Processing math: 9%

CBSE Questions for Class 12 Engineering Chemistry The Solid State Quiz 10 - MCQExams.com

The maximum radius of sphere that can be fitted in the octahedral hole of cubical closed packing of sphere of radius

$r$ is:

0%
$0.732\ r$
0%
$0.225\ r$
0%
$0.155\ r$
0%
$0.414\ r$

Explanation

For octahedral voids,

$r+r_1=2r\cdot \cfrac{1}{\sqrt{2}}$

$\implies \cfrac{r_1}{r}=\sqrt 2 r$

$\implies \cfrac{r_1}{r}=\sqrt{2}-1=0.414$

$\implies r_1=0.414r$

987550_1026415_ans_3dfda5bfbecc4ec2876993cf7db4d070.png

The unit cell length of

$NaCl$ is observed to be

$0.5627$ nm by X-ray difference studies; the measured density of

$NaCl$ is

$2.164$ g

$cm^{-3}$ / Calculate the difference of observed and calculated density. lso calculate

$\%$ is missing

$Na^+$ and

$Cl^-$ ions:

0%
$0.031, 0.775\%$
0%
$3.96, 0.031\%$
0%
$0.0031, 7.75\%$
0%
$1.7, 1.75\%$

Explanation

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

$\rho_{observed}=2.164$

$gcm^{-3}$

$a=0.5627$

$nm$

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

$Z_{observed}=?$

$2.164=\cfrac{58.5\times Z_{observed}}{6.023\times (5.627)^3\times 10^{-24}\times 10^{23}}$

$Z_{observed}=3.96$

$Z_{theoretical}=4$

$\rho_{theoretical}=2.180$

${g/cm^3}$

So,

$\rho=\cfrac{4\times M_A}{N_A\times a^3}$

$\rho_{theoretical}-\rho_{observed}=0.016$

${gm/cm^3}$

Percentage of missing ions

$=\cfrac{4-396}{4}\times 100=1\%$

A crystalline solid

$AB$ adopts sodium chloride type structure with edge length of the unit cell as 745 pm and formula mass of 74.5 a.m.u. The density of the crystalline compound is:

0%
$2.16 g cm^{-3}$
0%
$0.99g cm^{-3}$
0%
$1.88g cm^{-3}$
0%
$1.97 g cm^{-3}$

Explanation

Density

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

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

$Z_{NaCl}=4$ (FCC structure)

$M_{AB}=74.5$

$amu$

$a=7.45\mathring A$

$\rho=\cfrac{4\times 74.5}{6.023\times 10^{23}\times (7.45)^3\times 10^{-24}}=1.197$

${g/cm^3}$

The fraction of octahedral voids filled by

$A{l^{3 + }}$ ions in

$A{l_2}{O_3}\left( {{r_{Al^{3 \oplus }}}}/r_{O^{2 - }} = 0.43 \right)$ is:

0%
0.43
0%
0.287
0%
0.667
0%
1

Explanation

For

$A l_{2} O_{3}, \dfrac{\gamma_{A L}+3}{\gamma_{0}^{2}-}=0.43$ .

$\begin{array}{l} \text { It forms hexagonal closed packed lattice in which } A l^{+3} \text {ions } \\ \text { occupy } \dfrac{2}{3} \text { rd of } \text { octahedral voids. } \\ \text { so, option (c) is correct. } \end{array}$

The pattern in Hexagonal close packing and cubic close packing are same.

Is the statement true or false?

0%
True
0%
False

Silver crystallizes in fcc structure with edge length of unit cell,

$408.7pm$ . Hence density of silver is

$(Ag = 108\ g\ mol^{-1})$

0%
$10.5\ g \ cm^{-3}$
0%
$9.82\ g \ cm^{-3}$
0%
$12.06\ g \ cm^{-3}$
0%
$8.86\ g \ cm^{-3}$

Explanation

Density

$\displaystyle \rho = \dfrac {Z \times M}{N_0 \times a^3}$

$\displaystyle Z=4 \ atoms=$ number of atoms in fcc unit cell

$\displaystyle M=108 \ g/mol =$ atomic mass of silver

$\displaystyle N_0=6.023 \times 10^{23} \ atoms/mol=$ Avogadro's number

$\displaystyle a=408.7 \ pm = 408.7 \times 10^{-10} \ cm =$ edge length of unit cell

