MCQ Questions for Class 12 Engineering Chemistry The Solid State Quiz 9

The empty space left in hcp in three-dimensions is__________.

0%
$26$ %
0%
$74$ %
0%
$52.4$ %
0%
$80$ %

Explanation

The space left in hcp in three-dimensions is 26%.
The space occupied by spheres in the hcp arrangement is 74%.

Packing fraction is

$\displaystyle\dfrac {6 \times \dfrac {4}{3}\pi r^3}{ a^2csin60^o}= 0.74$

i.e. 74 %

The space left in hcp in three-dimensions is 26 %

a=2r, c=

$2\sqrt{\dfrac{2}{3}}a$

Hence, the correct option is

$\text{A}$

The flame colours of metal ions are due to:

0%
metal excess defect
0%
metal deficiency defect
0%
Schottky defect
0%
Frenkel defect

$KBr$ crystallises in

$NaCl$ type of crystal lattice and its density is

$2.75 g/cm^3$ . Number of unit cells of

$KBr$ present in a 1.00

$mm^3$ grain of

$KBr$ are:

0%
$3. 6 \times 10^{18}$
0%
$3. 6 \times 10^{16}$
0%
$2. 34 \times 10^{23}$
0%
$6.02 \times 10^{24}$

Explanation

In a unit cell of

$KBr,$ there are four formula units of

$KBr$ therefore,

Density

$(\rho)$

$=\dfrac {4 \times 119}{6.023\times 10^{23}a^3}= 2.75$

$a^3= 2.873 \times 10^{-22}\, cm^3$

$=2.873 \times 10^{-19}\, mm^3$

Number of units cells per

$mm^3$

$\dfrac{1}{2.873 \times 10^{-19}}=\dfrac {10^{19}}{2.873}= 3.5 \times 10^{18}$

Option A is correct.

Crystals are formed by ions, atoms, and molecules having an internal pattern that is ___________ and ___________.

0%
irregular, non-repeating
0%
regular, repeating
0%
regular, non-repeating
0%
none of the above

Which one of the following has a different crystal lattice from those of the rest?

0%
$Ag$
0%
$V$
0%
$Cu$
0%
$Pt$
0%
$Au$

Explanation

Vanadium crystallizes in bcc structures while

$Ag, Cu, Pt$ and

$Au$ crystallize in ccp structures.

In a face-centre cubic arrangement of A and B atoms whose A atoms are at the corner of the unit cell and B atoms at the face-centre, one of A atom is missing from one corner in the unit cell. The simplest formula of the compound is:

0%
$A_{24}B_7$
0%
$A_7B_{24}$
0%
$A_7B_4$
0%
$AB_2$

Explanation

No. of atoms of

$A$ from corners of unit cell

$=\displaystyle\frac{7}{8}$
No. of atoms of

$B$ from faces of unit cell

$=3$

$\therefore A:B=\displaystyle\frac{7}{8}:3=7:24$
Thus, formula of the compound is

$A_7B_{24}$ .

Arrangement of sulphide ions in zinc blende is

0%
Simple cubic
0%
HCP
0%
BCC
0%
FCC

Explanation

Arrangement of sulphide ions

$(S^{2-})$ in zinc blende

$(ZnS)$ is fcc (occupy all corners and face centres) while

$Zn^{2+}$ ion occupy alternate tetrahedral voids.

Iron crystallises in a bcc system with a lattice parameter of

$2.861\mathring { A }$ . Calculate the density of iron in the bcc system

[atomic weight of

$Fe=56,{ N }_{ A }=6.02\times { 10 }^{ 23 }{ mol }^{ -1 }$ ].

0%
$7.94g\ m{ L }^{ -1 }$
0%
$8.96g\ m{ L }^{ -1 }$
0%
$2.78g\ m{ L }^{ -1 }$
0%
$6.72g\ m{ L }^{ -1 }$

Explanation

$d=\cfrac { ZM }{ { N }_{ A }{ a }^{ 3 } }$ (for bcc,

$Z=2$ ),

$a= 2.861 \mathring {A}= 2.861 \times 10^{-8} cm$

${ d }_{ Fe }=\cfrac { (2)\times 56.0\ g\ { mol }^{ -1 } }{ \left( 6.02\times { 10 }^{ 23 }{ mol }^{ -1 } \right) { \left( 2.861\times { 10 }^{ -8 } \right) }^{ 3 }{ cm }^{ 3 } }$

