Xii Chemistry Numericals

A book for Std. XII/12th Science Chemistry Numericals/Problems

Written according to the New Text book (2012-2013) published by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune.

Std. XII Sci.

Chemistry Numericals Prof. Santosh B. Yadav (M. Sc., SET, NET) Department of Chemistry R. Jhunjunwala College, Ghatkopar

Salient Features: 9 9 9 9 9

Completely exam oriented solved problems. Formulae bank for every topic. Practice problems with hints for every subtopic. Problems from various competitive exams. 236 Solved problems, 637 Problems for practice and 104 Multiple Choice Questions. 9 Self evaluative in nature.

Target PUBLICATIONS PVT. LTD. Mumbai, Maharashtra Tel: 022 – 6551 6551 Website: www.targetpublications.in www.targetpublications.org email : [email protected]

Std. XII Sci. Chemistry Numericals

©

Target Publications Pvt Ltd.

Sixth Edition: November 2012

Price: ` 120/-

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PREFACE The desire to learn Chemistry remains diminished unless and until the student masters Physical chemistry. Physical chemistry is a field of science which mainly consists of problems and hence it calls for a deep knowledge of formulas and ability to solve numerical problems quickly and efficiently. Hence to ease this task we bring to you “Std. XII Sci. Chemistry Numericals” a book containing adequate solved problems for every chapter classified into subtopics that provides an indepth knowledge of the procedure to tackle the problems. At the end of each topic and sub-topic practice problems are provided to test the student’s preparation and increase his confidence. Additional and multiple choice questions are also provided to increase the knowledge and ability of the student. Board problems and various competitive exams problems of the last many years have been included to provide the importance of questions. To end on a candid note, I would like to make a humble request to each and every student: Preserve this book as a Holy Grail as it helps you in the complete and thorough preparation from the examination point of view. There is always a room for improvement, hence I welcome all suggestions and regret any errors that may have occurred in the making of this book.

Best of luck to all the aspirants! Your’s faithfully

Publisher

Contents No.

Topic Name

Page No.

1.

Solid State

1

2.

Solutions and Colligative Properties

20

3.

Chemical Thermodynamics and Energetics

62

4.

Electrochemistry

116

5.

Chemical Kinetics

163

6.

IUPAC Name and Nomenclature

234

Std. XII Sci.: Chemistry Numericals

TARGET Publications

Solid state

01 Formulae 1.

Density of unit cell: z.M d= 3 a .NA where, a is edge of unit cell NA = Avogadro number (6.023 × 1023) M = Molar mass z = number of atoms per unit cell For fcc, z = 4 for bcc, z = 2 for simple cubic, z = 1

2.

Radius rule and coordination number for ionic crystals: In simple ionic crystals, the cations commonly occupy the voids or holes. The voids are empty spaces left between anionic spheres. ⎛ r+ ⎞ i. Radius Ratio ⎜ − ⎟ : ⎝r ⎠ The critical radius ratio of the void (cation) and sphere (anion), is calculated by solid geometry. r+ Cation radius Radius ratio = − = r Anion radius

∴

ii.

3.

Coordination Number (CN) : The number of spheres (atoms, molecules or ions) directly surrounding a single sphere in a crystal, is called coordination number.

Crystal structures of some elements and their coordination number’s (CN): Crystal structure bcc fcc or ccp hcp (Hexagonal closed packed)

4.

Example Li, Na, K, Rb, Cs, Ba Ni, Cu, Ag, Au, Pt Zn, Mo, Cd, V, Be, Mg

Coordination No. 8 12 21

Relation between radius ratio, coordination number and geometry : ⎛ r+ ⎞ − ⎟ ⎝r ⎠

Radius ratio ⎜

0.155 to 0.225 0.225 to 0.414 0.414 to 0.732 0.732 to 1.0 Solid State

Coordination number

Geometry

Examples

3 4 6 8

Planar triangular Tetrahedral Octahedral Cubic

B2O3 ZnS NaCl CsCl 1

Std. XII Sci.: Chemistry Numericals 5.

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Characteristics of some typical crystal structure : Crystal

CsCl NaCl ZnS CaF2

Type of unit cell bcc fcc fcc fcc

Examples

CsCl, CsBr, TiCl AgCl, MgO ZnS CaF2, SrF2, CdF2

Solved Examples Type 1: Radius Ratio of ionic compound/ The Formula of compound Example 1.1 Barium has a radius of 224 pm and crystallizes in a body-centred cubic structure. What is the edge length of the unit cell? Solution: Given: Radius (r) = 224 pm To find: Edge length of unit cell (a) = ? 3a Formula: r = 4 Calculation: For BCC From formula, r × 4 224 × 4 = = 517.3 pm a= 1.7320 3 Example 1.2 Aluminium crystallizes in cubic close packed structure. Its metallic radius is 125 pm. What is the edge length of unit cell? Solution: Given: Radius (r) = 125 pm Edge length of unit cell (a) = ? To find: a Formula: r = 2× 2 Calculation: Since Al crystallizes in Face centred cubic (FCC) structure From formula, a = r×2× 2 = 125 × 2 × 1.4142 ∴ a = 353.5 pm 2

Radius ratio 0.93 0.52 0.40 0.73

CN Cation Anion 8 8 6 6 4 4 8 4

Example 1.3 In silicates the oxygen atom forms a tetrahedral void. The limiting radius ratio for tetrahedral void is 0.22. The radius of oxide is 1.4 Å. Find out the radius of cation. Solution: Given: Radius of oxide (r−) = 1.4 Å Radius ratio = 0.22 To find: Radius of cation (r+) = ? Formula: Radius of the cation Radius ratio = Radius of the anion Calculation: From formula,

Radius ratio =

r+ r−

r+ 1.4 + r = 0.22 × 1.4 r + = 0.308 Å

0.22 =

∴

Example 1.4 The radius of Be2+ cation is 59 pm and that of S2− is 170 pm. Find out the coordination number and structure of BeS. Solution: Given: Radius of cation Be2+(r+)= 59 pm Radius of anion S2− (r−) = 170 pm To find: i. The coordination number of Be2+ S2− = ? ii. Structure of BeS = ? Formula: Radius of the cation Radius ratio = Radius of the anion Solid State

TARGET Publications

Formula:

Calculation: From formula,

Radius ratio =

r + rBe2 + 59 = = = 0.347 − r 170 r 2− S

Since the radius ratio lies in between 0.225 – 0.414 The coordination number of Be2+ S2− is 4 And the structure of BeS is tetrahedral. Example 1.5 If the radius of cation is 96 pm and that of anion is 618 pm. Determine the coordination number and structure of the crystal lattice. Solution: Given: Radius of cation (r+) = 96 pm Radius of anion (r−) = 618 pm To find: i. Coordination number = ? ii. Structure of the crystal lattice =? Formula: Radius of the cation Radius ratio = Radius of the anion Calculation: From formula,

r+ 96 = = 0.1553 − r 618 radius ratio lies in

Radius ratio =

Since the between 0.155 – 0.225 The coordination number of crystal is 3 And the structure of crystal lattice is Trigonal planar. Example 1.6 The radius of calcium ion is 94 pm and that of an oxide ion is 146 pm. Find the coordination number of calcium. Solution: Given: Radius of cation (r+) = 94 pm Radius of anion (r–) = 146pm To find: The coordination number of calcium = ? Solid State

Std. XII Sci.: Chemistry Numericals Radius ratio =

Radius of the cation Radius of the anion

Calculation: From formula,

94 r+ = = 0.6438 − 146 r radius ratio lies in

Radius ratio =

Since the between 0.414 – 0.732 The coordination number of calcium is 6. Example 1.7 Sodium metal crystallizes in body centered cubic lattice with cell edge = 4.29 Å. What is the radius of sodium atom? Solution: Given: Edge length of unit cell (a) = 4.29 Å Radius (r) =? To find: 3a Formula: Radius (r) = 4 Calculation: For BCC From formula, 3a 1.7320 × 4.29 Radius (r) = = = 1.86 Å 4 4 Example 1.8 Br− ion forms a close packed structure. If the radius of Br− ions is 195 pm. Calculate the radius of the cation that just fits into the tetrahedral hole. Can a cation A+ having a radius of 82 pm be slipped into the octahedral hole of the crystal A+Br-? Solution: Given: Radius of anion Br– (r– ) = 195 pm Radius of cation ( rA+ ) = 82 pm To find: i. The radius of the cation that just fits into the tetrahedral hole (r+) = ? ii. Whether the cation A+ having a radius of 82 pm can be slipped into the octahedral hole of the crystal (A+ Br–) = ? 3

Std. XII Sci.: Chemistry Numericals Formula: Radius of the cation Radius ratio = Radius of the anion Calculation: i. For, r+ Limiting value for − for tetrahedral hole r is 0.225 – 0.414 From formula, Radius of the tetrahedral hole ( rA+ ) = Radius ratio × r–

= 0.225 × 195 = 43.875 pm For cation A+ with radius = 82 pm From formula, 82 r+ Radius ratio = − = = 0.4205 195 r As it lies in the range 0.414 – 0.732, hence the cation A+ can be slipped into the octahedral hole of the crystal A+Br−.

ii.

Example 1.9 A solid AB has ZnS type structure. If the radius of cation is 50 pm, calculate the maximum possible value of the radius of anion B−. Solution: Given: Radius of cation (r+) = 50 pm To find: Radius of anion (r−) = ? Formula:

Radius ratio =

Radius of the cation Radius of the anion

Calculation: ZnS has tetrahedral arrangement. r+ The range of for stable four fold r− coordination is 0.225 to 0.414 Hence the radius of anion can be calculated by r+ taking − = 0.225 r 50 r+ = ∴ r– = 0.225 0.225 = 222.22 pm 4

TARGET Publications

Example 1.10 Determine the structure and coordination number of MgS on the basis of radius ratio in which radius of Mg2+ and S2– is 65 pm and 184 pm respectively. Solution: Given: Radius of cation Mg2+ (r+) = 65 pm Radius of anion S2− (r−) = 184 pm To find: i. The coordination number of MgS = ? ii. Structure of MgS = ? Formula: Radius of the cation Radius ratio = Radius of the anion Calculation: From formula,

Radius ratio =

r 2+ 65 r+ Mg = = = 0.3533 − 184 r r 2− S

Since the radius ratio lies in between 0.225 – 0.414 The coordination number of MgS is 4. And the structure of MgS is Tetrahedral. Type 2: Density of the unit cell Example 2.1 Al crystallizes in FCC structure. Calculate the molar mass of Al atoms, if length of the unit cell is 404 pm and density of Al is 2.7 g/cm3. Solution: Density (d) = 2.7 g/cm3 Given: Length of unit cell (a) = 404 pm = 4.04 × 10−8 cm z = 4 (FCC) To find: Atomic mass of element (M) =? Formula: i. V = a3 z×M ii. Density (d) = NA × V Solid State


TARGET Publications

Calculation: From formula (i), V = (4.04 × 10−8 cm)3 = 6.594 × 10−23 cm3 From formula (ii), N ×V×d M= A z 6.023 × 1023 × 6.594 × 10−23 × 2.7 = 4 M = 26.81 amu Example 2.2 If the radius of palladium is 248 pm and the lattice type is body centered cubic, what is the theoretical density of palladium ? Solution: Radius (r) = 248 pm Given: = 2.48 × 10−8cm z = 2 (BCC) Atomic mass of Pd = 106 To find: Density (d) = ? Formula:

i.

Atomic Radius (r) =

ii.

V = a3

iii.

Density (d) =

Calculation: For BCC From formula (i),

1.732 × a 4 −8 2.48 × 10 cm × 4 a= 1.732 = 5.727 × 10−8 cm From formula (ii), V = (5.727 × 10−8 cm)3 = 18.78 × 10−23 cm3 From formula (iii), 2 × 106 d= 6.023 × 1023 × 18.78 × 10−23 = 1.87 g/cm3 2.48 × 10−8 cm =

Solid State

z×M NA × V

3a 4

Example 2.3 Polonium exist as a simple cube. The edge of its unit cell is 334.7 pm. Calculate its density. Solution: Given: Edge length (a) = 334.7 = 3.347 × 10−8 cm Atomic mass of Po = (M) = 210 z=1 (Simple cube) Avogadro’s number = NA = 6.023 × 1023 To find: Density (d) = ? Formula: i. V = a3 z×M ii. Density (d) = NA × V Calculation: From formula (i), V = (3.347 × 10−8 cm)3 = 3.7494 × 10−23 cm3 From formula (ii), 1 × 210 ⎛ ⎞ d= ⎜ 23 −23 ⎟ ⎝ 6.023 × 10 × 3.7494 × 10 ⎠ = 9.30 g/cm3 Example 2.4 Gallium crystallizes in a simple cubic lattice. The density of gallium is 5.904 g/cm3. Determine a value for atomic radius of gallium. Solution: Given: Density (d) = 5.904 g/cm3 Atomic mass of Ga (M) = 69.7 z = 1 (Simple cube) Avogadro’s number (NA) To find:

= 6.023 × 1023 Atomic radius (r) = ?