$\displaystyle \rho = \dfrac {Z \times M}{N_0 \times a^3}$

$\displaystyle \rho = \dfrac {4 \ atoms \times 108 \ g/mol}{6.023 \times 10^{23} \ atoms/mol \times (408.7 \times 10^{-10} \ cm)^3}$

$\displaystyle \rho = 10.5 \ g/cm^3$

Copper crystallizes in fcc structure and the edge length of the unit cell is

$3.61A^o$ . Hence the density of copper is:
(atomic mass of

$Cu = 63.5\ g \ mol^{-1}$ )

0%
$5.17\ g \ cm^{-3}$
0%
$0.72\ g \ cm^{-3}$
0%
$7.09\ g \ cm^{-3}$
0%
$8.97\ g \ cm^{-3}$

Explanation

The density

$\displaystyle \rho = \dfrac {Z \times M}{N_0 \times a^3}$

$\displaystyle Z=4 \ atoms =$ number of Cu atoms in one fcc unit cell.

$\displaystyle M=63.5 \ g/mol =$ atomic mass of copper.

$\displaystyle N_0 = 6.023 \times 10^{23} \ atoms/mole=$ Avogadro's number.

$\displaystyle a=3.61 \ A^o =3.61 \times 10^{-8} \ cm =$ edge length of the unit cell.

The density

$\displaystyle \rho = \dfrac {Z \times M}{N_0 \times a^3}$

The density

$\displaystyle \rho = \dfrac {4 \ atoms \times 63.5 \ g/mol }{6.023 \times 10^{23} \ atoms/mole \times (3.61 \times 10^{-8} \ cm)^3}$

The density

$\displaystyle \rho =8.97 \ g \ cm^{-3}$

The side length of diamond unit cell is (in pm):

0%
$154$
0%
$1422.63$
0%
$711.32$
0%
$355.66$

Explanation

Given,

$C-C$ bond length

$=154$

$pm$

Let, side length be

$a$

$\cfrac{a\sqrt{3}}{4}=154$

$\Rightarrow a=355.66$

$pm$

The crystal system possess by

${K}_{2}{Cr}_{2}{O}_{7}$ is:

0%
Tetragonal
0%
Cube
0%
Hexagonal
0%
Triclinic

Explanation

The crystal system possessed by

$K_2Cr_2O_7$ is Triclinic. It has been determined by single-crystal

$X-ray$ diffraction.

The coefficient of linear expansion of crystal in x-direction is

$6\times {10}^{-5}/^{o}C$ and that in y and z- directions is

$4\times {10}^{-5}/^{o}C$ . The coefficient of cubical expansion of the crystal is?

0%
$18\times {10}^{-5}/^{o}C$
0%
$16\times {10}^{-5}/^{o}C$
0%
$14\times {10}^{-5}/^{o}C$
0%
$32\times {10}^{-5}/^{o}C$

Explanation

$\gamma =\alpha _x+\alpha_y+\alpha_z$

$\implies \gamma=6\times 10^{-5}+4\times 10^{-5}+4\times 10^{-5}$

$\implies \gamma= 14\times 10^{-5}/^oC$

Number of unit cells in

$10\ g\ NaCl$ is:

0%
$\dfrac {1.5}{58.5}\times 10^{24}$
0%
$\dfrac {2.5}{58.5}\times 10^{23}$
0%
$\dfrac {5.6}{58.5}\times 10^{20}$
0%
$\dfrac {5.6}{58.5}\times 10^{21}$

Explanation

Moles of

$NaCl=\cfrac{10}{58.5}$

Number of molecules of

$NaCl=\cfrac{10}{58.5}\times 6.023\times 10^{23}$

In one unit cell

$4$ molecules of

$NaCl$ are present.

Number of unit cells

$=\cfrac{10\times 6.023\times 10^{23}}{58.5}\times \cfrac{1}{4}=\cfrac{1.5}{58.5}\times 10^{24}$

A metal

$M$ is crystallised in

$F.C.C$ lattice. The number of unit cells in it having

$2.4\times 10^{24}$ atoms are:

0%
$6.023\times 10^{23}$
0%
$\frac{6.023\times 10^{23}}{2}$
0%
$2\times 6.023\times 10^{23}$
0%
$4\times 6.023\times 10^{23}$

Explanation

In FCC lattice

$Z_M=4$ .

$4$ atoms of

$M$ is present in a unit cell of

$M$ .