$=7.94\ g/mL$

Hence, the correct option is

$\text{A}$

The pyknometer density of

$NaCl$ crystal is

$2.165\times { 10 }^{ 3 }kg\quad { m }^{ -3 }$ while its X-rays density is

$2.178\times { 10 }^{ 3 }kg\quad { m }^{ -3 }$ . The fraction of the unoccupied sites in

$NaCl$ crystal is:

0%
$5.968$
0%
$5.96\times { 10 }^{ -2 }$
0%
$5.96\times { 10 }^{ -3 }$
0%
$5.96\times { 10 }^{ -4 }$

Explanation

Fraction occupied =

$\dfrac{\text{X- ray density - pyknometer density}}{\text{X- ray density}}$

Fraction unoccupied

$= \dfrac{2.178\times 10^3 - 2.165\times 10^3}{2.178\times 10^3}$

$= 5.96\times 10^{-3}$

Hence, option C is correct.

If silver iodide crystallizes in a zinc blendek structure with

$I^-$ ions forming the lattice, then the fraction of the tetrahedral voids occupied by

$Ag^+$ ions will be:

0%
10%
0%
25%
0%
50%
0%
75%

Explanation

$Agl,$

$I^-$ ions form the lattice.

Let the number of

$I^-$ ion

$= 1$

There are two tetrahedral voids for 1 ion (sphere).

As there are one

$Ag^+$ ion in

$AgI$ so fraction of tetrahedral voids occupied by

$Ag^+$ ions

$=\cfrac {1}{2}$ or

$50\%$

Option C is correct.

When heated above

$916^o C ,$ iron changes its crystal structure from bcc to ccp structure without any change in the radius of atom. The ratio of density of the crystal before heating and after heating is :

0%
$1. 089$
0%
$0.918$
0%
$0.725$
0%
$1.231$

Explanation

I) for BCC ,

$Z=2$ ,

$4r=a\sqrt{3}$ ,

$M_A=56$ edge length=

$a_1$

$\rho_{BCC}=\cfrac {2\times 56}{N_A\times {a_1}^{3}}=\cfrac {112\times 3\sqrt {3}}{N_A\times 64r^{3}}=K\times 3\sqrt{3}$

II) for FCC,

$Z=4$ ,

$4r=a_2\sqrt{2}$ ,

$M_A=56$ edge length=

$a_2$

$\rho_{FCC}=\cfrac {4\times 56}{N_A\times {a_2}^{3}}=\cfrac {2\times 112\times 2\sqrt{2}}{N_A\times {64r}^{3}}=\cfrac {224\times 2\sqrt{2}}{N_A\times {64r}^{3}}=K\times 4\sqrt{2}$

$\cfrac {\rho_{BCC}}{\rho_{FCC}}=\cfrac {3\sqrt{3}}{4\sqrt{2}}=0.918$

In a crystal, atoms are located at the position of:

0%
Maximum potential energy
0%
Minimum potential energy
0%
Zero potential energy
0%
Infinite potential energy

Which of the following defect is seen in

$FeO$ ?

0%
Metal excess defect
0%
Metal deficiency defect
0%
Displacement defect
0%
Impurity defect

Explanation

$FeO$ has metal deficiency defect. Metal deficiency defect is supposed to arise when there are lesser number of positive ions than negative ions. In case of $FeO$ , the positive ions are missing from their lattice sites.

The additional negative charge is balanced by some nearby metal ion by acquiring one more positive charge. It happens in $FeO$ because $Fe$ has capacity of showing variable oxidation states.

An element occurs in two crystalline forms

$\alpha$ and

$\beta$ . The

$\alpha$ -form a fcc arrangement with

$a=3.68$

$\mathring{A}$ and the

$\beta$ -form has a bcc arrangement with

$a=2.92 \mathring { A}$ . Calculate the ratio of their densities.

0%
2 : 1
0%
1 : 2
0%
1 : 3
0%
1 : 1

Explanation

$Unit\; cell\; length\; in\; Fcc\; =\; 3.68{ A }^{ 0 }\; ({ a }_{ 1 })\\ Unit\; cell\; length\; in\; Bcc\; =\; 2.92{ A }^{ 0 }\; ({ a }_{ 2 })\\ Density\; \alpha \; (Fcc)\; =\; \cfrac { M\times { Z }_{ 1 } }{ { N }_{ 0 }\times { a }_{ 1 }^{ 3 } } \\ Density\; \beta \; (Bcc)\; =\; \cfrac { M\times { Z }_{ 2 } }{ { N }_{ 0 }\times { a }_{ 2 }^{ 3 } } \\ \cfrac { { d\alpha } }{ d\beta } =\cfrac { { Z }_{ 1 }\times ({ { a }_{ 2 }) }^{ 3 } }{ { Z }_{ 2 }\times { ({ a }_{ 1 } })^{ 3 } } =\cfrac { 4\times ({ 2.92) }^{ 3 } }{ 2\times { (3.68) }^{ 3 } } =\; 0.999\sim 1\\ Hence\; ratio\; of\; density\; =\; 1:1$

