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Chemistry 2 Marks Questions

The Solid State · Jharkhand Board (JAC) STD 12 Science (Jharkhand - English Medium). 102 questions.

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Question 12 Marks
Explain the following with suitable examples:
Antiferromagnetism.
Answer
Antiferromagnetism:Antiferromagnetic substanceshave domain structures similar to ferromagnetic substances, but are oppositely-oriented. The oppositely-oriented domains cancel out each other's magnetic moments.
Schematic alignment of magnetic moments in antiferromagnetic substances.
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Question 22 Marks
A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?
Answer
Number of closed packed particles $= 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$
$0.5 \times 6.022 \times 1023 = 3.011 \times 1023$
Therefore number of octhahedral voids $= 3.011 \times 10^{23}\times 1023$
and number of tetrahedral voids $= 2 \times 3.011 \times 10^{23}$
$= 6.022 \times 10^{23}2 \times 3.011 \times 1023$
$= 6.022 \times 1023$
Therefore total number of voids $= 3.011 \times 10^{23} + 6.022\times 10^{23}$
$ = 9.033 \times 10^{23}3.011 \times 1023 + 6.022 \times 1023$
$​​​​​​​= 9.033 \times 1023.$
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Question 32 Marks
Explain the following with suitable examples:
Ferromagnetism.
Answer
Ferromagnetism:Few substances like iron, cobalt, nickel, gadolinium and $CrO_2$ are strongly attracted by a magnetic field. Such substances are known as ferromagnetic substances. These substances can be permanently magnetised with the help of strong electrostatic field. The metal ions of ferromagnetic substances are grouped together into small regions called domains in solid state. So that each domain acts as a tiny magnet. The domains of un magnetized piece of a ferromagnetic are randomly oriented so that their magnetic moments get cancelled out. When such substance is placed in a magnetic field, all the domains get oriented in the direction of the magnetic field and a strong magnetic effect is produced. This ordering of domains remains same even when the magnetic field is removed and the ferromagnetic substance becomes a permanent magnet.

Fig. Schematic alignments of magnetic moments in ferromagnetism.
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Question 42 Marks
Distinguish between:
  1. Hexagonal and monoclinic unit cells.
  2. Face-centred and end-centred unit cells.
Answer
  1.  
Hexagonal unit cell Monoclinic unit cell
a=b≠c a≠b≠c
α = β = 90° α = γ = 90°
γ = 120° β ≠ 90°
  1.  
Face-centred unit cell End-centred unit cell
A Face-centred unit cell the constituent particles are present at the corners and one at the centre of each face. An End-centred unit cell contains particles at the corners and one at the centre of any two opposite faces.
Total no of particles in a face centered unit cell= 4 Total no. of particles in an end centered unit cell = 2
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Question 52 Marks
Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?
Answer
Radius, r = 0.144 nm, Unit cell is fcc, edge of unit cell, a = ? In face centered unit cell, diagonal of face = $4\text{r}=\text{a}\sqrt{2}\text{a}2$ $[\because\ \text{face diagonal}=\text{a}\sqrt{2}\text{a}2]$$\therefore\ \text{a}=4\text{r}\sqrt{2}=\sqrt{2}\text{r}\text{a}=4\text{r}2=2\text{r}$
$=2\times1.4142\times0.144\text{ nm}=0.4073\text{ nm.}$
Length of side of unit cell, a = 0.4073 nm.
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Question 62 Marks
Non-stoichiometric cuprous oxide, $Cu_2O$ can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?
Answer
In the cuprous oxide $(Cu_2O)$ prepared in the laboratory, copper to oxygen ratio is slightly less than $2 : 1.$ This means that the number of $Cu^+$ ions is slightly less than twice the number of $O_2$ −ions. This is because some $Cu^+$ ions have been replaced by $Cu^{2+}$ ions. Every $Cu^{2+}$ ion replaces two $Cu^+$ ions, thereby creating holes. As a result, the substance conducts electricity with the help of these positive holes. Hence, the substance is a p-type semiconductor.
