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CHXII01:THE SOLID STATE

318739 A crystalline solid $X Y_{3}$ has ccp arrangement for its element $Y$ . $X$ occupies

1

66 %

of tetrahedral voids

2

33 %

of tetrahedral voids

3

66 %

of octahedral voids

4

33 %

of octahedral voids

Explanation:

As ratio on of $X$ and $Y$ is $1 : 3, X$ will occupy ${\frac{1}{3}}^{r d}$ of octahedral voids.
$∴ %$ of octahedral voids occupied by $X$
$\frac{1}{3} \times 100 = 33.3 %$

KCET - 2014

CHXII01:THE SOLID STATE

318740 Consider a fcc lattice made of a metal cation $(M^{6 +})$ and three oxide anions per unit cell. The resultant structure would have

1 3 D network of edge shared octahedra

2 3 D network of corner shared octahedra

3 2 D network of edge shared octahedra

4 2 D network of face shared octahedra

CHXII01:THE SOLID STATE

318741 Structure of a mixed oxide is cubic close-packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One-fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal $B$ . The formula of the oxide is

1

{ABO}_{2}

2

A_{2} B O_{2}

3

A_{2} B_{3} O_{4}

4

A B_{2} O_{2}

Explanation:

We know, in $c c p$
If number of atoms $= n$
Then number of octahedral voids $= n$
and number of tetrahedral voids $= 2 n$
$\begin{array}{ccc} A^{2 +} & B^{+} & O^{2 -} \\ n 2 \times \frac{1}{4} & n & n \\ \frac{n}{2} & n & n \\ 1 & 2 & 2 \end{array}$
So, ${AB}_{2} O_{2}$

CHXII01:THE SOLID STATE

318742 Assertion :
The number of spheres are equal to the number of octahedral void as well as tetrahedral void.
Reason :
Number of octahedral void $= 2 \times$ Number of tetrahedral void

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

CHXII01:THE SOLID STATE

318743 In a compound, $n$ atoms of element $Y$ form ccp lattices and those of element $X$ occupy ${\frac{2}{3}}^{r d}$ of tetrahedral voids. The formula of the compound will be:

1

X_{2} Y

2

X_{3} Y_{4}

3

X_{4} Y_{3}

4

X_{2} Y_{3}

Explanation:

In ccp array, number of tetrahedral voids is twice the no. of atoms.
$\begin{aligned} X : Y & = [\frac{2}{3} \times 2 n] : n \\ = 4 : 3 \end{aligned}$
$∴$ Formula $= X_{4} Y_{3}$

CHXII01:THE SOLID STATE

318739 A crystalline solid $X Y_{3}$ has ccp arrangement for its element $Y$ . $X$ occupies

1

66 %

of tetrahedral voids

2

33 %

of tetrahedral voids

3

66 %

of octahedral voids

4

33 %

of octahedral voids

Explanation:

As ratio on of $X$ and $Y$ is $1 : 3, X$ will occupy ${\frac{1}{3}}^{r d}$ of octahedral voids.
$∴ %$ of octahedral voids occupied by $X$
$\frac{1}{3} \times 100 = 33.3 %$

KCET - 2014

CHXII01:THE SOLID STATE

318740 Consider a fcc lattice made of a metal cation $(M^{6 +})$ and three oxide anions per unit cell. The resultant structure would have

1 3 D network of edge shared octahedra

2 3 D network of corner shared octahedra

3 2 D network of edge shared octahedra

4 2 D network of face shared octahedra

CHXII01:THE SOLID STATE

318741 Structure of a mixed oxide is cubic close-packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One-fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal $B$ . The formula of the oxide is

1

{ABO}_{2}

2

A_{2} B O_{2}

3

A_{2} B_{3} O_{4}

4

A B_{2} O_{2}

Explanation:

We know, in $c c p$
If number of atoms $= n$
Then number of octahedral voids $= n$
and number of tetrahedral voids $= 2 n$
$\begin{array}{ccc} A^{2 +} & B^{+} & O^{2 -} \\ n 2 \times \frac{1}{4} & n & n \\ \frac{n}{2} & n & n \\ 1 & 2 & 2 \end{array}$
So, ${AB}_{2} O_{2}$

