these elements to occupy holes in the host lattice. Hydrogen occupies tetrahedral holes but carbon occupies octahedral holes the formula of titanium carbide and hydride are

1

{TiC}_{2}, TiH

2

{TiC}_{2}, {TiH}_{4}

3

TiC, {TiH}_{2}

4

{TiC}_{3}, {TiH}_{2}

Explanation:

Number of Ti atoms $= 4$
Number of $H$ atoms $= 2 \times 4 = 8$
Number of $C$ atoms $= 4$
Formula: $TiC, {TiH}_{2}$

CHXII01:THE SOLID STATE

318748 In a crystalline solid, having formula $X Y_{2} O_{4}$ , oxide ions are arranged in cubic close packed lattice while, cations $X$ are present in tetrahedral voids and cations $Y$ are present in octahedral voids. The percentage of tetrahedral voids occupied by $X$ is

1

12.5 %

2

25 %

3

50 %

4

75 %

Explanation:

In a cubic close packed lattice of oxide ions, there would be two tetrahedral and one octahedral void per oxide ion.Since, the formula shows the presence of 4 oxide ions, the number of tetrahedral voids is eight and that of octahedral voids is four. Out of the eight tetrahedral voids, one is occupied by $X$ .
$∴$ Percentage of tetrahedral voids
occupied $= \frac{1}{8} \times 100 = 12.5 %$

CHXII01:THE SOLID STATE

318749 Total number of voids in 0.5 mole of a compound forming hexagonal closed packed structure are

1

6.022 \times 10^{23}

2

3.011 \times 10^{23}

3

9.033 \times 10^{23}

4

4.516 \times 10^{23}

Explanation:

In hexagonal close packed structure, there are 6 atoms per unit cell.
$∴$ Number of octahedral voids $= 6$
$∴$ Number of tetrahedral voids
$= 2 \times 6 = 12$
Total number of voids $= 6 + 12 = 18$
$∴$ Total number of voids per atom
$= \frac{18}{6} = 3$
$∵$ In 1 mole of a compound, the total number of voids $= 3 \times 6.023 \times 10^{23}$
$∴$ In 0.5 mole compound, the total number of voids
$= 3 \times 0.5 \times 6.022 \times 10^{23} = 9.033 \times 10^{23}$

CHXII01:THE SOLID STATE

318750 In chrysoberyl, a compound containing Beryllium, Aluminium and Oxygen, oxide ions form cubic close packed structure. Aluminium ions occupy $\frac{1}{4}$ th of octahedral voids. The formula of the compound is:

1

{BeAlO}_{4}

2

{BeAl}_{2} O_{4}

3

{Be}_{2} {AlO}_{2}

4

BeAlO

Explanation:

$Be Al O N =$ No. of oxide ions involved in $CCP$
$\frac{N}{4} : \frac{2 N}{4} : N ∴ octahedral voids = N$
1:2:4 Tetrahedral voids $= 2 N$

CHXII01:THE SOLID STATE

318751 Titanium crystallises in a face centred cubic lattice. It reacts with carbon or hydrogen interstitially by allowing atoms of these elements to occupy holes in the host lattice. Hydrogen occupies tetrahedral holes but carbon occupies octahedral holes the formula of titanium carbide and hydride are

1

{TiC}_{2}, TiH

2

{TiC}_{2}, {TiH}_{4}

3

TiC, {TiH}_{2}

4

{TiC}_{3}, {TiH}_{2}

Explanation:

Number of Ti atoms $= 4$
Number of $H$ atoms $= 2 \times 4 = 8$
Number of $C$ atoms $= 4$
Formula: $TiC, {TiH}_{2}$