In a face centred cubic arrangement of

$A$ and

$B$ atoms when

$A$ atoms are at the corner of the unit cell and

$B$ atoms of the face centres. One of the

$A$ atom is missing from one corner in unit cell. The simplest formula of compound is:

0%
$A_{7}B_{3}$
0%
$AB_{3}$
0%
$A_{7}B_{24}$
0%
$A_{7}B_{8}$

Explanation

Hint: Consider the arrangement of atoms A and B in the lattice.

Explanation:

Step 1: Atoms calculation in the lattice.

Atoms are present at the face center of the lattice.

As there are

$8$ corners in a face-centered cubic arrangement, each contributes to

$\frac {1}{8}$ .

Since

$1$ atom is missing from lattice, then total atoms =

$7$ x

$\frac {1}{8}$ =

$\frac {7} {8}$

Therefore

$\frac{7}{8}$ atoms are present of atom A.

Now B atoms are present at the face center and there are

$6$ faces.

Each face center contributes half an atom to the cell.

Therefore total contribution of B =

$6$ x

$\frac {1} {2}$ =

$3$ .

Step 2: Calculation of ratio.

$A:B$ =

$\frac{7}{8}$ :

$3$ .

Therefore

$A:B$ is

$A_7B_{24}$

Hence B is the correct option.

If the height of

$HCP$ unit cell of identical particles is

$h$ , then height of octahedral voids from the base is________.

0%
$\dfrac {h}{2}$
0%
$\dfrac {h}{3}, \dfrac {2h}{3}$
0%
$\dfrac {h}{4}, \dfrac {3h}{4}$
0%
$\dfrac {h}{8}, \dfrac {7h}{8}$

Explanation

Since, octahedral voids are present at body center and edge centers. So, if $h$ is the height of HCP unit cell, the body center must be located at $\cfrac h2$ height from the base, where $a$ octahedral void is located.

Which type of 'defect' has the presence of cations in the interstitial sites ?

0%
Frenkel defect
0%
Metal deficiency defect
0%
Schottky defect
0%
Vacancy defect

Which among the following metals crystallise as a simple cube?

0%
Polonium
0%
Iron
0%
Copper
0%
Gold

A solid can be irreversibly deformed. The stamping of sheets of steel into 'fins' of motor cars is a case of :

0%
Vaccum deformation
0%
Plastic deformation
0%
Elastic deformation
0%
Flexible shaping

The major binding force in diamond, silicon and quartz is:

0%
electrostatic force
0%
electrical attraction
0%
covalent bond force
0%
van der waals force

A crystalline structure has radius ratio (

$r_+ /r_-$ ) in the range of

$0.225-0.414$ . The coordination number and arrangement of anions around the cations are:

0%
$3$ , plane triangular
0%
$6$ , octahedral
0%
$4$ , tetrahedral
0%
$8$ , cubic

Which type of crystals contains more than one Bravais lattice?

0%
Hexagonal
0%
Triclinic
0%
Rhombohedral
0%
Monoclinic

$ABC$ packing if the number of atoms in the unit cell is

$n$ then the number of tetrahedral voids in the unit cell is equal to:

0%
$n$
0%
$\dfrac{n}{2}$
0%
$\dfrac{n}{4}$
0%
$2n$

Explanation

In a closed packed structure, the number of tetrahedral voids in a single unit cell is twice the number of octahedral voids.

number of octahedral voids = number of atoms

So for

$n$ atoms, the number of tetrahedral voids will be

$2n$ .

Hence, the correct answer is option "D".

The unit cell shown in the figure belongs to :

0%
$NaCl$ type
0%
$ZnS$ type
0%
$CsCl$ type
0%
$CaF_2$ type

Explanation

In the given figure,

$Y$ is present at corners of the cube &

$X$ occupy some of the tetrahedral voids. This is

$ZnS$ type structure in which sulphide ions are present at corners and zinc ions occupy half of the tetrahedral voids.

