Solid State Test

Result

Solid State Test - 43

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The percentage of empty space in body centered cubic arrangement is _______.

A
74

B
68

C
32

D
26

Solution

Packing efficiency for bcc arrangement is 68% which represents total filled space in the unit cell. Hence, empty space in a body centered arrangement is 100 – 68 = 32%

Hence, Option "C" is the correct answers.
Question 2
1 / -0

The lattice site in a pure crystal cannot be occupied by ______.

A
molecule

B
ion

C
electron

D
atom

Solution

A pure crystal consists of either atoms, molecules or ions at lattice sites. Electrons can only occupy the lattice sites when there is any vacancy at that site which means it is not a pure crystal.

Hence, Option "C" is the correct answer.
Question 3
1 / -0

Number of tetrahedral void completely inside a BCC metallic crystal is:

A
0

B
2

C
3

D
4

Solution

In BCC cell, 4 tetrahedral voids are preset on each of the face . But none of the voids are present inside it.
Four tetrahedral sites on each of the six BCC cell faces $$\left(\dfrac{1}{2},\dfrac{1}{4},0\right)$$.
Therefore, no tetrahedral void is completely present inside a BCC metallic crystal.
Question 4
1 / -0

If the ratio of coordination no. of $$A$$ to that $$B$$ is $$x:y$$, then the ratio of no. of atoms of $$A$$ to that no. of atoms of $$B$$ in unit cell is ___.

A
$$x:y$$

B
$$y:x$$

C
$$x^2:y$$

D
none of the above

Solution

Coordination number of $$A$$ = Number of $$B$$ atoms present
& Coordination number of $$B$$ = Number of $$A$$ atoms present
Since, Ratio of coordination number of $$A$$ and $$B$$ = $$x:y$$
So, number of $$A$$ atoms = $$y$$
number of $$B$$ atoms = $$x$$
Ratio of number of atoms of $$A$$ and $$B$$ = $$y:x$$
So, correct answer is option B.
Question 5
1 / -0

for a certain crystal system the relation between edge lengths (a, b, and c) is
a = b = c
The relation between axial angles $$\left( {\alpha ,\beta \,\,{\text{and}}\,\gamma } \right)$$ is $$\alpha = \beta = \gamma $$
The pair of crystal system for which above relation holds true is:

A
cubic and triclinic

B
cubic and tetragonal

C
hexagonal and orthorhombic

D
cubic and rhombohedral

Solution

The relation $$a=b=c$$ and $$\alpha=\beta=\gamma$$ holds for cubic and rhombohedral systems.
Cubic
$$a=b=c$$
$$\alpha=\beta=\gamma=90^o$$
Rhombohedral
$$a=b=c$$
$$\alpha=\beta=\gamma\neq 90^o$$
Question 6
1 / -0

By X-ray diffraction it is found that nickel (at mass $$= 59 \,g \,mol^{-1}$$), crystallized with $$ccp$$. The edge length of the unit cell is $$3.52 \mathring{A}$$. If density of $$Ni$$ crystal is $$8.94 \,g/cm^3$$. Then value of Avogadro's number from the date is:

A
$$5.05 \times 10^{23}$$

B
$$4.11 \times 10^{23}$$

C
$$3.02 \times 10^{23}$$

D
$$6.023 \times 10^{23}$$

Solution

$$\rho= \cfrac {Z\times M_A}{N_A\times a^3}$$
$$M_{Ni}=59\ g/mol$$
$$Z_{ccp}=6\times\cfrac{1}{2}+1=4$$
$$a=3.52\mathring A$$
$$\rho=8.94$$ $${g/cm^3}$$
$$N_A=?$$
$$8.94=\cfrac{4\times 59}{N_A\times (3.52)^3\times 10^{-24}}$$
$$N_A=6.0\times 10^{23}$$
Question 7
1 / -0

The unit cell cube length of $$LiCl$$ ($$NaCl$$ structure) is $$5.14\overset{o}{A}$$. Assuming anion-anion contact, calculate the ionic radius for chloride ion.