Formula:

i.

Density (d) =

ii.

V = a3 a r= 2

iii.

z×M NA × V

5

Std. XII Sci.: Chemistry Numericals Calculation: From formula (i), 1 × 69.7 5.904 = 6.023 × 1023 × V V = 1.96 × 10−23 From formula (ii),

a = 3 1.96 × 10−23 = 2.7 × 10−8 For Simple cube structure From formula (iii), 2.7 × 10−8 r = 2 = 1.35 × 10−8 cm = 135 pm Example 2.5 You are given a small bar of an unknown metal. You find the density of the metal to be 11.5 g/cm3. An X-ray diffraction experiment measures the edge of the face-centred cubic unit cell as 4.06 × 10−10 m. Find the gram-atomic mass of this metal and tentatively identify it. Solution: Given: Density (d) = 11.5 g/cm3 z = 4 (FCC) Edge Length (a) = 4.06 × 10−10 m = 4.06 × 10−8 cm To find: Atomic mass (M) =? Formula: i. V = a3 z×M ii. Density (d) = NA × V Calculation: From formula (i), V = (4.06 × 10-8 cm)3 = 6.69234 × 10−23 cm3 From formula (ii), d × NA × V M= z 11.5 × 6.023 × 1023 × 6.69234 × 10−23 M= 4 = 115.88 amu This weight is close to that of Indium. 6

TARGET Publications

Example 2.6 The edge length of the unit cell of Ta, is 330.6 pm; the unit cell is body-centred cubic. Tantalum has a density of16.69 g/cm3 i. Calculate the mass of a tantalum atom. ii. Calculate the atomic mass of tantalum in g/mol. Solution: Given: Edge length of the unit cell (a) = 330.6 pm = 330.6 × 10−10 cm = 3.306 × 10−8 cm Density (d) = 16.69 g/cm3 z = 2 (FCC) To Find: i. Mass of a tantalum atom = ? ii. Atomic mass of tantalum in g/mol = ? Formula: i. V = a3 Mass ii. Density = Volume z×M ii. Density (d) = NA × V Calculation: From formula (i), V = (3.306 × 10−8 cm)3 = 3.6133 × 10−23 cm3 i. Mass of the 2 tantalum atoms in the bodycentered cubic unit cell From formula (ii), Mass = Density × Volume = 16.69 × 3.6133 × 10–23 = 6.0307 × 10−22 g The mass of one atom of Ta 6.0307 × 10−22 = 2 = 3.015 × 10−22 g ii. Atomic mass of tantalum in g/mol From formula (iii), N ×V×d M= A z 6.023 × 1023 × 3.6133 × 10−23 × 16.69 = 2 Atomic mass of Ta = 181.6 g/mol Solid State

TARGET Publications

Example 2.7 Nickel crystallizes in a face-centred cubic lattice. If the density of the metal is 8.908 g/cm3, what is the unit cell edge length in pm? Solution: Density (d) = 8.908 g/cm3 Given: z = 4 (FCC Lattice) Atomic mass of Ni (M) = 58.6934 NA = 6.023 × 1023 To find: Edge length of unit cell (a) = ? z×M Formula: i. Density (d) = NA × V ii. V = a3 Calculation: From formula (i), 4 × 58.6934 4 × 58.6934 V = = 6.023 × 1023 × 8.908 NA × d

= 4.376 × 10−23 cm3 From formula (ii), a = 4.376 × 10−23 = 3.524 × 10−8 cm = 352.4 pm Example 2.8 A metal crystallizes in a face-centred cubic lattice. The radius of the atom is 0.197 nm. The density of the element is 1.54 g/cm3. What is this metal? Solution: Radius of atom (r) = 0.197 nm Given: = 1.97 × 10−8 cm Density (d) = 1.54 g/cm3 z = 4 (FCC Lattice) NA = 6.023 × 1023 atoms To Find: Name of metal = ? a Formula: i. r= 2 2 ii. V = a3 z×M iii. Density (d) = NA × V Solid State

Std. XII Sci.: Chemistry Numericals Calculation: For FCC Lattice From formula (i), a 1.97 × 10−8 = 2 2 a = 5.572 × 10−8 cm From formula (ii), V = (5.572 × 10−8 cm)3 = 1.72995 × 10–22 cm3 From formula (iii), N ×V×d M= A z

=

6.023 × 1023 × 1.72995 × 10−22 × 1.54 4

M = 40.11 g/mol The metal is calcium. Example 2.9 Metallic iron crystallizes in a type of cubic unit cell. The unit cell edge length is 287 pm. The density of iron is 7.87 g/cm3. How many iron atoms are there within one unit cell? Solution: Given: Edge length of unit cell (a) = 287 pm = 287 × 10−10 cm = 2.87 × 10–8 cm Density of iron (d) = 7.87 g/cm3 NA = 6.023 × 1023 atoms mol−1 Atomic mass of iron (M) = 55.845 To find: Number of iron atoms (z) = ? Formula: i. V = a3 z×M ii. Density (d) = NA × V Calculation: From formula (i), V = (2.87 × 10−8 cm)3 = 2.364 × 10−23 cm3 From formula (ii), d × NA × V z= M 7.87 × 6.023 × 1023 × 2.364 × 10−23 = 55.845 z = 2.006 z = 2 atoms per unit cell. Hence it is Face centred cubic structure (FCC) 7

Std. XII Sci.: Chemistry Numericals Example 2.10 A metal crystallizes into two cubic system-face centred cubic (FCC) and body centred cubic (BCC) whose unit cell lengths are 3.5 and 3.0Å respectively. Calculate the ratio of densities of FCC and BCC. Solution: FCC unit cell length = 3.5Å Given: BCC unit cell length = 3.0Å z1 for FCC = 4 z2 for BCC = 2 To Find: Ratio of densities of FCC and BCC d = 1 =? d2 Formula: i.

ii.