Hence, option "B" is the correct answer.

A compound is formed by two elements

$Y$ and

$Z.$ The elements

$Z$ forms

$ccp$ and atoms

$Y$ at

$1/3$ rd of tetrahedral voids. The formula of the compound is:

0%
$Y_2Z_3$
0%
$YZ$
0%
$YZ_3$
0%
$Y_2Z$

Explanation

Since, in

$CCP$ , number of atoms = 4

So, the number of

$Z$ atoms = 4

Number of tetrahedral voids =

$4\times 2=8$

Number of

$Y$ atoms =

$\cfrac { 1 }{ 3 } \times 8=\cfrac { 8 }{ 3 }$

Ratio of

$Y$ and

$Z$ is =

$\cfrac { 8 }{ 3 } :4=2:3$

So, the formula is

${ Y }_{ 2 }{ Z }_{ 3 }$ .

Hence, the correct answer is option "A".

The radii of

$Na^+$ and

$Cl^-$ ions are

$95 pm$ and

$181 pm$ respectively. The edge length of

$NaCl$ unit cell is:

0%
276 pm
0%
138 PM
0%
552 pm
0%
415 pm

Explanation

For

$NaCl$ type crystal structure,

$a={ 2r }_{ { Na }^{ + } }+{ 2r }_{ { Cl }^{ - } }\\ a=2\times 95+2\times 181\\ a=552pm$ .

Hence, correct answer is option C.

Which of the following is true about the value of refractive index quartz glass?

0%
Same in all direction
0%
Different in different direction
0%
Cannot be measured
0%
Always zero

The sharp melting point of crystalline solids is due to______.

0%
a regular arrangement of constituent particles observed over a short distance in crystal lattice.
0%
a regular arrangement of constituent particles observed over a long distance in crystal lattice.
0%
same arrangement of constituent particles in different direction
0%
different arrangement of constituent particles in different directions

A unit of

$BaCl_2$ (fluorite structure) is made up of:

0%
four $Ba^{2+}$ ions and four $Cl^-$ ions
0%
four $Ba^{2+}$ ions and eight $Cl^-$ ions
0%
eight $Ba^{2+}$ ions and four $Cl^-$ ions
0%
four $Ba^{2+}$ ions and six $Cl^-$ ions

Explanation

${ BaCl }_{ 2 }$ has

${ CaF }_{ 2 }$ type structure. In

${ BaCl }_{ 2 }$ ,

The ${ Ba }^{ 2+ }$ ions form the ccp arrangement, i.e., these ions occupy all the corner positions and the center of each face of the cube.

The coordination of the compound is 8:4.

In a unit of

$BaCl_2$ , there are eight

$Ba^{2+}$ ions and four

$Cl^-$ ions.

So, option C is correct.

Complete the following table:

0%
(A)- Impurity defects, (B)- Stoichiometric defects, (C)- Non-stoichiomeric defects (D)- Anion excess defects
0%
(A)- Stoichiometric defects, (B)- Non-stoichiomeric defects, (C)- Impurity defects, (D)- Metal excess defects
0%
(A)- Non-stoichiomeric defects, (B)- Stoichiometric defects, (C)- Impurity defects, (D)- Cation vacancy
0%
(A)- Impurity defects, (B)- Stoichiometric defects, (C)- Metal excess defects, (D)- Non-stoichiomeric defects

Explanation

Points defects can be classified into three types:

1. Stoichiometric defects (stoichiometric ratios of cations and anions remains unchanged) (A).

2. Non-stoichiometric defects (ratios of anions and cations changes due to this type of defects) (B)

3. Impurity defects (foreign atoms replace some of the atoms in the lattice) (C)

The non-stoichiometric defects further classified into two types:

1. Metal Excess defect (due to anionic vacancies) (D)

2. Metal deficiency defect (due to cation vacancies)

Hence, the correct answer is option

$\text{B}$ .

Silver halides generally show

0%
Schottky defect
0%
Frenkel defect
0%
both Frenkel and Schottky defects
0%
cation excess defect

Which of the following is an amorphous solid?

0%
Graphite $(C)$
0%
Quartz glass $(SiO_2)$
0%
Chrome alum
0%
Silicon carbide $(SiC)$

Explanation

$\bf{Hint-}$ Amorphous solids have short-range order with irregular shapes of constituent particles.

$\bf{Explanation-}$

$\bullet$ An amorphous solid is a solid that has no ordered internal structure.