Number of unit cell

$=\cfrac{2.4}{4}\times 10^{24}$ atom

$=6\times 10^{23}$ atom

$\approx N_A$ (Avogadro's number)

$NaBr$ has same crystal structure as

$NaCl$ . If density is

$3.203g\ cm^{-3}$ . Calculate the unit cell side length. [

$Na=23, Br=79.9$ ]

0%
$5.976A^{o}$
0%
$6.976A^{o}$
0%
$7.976A^{o}$
0%
$4.976A^{o}$

Explanation

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

$Z=4$ ,

$\rho=3.203$

$g/cm^3$ ,

$M_A=102.9$

$amu$

$\implies 3.203=\cfrac{4\times 102.9}{6.023\times 10^{23}\times a^3}$

$\implies a^3=\cfrac{4\times 102.9}{6.023\times 3.203}\times 10^{-23}$

$\implies a^3=213.356\times 10^{-24}$

$\implies a=5.976\mathring A$

The rank of atoms in the hexagonal until cell is:

0%
4
0%
3
0%
5
0%
6

Explanation

Rank of atom is another term for effective number of atoms for a unit cell, i.e.,

$Z_{effective}$

Therefore, For HCP structure rank is

$6$ .

In fcc, the neighbouring number of atoms for any lattice points is:

0%
6
0%
8
0%
12
0%
14

Explanation

There are 12 nearest neighbors to a given lattice point in a FCC lattice. We say that the coordination number of a lattice point in the FCC lattice is 12. The distance of the nearest lattice point in terms of the lattice parameter (i.e, the edge length of the cubic unit cell) is

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

If the edge length of a KCl cell is 488 pm, what is the length of KCl bond if it crystallizes in the fcc structure?

0%
122 pm
0%
244 pm
0%
488 pm
0%
974 pm

Explanation

In FCC structure of NaH, Na atom is present at the corner and H is present at the edge center.

Hence,

$Bond\ length=\dfrac{Edge\ length}{2}=\dfrac{488}{2}=244\ pm$

$fcc$ unit cell is converted in to End centred unit cell, then its packing efficiency will decrease by:

0%
$25\%$
0%
$50\%$
0%
$40\%$
0%
$60\%$

Explanation

$Z=4$ for FCC unit cell

In end centered unit cell,

$8$ atoms at body corners,

$2$ atoms at any

$2$ of the face centres.

$Z_{eff}$ for end centred unit cell

$=\cfrac{1}{8}\times 8+2\times \cfrac{1}{2}=2$

$packaging\quad fraction\propto Z_{effective}$

$\%$ decrease in packaging fraction

$=\cfrac{4-2}{4}\times 100=50\%$

The unit cell cube length for

$LiCl$ (just like

$NaCl$ structure) is

$5.14 \ \mathring { A }$ . Assuming anion-anion contact, the ionic radius for chloride ion is:

0%
$1.815 \ \mathring { A }$
0%
$2.8 \ \mathring { A }$
0%
$3.8 \ \mathring { A }$
0%
$4.815 \ \mathring { A }$

Explanation

In NaCl structure, anions have face centered cubic arrangement.

In such a unit cell face diagonal is four times the radius of anions.

Face diagonal

$=\sqrt{2}a=\sqrt{2}\times5.14\mathring {A}$

Face diagonal

$=4r$

$4r=\sqrt{2}\times5.14$

$r=\dfrac{\sqrt{2}\times5.14}{4}=1.817\mathring A$

A compound

$CuCl$ has face centered cubic structure. Its density is

$3.4 \ g \ cm^{-3}$ . The length of unit cell is :

0%
$5.783 \ \mathring { A }$
0%
$6.783 \ \mathring { A }$
0%
$7.783 \ \mathring { A }$
0%
$8.783 \ \mathring { A }$

Explanation

$Z_{eff}=4; \ \rho=3.4 \ g/cm^3; \$ Molecular weight

$=99; \ N_A =$ Avogadro Number

$= 6.02 \times 10^{23}$

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

$a^3= \cfrac {99 \times 4}{6.02 \times 10^23 \times 3.4}=19.34 \times 10^{-23}$

$a^3=193.4 \times 10^{-24} \ cm$

$a=5.783 \ \mathring A$

The Aluminium chloride crystallizes in BCC lattice with an edge length of unit cell equal to

$387\ pm$ . If the size of

$Cl^-$ ion is

$181\ pm$ , the size of

$NH_4^{+}$ ion would be:

0%
$116\ pm$
0%
$154\ pm$
0%
$174\ pm$
0%
$206\ pm$

Explanation

In BCC lattice,

$\sqrt{3}\times edge\ length = 2(r_{+} + r_{-})$

$\Rightarrow \dfrac{\sqrt{3}\times 387}{2 } = {^{r}NH_{4}^{+}} +$

$^{r}Cl^{-}$

$\Rightarrow {^{r}NH_{4}^{+}} = \dfrac{387\times \sqrt{3}}{2} - 181\ pm$

$\Rightarrow{^{r}NH_{4}^{+}} = 154\ pm$

Hence, option

$(2)$ is correct.

