Solid State Test

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Solid State Test - 58

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Weekly Quiz Competition

Question 1
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Directions For Questions

Density of a unit cell is represented as
$$\rho = \dfrac{Effective \,no. \,of \,atom(s) \times Mass \,of \,a \,unit \,cell}{Volume \,of \,a \,unit cell} = \dfrac{Z.M.}{N_{A}.a^{3}}$$
where, mass of unit cell = mass of effective no. of atom(s) or ion(s).
M = At. wt. / formula wt.
$$N_{A}$$ = Avogadro's no. $$\Rightarrow 6.023 \times 10^{23}$$
a = edge length of unit cell

...view full instructions

Silver crystallizes in a fcc lattice and has a density of $$10.6 \,g/cm^3$$. What is the length of an edge of the unit cell ?

A
$$0.407 \,nm$$

B
$$0.2035 \,nm$$

C
$$0.101 \,nm$$

D
$$4.07 \,nm$$

Solution

$$10.6 = \dfrac{4 \times 108}{a^{3} \times 6.023 \times 10^{23}}$$
$$\therefore a = 4.07 \,nm$$
Question 2
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An element crystallizes in a structure having FCC unit cell of an edge length $$200$$ pm. If $$200$$ g of this element contains $$24\times 10$$ $$^{23}$$ atoms, the density (in g/cc) of the element is :

A
$$50.3$$

B
$$63.4$$

C
$$41.6$$

D
$$34.8$$

Solution

It is given that an element crystallizes in a structure having FCC unit cell of an edge length $$200$$ pm.

$$200$$ gm of the element contains $$24 \times 10^{23}$$ atoms, which implies $$6.022 \times 10^{23}$$ atoms are present in $$50$$ gm, which is its molecular weight.

Since, the lattice is FCC, $$z=4$$

$$d=\dfrac { zM }{ { N }_{ A }{ a }^{ 3 } } $$

$$d=\dfrac { 4\times50}{ { 6\times10 }^{ 23 }\times{ (200\times{ 10 }^{ -10 }) }^{ 3 } } = 41.6$$ g/cc
Question 3
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A solid made up of ions of A and B possesses an edge length of unit cell equal to $$0.564$$ nm has four formula units. Among the two ions, the smaller one occupies the interstitial void and the larger ions occupy the space lattice with ccp type of arrangement. One molecule of solid has mass as $$9.712\times10^{-23}$$ g. The density (in g cm$$^{-3}$$) of solid is:

A
$$2.16$$

B
$$0.54$$

C
$$1.08$$

D
$$1.562$$

Solution

The expression for density of the unit cell is given below:

$$a= 0.564\ nm = 564\ pm= 564 \times 10^-{10} cm$$

$$d= \dfrac{ZM}{N_A\times a^3}$$

where Z is effective no. of atoms, for CCP Z=4

Molar mass of $$AB$$:

1 molecule has a mass = $$9.712 \times 10^{-23}$$ g

1 mole has a mass, $$M $$ = $$9.712 \times 10^{-23} \times N_A$$ g

$$d= \dfrac{4\times9.712\times10^{-23}\times N_A}{N_A\times(564)^3\times10^{-30}} = 2.16$$ g cm$$^{-3}$$

Hence, the correct option is A.
Question 4
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An element crystallizes in a structure having FCC unit cell of an edge length $$200$$ pm. Calculate the density (in g cm$$^{-3}$$) if $$200$$ grams of it contains $$24\times 10^{23}$$ atoms.

A
$$41.6$$

B
$$42.6$$

C
$$43.6$$

D
$$44.6$$

Solution

The formula of density is as follows:

$$d=\dfrac { z\times w }{ { N }\times { a }^{ 3 } } $$

For FCC unit cell, $$z = 4$$

$$N = 24\times 10^{23} $$ atoms

$$a=200 pm= 2\times 10^{−8} cm$$

Now,
$$d = \dfrac{4\times 200}{24\times 10^{23}\times 8\times 10^{−24} }$$

$$ = \dfrac{1000}{24}$$

$$x = 41.67\ g.cm^{−3}$$

So, the correct option is A.
Question 5
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Assertion: The close packing of atoms in cubic structure is in the order: scp > bcc > ccp.

Reason: $$\text{Packing density} = \dfrac{\text{Volume of unit cell}}{a^{3}} $$

A
Both Assertion and Reason are correct and Reason is correct explanation of Assertion.

B
Both Assertion and Reason are correct but Reason is not a correct explanation of Assertion.

C
Assertion is wrong but Reason is correct.

D
Assertion is correct but Reason is not correct.

Solution

Packing efficiency of SCP(simple cubic) $$= 52\%$$

Packing efficiency for BCC(body centred) $$= 68\%$$

Packing efficiency of CCP(cubic-closed/face-centred) $$= 74\%$$

The order of close packing would be, FCC<BCC<SCP

$$\text{Packing density} = \dfrac{\text{Volume of unit cell}}{a^{3}} $$
Question 6
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Match the elements in list I with the shape of its crystal in list II.
List I List II
A) $$Be$$ 1) Body-centred cubic
B) $$Ca$$ 2) Simple cubic
C) $$Ba$$ 3) Face-centred cubic
D) $$Po$$ 4) Hexagonal close-packed

A
A - 4, B-3, C-1, D-2

B
A - 4, B-3, C-2, D-1

C
A-2, B-4, C-1, D-3

D
A-4, B-1, C-3, D-2

Solution

A. Beryllium is a steel gray and hard metal that is brittle at room temperature and has a close-packed hexagonal crystal structure.

