Solid State Test

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Solid State Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Select the correct order for covalent radii :

A
Octahedral radii > linear radii > tetrahedral radii

B
Octahedral radii > tetrahedral radii > linear radii

C
Linear radii > tetrahedral radii > octahedral radii

D
Tetrahedral radii > octahedral radii > linear radii

Solution

1. If the triangular void in a close-packed layer has a sphere directly over it, there results a void with four spheres around it, as shown Such a void is called a tetrahedral void since the four spheres surrounding it are arranged on the corners of a regular tetrahedron.
If R denotes the radius of the four spheres surrounding a tetrahedral void, the radius of the spheres that would just fit into this void is 0.225 R.
(ii) If a triangular void pointing up in one close-packed layer is covered by a triangular void pointing down in the adjacent layer, then a void surrounded by six spheres results . Such a void is called an octahedral void since the six spheres surrounding it lie at the corners of a regular octahedron.
The radius of the sphere that would just fit into an octahedral void in a close-packing is 0.414 R.
Ocathedral radii > tetrahedral radii > linear radii
Question 2
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Metallic copper crystallises in bcc lattice. The length of the cubic unit cell is $$362$$ pm. The density (in g cm$$^{-3}$$) of crystalline copper is:

A
$$2.78$$

B
$$3.51$$

C
$$4.45$$

D
none of the above

Solution

For a BCC lattice, $$ z =2 $$
density = $$ \dfrac{zM}{N_a \times (a \times 10^{-10})^3} $$
density = $$ \dfrac{2 \times 65}{6 \times 10^{23}\times (362 \times 10^{-10})^3} $$
density = $$ 4.45 \text{ gcm}^{-3}$$
Question 3
1 / -0

The ratio of packing density in fcc, bcc and cubic structure is respectively_________.

A
$$1:0.92 :0.70$$

B
$$0.70 :0.92 :1$$

C
$$1:0.70:0.92$$

D
$$0.92:0.70:1$$

Solution

As we know,
Packing fraction in fcc $$= 74\%$$, bcc $$= 68\%$$ and sc $$= 52\%$$.
Hence, their ratio is $$74:68:52=1:0.92:0.70$$.

Question 4

1 / -0

The packing fraction for a body-centered cube is________.

$$0.42$$

$$0.53$$

$$0.68$$

$$0.82$$

Solution

Unit cell	Number of atoms at			No. of atoms per unit cell	Volume occupied by particles (%)
Unit cell	Corners	Centres	Faces	No. of atoms per unit cell	Volume occupied by particles (%)
Simple cube	$$8\times \dfrac{1}{8}=1$$	0	0	1	52.4
Body centred cube (BCC)	$$8\times \dfrac{1}{8}=1$$	1	0	2	68
Face centred cube (FCC)	$$8\times \dfrac{1}{8}=1$$	0	$$6\times \dfrac{1}{2}=3$$	4	74

Question 5

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Assertion (A): Triclinic system is the most unsymmetrical system.

Reason (R): No axial angle is equal to $$90^o$$ in the triclinic system.

Choose the correct option.

Both (A) and (R) are correct and (R) is the correct explanation of (A)

Both (A) and (R) are correct but (R) is not the correct explanation of (A)

(A) is correct but (R) is incorrect

(A) is incorrect but (R) is correct

Both (A) and (R) are incorrect

Solution

The edges of the triclinic system are not equal as $$a \neq b \neq c$$. Hence, it is known as the most unsymmetrical system.

Name of System	Axes	Angles	Bravais Lattices
Cubic	$$a=b=c$$	$$\alpha = \beta = \gamma = 90^o$$	Primitive, Face -centred, Body centred$$=3$$
Tetragonal	$$a=b\neq c$$	$$\alpha = \beta = \gamma = 90^o$$	Primitive, Face -centred, Body centred$$=2$$
Rhombohombic or trigonal	$$a=b= c$$	$$\alpha = \beta = \gamma \neq 90^o$$	Primitive$$=1$$
Orthorhombic or Rhombic	$$a\neq b\neq c$$	$$\alpha = \beta = \gamma = 90^o$$	Primitive, Face -centred, Body centred End centred$$=4$$
Monoclinic	$$a\neq b\neq c$$	$$\alpha =\gamma=90^o\, ;\beta\neq 90^o$$	Primitive, End-centred$$=2$$
Triclinic	$$a\neq b\neq c$$	$$\alpha \neq \beta \neq \gamma = 90^o$$	Primitive$$=1$$
Hexagonal	$$a=b\neq c$$	$$\alpha = \beta =90^o\, \gamma = 90^o$$	Primitive $$=1$$ total $$=14$$

Question 6
1 / -0

KF has a NaCl structure. What is the distance (in pm) between $$K^+$$ and $$F^-$$ in KF if the density is $$2.48$$ g cm$$^{-3}$ $?