How many unit cells are there in

$1.00$ gm cube shaped ideal crystal

$AB (MF = 60)$ having

$NaCl$ type lattice?

0%
$6.02 \times 10^{23}$
0%
$2.50 \times 10^{21}$
0%
$1.00 \times 10^{22}$
0%
$6.02 \times 10^{24}$

Explanation

$NaCl$ type lattice,

$Na^{+} \longrightarrow$ present octahedral voids

Thus, no. of

$Na^{+}$ ions=4

$Cl^{-}$ is present in fcc structure.

Thus, no. of

$Cl^{-}$ ions=4

1 unit cell contains four

$NaCl$ molecules.

Thus, 1 unit cell contains 4

$Na^{+}$ ion and 4

$Cl^{-}$ ions.

No. of moles in 1 g of crystal of AB=

$\cfrac{1}{60}$ moles

1 mole contains=

$6.023 \times 10^{23}$ ions

$\cfrac{1}{60}$ moles contains=

$\cfrac{6.023 \times 10^{23}}{60}$ ions or molecules.

1 unit cell has=8 ions or 4 molecules

or 1 ion=

$\cfrac{1}{8}$ unit cell

$\cfrac{6.023 \times 10^{23}}{60}$ ions=

$\cfrac{1}{8} \times \cfrac{6.023 \times 10^{23}}{60}$ unit cell

$\cfrac{6.023 \times 10^{23}}{60}$ molecules=

$\cfrac{1}{4} \times \cfrac{6.023 \times 10^{23}}{60}$ unit cell

Thus, 1 g of crystal contains

$2.5 \times 10^{21}$ unit cell.

In a cubic close packing of spheres in three dimensions, the co-ordination number of each sphere?

0%
6
0%
9
0%
3
0%
12

Explanation

In cubic close packing the atoms are present at corners and face center of the unit cell.
As the atom present at face center touches

$12$ atoms of face centers .

$4$ in its plane ,

$8$ outside the plane..
hence the co- ordination number is

$12$ .

1089258_1089486_ans_b1d9525829b2401896317b81a4fe06b2.png

In a hypothetical solid, C atoms are found to form cubical close-packed lattice . A atoms occupy att tetrahedral voids and B atoms occupy all octahedral voids.
A and B atoms are of appropriate size, so that there is no distortion in the ccp lattice of C atoms .Now if a plane as shown in the following figure is cut, then the cross section of this plane will look like?

For bcc structure, the coordination number and packing are ______ respectively.

0%
$8, \cfrac {\pi \sqrt 3}{8}$
0%
$6, \cfrac {\pi \sqrt 2}{6}$
0%
$8, \cfrac {\pi \sqrt 2}{6}$
0%
none of the above

Explanation

For bcc structure, $4r= a \sqrt 3 \Rightarrow a = \cfrac {4r}{\sqrt 3}$ (along body diagonal)

$\text{Packing Fraction}= \cfrac {2 \times \cfrac 43 \pi r^3}{a^3}= \cfrac {2 \times \cfrac 43 \pi r^3}{\cfrac {64 r^3}{3 \sqrt 3}}= \cfrac {\pi \sqrt 3}{8}$

Since, the body center atom is surrounded by $8$ other atoms. Therefore coordination number is $8$ .

The number of unit cells in 58.5 g of

$NaCl$ is approximately:

0%
$6 \times 10^{20}$
0%
$1.5 \times 10^{23}$
0%
$9 \times 10^{23}$
0%
None of these

Explanation

Atomic mass of sodium chloride

$= 23 + 35.5 = 58.5$ g

this means 23 g has no. of toms

$=6.022 \times 10^{23}$

therefore 58.5 g has no. of atoms

$= 6.022 \times 10^{23}\times 58.5/23$

no. of atoms in a unit cell of fcc structure=4

therefore no. of unit cell which contains no. of atoms

$= 6.022 \times 10^{23} \times 58.5/23\times 4 = 1.5 \times 6\times 10^{23}$

Which is correct statement?