3

V=a

Density (d) =

Calculation: FCC unit cell length = 3.5 Å BCC unit cell length = 3.0 Å From formula (i), V1 = (3.5 × 10−8)3 V2 = a3 = (3.0 × 10−8)3 From formula (ii), z ×M Density in FCC (d1) = 1 N A × V1

Density in BCC (d2) =

TARGET Publications

Problems for Practice Type 1: Radius Ratio of ionic compound/ The Formula of compound

1.

A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and that of P are at the bodycentre. What is the formula of the compound? What are the coordination numbers of P and Q?

2.

The two ions A+ and B– have radius 58 and 210 pm respectively in closed packed crystal of compound AB. Predict the coordination number of A+.

3.

The ionic radii of Rb+, Br– are 1.47 and 1.95 respectively. Predict the most probable type of geometry exhibited by RbBr on the basis of radius ratio rule.

4.

A solid has NaCl structure. If radius of the cation is 150 pm. Calculate the maximum possible value of the radius of the anion.

5.

Why is coordination number of 12 not found in ionic crystals?

6.

Gold crystallizes in a FCC lattice, the observed unit cell length is 4.070 Å. Calculate the radius of a gold atom.

7.

A compound is formed by two elements M and N. The element N forms CCP and rd 1 of tetrahedral atoms of M occupy 3 voids. What is the formula of the compound?

8.

Ferric oxide crystallizes 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.

9.

A compound forms hexagonal closepacked structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?

z×M NA × V

z2 × M N A × V2

⎛z ⎞ ⎛V ⎞ d1 = ⎜ 1 ⎟× ⎜ 2 ⎟ d2 ⎝ z 2 ⎠ ⎝ V1 ⎠ ⎛4⎞ = ⎜ ⎟× ⎝2⎠

⎛ (3.0 × 10−8 )3 ⎞ ⎜ −8 3 ⎟ ⎝ (3.5 × 10 ) ⎠

⎛ 2.7 ×10−23 ⎞ −23 ⎟ ⎝ 4.2875 ×10 ⎠

=2× ⎜

= 2 × 0.6297 = 1.259 d1 = 1.259 ∴ d2 8

Solid State


TARGET Publications

Type 2: Density of the unit cell

10.

11.

12.

13.

Thallium(I) chloride crystallizes in either a simple cubic lattice or FCC lattice of Cl− ion. The density of a given sample of solid is 9.0 g cm−3 and edge of the unit cell is 3.95 × 10−8 cm. Predict the category of unit cell. Tungsten has a BCC lattice and each lattice point is occupied by one atom. Calculate the metallic radius of the tungsten atom if density of tungsten is 19.30 g/cm3 and its atomic mass is 183.9.

Calculate the X ray density of Aluminium which forms FCC crystal lattice, if edge length of unit cell is 4.049 Å. (Atomic mass of Al = 26.98 g/mol. Avogadro’s number = 6.023 × 1023)

18.

Platinum crystallizes in FCC crystal with unit length of 3.9231 Å. Calculate the density and atomic radius of platinum. (Atomic mass of Pt = 195.08) Additional Problems for Practice

1.

Europium crystallizes in a BCC lattice. The density of europium is 5.26 g/cm3. Calculate the unit cell edge length. (Atomic mass = 152)

Metallic uranium crystallizes in a body‐centered cubic lattice, with one U atom per lattice point. How many atoms are there per unit cell? If the edge length of the unit cell is found to be 343 pm, what is the metallic radius of U in pm?

2.

Al crystallizes in FCC structure. Its metallic radius is 125 pm. What is the edge length of unit cell? How many unit cells are there in 1 cm3 of Al.

A solid is made up of two elements P and Q. Atoms Q are in FCC arrangement, while P occupy all the tetrahedral sites. What is the formula of the compound ?

3.

In FCC structure of mixed oxide, the lattice is made up of oxide ions, one eighth of tetrahedral voids are occupied by divalent ions (A2+) while one half of octahedral voids are occupied by trivalent ions (B+). What is the formula of the oxide?

4.

Niobium is found to crystallize with BCC structure and found to have density of 8.55 g/cm3. Determine the edge length of unit cell.

5.

A metallic crystal has FCC lattice structure. Its edge length is 360 pm. What is the distance of closest approach for two atoms?

6.

Gold (atomic radius = 0.144 nm) crystallizes in a face-centred unit cell. What is the length of a side of the cell?

7.

Given that a solid crystallizes in a bodycentred cubic structure that is 3.05 Å on each side. What is the volume of one unit cell in Å? 9

14.

Copper crystal has a face centred cubic structure. Atomic radius of copper atom is 128 pm. What is the density of copper metal? (Atomic mass of copper is 63.5)

15.

Krypton crystallizes with a face-centered cubic unit cell of edge 559 pm. i. What is the density of solid krypton? ii. What is the atomic radius of krypton? iii. What is the volume of one krypton atom? iv. What percentage of the unit cell is empty space if each atom is treated as a hard sphere?

16.

17.

At a certain temperature and pressure an element has a simple body-centred cubic unit cell. The corresponding density is 4.253 g/cm3 and the atomic radius is 9.492 Å. Calculate the atomic mass for this element.

Solid State


Many metals pack in cubic unit cells. The density of a metal and length of the unit cell can be used to determine the type for packing. For example, gold has a density of 19.32 g/cm3 and a unit cell side length of 4.08 Å. (1 Å = 1 × 10–8 cm.) i. How many gold atoms are in exactly 1 cm3? ii. How many unit cells are in exactly 1 cm3? iii. How many gold atoms are there per unit cell? iv. The atoms/unit cell suggests that gold packs as a (a) simple, (b) body-centered or (c) face-centered unit cell.

9.

Niobium with atomic crystallizes in body structure. If density 85.5 g/cm3. Calculate Niobium

10.

11.

12.