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Question 72 Marks
Explain the following with suitable examples:
Paramagnetism.
Answer
Paramagnetism:Few substances like $O_2, Cu^{2+}, Fe^{3+}, Cr^{3+}$ are weekly attracted by a magnetic field.
These substabces are magnetised in a magnetic field in the same direction. When we remove magnetic field ,they lose their magnetism. paramagnetism is takes place due to presence of one or more unpaired electrons. These unpaired electron are attracted by the magnetic field.
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Question 82 Marks
Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.
Answer
If we mix a small amount of molten strontium chloride $(SrCl_2)$ with molten sodium chloride (NaCl) and the resulting solution is cooled, in the crystals of NaCl some $Na^+$ ions will get replaced by $Sr^{2+}$ ions. In order to maintain electron neutrality too $Na^+$ ions are to leave their respective lattice sites. However, one out of them will be occupied by the $Sr^{2+}$ ion while the other will remain unoccupied or it will be vacant. Thus vacancies are introduced in an ionic solid.
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Question 92 Marks
Name the parameters that characterise a unit cell.
Answer
A unit cell is characterised by following parameters:
  1. The dimensions of unit cell along three edges: a, b and c.
  2. The angles between the edges: α (between b and c); β (between a and c) and γ (between a and b).
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Question 102 Marks
A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?
Answer
As atoms Q are present at the eight comers of the cube, therefore, the contribution of atoms of
Q in the unit cell = 1/8 x 8 = 1.
As atom P is present at the body centre, therefore, the contribution of atoms of P in the unit cell = 1.
$\therefore$ Ratio of atoms of P: Q = 1:1
Hence, the formula of the compound = PQ The coordination number of each P and Q = 8.
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Question 112 Marks
Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
  1. Tetra phosphorus decoxide $(P_4O_{10})$
  2. Ammonium phosphate $(NH_4)_3PO_4$
  3. SiC
  4. $I_2$
  5. $P_4$
  6. Plastic
  7. Graphite
  8. Brass
  9. Rb
  10. LiBr
  11. Si
Answer
  1. Tetra phosphorous decoxide $(P_4O_{10}) –$ Molecular
  2. Ammonium phosphate $(NH_4)_3PO_4 –$ Ionic
  3. SiC - Covalent (network)
  4. $I_2$ - Molecular
  5. $P_4$ – Molecular
  6. Plastic – Amorphous
  7. Graphite – Covalent (network)
  8. Brass – Metallic
  9. Rb – Metallic
  10. LiBr – Ionic
  11. Si – Covalent (network)
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Question 122 Marks
Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
Answer
The colour develops because of the presence of electrons in the anionic sites. These electrons absorb energy from the visible part of radiation and get excited.For example, when crystals of NaCl are heated in an atmosphere of sodium vapours, the sodium atoms get deposited on the surface of the crystal and the chloride ions from the crystal diffuse to the surface to form NaCl with the deposited Na atoms. During this process, the Na atoms on the surface lose electrons to form Na ions and the released electrons diffuse into the crystal to occupy the vacant anionic sites. These electrons get excited by absorbing energy from the visible light and impart yellow colour to the crystals.
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Question 132 Marks
  1. What is meant by the term 'coordination number'?
  2. What is the coordination number of atoms:
  1. In a cubic close-packed structure?
  2. In a body-centred cubic structure?
Answer
  1. The number of nearest neighbours of any constituent particle present in the crystal lattice is called its coordination number.
  2. The coordination number of atoms:
  1. In a cubic close-packed structure is 12, and
  2. In a body-centred cubic structure is 8.
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Question 142 Marks
Explain the following term with suitable examples:
Frenkel defect.
Answer
Frenkel defect:It is a type of vacancy defect. In ionic compounds, some of the ions (usually smaller in size) get dislocated from their original site and create defect. This defect is known as Frenkel Defects. Since this defect arises because of dislocation of ions, thus it is also known as Dislocation Defects. As there are a number of cations and anions (which remain equal even because of defect); the density of the substance does not increase or decrease.