$\bullet$ Examples of amorphous solids include glass, rubber, plastic, and gels.

$\bullet$ Amorphous solids break into uneven pieces with irregular edges.

$\bullet$ Graphite, alum, and silicon carbide are crystalline solids.

$\bf{Conclusion-}$ Hence, quartz glass

$(SiO_2)$ is an amorphous solid due to its short-range order of constituent particles. Option B is correct.

Iodine molecules are held in the crystals lattice by ______.

0%
london forces
0%
dipole-dipole interaction
0%
covalent bonds
0%
coulombic forces

Explanation

The London dispersion force is the weakest intermolecular force. The London dispersion force is a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles. This force is sometimes called an induced dipole-induced dipole attraction.

Iodine molecules are a class of non- polar molecular solid in which constitute molecules are held together by London or dispersion forces. These solids are soft and non-conductor of electricity.

So, the correct answer is option

$A$ .

The correct order of the packing efficiency in different types of the unit cell is:

0%
fcc < bcc <simple cubic
0%
fcc > bcc >simple cubic
0%
fcc < bcc >simple cubic
0%
bcc < fcc >simple cubic

Explanation

Packing efficiencies of different cells are shown in the above table.

Hence, correct order is fcc (74%) > bcc(68%) > simple cubic (52%).

Hence, the correct option is

$\text{B}$

The total number of tetrahedral voids in face centered unit cell is ______.

0%
6
0%
8
0%
10
0%
12

Explanation

In face centered cubic unit cell (FCC), tetrahedral voids are present on body diagonals. There are two tetrahedral voids on each body diagonal at $\left (\cfrac 14 \right)^{th}$ of the distance from each corner.

So, total number of tetrahedral voids = (no. of tetrahedral voids per body diagonal) x (no. of body diagonals) = $2 \times 4 = 8$

Hence, Option "B" is the correct answer.

In which pair most efficient packing in present?

0%
$hcp$ and $bcc$
0%
$hcp$ and $ccp$
0%
$bcc$ and $ccp$
0%
$bcc$ and simple cubic cell

Explanation

The packing efficiency of the lattices are given below.

$hcp$ =

$74\%$

$ccp$ =

$74\%$

$bcc$ =

$68\%$

Simple cubic =

$54\%$ .

So, the pair with most efficient packing is

$hcp$ and

$ccp$ .

Hence, Option "B" is the correct answer.

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:

0%
cubic and triclinic
0%
cubic and tetragonal
0%
hexagonal and orthorhombic
0%
cubic and rhombohedral

Explanation

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

$a=b=c$

$\alpha=\beta=\gamma\neq 90^o$

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

0%
molecule
0%
ion
0%
electron
0%
atom

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

0%
74
0%
68
0%
32
0%
26

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

0%
0
0%
2
0%
3
0%
4

Explanation

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.

987586_1025025_ans_b29c106c7c31445a997c8afe7668d375.png

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 ___.

0%
$x:y$
0%
$y:x$
0%
$x^2:y$
0%
none of the above

Explanation

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.

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:

0%
$0.98A^0$
0%
$0.80A^0$
0%
$1.33A^0$
0%
$2.29A^0$

Explanation

Given:

$r_{Cl^-}=1.81\mathring A$

$r_{Na^+}+r_{Cl^-}=2.79$

$\therefore r_{Na^+}=0.98\mathring A$

Now,

$r_{Na^+}+r_{F^-}=2.31$

$\therefore r_{F^-}=1.33\mathring A$

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:

0%
$5.05 \times 10^{23}$
0%
$4.11 \times 10^{23}$
0%
$3.02 \times 10^{23}$
0%
$6.023 \times 10^{23}$

Explanation

$\rho= \cfrac {Z\times M_A}{N_A\times a^3}$

$M_{Ni}=59\ g/mol$

$Z_{ccp}=6\times\cfrac{1}{2}+1=4$

$a=3.52\mathring A$

$\rho=8.94$

${g/cm^3}$

$N_A=?$

$8.94=\cfrac{4\times 59}{N_A\times (3.52)^3\times 10^{-24}}$

$N_A=6.0\times 10^{23}$

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.

0%
$2.05$
0%
$1.82$
0%
$2.92$
0%
$3.92$

Explanation

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

$r = \dfrac{{\sqrt 2 \times 5.14}}{4} = 1.82\:{A^ \circ }$

968780_1026721_ans_8d519a990c5d4894ac89d46aac95ce5d.jpg

The number of unit cells in

$58.5$ g of

$NaCl$ is nearly ?