0%
Schottky defect occurs when radius of cation is smaller
0%
When temperature increases then number of defects decreases
0%
Frenkel defect occurs when radius of cation is smaller
0%
None of these

How many units cells are there in

$1.00 g$ cube shaped ideal crystal of

$AB (Mw = 60 )$ which has a

$NaCl$ type lattice?

0%
$6.02 \times 10^{23}$
0%
$1.00 \times 20^{22}$
0%
$2.50 \times 10^{21}$
0%
$6.02 \times 10^{24}$

Explanation

$NaCl$ has Rock salt type lattice, so Numbers of atoms per unit cell = 4
That means 4

$NaCl$ farms = 1 unit cell
So in

$AB$ (mw = 60)

$= 4 AB$ forms one unit cell.
Now 1.0 gram of

$AB$ has Numbers of moles =

$\dfrac{Given mass}{molar mass}$
=

$\dfrac{1}{60}$ mole
1 mole of

$AB =$

$6.022 \times 10^{23}$

$\dfrac{1}{60}$ mole of

$AB = \dfrac{1}{60} \times 6.022 \times 10^{23}$
=

$1.004 \times 10^{22} AB$

$4 AB$ forms = 1 unit all

$1 AB$ forms =

$\dfrac{1}{4}$ unit all

$1.004 \times 10^{22}$ AB form =

$\dfrac{1}{4} \times 1.004 \times 10^{22}$

$9.50 \times 10^{21}$

$0.250 \times 10^{22}$

$2.50 \times 10^{21}$

So, the answer is

$2.50 \times 10^{21}$

The density of crystalline sodium chloride is 5.85 g

${cm}^{-3}$ . What is the edge length of the unit cell?

0%
$4.04 \times 10^{-9}$ cm
0%
$1.32 \times 10^{-14}$ cm
0%
$7.8 \times 10^{-23}$ cm
0%
$9.6 \times 10^{-24}$ cm

Explanation

Density

$(d) = 5.85\ g\ {cm}^{-3}$

Molar Mass of Sodium Chloride

$(m) = 58.5 g/mol$

$N_A = 6.022 \times 10^{23}$

Unit Cell Type is

$FCC,$

$Z = 4$

We know,

$d = \cfrac{(Z \times M)}{(N_A \times a^3)}$

$=> a^3 = \cfrac{(Z \times m)}{(N_A \times d)}$

$=> a^3 = \cfrac{(4 \times 58.5)}{(6.022 \times 10^{23} \times 5.85)}$

$=> a^3 = 0.066 \times 10^{-24}$

$=> a = 0.404 \times 10^{-8}$

So, edge length of the unit cell of NaCl is

$0.404 \times 10^{-8}\ cm.$

An element X (At. wt. = 80 g / mol ) having fcc structure, calculate no. of unit cells in 8 gm of X:

0%
$0.4 \times N_A$
0%
$0.1 \times N_A$
0%
$4 \times N_A$
0%
None of these

Explanation

Effective no. of atoms in a unit cell

$= 4$ N
No. of atoms

$= \dfrac{8}{80} \times N_{A}$

$\therefore$ No. of unit cell

$= \dfrac{N_{A}}{10} \times \dfrac{1}{4}$

Certain quantity of metal is crystallised

$60\%$ in ccp structure

$40\%$ in hcp structure. What is the ratio of number of unit cells in ccp and hcp?

0%
$3:2$
0%
$1:1$
0%
$9:4$
0%
$4:9$

Explanation

Let number of moles of metal crystallized=

$n$

Number of moles crystallized in

$ccp=n \times \cfrac {60}{100}=0.6n$

Number of moles crystallized in

$hcp= n \times \cfrac {40}{100}=0.4n$

$1$ unit of

$ccp$ contains=

$4$ atoms of metal or

$\cfrac {4}{N_a}$ moles of metal

Thus

$0.6n$ moles contains=

$\cfrac {0.6 n \times N_a}{4}$ unit cell of

$ccp$

Similarly,

$1$ unit cell of

$hcp$ contains=

$6$ atoms of metal or

$\cfrac {6}{N_a}$ moles of metal.