10

mass 92.9 amu centered cubic of Niobium is atomic radius of

If the length of body diagonal for CsCl which into a cubic structure with Cl– ions at the corners and Cs+ ions at centre of unit cell is 7Å and the radius is 1.69 Å What is the radius of Cl– ? Many metals pack in cubic unit cells. The density of a metal and length of the unit cell can be used to determine the type for packing. For example, sodium has a density of 0.968 g/cm3 and a unit cell side length (a) of 4.29 i. How many sodium atoms are in 1 cm3? ii. How many unit cells are in 1 cm3? iii. How many sodium atoms are there per unit cell? Chromium crystallizes in a body-centred cubic structure. The unit cell volume is 2.583 × 10−23 cm3. Determine the atomic radius os Cr in pm.

TARGET Publications

13.

Sodium has a density of 0.971 g/cm3 and crystallizes with a body-centred cubic unit cell. i. What is the radius of a sodium atom? ii. What is the edge length of the cell? Give answers in picometers.

14.

Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of 197 pm, calculate the density of solid calcium.

15.

Calculate the length of edge of unit cell for α-iron belonging to BCC structure. Take the density of α-iron as 7.86 × 103 kg/m3. (Atomic mass of iron = 55.85)

16.

Metallic copper crystallizes in BCC lattice. If the length of cubic unit cell is 362 pm then calculate the closest distance between two copper atoms, also calculate the density of crystalline copper.

17.

Copper has FCC structure and its atomic radius is 0.1278 nm. Calculate its density. (Atomic mass of copper = 63.5)

18.

Vanadium has the iron (monoatomic FCC) structure. If the length of unit cell edge is 305 pm, calculate the density of vanadium. (Atomic mass of V = 50.94 g/mol) Questions From Various Exams

1.

The ionic radius of an anion is 2.11 Å . Find the radius of the smallest cation that can have stable eight fold coordination with the above anions. [GATE-1987]

2.

The chloride ion has a radius of 0.181 nm. Calculate the radius of smallest cation which can be coordinated with eight neighbouring chloride ions. [GATE-1989]

3.

A solid has NaCl structure. If the radius of the cation is 100 pm, what is the radius [CBSE 1985] of the anion? Solid State

TARGET Publications


4.

Predict the closed packed structure of an ionic compound A+B– in which the radius of cation = 148 pm and radius of anion = 195 pm. What is the coordination number of cation? [CBSE-1998]

12.

A unit cell of sodium chloride has four formula unit. The edge length of the unit cell is 0.564 nm. What is the density of sodium chloride? [IIT May 1997]

5.

Predict the structure of MgO crystal and coordination number of its cation in which radii of cation and anion are equal to 65 pm and 140 pm respectively. [CBSE 1998]

13.

6.

The two ions A+ and B– have radius 88 and 200 pm respectively in closed packed crystal of compound AB. Predict the coordination number of A+ [CBSE 1990]

The unit cell of an element of atomic mass 96 and density 10.3 g cm−3 is cube with edge length 314 pm. Find the structure of the crystal lattice. (Simple cubic, FCC , BCC) (Avogadro constant = 6.023 × 1023 mol−1) [CBSE 1995]

14.

The unit cell of an element of atomic mass 108 and density 10.5 g/cm–3 is a cube with edge length 409 pm. Find the structure of the crystal lattice (Simple cubic, FCC, BCC) (Avogadro constant(NA) = 6.023 × 1023 mol−1) [CBSE 1995] An element (Atomic mass = 60) having FCC unit cell, has density of 6.23 g cm−3. What is the edge length of the unit cell?

7.

An ionic compound has unit cell consisting of A ions at the corners of a cube and B ions on the centres of face after cube. What would be the empirical formula of this compound? [AIEEE 2005]

8.

In a solid AB having the NaCl structure A atoms occupies the corners of the cubic unit cell. If all the face centered atoms along one of the axes are removed then the resultant stoichiometry of the solid is [IIT 2001] A metallic element crystallizes into lattice containing a sequence of layers of ABABAB…. Any packing of spheres leaves out voids in the lattice. Then calculate the empty space in percentage by volume in this lattice. [IIT 1996]

9.

10.

11.

A substance Ax By crystallizes in FCC lattice in which atoms A occupy each corner of the cube and atom B occupy the centers of each face of the cube. Identify the composition of AxBy [IIT 2002] Chromium metal crystallizes with BCC lattice. The length of the unit cell edge is found to be 287 pm. Calculate the atomic radius. What would be the density of chromium in g/cm3. (Atomic mass of Cr = 51.99) [IIT July 1997]

Solid State

15.

16.

The compound CuCl has ZnS structure and the edge length of the unit cell is 500 pm. Calculate the density. (Atomic mass of Cu = 63, Cl = 35.5 Avogadro constant = 6.023 × 1023 mol−1) [CBSE 1997]

17.

An element A (atomic mass 100) having BCC structure has unit cell edge 400 pm. Calculate the density of A and the number of unit cells for 10 g of A. (Avogadro Number = 6.023 × 1023 ) [CBSE 1990]

18.

An element of atomic mass 98.5 g/mol occurs in FCC structure. If its unit cell edge length is 500 pm and its density is 5.22 g/cm3. What is the value of Avogadro constant? [CBSC 1997]

19.

A face centred cubic element (atomic mass = 60) has a unit cell 400 pm. What is its density?(N = 6.023 × 1023 mol−1) [CBSE 1992] 11


21.

22.

Copper crystal has face centred cubic lattice structure. Its density is 3 8.93 g/cm .What is the length of the unit cell? (NA = 6.023 × 1023 mol−1; Atomic mass of Cu = 63.5) [CBSE 1992] A metal (At mass = 50) has a BCC crystal structure. The density of the metal is 5.96 g/cm3. Find the volume of unit cell. (NA = 6.023 × 1023 mol−1 ) [CBSE 1993] The density of chromium metal is 7.2 cm3. If unit cell is cubic with edge length of 289 pm, determine the type of unit cell (Simple/BCC/FCC) At mass of Cr = 52 amu [CBSE 1997]

23.

An element crystallizes in a structure having FCC unit cell of an edge of 200 pm. Calculate its density if 200 g of this element contains 24 × 1023 atoms [CBSE 1991]

24.

A metal has FCC crystal structure. The length of its unit cell is 404 pm. What is the molar mass of metal atoms if the density of the metal is 2.72 g/cm3 (NA = 6.023 × 1023) [CBSE 1993]

25.