Ionic compounds; having large difference in the size between their cations and anions; show Frenkel Defects, such as ZnS, AgCl, AgBr, AgI, etc. These compounds have smaller size of cations compared to anions.
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Question 152 Marks
What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
Answer
The arrangement of the constituent particles makes glass different from quartz. In glass, the constituent particles have short-range order, but in quartz, the constituent particles have both long range and short range orders. Quartz can be converted into glass by heating and then cooling it rapidly.
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Question 162 Marks
Calculate the efficiency of packing in case of a metal crystal for:
Face-centred cubic (with the assumptions that atoms are touching each other).
Answer
In fcc: Let the edge length of unit cell = a Let the radius of each sphere = r $\therefore\ \text{AC}=4\text{r}$ From right angle triangle ABC,$\text{AC}=\sqrt{\text{AB}^2+\text{BC}^2}=\sqrt{\text{a}^2+\text{a}^2}=\sqrt{2\text{a}^2}=\sqrt{2}\text{a}$

$\therefore\ \sqrt{2}\text{a}=4\text{r}$
$\therefore\ \text{a}=\frac{4\text{r}}{\sqrt{2}}$
$\therefore$ Volume of the unit cell
$=\text{a}^3=\Big(\frac{4}{\sqrt{2}}\text{r}\Big)^3=\frac{64\text{r}^3}{2\sqrt{2}}=\frac{32\text{r}^3}{\sqrt{2}}$
No. of unit cell in fcc = 4 $\therefore\ \text{Volume of four spheres}=4\times\frac{4}{3}\pi\text{r}^3=\frac{16}{3}\pi\text{r}^3$ $\therefore$ Packing efficiency$=\frac{\frac{16\pi\text{r}^3}{3}}{\frac{64\text{r}^3}{3}\sqrt{3}}=0.74, \text{i.e.}, 74\%$
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Question 172 Marks
Explain the following term with suitable examples:
Schottky defect.
Answer
Schottky defects:When cations and anions both are missing from regular sites, the defect is called Schottky Defect. In Schottky Defects, the number of missing cations is equal to the number of missing anions in order to maintain the electrical neutrality of the ionic compound.
Schottky Defect is type of simple vacancy defect and shown by ionic solids having cations and anions; almost similar in size, such as NaCl, KCl, CsCl, etc. AgBr shows both types of defects, i.e. Schottky and Frenkel Defects.
Since, Schottky Defects arises because of mission of constituent particles, thus it decreases the density of ionic compound.
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Question 182 Marks
Explain the following with suitable examples:
12-16 and 13-15 group compounds.
Answer
12-16 and 13-15 group compounds:The 12-16 group compounds are prepared by combining group 12 and group 16 elements and the 13-15 group compounds are prepared by combining group 13 and group15 elements. These compounds are prepared to stimulate average valence of four as in Ge or Si. Indium (III) antimonide (IrSb), aluminium phosphide (AlP), and gallium arsenide (GaAS) are typical compounds of groups 13-15. GaAs semiconductors have a very fast response time and have revolutionised the designing of semiconductor devices. Examples of group 12-16 compounds include zinc sulphide (ZnS), cadmium sulphide (CdS), cadmium selenide (CdSe), and mercury (II) telluride (HgTe). The bonds in these compounds are not perfectly covalent. The ionic character of the bonds depends on the electronegativities of the two elements.
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Question 192 Marks
Calculate the efficiency of packing in case of a metal crystal for:
Simple cubic.