Thus

$0.4n$ moles contains=

$\cfrac {0.4n \times N_a}{6}$ unit cell of

$hcp$

Thus ratio of number of unit cell of

$ccp$ and

$hcp$ is

$R=\cfrac {\text{unit cell of ccp}}{\text {unit cell of hcp}}$

$=\cfrac {\cfrac {0.6 \times n \times N_a}{4}}{\cfrac {0.4n \times N_a}{6}}=\cfrac {9}{4}$

So, the ratio is

$9:4$

The density of solid argon (Ar = 40 g / mol) is 1.68 g/mL at 40 K. If the argon atom is assumed to be a sphere of radius 1.50 x

$10^{-8}$ cm, what % of solid Ar is apparently empty space? (use

$N_A = 6 \times10^{23}$ )

0%
35.64
0%
64.36
0%
74%
0%
none of these

Explanation

Volume of one atom of

$Ar =\frac{4}{3}πr^3$

Also, number of atoms in 1.65 g/mL

$= \frac{1.65}{40}\times 6.023 \times {10}^{23}$

∴ Total volume of all atoms of Ar in solid state

$=\frac{4}{3}πr^3 \times \frac{1.65}{40}\times 6.023 \times {10}^{23}$

$=\frac{4}{3}\times 3.14\times (1.54\times 10^{-8})^3\times \frac{1.65}{40}\times 6.023 \times {10}^{23} =0.380 cm^3$

Volume of solid argon

$= 1 cm^3$

∴% Empty space

$=\frac{1−0.380}{1}\times 100 = 62$ %

The compound

${ MX }_{ 4 }$ is tetrahedral . The number of

$X-M-X$ angles in the compound is:

0%
three
0%
four
0%
five
0%
six

Explanation

$MX_4$ is tetrahedral, therefore total number of angles

$XMX$ is six.

Select the wrong statement (s) :-

0%
A NaCl type AB crystal lattice can be interpreted to be made up of two individual fcc type unit lattice of $A^+$ and $B^-$ fused together in such a manner that the corner of one unit lattice becomes the edge centre of the other
0%
In a fcc unit cell, the body center is an octahedral void
0%
In an scc lattice, there can be no octahedral void
0%
In an scc lattice, the body center is the octahedral void

Packing efficiency of body centred unit cell is:

0%
85%
0%
68%
0%
33.33%
0%
45%

Explanation

Total number of atoms per unit cell in bcc=

$2$

From

$\Delta EFD$ ,

$\Rightarrow b^2= a^2+a^2$

$\Rightarrow b= \sqrt {2} a$

$\Delta AFD$ ,

$\Rightarrow c^2= a^2+b^2=a^2+2a^2=3a^2$

$\Rightarrow c= \sqrt {3}a$

But

$c=4r$ (

$r$ = radius of sphere)

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

$\Rightarrow a=\cfrac {4r}{\sqrt {3}}$

Total number of atoms per unit cell in bcc=

$2$

Packing efficiency=

$\cfrac {\text {Volume occupied by 2 spheres in unit cell }\times 100}{\text{Total volume of unit cell}}$

$=\cfrac {2 \times \cfrac {4}{3}\Pi r^3\times 100}{a^3}$

$=\cfrac {2 \times \cfrac {4}{3}\Pi r^3 \times 100}{\left[\cfrac {4}{\sqrt {3}}r\right]^3}=68$ %

$\therefore$ Packing efficiency of bcc

$=$

$68$ %

1185726_1120564_ans_f526933a65f34c2d8b9204ec46c5093d.JPG

$A_2B$ molecules (molar mass =259.8 g/mol) crystallises in a hexagonal lattice as shown in figure. The lattice constants were

$a = 5 \mathring{A}$ and

$b= 8 \mathring{A}$ . If density of crystal is 5

$g/cm^3$ then how many molecules are contained in given unit cell? (use

$N_A= 6 \times 10^{23}$ )

0%
6
0%
4
0%
3
0%
2

Explanation

Volume of unit cell

$= a^2\sin 60^o\times b$

$=173.2\times 10^{-24}$

$\Rightarrow a=5\overset {o}{A}$

Mass of unit cell=

$173.2\times 10^{-24}\times 5\times 6 \times 10^{23}= 519.6 g$

Number of molecules presenr in given cell=

$\cfrac {519.6}{259.8}=2$ .

Total number of unit cells present in a cube shaped ideal crystal of

$1g \ NaCl(s)$ is (Avogdro's number =

$6\times10^{23}$ ).

0%
$2.56\times10^6$
0%
$2.56\times10^{21}$
0%
$2.56\times10^{23}$
0%
$4.2\times10^{22}$

Explanation

The number of

${Na}^+$ ions present in one unit cell

$= 12 \times \frac{1}{4} + 1 = 4$ atoms per unit cell.

The number of

${Cl}^-$ ions present in one unit cell

$= 6 \times \frac{1}{2} + 8 \times \frac{1}{8} = 4$ atoms per unit cell.