B. Crystal Structure of Calcium. —The X-ray pattern obtained with powdered calcium shows that the atoms are arranged in a face-centered cubic lattice. The side of the elementary cube is 5.56 Å.

C. At room temperature and pressure, barium has a body-centered cubic structure, with a barium–barium distance of 503 picometers.

D. Polonium is a radioactive element that exists in two metallic allotropes. The alpha form is the only known example of a simple cubic crystal structure in a single atom basis at STP, with an edge length of 335.2 picometers.

$$Be$$ (Beryllium) - HCP packing
$$Ca$$ (Calcium) - FCC packing
$$Ba$$ (Barium)- BCC packing
$$Po$$ (Polonium) - Simple cubic arrangement
Question 7
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If the edge length of an NaH unit cell is $$488$$ pm, what is the length of an Na-H bond if its crystal has fcc structure?

A
$$488$$ pm

B
$$122$$ pm

C
$$244$$ pm

D
$$976$$ pm

Solution

Length of an $$ Na-H $$ bond $$ = r_+ + r_- $$

$$\because 2({ r }_{ + }{ +r }_{ - })=a$$(edge length)

$$ \because 2({ r }_{ + }{ +r }_{ - })=488\ pm$$

The length of Na-H bond is the interatomic separation between $$Na^+$$ and $$H^-$$ ions

$$ \therefore { r }_{ + } + {r }_{ - }=244\quad pm$$
Question 8
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Statement 1: In a crystal of $$Ca$$, the separation of $$(1,1,1)$$ planes is twice as great as that of $$(2,2,2)$$ planes.
Statement 2: The length of the side of crystal lattice is $$0.556$$ nm $$(\sqrt{12}=3.46)$$.

A
Statement 1 is True, Statement 2 is True, Statement 2 is a correct explanation of Statement 1.

B
Statement 1 is True, Statement 2 is True, Statement 2 is not a correct explanation of Statement 1.

C
Statement 1 is True, Statement 2 is False.

D
Statement 1 is False, Statement 2 is True.

Solution

The expression for the distance between two planes is given below:

$$d=\dfrac{a}{\sqrt{h^{2}+k^{2}+ \ell ^{2}}}$$

$$d_{111}\dfrac{0.556}{\sqrt{1^{2}+1^{2}+1^{2}}}=0.321$$

$$d_{222}=\dfrac{0.556}{\sqrt{2^{2}+2^{2}+2^{2}}}=\dfrac{0.556}{3.46}=0.161$$

$$d_{111} = 2\times d_{222} $$
Question 9
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In a multi layered close-packed structure :

A
there are twice as many tetrahedral holes as many close-packed atoms

B
there are as many tetrahedral holes as many closed packed atoms

C
there are twice as many octahedral holes as many close-packed atoms

D
there are as many tetrahedral holes as many octahedral holes

Solution

No. of octahedral holes $$=$$ No. of close packed atoms
No. of Tetrahedral holes $$= 2 \times$$ No. of close packed atoms
Hence, in a multi-layered close-packed structure, there are twice as many tetrahedral holes as many close-packed atoms.
Question 10
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Lithium metal crystallizes in a body-centered cubic crystal. If the length of the side of the unit cell of lithium is 351 pm, the atomic radius of lithium will be:

A
300.5 pm

B
240.8 pm

C
151.98 pm

D
75.5 pm

Solution

In case of BCC crystal , $$ a \times \sqrt 3 =4r $$
Hence , atomic radius of lithium,
$$ r=\dfrac {a\times\sqrt3}{4}$$

= $$ \dfrac{ 351\times 1.732}{4}$$
= 151.98 pm
So option C is correct.

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Re-Attempt Weekly Quiz Competition

List I	List II
A) $$Be$$	1) Body-centred cubic
B) $$Ca$$	2) Simple cubic
C) $$Ba$$	3) Face-centred cubic
D) $$Po$$	4) Hexagonal close-packed

Solid State Test - 58

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Solid State Test - 58

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SHARING IS CARING

Question 1

$$0.407 \,nm$$

$$0.2035 \,nm$$

$$0.101 \,nm$$

$$4.07 \,nm$$

Solution

Question 2

$$50.3$$

$$63.4$$

$$41.6$$

$$34.8$$

Solution

Question 3

$$2.16$$

$$0.54$$

$$1.08$$

$$1.562$$

Solution

Question 4

$$41.6$$

$$42.6$$

$$43.6$$

$$44.6$$

Solution

Question 5

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

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

Assertion is wrong but Reason is correct.

Assertion is correct but Reason is not correct.

Solution

Question 6

A - 4, B-3, C-1, D-2

A - 4, B-3, C-2, D-1

A-2, B-4, C-1, D-3

A-4, B-1, C-3, D-2

Solution

Question 7

$$488$$ pm

$$122$$ pm

$$244$$ pm

$$976$$ pm

Solution

Question 8

Statement 1 is True, Statement 2 is True, Statement 2 is a correct explanation of Statement 1.

Statement 1 is True, Statement 2 is True, Statement 2 is not a correct explanation of Statement 1.

Statement 1 is True, Statement 2 is False.

Statement 1 is False, Statement 2 is True.

Solution

Question 9

there are twice as many tetrahedral holes as many close-packed atoms

there are as many tetrahedral holes as many closed packed atoms

there are twice as many octahedral holes as many close-packed atoms

there are as many tetrahedral holes as many octahedral holes

Solution

Question 10

300.5 pm

240.8 pm

151.98 pm

75.5 pm

Solution

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