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CHXII01:THE SOLID STATE

318868 The contribution of particle at the edge centre of a particular unit cell is

1

\frac{1}{2}

2

\frac{1}{4}

3 1

4

\frac{1}{8}

Explanation:

The atom present at the edge centre of a unit cell is shared by a total of 4 unit cells. Hence, the contribution of the edge centre atom $= 1 / 4$

KCET - 2016

CHXII01:THE SOLID STATE

318869 A solid has a structure in which $X$ atoms are located at cubic corners of unit cell, $O$ atoms are at the edge centres and $Y$ atoms are at cube centre. Then the formula of compound is $X_{a} Y_{b} O_{c}$ . If two atoms of $O$ are missing from any of two edge centres per unit cell, then the molecular formula is $X_{p} Y_{q} O_{r}$ . Then, find the value of $(p + q + r) - (a + b + c)$ .

1 2

2 4

3 5

4 8

Explanation:

$X_{a} Y_{b} O_{c} = X_{1} Y_{1} O_{3}$
$X_{p} Y_{q} O_{r} = X_{2} Y_{2} O_{5}$
$(p + q + r) - (a + b + c) = 4$

CHXII01:THE SOLID STATE

318870 A fcc unit cell of aluminium contains the equivalent of how many atoms?

1 1

2 2

3 3

4 4

Explanation:

For fcc, number of atoms at corner
$= \frac{1}{8} \times 8 = 1$
Number of atoms at face centre
$= \frac{1}{2} \times 6 = 3$
$∴$ Total number of atoms $= 1 + 3 = 4$

CHXII01:THE SOLID STATE

318871 A compound formed by elements $X$ and $Y$ has a cubic structure in which $X$ atoms are at the corner of cube and also at the face centres. Y atoms are present at the body centre and at the edge centres of the cube. The total number of atoms in a cube are

1 12

2 20

3 27

4 15

Explanation:

Total number of atoms in a cube $= 8$ (corner) +6 (face centre) +1 (body centre) +12 (edge centre) $= 27$

CHXII01:THE SOLID STATE

318872 In a face centred cubic lattice atoms $A$ are at the corner points and atoms $B$ at the face centred points. If atoms B is missing from one of the face centred points, the formula of the ionic compound is :

1

{AB}_{2}

2

A_{2} B_{3}

3

A_{5} B_{2}

4

A_{2} B_{5}

Explanation:

$A = 8 \times \frac{1}{8} = 1$
$\begin{aligned} B = 6 \times \frac{1}{2} - 1 \times \frac{1}{2} = \frac{5}{2} \\ A : B = 1 : \frac{5}{2} \Rightarrow 2 : 5 \end{aligned}$

JEE - 2014

CHXII01:THE SOLID STATE

318868 The contribution of particle at the edge centre of a particular unit cell is

1

\frac{1}{2}

2

\frac{1}{4}

3 1

4

\frac{1}{8}

Explanation:

The atom present at the edge centre of a unit cell is shared by a total of 4 unit cells. Hence, the contribution of the edge centre atom $= 1 / 4$

KCET - 2016

CHXII01:THE SOLID STATE

318869 A solid has a structure in which $X$ atoms are located at cubic corners of unit cell, $O$ atoms are at the edge centres and $Y$ atoms are at cube centre. Then the formula of compound is $X_{a} Y_{b} O_{c}$ . If two atoms of $O$ are missing from any of two edge centres per unit cell, then the molecular formula is $X_{p} Y_{q} O_{r}$ . Then, find the value of $(p + q + r) - (a + b + c)$ .

1 2

2 4

3 5

4 8

Explanation:

$X_{a} Y_{b} O_{c} = X_{1} Y_{1} O_{3}$
$X_{p} Y_{q} O_{r} = X_{2} Y_{2} O_{5}$
$(p + q + r) - (a + b + c) = 4$

CHXII01:THE SOLID STATE

318870 A fcc unit cell of aluminium contains the equivalent of how many atoms?

1 1

2 2

3 3

4 4

Explanation:

For fcc, number of atoms at corner
$= \frac{1}{8} \times 8 = 1$
Number of atoms at face centre
$= \frac{1}{2} \times 6 = 3$
$∴$ Total number of atoms $= 1 + 3 = 4$

CHXII01:THE SOLID STATE

318871 A compound formed by elements $X$ and $Y$ has a cubic structure in which $X$ atoms are at the corner of cube and also at the face centres. Y atoms are present at the body centre and at the edge centres of the cube. The total number of atoms in a cube are