CHXII01:THE SOLID STATE

318742 Assertion :
The number of spheres are equal to the number of octahedral void as well as tetrahedral void.
Reason :
Number of octahedral void $= 2 \times$ Number of tetrahedral void

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

CHXII01:THE SOLID STATE

318743 In a compound, $n$ atoms of element $Y$ form ccp lattices and those of element $X$ occupy ${\frac{2}{3}}^{r d}$ of tetrahedral voids. The formula of the compound will be:

1

X_{2} Y

2

X_{3} Y_{4}

3

X_{4} Y_{3}

4

X_{2} Y_{3}

Explanation:

In ccp array, number of tetrahedral voids is twice the no. of atoms.
$\begin{aligned} X : Y & = [\frac{2}{3} \times 2 n] : n \\ = 4 : 3 \end{aligned}$
$∴$ Formula $= X_{4} Y_{3}$

CHXII01:THE SOLID STATE

318739 A crystalline solid $X Y_{3}$ has ccp arrangement for its element $Y$ . $X$ occupies

1

66 %

of tetrahedral voids

2

33 %

of tetrahedral voids

3

66 %

of octahedral voids

4

33 %

of octahedral voids

Explanation:

As ratio on of $X$ and $Y$ is $1 : 3, X$ will occupy ${\frac{1}{3}}^{r d}$ of octahedral voids.
$∴ %$ of octahedral voids occupied by $X$
$\frac{1}{3} \times 100 = 33.3 %$

KCET - 2014

CHXII01:THE SOLID STATE

318740 Consider a fcc lattice made of a metal cation $(M^{6 +})$ and three oxide anions per unit cell. The resultant structure would have

1 3 D network of edge shared octahedra

2 3 D network of corner shared octahedra

3 2 D network of edge shared octahedra

4 2 D network of face shared octahedra

CHXII01:THE SOLID STATE

318741 Structure of a mixed oxide is cubic close-packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One-fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal $B$ . The formula of the oxide is

1

{ABO}_{2}

2

A_{2} B O_{2}

3

A_{2} B_{3} O_{4}

4

A B_{2} O_{2}

Explanation:

We know, in $c c p$
If number of atoms $= n$
Then number of octahedral voids $= n$
and number of tetrahedral voids $= 2 n$
$\begin{array}{ccc} A^{2 +} & B^{+} & O^{2 -} \\ n 2 \times \frac{1}{4} & n & n \\ \frac{n}{2} & n & n \\ 1 & 2 & 2 \end{array}$
So, ${AB}_{2} O_{2}$

CHXII01:THE SOLID STATE

318742 Assertion :
The number of spheres are equal to the number of octahedral void as well as tetrahedral void.
Reason :
Number of octahedral void $= 2 \times$ Number of tetrahedral void

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

CHXII01:THE SOLID STATE

318743 In a compound, $n$ atoms of element $Y$ form ccp lattices and those of element $X$ occupy ${\frac{2}{3}}^{r d}$ of tetrahedral voids. The formula of the compound will be:

1

X_{2} Y

2

X_{3} Y_{4}

3

X_{4} Y_{3}

4

X_{2} Y_{3}

Explanation:

In ccp array, number of tetrahedral voids is twice the no. of atoms.
$\begin{aligned} X : Y & = [\frac{2}{3} \times 2 n] : n \\ = 4 : 3 \end{aligned}$
$∴$ Formula $= X_{4} Y_{3}$

CHXII01:THE SOLID STATE

318739 A crystalline solid $X Y_{3}$ has ccp arrangement for its element $Y$ . $X$ occupies

1

66 %

of tetrahedral voids

2

33 %

of tetrahedral voids

3

66 %

of octahedral voids

4

33 %

of octahedral voids

Explanation:

As ratio on of $X$ and $Y$ is $1 : 3, X$ will occupy ${\frac{1}{3}}^{r d}$ of octahedral voids.
$∴ %$ of octahedral voids occupied by $X$
$\frac{1}{3} \times 100 = 33.3 %$

KCET - 2014

CHXII01:THE SOLID STATE

318740 Consider a fcc lattice made of a metal cation $(M^{6 +})$ and three oxide anions per unit cell. The resultant structure would have