Answer
In simple cubic lattice:
Here $a = 2r$
No. of spheres per unit cell $= 1$
Volume of spheres = $\frac{4}{3}\pi\text{r}^3$
Volume of cube $= a^3= (2r)^3= 8r^3$
$\therefore$ Pacing efficiency
$= \frac{\frac{4}{3}\pi\text{r}^3}{8\text{r}^3}=0.524, \text{i.e.}, 52.4\%.$
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Question 202 Marks
Classify each of the following as being either a p-type or a n-type semiconductor:
  1. Ge doped with In.
  2. B doped with Si.
Answer
  1. Ge (a group 14 element) is doped with In (a group 13 element). Therefore, a hole will be created and the semiconductor generated will be a p-type semiconductor.
  2. B (a group 13 element) is doped with Si (a group 14 element). Thus, a hole will be created and the semiconductor generated will be a p-type semiconductor.
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Question 212 Marks
Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.
Answer
Amorphous solids: Polyurethane, teflon, cellophane, polyvinyal chloride, fibre glass.Crystalline solids: naphthlene, benzoic acid, potassium nitrate, copper.
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Question 222 Marks
Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.
Answer
Let the number of oxide $(O_{2-})$ ions be x.
So, number of octahedral voids $= x.$
It is given that two out of every three octahedral holes are occupied by ferric ions. So, number of ferric $(Fe^{3+})$ ions $= 23x$
Therefore, ratio of the number of $Fe^{3+}$ ions to the number of $O_{2−}$ ions, $Fe^{3+}Fe^{3+} : O_{2-} = 23 x : x$
$= 23 : 1$
$= 2 : 3$
Hence, the formula of the ferric oxide is $Fe_2O_3.$
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Question 232 Marks
Explain the following term with suitable examples:
Interstitials.
Answer
Interstitials:Interstitials – Sometime in the formation of lattice structure some of the atoms or ions occupy vacant interstitial site, and are known as interstitials. These interstitials are generally small size non-metals, such as H, B, C, etc. Defect arises because of interstitials is called interstitial defect.
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Question 242 Marks
Calculate the efficiency of packing in case of a metal crystal for:
Body-centred cubic.
Answer
In bcc: AD = 4r
From right angled triangle ABC,$\text{AC}=\sqrt{\text{AB}^2+\text{BC}^2}=\sqrt{\text{a}^2+\text{a}^2}=\sqrt{2}\text{a}$
Body diagonal, AD $=\sqrt{\text{AC}^2+\text{CD}^2}=\sqrt{2\text{a}^2+\text{a}^2}=\sqrt{3\text{a}^2}=\sqrt{3}\text{a}$ $\therefore\ \sqrt{3}\text{a}=4\text{r}\ \ \ \text{a}=\frac{4\text{r}}{\sqrt{3}}$ $\therefore\ \text{Volume of unit cell}=\text{a}^3=\Big(\frac{4\text{r}}{\sqrt{3}}\Big)^3=\frac{64\text{r}^3}{3\sqrt{3}}$ No. of spheres per unit cell = 2 $\therefore\ \text{Volume of two spheres}=2\times\frac{4}{3}\pi\text{r}^3=\frac{8}{3}\pi\text{r}^3$ $\therefore$ Packing efficiency $=\frac{\frac{8}{3}\pi\text{r}^3}{\frac{64\text{r}^3}{3}\sqrt{3}}=0.68, \text{i.e.}, 68\%$
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Question 252 Marks
In term of band theory, what is the difference:
Between a conductor and a semiconductor.
Answer
Difference between conductors and semiconductor:In conductors there is no energy gap between the valence band and conduction band, which facilitates the flow of electrons easily under an applied electric field and metals show conductivity.
While in semi conductors, there is small energy gap between valence bond and conduction band. The small gap between band facilitates some electrons to jump to the conduction band by acquiring extra energy.
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Question 262 Marks
Explain the following with suitable examples:
Ferrimagnetism.
Answer
Ferrimagnetism:
This effect is observed when the magnetic moments of the domains in the substance are aligned in parallel and anti–parallel directions in unequal numbers. Ferrimagnetic substances are weakly attracted by magnetic field as compared to ferromagnetic substances. Magnetite like $Fe_3O_4$ and ferrites like $MgFe_2O_4$ and $ZnFe_2O_4$ are examples.