1 12

2 20

3 27

4 15

Explanation:

Total number of atoms in a cube $= 8$ (corner) +6 (face centre) +1 (body centre) +12 (edge centre) $= 27$

CHXII01:THE SOLID STATE

318872 In a face centred cubic lattice atoms $A$ are at the corner points and atoms $B$ at the face centred points. If atoms B is missing from one of the face centred points, the formula of the ionic compound is :

1

{AB}_{2}

2

A_{2} B_{3}

3

A_{5} B_{2}

4

A_{2} B_{5}

Explanation:

$A = 8 \times \frac{1}{8} = 1$
$\begin{aligned} B = 6 \times \frac{1}{2} - 1 \times \frac{1}{2} = \frac{5}{2} \\ A : B = 1 : \frac{5}{2} \Rightarrow 2 : 5 \end{aligned}$

JEE - 2014

CHXII01:THE SOLID STATE

318868 The contribution of particle at the edge centre of a particular unit cell is

1

\frac{1}{2}

2

\frac{1}{4}

3 1

4

\frac{1}{8}

Explanation:

The atom present at the edge centre of a unit cell is shared by a total of 4 unit cells. Hence, the contribution of the edge centre atom $= 1 / 4$

KCET - 2016

CHXII01:THE SOLID STATE

318869 A solid has a structure in which $X$ atoms are located at cubic corners of unit cell, $O$ atoms are at the edge centres and $Y$ atoms are at cube centre. Then the formula of compound is $X_{a} Y_{b} O_{c}$ . If two atoms of $O$ are missing from any of two edge centres per unit cell, then the molecular formula is $X_{p} Y_{q} O_{r}$ . Then, find the value of $(p + q + r) - (a + b + c)$ .

1 2

2 4

3 5

4 8

Explanation:

$X_{a} Y_{b} O_{c} = X_{1} Y_{1} O_{3}$
$X_{p} Y_{q} O_{r} = X_{2} Y_{2} O_{5}$
$(p + q + r) - (a + b + c) = 4$

CHXII01:THE SOLID STATE

318870 A fcc unit cell of aluminium contains the equivalent of how many atoms?

1 1

2 2

3 3

4 4

Explanation:

For fcc, number of atoms at corner
$= \frac{1}{8} \times 8 = 1$
Number of atoms at face centre
$= \frac{1}{2} \times 6 = 3$
$∴$ Total number of atoms $= 1 + 3 = 4$

CHXII01:THE SOLID STATE

318871 A compound formed by elements $X$ and $Y$ has a cubic structure in which $X$ atoms are at the corner of cube and also at the face centres. Y atoms are present at the body centre and at the edge centres of the cube. The total number of atoms in a cube are

1 12

2 20

3 27

4 15

Explanation:

Total number of atoms in a cube $= 8$ (corner) +6 (face centre) +1 (body centre) +12 (edge centre) $= 27$

CHXII01:THE SOLID STATE

318872 In a face centred cubic lattice atoms $A$ are at the corner points and atoms $B$ at the face centred points. If atoms B is missing from one of the face centred points, the formula of the ionic compound is :

1

{AB}_{2}

2

A_{2} B_{3}

3

A_{5} B_{2}

4

A_{2} B_{5}

Explanation:

$A = 8 \times \frac{1}{8} = 1$
$\begin{aligned} B = 6 \times \frac{1}{2} - 1 \times \frac{1}{2} = \frac{5}{2} \\ A : B = 1 : \frac{5}{2} \Rightarrow 2 : 5 \end{aligned}$

JEE - 2014

CHXII01:THE SOLID STATE

318868 The contribution of particle at the edge centre of a particular unit cell is

1

\frac{1}{2}

2

\frac{1}{4}

3 1

4

\frac{1}{8}

Explanation:

The atom present at the edge centre of a unit cell is shared by a total of 4 unit cells. Hence, the contribution of the edge centre atom $= 1 / 4$

KCET - 2016

CHXII01:THE SOLID STATE

318869 A solid has a structure in which $X$ atoms are located at cubic corners of unit cell, $O$ atoms are at the edge centres and $Y$ atoms are at cube centre. Then the formula of compound is $X_{a} Y_{b} O_{c}$ . If two atoms of $O$ are missing from any of two edge centres per unit cell, then the molecular formula is $X_{p} Y_{q} O_{r}$ . Then, find the value of $(p + q + r) - (a + b + c)$ .