26.

12

The density of CsBr which has CsCl (BCC) structure is 4.4 g/cm3. The unit cell edge length is 400 pm. Calculate the interionic distance in crystal of CsBr. (NA = 6.023 × 1023. At mass of Cs = 133, [CBSE 1993] Br = 80) Potassium fluoride has the NaCl type structure. The density of KF is 2.48 g/cm3 at 20 °C. i. Calculate the unit cell length ii. Calculate the nearest neighbour distance in KF [CBSE 1999]

TARGET Publications

Multiple Choice Questions

1.

The space occupied by b.c.c. arrangement is approximately (A) 50% (B) 68% (C) 74% (D) 56%

2.

The maximum percentage of available volume that can be filled in a face centered cubic system by an atom is (A) 74% (B) 68% (C) 34% (D) 26%

3.

In NaCl lattice, the radius ratio is

r

Na +

r

=

Cl−

(A) (C)

0.225 0.5414

(B) (D)

0.115 0.471

4.

Xenon crystallizes in face centre cubic lattice and the edge of the unit cell is 620 pm, then the radius of Xenon atom is (A) 219.20 pm (B) 438.5 pm (C) 265.5 pm (D) 536.94 pm

5.

A metallic element crystallizes in simple cubic lattice. Each edge length of the unit cell is 3 Å. The density of the element is 8 g / cc. Number of unit cells in 108 g of the metal is (A) (C)

6.

1.33 × 1020 5 × 1023

(B) (D)

2.7 × 1022 2 × 1024

The density of KBr is 2.75 gm cm−3. Length of the unit cell is 654 pm. K = 39, Br = 80. Then what is TRUE about the predicted nature of the solid. (A) Solid has face centered cubic system with z = 4. (B) Solid has simple cubic system with z = 4. (C) Solid has face centered cubic system with z = 1 (D) Solid has body centered cubic system with z = 2 Solid State

TARGET Publications

7.

8.

9.

10.

11.

12.

13.

A compound CuCl has face centered cubic structure. Its density is 3.4 g cm–3. The length of unit cell is. (At mass of Cu = 63.54 and Cl = 35.5) (A) 5.783 Å (B) 6.783 Å (C) 7.783 Å (D) 8.783 Å At room temperature, sodium crystallizes in a body centered cubic lattice with a = 4.24 Å. The theoretical density of sodium (At. mass of Na = 23) is (A) 1.002 g cm–3 (B) 2.002 g cm–3 (C) 3.002 g cm–3 (D) 4.002 g cm−3


15.

The edge length of the unit cell of NaCl crystal lattice is 552 pm. If ionic radius of sodium ion is 95 pm, what is the ionic radius of chloride ion? (A) 190 pm (B) 368 pm (C) 181 pm (D) 276 pm The radius of the Na+ is 95 pm and that of Cl– ion is 181 pm. Predict the coordination number of Na+. (A) 4 (B) 6 (C) 8 (D) Unpredictable A solid AB has rock salt structure. If the edge length is 520 pm and radius of A+ is 80 pm, the radius of anion B– would be (A) 440 pm (B) 220 pm (C) 360 pm (D) 180 pm NH4Cl crystallizes in bcc lattice with edge length of unit cell equal to 387 pm. If – radius of Cl is 181 pm, the radius of NH +4 will be (A) 174 pm (B) 154 pm (C) 116 pm (D) 206 pm What is the simplest formula of a solid whose cubic unit cell has the atom A at each corner, the atom B at each face centre and C atom at the body centre (B) A2BC (A) AB2 C (D) ABC3 (C) AB3C

Solid State

The packing efficiency of the two− dimensional square unit cell shown below is (A) 39.27 % (B)

68.02 %

(C)

74.05 %

(D)

78.54 %

L

If ‘a’ stands for the edge length of the cubic systems: simple cubic, body centered cubic and face centered cubic, then the ratio of radii of the spheres in these systems will be respectively. (A)

1 3 a: a: 2 2

(B)

1a :

(C) (D)

3a:

3 a 2 2a

1 3 1 a: a: a 2 4 2 2 1 1 a: 3a: a 2 2

16.

CsBr crystal has bcc structure. It has an edge length of 4.3 Å. The shortest interionic distance between Cs+ and Br− ions is (A) 1.86 Å (B) 3.72 Å (C) 4.3 Å (D) 7.44 Å

17.

The number of atoms in 100 g of an fcc crystal with density d = 10 g / cm3 and cell edge equal to 100 pm, is equal to (A) 4 × 1025 (B) 3 × 1025 (C) 2 × 1025 (D) 1 × 1025

18.

An element (atomic mass 100 g / mol ) having bcc structure has unit cell edge 400 pm. Then density of the element is (A) 10.376 g / cm3 (B) 5.188 g / cm3 (C) 7.289 g / cm3 (D) 2.144 g / cm3 13


Copper crystallizes in fcc with a unit cell length of 361 pm. What is the radius of copper atom ? (A) 108 pm (B) 127 pm (C) 157 pm (D) 181 pm

20.

AB crystallizes in a body centered cubic lattice with edge length ‘a’ equal to 387 pm. The distance between two oppositely charged ions in the lattice is (A) 335 pm (B) 250 pm (C) 200 pm (D) 300 pm

21.

A solid has a structure in which ‘W’ atoms are located at the corners of a cubic lattice, ‘O’ atoms at the centre of edges and ‘Na’ atoms at the centre of the cube. The formula for the compound is (B) NaWO3 (A) NaWO2 (D) NaWO4 (C) Na2WO3

Answers to Additional Problems for Practice

1. 2. 3. 4. 5. 6. 7. 8.

9. 10. 11.

12. 13. 14. 15. 16. 17. 18. 14

2 atoms, 8.9 pm P2Q AB2O 303.5 pm 255 pm 0.407 nm 28.372 Å i. 5.9058 × 1022 atoms ii. 1.47238 × 1022 unit cells iii. 4 atom/unit cell iv. FCC 1.43 × 102 pm 181 pm i. 2.54 × 1022 atoms in 1 cm3 ii. 1.27 × 1022 unit cells iii. 2 atoms per unit cell 128 × 10−10 pm i. 185.5 pm ii. 428.4 pm 1.54 g/cm3 0.124 nm 313 pm, 4.45 g/cm3 8.98 kg/m3 5.96 g/cm3

TARGET Publications

Answers to Questions from Various Exams

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.