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Question 272 Marks
Explain the following term with suitable examples:
F-centres.
Answer
F-centres:F-centres – This is type of defect and called metal excess defect. These type of defects seen because of missing of anions from regular site leaving a hole which is occupied by electron to maintain the neutrality of the compound. Hole occupied by electron is called F-centre and responsible for showing colour by the compound. example sodium chloride impart yellow colour beacuse of F centre.
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Question 282 Marks
Silver crystallises in fcc lattice. If edge length of the cell is $4.07 \times 10^{-8} \mathrm{~cm}$ and density is $10.5 \mathrm{~g} \mathrm{~cm}^{-3}$, calculate the atomic mass of silver.
Answer
$\text{M}=\frac{\text{d.a}^3\text{.N}_a}{\text{Z}}=\frac{10.5\times(4.07\times10^{-8})^3\times6.023\times10^{23}}{4}$$=107.09 \text{ g mol}^{-1}.$
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Question 292 Marks
Niobium crystallises in body-centred cubic structure. If density is $8.55 \mathrm{~g} \mathrm{~cm}^{-3}$, calculate atomic radius of niobium using its atomic mass 93 u.
Answer
$\text{a}^3=\frac{\text{M.Z}}{\text{d.N}_a\times10^{-30}}=\frac{93\times2}{8.55\times6.02\times10^{23}\times10^{-30}}$$=3.61\times10^7$
$\therefore\ \text{a}=(3.61\times10^7)^\frac{1}{3}=(36.1\times10^6)\frac{1}{3}$
$=3.304\times10^2\text{pm}=330.4\text{pm}$
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Question 302 Marks
Copper crystallises into a fcc lattice with edge length $3.61 \times 10^{-8} \mathrm{~cm}$. Show that the calculated density is in agreement with its measured value of $8.92 \mathrm{~g} \mathrm{~cm}^{-3}$.
Answer
$\text{d}=\frac{\text{Z}\times\text{M}}{\text{a}^3\times\text{N}_\text{a}}$$=\frac{4\times63.5}{(3.61\times10^{-8})^3\times6.023\times10^{23}}\ \{{\text{M}=63.5 \text{ for Cu}}\}$
$=8.96 \text{ g cm}^{-3}$
This calculated value of density is closely in agreement with its measured value of $=8.96 \text{ g cm}^{-3}$.
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Question 312 Marks
Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property?
Answer
As the solid has same value of refractive index along all directions, it is isotropic in nature and hence amorphous. Being amorphous solid, it will not show a clean cleavage and when cut, it will break into pieces with irregular surfaces.
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Question 322 Marks
In term of band theory, what is the difference:
Between a conductor and an insulator.
Answer
Difference between conductor and insulator:In conductors there is no energy gap between the valence band, which facilitates the flow of electrons easily under an applied electric field and metals show conductivity.
While in insulators there is large energy gap between the valence band and electrons cannot jump to it i.e. large energy gap prevents the flow of electricity.
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Question 332 Marks
Classify the following solids in different categories based on the nature of intermolecular forces operating in them:
Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide.
Answer
Molecular solids: Water, argon.
Ionic solids: Potassium sulphate, zinc sulphate.
Metallic solids: Tin, rubidium.
Covalent or network: Benzene, urea, ammonia, graphite, silicon carbide.
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Question 342 Marks
Calculate the number of unit cells in 8·1 g of aluminium if it crystallizes in a face-centred cubic (f.c.c.) structure. $($Atomic mass of $Al = 27$ g mol$^{-1})$
Answer
n = given mass/molar mass
$= 8.1/27$ mol
Number of atoms = $\frac{8.1}{27}\times6.022\times10^{23}$
Number of atoms in one unit cell = 4(fcc)
Number of unit cells $=\frac{\big[\frac{8.1}{27}\times6.022\times10^{23}\big]}{4}\\=4.5\times10^{22}$
Or
27g of Al contains $= 6.022x10^{23}$ atoms
8.1g of Al contains $= (6.022x10^{23} / 27) \times 8.1$
No of unit cells = total no of atoms /4
$=\frac{\big[\frac{8.1}{27}\times6.022\times10^{23}\big]}{4}\\=4.5\times10^{22}$
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Question 352 Marks
An element with density $2.8\ g\ cm^{–3}$ forms a f.c.c. unit cell with edge length $4\times 10^{–8}\ cm.$ Calculate the molar mass of the element.
