chemistry

EAMCET PROBLEMS WITH SOLUTIONS FROM SOLIDSTATE CHAPTER PART-1

Example 1. A compound formed by elements A and B has a cubic structure in which A atoms are at the corners of the cube and B atoms are at face centres. Derive the formula of the compound.

Solution: As ‘A’ atom are present at the 8 corners of the cube therefore no of atoms of A in the unit cell = 1/8 × 8 = 1

As B atoms present at the face centres of the cube, therefore no of atoms of B in the unit cell = 1/2 × 6 = 3

Hence the formula of compound is AB3.

Example 2. Potassium crystallizes in a body centred cubic lattice. What is the approximate number of until cells in 4.0 g of potassium? Atomic mass of potassium = 39.

Solution: In bcc unit cell there are 8 atoms at the corners of the cube and one atom at the body centre

∴ No of atoms per unit = 8 × 1/8 + 1 = 2

No of atoms is 4.0 g of potassium = 4/39 × 6.023 ×1023

∴ No of unit cells in 4.0 g potassium = 4/39 × 6.023 × 1023 / 2 = 3.09 ×1022

Example 3. A compound made up of A and B atoms have a crystalline structure, in which A forms Hexagonal close packed structure and B occupies 2/3 of octahedral holes. What will be the simplest molecular formula?

Solution: Effective number of A atoms forming HCP = 6

Effective number of octahedral holes in HCP = 6

So molecular formula will

A6B6×2/3 = A6B4 = A3B2

Example 4. An ionic compound made up of atoms A and B has a face-centred cubic arrangement in which atoms A are at the corners and atoms B are at the face-centres. If one of the atoms is missing from the corner, what is the simplest formula of the compound?

Solution: Number of atoms of A at the corners = 7 (because one A is missing)

∴ Contribution atoms of A towards unit cell = 7 × 1/8 = 7/8

Number of atoms B at face-centres = 6

∴ Contribution of atom B towards unit cell = 6 × 1/2 = 3

Ratio of A : B = 7/8 : 3 = 7 : 24

∴ Formula is A7B24

Example 5. In a face centered cubic arrangement of X and Y atoms, whose Y atoms are at the corner of the unit cell and X-atoms at the face centers. One of the X-atoms is missing from one of the faces in the unit cell. The simplest formula of the compound is

(A) X5Y2 (B) X2Y5

Solution: (A)

Example 6. A solid crystal is composed of X, Y and Z atoms. Y atoms are occupying 50% of octahedral voids, where as X atoms are occupying the 100% tetrahedral void with Z atoms in ccp array arrangement, then the rational formula of the compound in the given crystal is

(A) X8Y2Z4 (B) X5Y10Z8

Solution: (A)

Example 7. In FCC lattice of NaCl structure, if the diameter of Na+ is x, and the radius of Cl–is y, then the bond length of NaCl in the crystal is

(A) 2x + 2y (B) x + y

Solution: (C)

Example 8. In a cubic packed structure of mixed oxides, the lattice is made up of oxide ions, one fifth of tetrahedral voids are occupied by divalent (X++) ions, while one-half of the octahedral voids are occupied by trivalent ions (Y+3), then the formula of the oxide is.

(A) XY2O4 (B) X2YO4

Solution: (C)In ccp anions occupy primitives of the cube while cations occupied voids. In ccp there are two tetrahedral voids and one octahedral holes per anion. For one oxygen atom there are two tetrahedral holes and one octahedral hole.

Since one fifth of the tetrahedral voids are occupied by divalent cations (X2+)

∴ number of divalent cations in tetrahedral voids = 2 × 1/5.

Since half of the octahedral voids are occupied by trivalent cations (Y3+)

∴ number of trivalent cations = 1 × 1/2.

So the formula is the compounds is (X)2×1/5 (Y)1/2 (O)1

or X2/5 Y1/2 O1,

or X4Y5O10

Example 9. If three elements X, Y & Z crystallized in cubic solid lattice with X atoms at corners, Y atoms at cube centre & Z-atoms at the edges, then the formula of the compound is:

(A) XYZ (B) XY3Z

Solution: (C)Atom X is shared by 8 corners

1 atoms of Z is shared by 4 unit cells

Atom Y is present at centre of the unit cell

Hence, effective number of atoms of X per unit cell = 8 × 1/8 = 1

Effective number of atoms of Z per unit cell = 12/4 = 3

Effective number of atoms of Y per unit cell = 1

Hence, the formula of the compound is XYZ3

Example 10. An element crystallizes into a structure which may be described by a cubic type of unit cell having one atom in each corner of the cube and two atoms on one of its face diagonals. If the volume of this unit cell is 24 x 10-24cm3 and density of the element is 7.20gm/cm3, calculate no. of atoms present in 200gm of the element.