A
$$2.05$$

B
$$1.82$$

C
$$2.92$$

D
$$3.92$$

Solution

In NaCl structure, anions have a face-centered cubic arrangement.
In such a unit cell face diagonally in $$4$$ times the radius of the anion(as shown in the figure).
Face diagona$$[ = \sqrt 2 .a = \sqrt 2 \times 5.14{A^ \circ }$$
Face diagonal$$=4r$$.
$$4r = \sqrt 2 \times 5.14$$
Implies that
$$r = \dfrac{{\sqrt 2 \times 5.14}}{4} = 1.82\:{A^ \circ }$$
Question 8
1 / -0

Copper has fcc crystal structural. Assuming an atomic radius of $$130$$ pm for copper atom. What is the edge length of the unit cell? $$(Cu=63.54)$$

A
$$3680$$ pm

B
$$368$$ pm

C
$$3.68$$ pm

D
$$268$$ pm

Solution

For $$f.c.c$$ structure.
$$4r = \sqrt 2 a$$
$$\therefore a=2\sqrt 2 r$$
Here $$r=130\ pm$$
$$\therefore a=2\sqrt 2 \times 130pm = 367.69\ pm$$.
Question 9
1 / -0

A unit cell of sodium chloride has four formula units. The edge length of the unit cell is $$0.564$$ nm. What is the density of sodium chloride? $$[Na=23, Cl=35.5]$$

A
$$1.16$$ g$$/cm^3$$

B
$$2.62$$ g$$/cm^3$$

C
$$2.16$$ g$$/cm^3$$

D
$$3.16$$ g$$/cm^3$$

Solution

Let us consider the problem.
Using the formula
$$\rho = \dfrac{N}{{{a^3}}}\frac{M}{{{N_A}}}$$
We have
$$\rho = \dfrac{4}{{{{\left( {0.564 \times {{10}^{ - 7}}cm} \right)}^3}}}\times \dfrac{{\left( {58.5gmo{l^{ - 1}}} \right)}}{{\left( {6.023 \times {{10}^{23}}mo{l^{ - 1}}} \right)}} = 2.166gc{m^{ - 3}}$$
Question 10
1 / -0

The ionic radius of $$Cl^-$$ ion is $$1.81A^0$$. The interionic distance of $$NaCl$$ and $$NaF$$ are $$2.79A^0$$ and $$2.31A^0$$ respectively. The ionic radius of $$F^-$$ ion will be:

A
$$0.98A^0$$

B
$$0.80A^0$$

C
$$1.33A^0$$

D
$$2.29A^0$$

Solution

Given:
$$r_{Cl^-}=1.81\mathring A$$
$$r_{Na^+}+r_{Cl^-}=2.79$$
$$\therefore r_{Na^+}=0.98\mathring A$$
Now,
$$r_{Na^+}+r_{F^-}=2.31$$
$$\therefore r_{F^-}=1.33\mathring A$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Solid State Test - 43

Result

Solid State Test - 43

Score

-

Rank

-

SHARING IS CARING

Question 1

74

68

32

26

Solution

Question 2

molecule

ion

electron

atom

Solution

Question 3

0

2

3

4

Solution

Question 4

$$x:y$$

$$y:x$$

$$x^2:y$$

none of the above

Solution

Question 5

cubic and triclinic

cubic and tetragonal

hexagonal and orthorhombic

cubic and rhombohedral

Solution

Question 6

$$5.05 \times 10^{23}$$

$$4.11 \times 10^{23}$$

$$3.02 \times 10^{23}$$

$$6.023 \times 10^{23}$$

Solution

Question 7

$$2.05$$

$$1.82$$

$$2.92$$

$$3.92$$

Solution

Question 8

$$3680$$ pm

$$368$$ pm

$$3.68$$ pm

$$268$$ pm

Solution

Question 9

$$1.16$$ g$$/cm^3$$

$$2.62$$ g$$/cm^3$$

$$2.16$$ g$$/cm^3$$

$$3.16$$ g$$/cm^3$$

Solution

Question 10

$$0.98A^0$$

$$0.80A^0$$

$$1.33A^0$$

$$2.29A^0$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.