$($Given: $N_A = 6.022 \times 10^{23}\ mol ^{–1})$
Answer
Given; $d = 2.8g/cm^3;$ $Z = 4; a = 4 x 10^{–8}$ cm $N_A= 6.022 \times 10^{23}$ per mol
$\text{d}=\frac{\text{Z}\times\text{M}}{\text{a}^{3}\times\text{N}_{A}}$ OR
$\text{M}=\frac{\text{d}\times\text{a}^{3}\times\text{N}_{A}}{\text{z}}$
$\Rightarrow$$\text{M}=\frac{\text{2.8 g cm}^{-3}\times\text{(4}\times\text{10}^{-8}\text{cm})^{3}\times\text{6.022}\times\text{10}^{23}}{\text{4}}$
$M = 2.8 \times 16 \times 10^{–1} \times 6.022 = 26.97$ g/mol.
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Question 362 Marks
  1. Write the type of magnetism observed when the magnetic moments are aligned in parallel and anti-parallel directions in unequal numbers.
  2. Which stoichiometric defect decreases the density of the crystal?
Answer
  1. Ferrimagnetism.
  2. Schottky defect.
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Question 372 Marks
Calculate the number of unit cells in 8·1 g of aluminium if it crystallises in a face-centered cubic (f.c.c.) structure. $($Atomic mass of Al $= 27$ g mol$^{-1})$
Answer
$n =$ given mass/molar mass
$= 8.1/27$ mol
Number of atoms $=\frac{8.1}{27}\times6.022\times10^{23}$
Number of atoms in one unit cell = 4 (fcc)
Number of unit cells $=[\frac{8.1}{27}\times6.022\times10^{23}]/4$
$=4.5\times10^{22}$
Alternate Answer
27g of $\text{A}l$ contains $=6.022\times10^{23}$ atoms
8.1g of $\text{A}l$ contains $=(6.022\times10^{23}/27)\times8.1$
No of unit cells = total no of atoms /4
$=[\frac{8.1}{27}\times6.022\times10^{23}]/4$
$=4.5\times10^{22}$
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Question 382 Marks
  1. Write the type of magnetism observed when the magnetic moments are oppositively aligned and cancel out each other.
  2. Which stoichiometric defect does not change the density of the crystal?
Answer
  1. Antiferromagnetism.
  2. Frenkel defect.
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Question 392 Marks
An element with density $2.8\ g\ cm^{–3}$ forms a f.c.c. unit cell with edge length $4 \times 10^{–8}\ cm.$ Calculate the molar mass of the element.
$($Given: $N_A = 6.022 \times 10^{23}\ mol ^{–1})$
Answer
Given; $d = 2.8g/cm^3;\ Z = 4;\ a = 4 x 10^{–8}\ cm\ \  N_A= 6.022 x 10^{23}$ per mol
$\text{d}=\frac{\text{Z}\times\text{M}}{\text{a}^{3}\times\text{N}_{A}}$ OR
$\text{M}=\frac{\text{d}\times\text{a}^{3}\times\text{N}_{A}}{\text{z}}$
$\Rightarrow$$\text{M}=\frac{\text{2.8 g cm}^{-3}\times\text{(4}\times\text{10}^{-8}\text{cm})^{3}\times\text{6.022}\times\text{10}^{23}}{\text{4}}$
$M = 2.8 \times 16 \times 10^{–1} \times 6.022 = 26.97$ g/mol.