So, each unit cell has 4 atoms of each

${Na}^+$ and

${Cl}^-$ . Hence there are total of 4 molecules of NaCl in each unit cell.

As the mass of -Avogadro number of molecules of NaCl is 58.5g.

Then, number of molecules in one gram

$= \frac{6.022 \times {10}^{23}} {58.5} = 1.025 \times {10}^{22}$ molecules

As we found above that each unit cell contains 4 molecules of NaCl hence we can now easily find the number of unit cells containing

$1.025 \times {10}^{22}$ molecules

$= \frac{1.03 \times {10}^{22}} {4}$

$=$

$2.56 \times {10}^{21}$ unit cells

So,

$2.56 \times {10}^{21}$ unit cells contains 1gm of NaCl molecules.

The magnitude of the lattice energy of a solid increases if:

0%
the ions are large
0%
the ions are small
0%
the ions are of equal size
0%
charges on the ions are large

Explanation

Ans : D, B.

Lattice energy is defined as the energy required to separate a mole of an ionic solid into gaseous ions.

The

$2$ factors that contributes to the lattice energy of an ionic solid are charge on the ions & the size of the ions.

Lattice energy increases with the increase in ionic charge and decreases with the increase in the ionic size.

Which forms a crystal of NaCl?

0%
$NaCl\ molecules$
0%
$Na^{+}$ and $Cl^{-}ions$
0%
$Na\ and\ Cl\ atoms$
0%
$None\ of\ these$

Explanation

NaCl crystal structure is formed by

$Na^+$ cation and

$Cl^-$ anion.

$Cl^-$ forms a FCC lattice and

$Na^+$ occupies all the octahedral voids.

It is also called as Rock Salt type structure.

Hecne, the correct option is B

A metallic crystal has the bcc type stacking pattern. What percentage of volume of this lattice is empty space?

0%

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

In a unit cell, atoms A, B, C and D are present at half of total corners, all face-centres, body-centre and one third of all edge-centres respectively. Then formula of unit cell is ?

0%
$AB_{3}CD_{3}$
0%
ABCD
0%
$AB_{6}C_{2}D_{4}$
0%
$AB_{6}C_{2}D_{2}$

Explanation

Atoms at the corner in a unit cell is 1
No of atoms of A in one unit cell =

$\frac{1}{2}$

Atoms at the corner in a unit cell is 3
No of atoms of B in one unit cell = 3

Atoms at the body center in a unit cell is 1
No of atoms of C in one unit cell = 1

Atoms at the edge center in a unit cell is 3
No of atoms of D in one unit cell = 1

Formula is

$A_{\frac{1}{2}}B_3CD or AB_6C_2D_2$

If the radius of

${Cl}^{-}$ ion is

$181pm$ , and the radius of

${Na}^{+}$ ion is

$101pm$ then the edge length of unit cell is:

0%
$282pm$
0%
$285.71pm$
0%
$512pm$
0%
$564pm$

Explanation

Edge length of unit cell,

$=2\times (\text{radius of }Cl^-+\text{radius of } Na^+)\\=2\times (181+ 101)\\=2 \times 282\\=564pm$

How many molecules are there in the unit cell of sodium chloride?

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

$NaCl$ (rock salt):

Number of

$Na^{+}$ ions

$= 12$ (at edge centers)

$\times \dfrac {1}{4} + 1$ (at body centre)

$\times 1 = 4$ .

Number of

$Cl^{-} ions = 8$ (at corners)

$\times \dfrac {1}{8} + 6$ (at face centre)

$\times \dfrac {1}{2} = 4$ .

Thus

$4$ formula units per unit cell.

(i) Frenkel defect create a vacancy defect at its original site and an interstitial defect at its new position.
(ii) Frenkel defect is stoichiometric point defect.
(iii) Frenkel defect is also called dislocation defect.
(iv) Silver halide shows Frenkel defect
Select the correct statements :

0%
$(ii),(iii),(iv)$ only
0%
$(i),(ii),(iii)$ only
0%
$(i),(ii),(iii),(iv)$
0%
$(i),(ii),(iv)$ only

Lattice energy of an ionic compound depends upon?