1 3 D network of edge shared octahedra

2 3 D network of corner shared octahedra

3 2 D network of edge shared octahedra

4 2 D network of face shared octahedra

CHXII01:THE SOLID STATE

318741 Structure of a mixed oxide is cubic close-packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One-fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal $B$ . The formula of the oxide is

1

{ABO}_{2}

2

A_{2} B O_{2}

3

A_{2} B_{3} O_{4}

4

A B_{2} O_{2}

Explanation:

We know, in $c c p$
If number of atoms $= n$
Then number of octahedral voids $= n$
and number of tetrahedral voids $= 2 n$
$\begin{array}{ccc} A^{2 +} & B^{+} & O^{2 -} \\ n 2 \times \frac{1}{4} & n & n \\ \frac{n}{2} & n & n \\ 1 & 2 & 2 \end{array}$
So, ${AB}_{2} O_{2}$

CHXII01:THE SOLID STATE

318742 Assertion :
The number of spheres are equal to the number of octahedral void as well as tetrahedral void.
Reason :
Number of octahedral void $= 2 \times$ Number of tetrahedral void

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

CHXII01:THE SOLID STATE

318743 In a compound, $n$ atoms of element $Y$ form ccp lattices and those of element $X$ occupy ${\frac{2}{3}}^{r d}$ of tetrahedral voids. The formula of the compound will be:

1

X_{2} Y

2

X_{3} Y_{4}

3

X_{4} Y_{3}

4

X_{2} Y_{3}

Explanation:

In ccp array, number of tetrahedral voids is twice the no. of atoms.
$\begin{aligned} X : Y & = [\frac{2}{3} \times 2 n] : n \\ = 4 : 3 \end{aligned}$
$∴$ Formula $= X_{4} Y_{3}$

CHXII01:THE SOLID STATE

318739 A crystalline solid $X Y_{3}$ has ccp arrangement for its element $Y$ . $X$ occupies

1

66 %

of tetrahedral voids

2

33 %

of tetrahedral voids

3

66 %

of octahedral voids

4

33 %

of octahedral voids

Explanation:

As ratio on of $X$ and $Y$ is $1 : 3, X$ will occupy ${\frac{1}{3}}^{r d}$ of octahedral voids.
$∴ %$ of octahedral voids occupied by $X$
$\frac{1}{3} \times 100 = 33.3 %$

KCET - 2014

CHXII01:THE SOLID STATE

318740 Consider a fcc lattice made of a metal cation $(M^{6 +})$ and three oxide anions per unit cell. The resultant structure would have

1 3 D network of edge shared octahedra

2 3 D network of corner shared octahedra

3 2 D network of edge shared octahedra

4 2 D network of face shared octahedra

CHXII01:THE SOLID STATE

318741 Structure of a mixed oxide is cubic close-packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One-fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal $B$ . The formula of the oxide is

1

{ABO}_{2}

2

A_{2} B O_{2}

3

A_{2} B_{3} O_{4}

4

A B_{2} O_{2}

Explanation:

We know, in $c c p$
If number of atoms $= n$
Then number of octahedral voids $= n$
and number of tetrahedral voids $= 2 n$
$\begin{array}{ccc} A^{2 +} & B^{+} & O^{2 -} \\ n 2 \times \frac{1}{4} & n & n \\ \frac{n}{2} & n & n \\ 1 & 2 & 2 \end{array}$
So, ${AB}_{2} O_{2}$

CHXII01:THE SOLID STATE

318742 Assertion :
The number of spheres are equal to the number of octahedral void as well as tetrahedral void.
Reason :
Number of octahedral void $= 2 \times$ Number of tetrahedral void

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

CHXII01:THE SOLID STATE

318743 In a compound, $n$ atoms of element $Y$ form ccp lattices and those of element $X$ occupy ${\frac{2}{3}}^{r d}$ of tetrahedral voids. The formula of the compound will be:

1

X_{2} Y

2

X_{3} Y_{4}

3

X_{4} Y_{3}

4

X_{2} Y_{3}

Explanation:

In ccp array, number of tetrahedral voids is twice the no. of atoms.
$\begin{aligned} X : Y & = [\frac{2}{3} \times 2 n] : n \\ = 4 : 3 \end{aligned}$
$∴$ Formula $= X_{4} Y_{3}$