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Question 402 Marks
Calculate the number of unit cells in 8·1 g of aluminium if it crystallises in a face-centered cubic (f.c.c.) structure. $($Atomic mass of $Al = 27$ g mol$^{-1})$
Answer
n = given mass/molar mass
= 8.1/27 mol
Number of atoms $=\frac{8.1}{27}\times6.022\times10^{23}$
Number of atoms in one unit cell = 4 (fcc)
Number of unit cells $=[\frac{8.1}{27}\times6.022\times10^{23}]/4$
$=4.5\times10^{22}$
Alternate Answer
27g of $\text{A}l$ contains $=6.022\times10^{23}$ atoms
8.1g of $\text{A}l$ contains $=(6.022\times10^{23}/27)\times8.1$
No of unit cells = total no of atoms /4
$=[\frac{8.1}{27}\times6.022\times10^{23}]/4$
$=4.5\times10^{22}$
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Question 412 Marks
An element with density $2.8\ g\ cm^{–3}$ forms a f.c.c. unit cell with edge length $4 $\times$ 10^{–8} cm.$ Calculate the molar mass of the element.
$($Given: $N_A = 6.022 \times 10^{23}\ mol^{–1})$
Answer
Given; $d = 2.8g/cm^3; Z = 4; a = 4 \times 10^{–8}$ cm $N_A= 6.022 \times 10^{23}$ per mol
$\text{d}=\frac{\text{Z}\times\text{M}}{\text{a}^{3}\times\text{N}_{A}}$ OR
$\text{M}=\frac{\text{d}\times\text{a}^{3}\times\text{N}_{A}}{\text{z}}$
$\Rightarrow$$\text{M}=\frac{\text{2.8 g cm}^{-3}\times\text{(4}\times\text{10}^{-8}\text{cm})^{3}\times\text{6.022}\times\text{10}^{23}}{\text{4}}$
$M = 2.8\times 16\times 10^{–1} \times 6.022 = 26.97$ g/mol.
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Question 422 Marks
Account for the following:
  1. Schottky defects lower the density of related solids.
  2. Conductivity of silicon increases on doping it with phosphorus.
Answer
  1. In Schottky defect some ions are missing (or due to vacancies) from their normal lattice sites due to which density decreases.
  2. This is due to availability of unpaired or odd electron provided by P.
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Question 432 Marks
Aluminium crystallises in an fcc structure. Atomic radius of the metal is 125 in. What is the length of the side of the unit cell of the metal?
Answer
$\text{For f.c.c. unit cell r=}\frac{\text{a}}{\text{2}\sqrt{2}}$
$\text{a = 2 r}\times\sqrt{2}$

= 2 x 125 pm x 1.414

=353.5 pm
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Question 442 Marks
Calculate the packing efficiency of a metal crystal for a simple cubic lattice.
Answer
Packing efficiency
$=\frac{\text{Z}\ \times\ \text{volume of one atom}}{\text{volume of cubic cell unit}}$
$=\frac{\text{1}\ \times\ \text{4/3}\pi\text{r}^{3}}{\text{a}^{3}}$
For simple cubic lattice a = 2r
Therefore packing effieciency $=\frac{\text{1}\ \times\ \text{4/3}\pi\text{r}^{3}}{\text{8r}^{3}}$
= 0.524 or 52.4%.
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Question 452 Marks
Explain how you can determine the atomic mass of an unknown metal if you know its mass density and the dimensions of unit cell of its crystal.
Answer
We can determine the atomic mass of an unknown by using the formula.
$\text{M}=\frac{\text{d}\times\text{a}^{3}\times\text{N}_{A}}{Z}$
By knowing $d, a, N_A\ \&\ Z$ We can calculate the M
Where
$d =$ Density of the element.
$N_A =$ Avogadro number.
$a =$ Cell edge or edge length.
$Z = $ No. of atoms present in one unit cell.
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Question 462 Marks
An element with density $11.2\ g\ cm^{-3}$ forms a f.c.c. lattice with edge length of $4 \times 10^{-8} $ cm.