1.545 Å r+ = 1.32 Å 241.5 pm Cubic, 8 Octahedral, 6 6 AB3 A3B4 26 % AB3 124.27 pm, 7.30 g/mL 2.16 g/cm3 Body centred cubic (BCC) lattice. Face centred cubic (FCC) lattice. 400 pm 5.22 g/cm3 5.188 g/cm3, 3.0 × 1022 unit cells 6.03 × 10+23 mol−1 6.226 g/cm3 361.5 pm 2.7857 × 10−23 cm3 Body centred cubic (BCC) lattice. 41.7 g/cm3 27 g/mol 346.4 pm i. 537.7 pm ii. 268.9 pm

Answer Key to Multiple Choice Questions

1.

(B)

2.

(A)

3.

(C)

4.

(A)

5.

(C)

6.

(A)

7.

(A)

8.

(A)

9.

(C)

10. (B)

11. (D)

12. (B)

13. (C)

14. (D)

15. (C)

16. (B)

17. (A)

18. (B)

19. (B)

20. (A)

21. (B) Solid State

TARGET Publications

Hints to Problems for Practice Problem 1: Atoms of P are present at the bodyGiven : centre Atoms of Q are present at the corners of the cube To find: Formula of the compound = ? Co-ordination numbers of P and Q=? Calculation: It is given that the atoms of Q are present at the corners of the cube. ∴ Number of atoms contributed by a corner 1 of atom Q per unit cell = atoms 8 Number of atoms contributed by 8 1 corners of atom Q per unit cell = × 8 8 = 1 atom It is also given that the atoms of P are present at the body-centre. Therefore, number of atoms of P in one unit cell = 1 atom This means that the ratio of the number of P atoms to the number of Q atoms, P:Q =1:1 Hence, the formula of the compound is PQ. The coordination number of both P and Q is 8 Problem 2: Radius of Cation A+ (r+) = 58 pm Given: Radius of anion B– (r–) = 210 pm To find: The coordination number of AB = ? Formula: Radius of thecation Radius ratio = Radius of theanion Calculation: From formula, r+ 58 r+ Radius ratio = − = A = = 0.276 r− r 210 B

Since the radius ratio lies in between 0.225 – 0.414 The coordination number of AB is 4 And the structure of AB is Tetrahedral Solid State

Std. XII Sci.: Chemistry Numericals Problem 3: Radius of Cation Rb+ (r+) = 1.47Å Given: Radius of anion Br– (r–) = 1.95 Å To find: Structure of RbBr = ? Formula: Radius of thecation Radius ratio = Radius of theanion Calculation: From formula,

Radius ratio =

r + 1.47 r+ = Rb = = 0.7538 − r − 1.95 r Br

Since the radius ratio lies in between 0.732 – 1.0 The coordination number of RbBr is 8 And the structure of RbBr is Cubic. Problem 4: Given: Radius of cation Na+ (r+) =150 pm To find: Radius of anion Cl– (r–) =? Formula: Radius of thecation Radius ratio = Radius of theanion Calculation: NaCl has octahedral structural arrangement r+ The range of for stable six fold r− coordination is 0.414 to 0.732 Hence the radius of cation can be calculated by r+ taking − = 0.414 r From formula, r+ 0.414 = − r + 150 r = = 362.32 pm r– = 0.414 0.414 Problem 5: Maximum radius ratio in ionic crystals lies in the range 0.732 – 1 which corresponds to a coordination number of 8. Hence coordination number greater than 8 is not possible in ionic crystals. 15


TARGET Publications

Problem 6: Given: Edge length of unit cell (a) = 4.070 Å To find: Radius (r) =? a Formula: r = 2 2 Calculation: Since Au crystallizes in face centred cubic (FCC) structure From formula, 4.070 r= 2 × 1.4142 = 1.44 Å

It is given that two out of every three octahedral holes are occupied by ferric ions. So, number of 2 ferric (Fe3+) ions = x 3 Therefore, ratio of the number of Fe3+ ions to the number of O2− ions, 2 Fe3+ : O2− = x : x 3 2 = :1 3 =2:3 Hence, the formula of the ferric oxide is Fe2O3.

Problem 7:

Problem 9: Compound has hexagonal closeGiven : packed structure Avogadro’s Number = NA = 6.023 × 1023 To find: Total number of voids = ? Number of tetrahedral voids = ? Calculation: Number of close-packed particles = 0.5 × NA = 0.5 × 6.023 × 1023 = 3.011 × 1023 Therefore, number of octahedral voids = 3.011 × 1023 And, number of tetrahedral voids = 2 × 3.011 × 1023 = 6.022 ×1023 Therefore, total number of voids = (3.011 × 1023) + (6.023 × 1023) = 9.034 × 1023

rd

1 of tetrahedral voids 3 To find : Formula of the compound = ? Calculation: The CCP lattice is formed by the atoms of the element N. Here, the number of tetrahedral voids generated is equal to twice the number of atoms of the element N. rd 1 The atoms of element M occupy of the 3 tetrahedral voids. 1 Therefore, the number of atoms of M = 2 × = 3 2 of the number of atoms of N. 3 Therefore, ratio of the number of atoms of M to 2 that of N is M: N = : 1 = 2:3 3 Thus, the formula of the compound is M2N3 Given:

M occupy

Problem 8: Ferric oxide has hexagonal closeGiven: packed array. Every three octahedral holes are occupied by ferric ions. To find: Formula of the ferric oxide = ? Calculation: Let the number of oxide (O2−) ions be x. So, number of octahedral voids = x 16

Problem 10: Given: Density = 9.0 g cm–3 Edge length (a) = 3.95 × 10−8 cm Atomic mass of Th (M) = 232 To find: Category of unit cell = ? Formula: i. V = a3 z×M ii. Density (d) = NA × V Calculation: From formula (i), V = (3.95 × 10−8 cm)3 = 6.163 × 10−23 cm3 Solid State


TARGET Publications

From formula (ii), d × NA × V z= M 9 × 6.023 × 1023 × 6.163 × 10−23 = 232 = 1.4 =1 z =1 atom per unit Hence it is Simple cubic structure (SC)

Calculation: From formula (i), 152 Eu = 6.023 × 1023 From formula (ii), 152 V= 6.023 × 1023 × 5.26 = 4.7978 × 10–23 cm3 From formula (iii),

Problem 11: Given: Atomic mass of Tungsten (M) = 183.9 Density (d) = 19.30 g/cm3 z = 2 (For BCC) To find: Metallic radius (a) = ? z×M Formula: i. Density (d) = NA × V

a=

ii. V = a3 Calculation: From formula (i), z×M V= NA × d

2 × 183.9 6.023 × 1023 × 19.30 = 3.1640 × 10−23 cm3 From formula (ii), =

a = 3 3.1640 × 10−23 = 3.1628 × 10−8 cm Problem 12: Given: Density of Europium (d) = 5.26 g/cm3 Atomic mass (M) = 152 To find: Edge length of unit cell (a) = ? Formula: Atomic mass i. Mass of 1 atom = Avogadro 's number

Mass Density

ii.