0%
Charge on the ion and size of the ion
0%
Packing of Ions only
0%
Size of the ion
0%
Charge on the Ion only

An element having a face-centered cubic structure has a density of

$6.23g\ cm^{-3}$ . The atomic mass of elements is

$60$ . The edge length of the unit cell is:

0%
$400\ pm$
0%
$300\ pm$
0%
$425\ pm$
0%
$370\ pm$

Explanation

$a^3=\large\dfrac{z\times M}{N_A\times d}$

$\;\;\;\;=\large\dfrac{4\times 60}{6.023\times 10^{23}\times 6.23}$

$\Rightarrow a^3=64\times 10^{-24}cm^3$

$\Rightarrow a=(4\times 10^{-8}\times 10^{10})pm=400pm$

Lattice energy of

$NaCl$ is

$'X'$ . If the ionic size of

$A^{ +2 }$ is equal to that of

$Na^{ + }$ and

$B^{ -2 }$ is equal to

$Cl^{ - }$ , then lattice energy associated with the crystal

$AB$ is:

0%
$X$
0%
$2X$
0%
$4X$
0%
$8X$

Explanation

The lattice energy of crystal $AB$ is $4X$ .

Lattice energy is the energy associated to the crystal lattice of a compound.

Mathematically,

$E_{lattice}=k\dfrac{Q_1Q_2}{d}$

where, $k =$ proportionality constant

$Q_1$ and $Q_2$ are the charges on the ions in coulombs.

$d =$ distance between the ion charges.

Lattice energy of $NaCl = X$

$E_{lattice}(NaCl)=\dfrac{Q_1Q_2}{d}=X$

As the ionic size remains same when $AB$ crystal is formed, d remains same

Lattice energy of $AB$ ,

$E_{lattice}(AB)=\dfrac{(2Q_1)(2Q_2)}{d}$

$E_{lattice}(AB)=4\left(\dfrac{Q_1Q_2}{d}\right )=4X$

The lattice energy of crystal $AB$ is $4X$ .

Which of the following expression is correct for packing fraction of

$NaCI$ if the ions along with face are diagonally removed ?

0%
$\dfrac{\dfrac{13}{3} \pi r^{3}_{-}+\dfrac{16}{3} \pi r_{+}^{3}}{8(r_{+}+r_{-})^{3}}$
0%
$\dfrac{\dfrac{13}{3} \pi r^{3}_{-}+\dfrac{4}{3} \pi r_{+}^{3}}{8(r_{+}+r_{-})^{3}}$
0%
$\dfrac{\dfrac{16}{3} \pi r^{3}_{-}+\dfrac{13}{3} \pi r_{+}^{3}}{8(r_{+}+r_{-})^{3}}$
0%
$\dfrac{\dfrac{3}{4} \pi r^{3}_{-}+\dfrac{13}{3} \pi r_{+}^{3}}{8(r_{+}+r_{-})^{3}}$

Which of the following pairs have layer lattice structure in solid state chemistry-

0%
$Sr{Cl}_{2}$ and $C{dI}_{2}$
0%
Diamond and Graphite
0%
Graphite and $Cd{I}_{2}$
0%
$Mg{SO}_{4}.7{H}_{2}O$ and $Fe{SO}_{4}.7{H}_{2}O$

In the given crystal what should be the cation

$X$ which replaces to create a

${ Na }^{ + }$ cation vacancy?

0%
${ Sr }^{ 2+ }$
0%
${ K }^{ + }$
0%
${ Li }^{ + }$
0%
${ Br }^{ - }$

Explanation

The correct option is (A).

Substitutional impurity in $NaCl$

1) Each

$Sr^{2+}$ replaces two

$Na^+$ ions( to maintain charge neutrality). It occupies the site of one ion and other site remains vacant.

2) The cationic vacancies thus produced are equal in number to that of

$Sr^{2+}$ ions.

Additional Information:
This defect occurs when regular cation is replaced by some different cation. The different cation may occupy a regular lattice site or interstitial site.

Thus we can say that when cation of higher valence is added as an impurity to an ionic solid, two or more cations of lower valency are replaced by cation of higher valency to maintain electrical neutrality. Hence, some cationic vacancies are created.

Which of the following solid substance(s) will have the same refractive index when measured in different directions?

0%
Rubber
0%
$NaCl$
0%
Plastic
0%
Graphite

Explanation

Isotropic solids have same refractive index in all the directions. As amorphous solids are isotropic in nature. Hence, amorphous solids like

$NaCl$ will show same refractive index when measured in different directions.

Hence, the correct option is

$B$

0:0:1

Answered

Not Answered

Not Visited

Correct : 0

Incorrect : 0

CBSE Questions for Class 12 Engineering Chemistry The Solid State Quiz 10 - MCQExams.com

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.