Calculate the atomic mass of the element.
$($Given: $N_A= 6.022\times 10^{23}\ mol^{-1})$
Answer
$d = 11.2\ g/cm^3$
$z = 4$
$a = 4x10^{-8}\ cm$
$\text{d}=\frac{\text{Z}\times{M}}{\text{N}_{a}\times\text{a}^{3}}$
$\text{11.2}=\frac{\text{4}\times{M}}{\text{6.022}\times\text{10}^{23}}\times{(4}\times{10}^{-8})^{3}$
$\text{M}=\frac{\text{11.2}\times\text{6.022}\times\text{10}^{23}\times\text{4}\times\text{10}^{-8}\times\text{4}\times\text{10}^{-8}\times\text{4}\times\text{10}^{-8}}{\text{4}}$
$M = 11.2\times 6.022\times 16\times 10^{-1}$
$M = 107.9\ g\ mol^{-1}$ or $107.9$ u
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Question 472 Marks
Examine the given defective crystal:

Answer the following questions:
  1. What type of stoichiometric defect is shown by the crystal?
  2. How is the density of the crystal affected by this defect?
  3. What type of ionic substances show such defect?
Answer
  1. Schottky defect.
  2. Decreases.
  3. Alkali metal halides/Ionic substances having almost similar size of cations and anions (NaCl/KCl).
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Question 482 Marks
Examine the given defective crystal:

Answer the following questions:
  1. What type of stoichiometric defect is shown by the crystal?
  2. How is the density of the crystal affected by this defect?
  3. What type of ionic substances show such defect?
Answer
  1. Schottky defect.
  2. Decreases.
  3. Alkali metal halides/Ionic substances having almost similar size of cations and anions (NaCl/KCl).
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Question 492 Marks
An element with density $11.2\ g\ cm^{-3}$ forms a f.c.c. lattice with edge length of $4 \times 10^{-8}$ cm.
Calculate the atomic mass of the element.
$($Given:$ N_A= 6.022\times 10^{23}\ mol^{-1})$
Answer
$d = 11.2\ g/cm^3$
$z = 4$
$a = 4x10^{-8}$ cm
$\text{d}=\frac{\text{Z}\times{M}}{\text{N}_{a}\times\text{a}^{3}}$
$\text{11.2}=\frac{\text{4}\times{M}}{\text{6.022}\times\text{10}^{23}}\times{(4}\times{10}^{-8})^{3}$
$\text{M}=\frac{\text{11.2}\times\text{6.022}\times\text{10}^{23}\times\text{4}\times\text{10}^{-8}\times\text{4}\times\text{10}^{-8}\times\text{4}\times\text{10}^{-8}}{\text{4}}$
$M = 11.2\times 6.022\times 16\times 10^{-1}$
$M = 107.9\ g\ mol^{-1}$ or $107.9$ u
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Question 502 Marks
An element with density $11.2\ g\ cm^{-3}$ forms a f.c.c. lattice with edge length of $4\times  10^{-8}\ cm.$
Calculate the atomic mass of the element.
$($Given: $N_A= 6.022\times 10^{23}\ mol^{-1})$
Answer
$d = 11.2 g/cm^3$
$z = 4$
$a = 4\times 10^{-8} cm$
$\text{d}=\frac{\text{Z}\times{M}}{\text{N}_{a}\times\text{a}^{3}}$
$\text{11.2}=\frac{\text{4}\times{M}}{\text{6.022}\times\text{10}^{23}}\times{(4}\times{10}^{-8})^{3}$
$\text{M}=\frac{\text{11.2}\times\text{6.022}\times\text{10}^{23}\times\text{4}\times\text{10}^{-8}\times\text{4}\times\text{10}^{-8}\times\text{4}\times\text{10}^{-8}}{\text{4}}$
$M = 11.2\times 6.022\times 16\times 10^{-1}$
$M = 107.9\ g\ mol^{-1}$or $107.9\  u$
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