0%
$6 \times 10^{23}$
0%
$3\times 10^{22}$
0%
$1.5\times 10^{23}$
0%
$0.5\times 10^{24}$

Explanation

$NaCl$ crystalline in FCC structure i.e.,

$4$ molecules of

$NaCl$ is present in one unit cell.

Moles of

$NaCl=\cfrac{58.5}{58.5}=1$

Moles of unit cell

$=\cfrac{1}{4}=0.25$

Total number of unit cells

$=\cfrac{1}{4}\times 6\times 10^{23}=1.5\times 10^{23}$ cells

Copper has fcc crystal structural. Assuming an atomic radius of

$130$ pm for copper atom. What is the edge length of the unit cell?

$(Cu=63.54)$

0%
$3680$ pm
0%
$368$ pm
0%
$3.68$ pm
0%
$268$ pm

Explanation

For

$f.c.c$ structure.

$4r = \sqrt 2 a$

$\therefore a=2\sqrt 2 r$

Here

$r=130\ pm$

$\therefore a=2\sqrt 2 \times 130pm = 367.69\ pm$ .

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?

$[Na=23, Cl=35.5]$

0%
$1.16$ g $/cm^3$
0%
$2.62$ g $/cm^3$
0%
$2.16$ g $/cm^3$
0%
$3.16$ g $/cm^3$

Explanation

Let us consider the problem.

Using the formula

$\rho = \dfrac{N}{{{a^3}}}\frac{M}{{{N_A}}}$

We have

$\rho = \dfrac{4}{{{{\left( {0.564 \times {{10}^{ - 7}}cm} \right)}^3}}}\times \dfrac{{\left( {58.5gmo{l^{ - 1}}} \right)}}{{\left( {6.023 \times {{10}^{23}}mo{l^{ - 1}}} \right)}} = 2.166gc{m^{ - 3}}$

If the ratio of the coordination number of

$A$ to that of

$B$ is

$x:y$ , then the ratio of number of atoms of

$A$ to that of

$B$ in the unit cell is:

0%
$x:y$
0%
$y:x$
0%
${ x }^{ 2 }:y$
0%
$y:{ x }^{ 2 }$

Explanation

Coordination number of A = Number of atoms of B

Coordination number of B = Number of atoms of A

So, The Formula is

${A}_{y}{B}_{x}$ and the ratio of number of atoms is y : x

So option B is correct

A compound

$AB$ has rock salt type structure. The formula weight of

$AB$ is

$6.023 \ Y$ amu, and closest

$A-B$ distance is

$Y^{1/3}$ nm, where

$Y$ is an arbitrary number, find the density of the lattice.

0%
$5$ kg $/m^3$
0%
$5\times 10^{-3}$ g $/m^3$
0%
$10^{-3}$ g $/m^3$
0%
$15\times 10^{-2}$ g $/m^3$

Explanation

Let

$AB$ has rock salt structure

$(A:B::1:1)$

For

$F.C.C$ structure

$(n=4)$ and formula weight of

$AB$ is $$6.023 y g

Closest distance

$A - B = y\frac{1}{3}nm$

Therefore,

Edge length of unit cell,

$a = 2\left( {{A^ + }{B^ - }} \right) = 2 \times \frac{y}{3} \times {10^{ - 9}}m$

Therefore ,

Density of

$AB = \frac{{n \times mol.wt}}{{{N_A}V}}$

$= \frac{{4 \times 6.023 \times y \times {{10}^{ - 3}}}}{{6.023 \times {{10}^{23}} \times {{\left( {2 \times \frac{y}{3}{{10}^{ - 9}}} \right)}^3}}}$

$= 5.0kg\,{m^{ - 3}}$ .

In the crystal

$A^{2+}B^{2-}$ , having anions in the face centred cubic packing if the radius of the anion if

$1.84A^0$ , ideal radius of the cation present in the tetrahedral hole will be:

0%
$0.225A^0$
0%
$0.414A^0$
0%
$0.732A^0$
0%
none of these

Explanation

For tetrahedral holes, ideal ratio of

$\cfrac{r_+}{r_-}$ is

$0.225$

Given:

$r_-=1.84\mathring A$

Then,

$\cfrac{r_+}{1.84}=0.225$

$r_+=0.414 \mathring A$

CBSE Questions for Class 12 Engineering Chemistry The Solid State Quiz 9 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The Solid State Quiz 9 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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