1 2

2 4

3 5

4 8

Explanation:

$X_{a} Y_{b} O_{c} = X_{1} Y_{1} O_{3}$
$X_{p} Y_{q} O_{r} = X_{2} Y_{2} O_{5}$
$(p + q + r) - (a + b + c) = 4$

CHXII01:THE SOLID STATE

318870 A fcc unit cell of aluminium contains the equivalent of how many atoms?

1 1

2 2

3 3

4 4

Explanation:

For fcc, number of atoms at corner
$= \frac{1}{8} \times 8 = 1$
Number of atoms at face centre
$= \frac{1}{2} \times 6 = 3$
$∴$ Total number of atoms $= 1 + 3 = 4$

CHXII01:THE SOLID STATE

318871 A compound formed by elements $X$ and $Y$ has a cubic structure in which $X$ atoms are at the corner of cube and also at the face centres. Y atoms are present at the body centre and at the edge centres of the cube. The total number of atoms in a cube are

1 12

2 20

3 27

4 15

Explanation:

Total number of atoms in a cube $= 8$ (corner) +6 (face centre) +1 (body centre) +12 (edge centre) $= 27$

CHXII01:THE SOLID STATE

318872 In a face centred cubic lattice atoms $A$ are at the corner points and atoms $B$ at the face centred points. If atoms B is missing from one of the face centred points, the formula of the ionic compound is :

1

{AB}_{2}

2

A_{2} B_{3}

3

A_{5} B_{2}

4

A_{2} B_{5}

Explanation:

$A = 8 \times \frac{1}{8} = 1$
$\begin{aligned} B = 6 \times \frac{1}{2} - 1 \times \frac{1}{2} = \frac{5}{2} \\ A : B = 1 : \frac{5}{2} \Rightarrow 2 : 5 \end{aligned}$

JEE - 2014

CHXII01:THE SOLID STATE

318868 The contribution of particle at the edge centre of a particular unit cell is

1

\frac{1}{2}

2

\frac{1}{4}

3 1

4

\frac{1}{8}

Explanation:

The atom present at the edge centre of a unit cell is shared by a total of 4 unit cells. Hence, the contribution of the edge centre atom $= 1 / 4$

KCET - 2016

CHXII01:THE SOLID STATE

318869 A solid has a structure in which $X$ atoms are located at cubic corners of unit cell, $O$ atoms are at the edge centres and $Y$ atoms are at cube centre. Then the formula of compound is $X_{a} Y_{b} O_{c}$ . If two atoms of $O$ are missing from any of two edge centres per unit cell, then the molecular formula is $X_{p} Y_{q} O_{r}$ . Then, find the value of $(p + q + r) - (a + b + c)$ .

1 2

2 4

3 5

4 8

Explanation:

$X_{a} Y_{b} O_{c} = X_{1} Y_{1} O_{3}$
$X_{p} Y_{q} O_{r} = X_{2} Y_{2} O_{5}$
$(p + q + r) - (a + b + c) = 4$

CHXII01:THE SOLID STATE

318870 A fcc unit cell of aluminium contains the equivalent of how many atoms?

1 1

2 2

3 3

4 4

Explanation:

For fcc, number of atoms at corner
$= \frac{1}{8} \times 8 = 1$
Number of atoms at face centre
$= \frac{1}{2} \times 6 = 3$
$∴$ Total number of atoms $= 1 + 3 = 4$

CHXII01:THE SOLID STATE

318871 A compound formed by elements $X$ and $Y$ has a cubic structure in which $X$ atoms are at the corner of cube and also at the face centres. Y atoms are present at the body centre and at the edge centres of the cube. The total number of atoms in a cube are

1 12

2 20

3 27

4 15

Explanation:

Total number of atoms in a cube $= 8$ (corner) +6 (face centre) +1 (body centre) +12 (edge centre) $= 27$

CHXII01:THE SOLID STATE

318872 In a face centred cubic lattice atoms $A$ are at the corner points and atoms $B$ at the face centred points. If atoms B is missing from one of the face centred points, the formula of the ionic compound is :

1

{AB}_{2}

2

A_{2} B_{3}

3

A_{5} B_{2}

4

A_{2} B_{5}

Explanation:

$A = 8 \times \frac{1}{8} = 1$
$\begin{aligned} B = 6 \times \frac{1}{2} - 1 \times \frac{1}{2} = \frac{5}{2} \\ A : B = 1 : \frac{5}{2} \Rightarrow 2 : 5 \end{aligned}$

JEE - 2014

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