A
$$220$$

B
$$135$$

C
$$540$$

D
$$269$$

Solution

As we know,
for such structure
$$2r^+ + 2r^ - = a$$
Also, as we know,
density $$ = nM/VN_o$$
here, n= 4
$$2.48 = 4*58/(a)^3(N_o)$$
distance between $$K^+\, and\, F^-$$ in KF = a/2 = 269 pm
Question 7
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Which one of the following schemes of ordering closed packed sheets of equal sized spheres do not generate close packed lattice?

A
$$ABCABC$$

B
$$ABACABAC$$

C
$$ABBAABBA$$

D
$$ABCBCABCBC$$

Solution

The scheme ABBAABBA of ordering closed packed sheets of equal sized spheres does not generate close packed lattice. Hexagonal close packing (ABAB type) generates close packed lattice in which 74% lattice is occupied.
Question 8
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Directions For Questions

Packing refers to the arrangement of constituent units in such a way that the forces of attraction among the constituent particles is maximum and the constituents occupy the maximum available space. In two dimensions, there are square close packing and hexagonal close packing. In three dimensions, however, there are hexagonal close packing, cubic close packing and body-centred cubic packing.
(i) HCP : AB AB AB AB . . . arrangement
Coordination number $$= 12$$
Percentage occupied space $$= 74\%$$
(ii) CCP : ABC ABC . . . arrangement
Coordination number $$= 12$$
Percentage occupied space $$= 74\%$$
(iii) BCC : $$68\%$$ space is occupied
Coordination number $$= 8$$
Answer the following questions:

...view full instructions

The empty space left in hcp in three dimensions is :

A
$$26\%$$

B
$$74\%$$

C
$$52.4\%$$

D
$$80\%$$

Solution

Packing efficiency is the ratio of the volume occupied by spheres in unit cell to the volume of hcp unit cell multiplied by 100.

For hcp unit cell, packing efficiency (or volume occupied) is 74%.

The empty space left in hcp in three dimensions is $$100-74=26$$%.
Question 9
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Chromium metal crystallizes as a body-centered cubic lattice. The length of the unit cell edge is found to be $$287$$ pm. What would be the density (in g cm$$^{-3}$$) of chromium?

A
$$3.40$$

B
$$5.20$$

C
$$12.10$$

D
$$7.30$$

Solution

If $$a$$ is the edge length, then for bcc lattice, we have

$$r=\dfrac {\sqrt3 a}{4}=\dfrac{\sqrt 3 \times 287}{4} =124.27 \mathring {A} $$
Density $$(\rho )=\dfrac {Z_{eff}\times Mw}{a^3 \times N_A}$$ $$=\dfrac{2\times 51.99}{(287 \times 10^{-10})^3 \times 6.023 \times 10^{23}}$$ $$=7.30$$ g cm$$^{-3}$$

Z for BCC is 2.

Hence, the correct option is $$\text{D}$$
Question 10
1 / -0

Ice crystallizes in a hexagonal lattice. At low temperature, at which the structure was determined, the lattice constants were $$a\, =\, 4.53\ \mathring A$$ and $$b = 7.60\ \mathring A$$ (see figure). How many molecules of water are contained in a given unit cell? [Density of ice is $$0.92$$ g cm$$^{-3}$$.]

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

As we know, density $$ = \dfrac{nM}{VN_o}$$, where n is the number of atoms per unit cell.
Density $$=0.92 = \dfrac{n\times 18}{(a^2b)\times (6\times 10^{23})} $$ $$\implies n =4$$
So, one unit of this structure contains $$4$$ molecules of water.

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Solid State Test - 28

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Solid State Test - 28

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Question 1

Octahedral radii > linear radii > tetrahedral radii

Octahedral radii > tetrahedral radii > linear radii

Linear radii > tetrahedral radii > octahedral radii

Tetrahedral radii > octahedral radii > linear radii

Solution

Question 2

$$2.78$$

$$3.51$$

$$4.45$$

none of the above

Solution

Question 3

$$1:0.92 :0.70$$

$$0.70 :0.92 :1$$

$$1:0.70:0.92$$

$$0.92:0.70:1$$

Solution

Question 4

$$0.42$$

$$0.53$$

$$0.68$$

$$0.82$$

Solution

Question 5

Both (A) and (R) are correct and (R) is the correct explanation of (A)

Both (A) and (R) are correct but (R) is not the correct explanation of (A)

(A) is correct but (R) is incorrect

(A) is incorrect but (R) is correct

Both (A) and (R) are incorrect

Solution

Question 6

$$220$$

$$135$$

$$540$$

$$269$$

Solution

Question 7

$$ABCABC$$

$$ABACABAC$$

$$ABBAABBA$$

$$ABCBCABCBC$$

Solution

Question 8

$$26\%$$

$$74\%$$

$$52.4\%$$

$$80\%$$

Solution

Question 9

$$3.40$$

$$5.20$$

$$12.10$$

$$7.30$$

Solution

Question 10

$$2$$

$$3$$

$$4$$

$$5$$

Solution

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