Volume =

iii.

Volume = a 3

Solid State

3

4.7978 × 10−23

a = 3.63 × 10−8 cm a = 363 pm Problem 13: Radius (r) = 125 pm Given: = 1.25 × 10−8 cm z = 4 (FCC) To find: Edge length of unit cell (a) = ? Number of unit cells in 1 cm3 of Al = ? a Formula: i. r= 2 2 ii. V = a3 Calculation: From formula (i), a 1.25 × 10-8 = 2 × 1.414 −8 a = 1.25 × 10 × 2 × 1.414 = 3.535 × 10−8 cm = 353.5 pm From formula (ii), V = (3.535 × 10−8)3 = 4.418 × 10–23 cm3 Number of unit cells in 1 cm3 of Al = 1 cm3/V 1 = 4.418 × 1023 = 2.266 × 1024 unit cells 17

Std. XII Sci.: Chemistry Numericals Problem 14: Given: Atomic radius of Cu atom = 128 pm = 128 × 10−10 cm z = 4 (FCC) Atomic mass of Cu = 63.5 To find: Density of Cu (d) = ? Formula: i. Face diagonal = 2 × edge length ii. Volume (V) = a3 z×M iii. Density (d) = NA × V Calculation: For FCC Lattice In face centred cubic arrangement face diagonal is four times the radius of atoms face diagonal = 4 × 128 = 512 pm = 512 × 10−10cm From formula (i), 512 Edge length (a) = 2 = 362 × 10–10 cm From formula (ii), V = (3.62 × 10−8 cm)3 = 47.4 × 10−24 cm3 From formula (iii), 4 × 63.5 d= (6.023 × 1023 × 47.4 × 10−24 ) = 8.897 g/cm3 Problem 15: Given: Edge length (a) = 559 pm = 5.59 × 10−8 cm z = 4 (FCC) Atomic mass of Krypton (M) = 83.798 To find: i. Density of solid krypton = ? ii. Atomic radius of krypton =? iii. Volume of one krypton atom =? iv. % of the unit cell which is empty space = ? 18

TARGET Publications

Formula:

3

i.

V=a

ii.

Density (d) =

iii.

r=

iv.

z×M NA × V

a

2 2 4 V = πr3 3

Calculation: From formula (i), V = (5.59 × 10−8 cm)3 = 1.7468 × 10−22 cm3 From formula (ii), 4 × 83.798 d= 6.023 × 1023 × 1.7468 × 10−22 = 3.19 g/cm3 For FCC From formula (iii), 5.59 × 10−8 cm 5.59 × 10−8 cm = r= 2 × 1.414 2 2 −8 r = 1.98 × 10 cm From formula (iv), 4 V = × 3.142 × (1.98 × 10–8)3 3 4 = × 3.142 × 7.762 × 10–24 3 9.756 × 10−23 = 3 V = 3.25 × 10−23 cm3 Volume of the 4 atoms in the unit cell: 3.25 × 10−23 cm3 × 4 = 1.292 × 10−22 cm3 Volume of cell not filled with Kr: (1.7468 × 10−22) − (1.292 × 10−22) = 4.568 × 10−23 cm3 % of empty space: 4.568 × 10−23 = 0.2615 1.7468 × 10−22 = 26.15 % Problem 16: Given: Density d = 4.253 g/cm3 Atomic radius (a) = 9.492Å = 9.492 × 10−8 cm z = 2 (BCC) To Find: Atomic mass of element (M) =? Solid State

TARGET Publications

Formula:

i. ii. iii.

3a 4 Volume of the unit cell V = a3 z×M Density (d) = N0 × V r=

Calculation: For BCC From formula (i),

3 × 9.492 ×10−8 = 4.11 × 10−8 cm r= 4 From formula (ii), V = (4.11 × 10−8 cm)3 = 6.94 × 10−23 cm3 From formula (iii), d M = N0 × V × z 6.023 × 1023 × 6.94 × 10−23 × 4.253 = 2 M = 88.89 amu Problem 17: Given: Edge length (a) = 4.049 Å = 4.049 × 10−8 cm Atomic mass of Al (M) = 26.98 g/mol z = 4 (FCC) Avogadro’s number = NA = 6.023 × 1023 To find: Density (d) = ? Formula: i. V = a3 z×M ii. Density (d) = NA × V

Std. XII Sci.: Chemistry Numericals Problem 18: Edge length (a) = 3.9231 Å Given: = 3.9231 × 10−8 cm Atomic weight of Pt (M)= 195.08 z = 4 (FCC) Avogadro’s number = NA = 6.023 × 1023 To find: Density (d) = ? Atomic radius (r) = ? Formula: i. V = a3 z×M ii. Density (d) = NA × V

iii.

Atomic Radius (r) =

a 2 4

Calculation: From formula (i), V = (3.9231 × 10−8 cm)3 = 6.038 × 10−23 cm3 From formula (ii), 4 × 195.8 d= 6.023 × 1023 × 6.038 × 10−23 = 21.53 g/cm3 From formula (iii), 3.9231 × 10−8 × 2 r= 4 = 138 .7 pm

Calculation: From formula (i), V = (4.049 × 10−8 cm)3 = 6.6381 × 10−23 cm3 From formula (ii), 4 × 26.98 d = 6.023 × 1023 × 6.6381 × 10−23 = 2.699 g/cm3 Solid State

19

Xii Chemistry Numericals

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