Solution: Number of atoms contributed in one unit cell

= one atom from the eight corners + one atom from the two face diagonals

= 1+1 = 2 atoms

Mass of one unit vell = volume × its density

= 24 × 10–24 cm3 × 7.2 gm cm3

= 172.8 × 10–24 gm

∴ 172.8 10–24 gm is the mass of one – unit cell i.e., 2 atoms

∴ 200 gm is the mass = 2 × 200 / 172.8 × 10–24 atoms = 2.3148 × 1024 atoms

Example 11. Calculate the void fraction for the structure formed by A and B atoms such that A form hexagonal closed packed structure and B occupies 2/3 of octahedral voids. Assuming that B atoms exactly fitting into octahedral voids in the HCP formed
by A

Solution: Total volume of A atom = 6 × 4 / 3 πrA3

Total volume of B atoms = 4 × 4/3 πrA3 4 × 4/3 π(0.414rA)3

Since rB/rA as B is in octahedral void of A

Volume of HCP = 24√2rA3

Packing fraction = 6 × 4/3 πrA3 + 4 × 4/3 π (0.414rA)3 / 24√2rA3 = 0.7756

Void fraction = 1-0.7756 = 0.2244

Example 12. Lithium iodide crystal has a face-centred cubic unit cell. If the edge length of the unit cell is 620 pm, determine the ionic radius of I- ion.

Solution: As LiI has fcc arrangement, I- ions will occupy the corners and face-centres. These ions will touch each other along the face diagonal.

∴ Face diagonal AB = 4r–1 = √a2 + a2 = √2a

∴ r1 = √2a / 4 = 1.414 × 620 pm / 4 = 219.17 pm

Example 13. When heated above 916°C, iron changes its crystal structure from bcc to ccp structure without any change in the radius of atom. The ratio of density of the crystal before heating and after heating is:

(A) 1.089 (B) 0.918

Solution: (B) dbcc / dccp = packing efficiency of bcc / packing efficiency of ccp = 67.92/74.02 = 0.918

Example 14. Select the correct statements:-

(A) For CsCl unit cell (edge-length = a), rc + ra = √3/2 a

For NaCl unit cell (edge-length =), rc + ra = l/2

The void space in a b.c.c. unit cell is 0.68

The void space % in a face-centered unit cell is 26%

Solution: (A), (B), (C), (D). In bcc structure are r+ + r– = √3/2 a

Hence, for CsCl, rC + ra = √3/2 a

∴ (A) is correct

Since, NaCl crystalise in fcc structure

∴ 2rC + 2ra = edge length of the unit cell

Hence, rC + ra = l/2

∴ (B) is correct

Since packing fraction of a bcc unit cell is 0.68

∴ void space = 1–0.68 = 0.32

∴ (C) is incorrect

In fcc unit cell PF = 74%

∴ VF = 100–74 = 26%

∴ (D) is correct

Example 15. In a solid, oxide ions are arranged in ccp. Cations A occupy one – sixth of the tetrahedral voids and cations B occupy one third of the octahedral voids. What is the formula of the compound?

Solution: In ccp with each oxide there would be 2 tetrahedral voids and one octahedral voids 1/3rd octahedral voids is occupied by B and 1/6th tetrahedral void by A. Therefore the compound can be

Example 16. In a crystalline solid, having formula AB2O4, oxide ions are arranged in cubic close packed lattice while cations A are present in tetrahedral voids and cations B are present in octahedral voids

(i) What percentage of the tetrahedral voids is occupied by A?

(ii) What percentage of the octahedral voids is occupied by B?

Solution: In a cubic close packed lattice of oxide ions there would be two tetrahedral voids and one octahedral void for each oxide ion.

∴ For four oxide ions there would be 8 tetrahedral and four octahedral voids two are occupied by B.

Percentage of tetrahedral voids occupied by A = 1/8 × 100 = 12.5%

Percentage of tetrahedral voids occupied by B = 2/4 × 100 = 50%

Example 17. A binary solid has zinc blend structure with ions constituting the lattice and ions occupying 25% tetrahedral voids. The formula of solid is

(A) AB (B) A2B

Solution: (C) No. of ions in unit cell = 8 × 1/8 + 1/2 × 6 = 4

Now ion occupies 25% of tetrahedral voids

∴ No. of A+ = 8 × 25 / 100 = 2

Thus ratio of A+ to B– is 1:2

Hence formula is AB2

Example 18. A compound crystallises as follows:

Ions “A“ are at corners of a cubic unit cell and “B“ ions at face centres of a cubic unit cell and “C“ ions in 1/4th of the total tetrahedral void. Assuming if this is dissolved, only the ions in the tetrahedral voids are dissociated completely in water, which one of the following statements is true. (Assuming all are univalent ions) and also A and C are cations and B is an anions.

(A) Boiling point of same concentration of solution (100% dissociation) will be greater than that of this solution.

(B) Boiling point of same concentration of (100% dissociation) will be greater than that of this solution.

(D) Data insufficient to predict.

Solution: (B) Compound is

So if C alone dissociates

AB3C2 ———> [AB3]2– + 2[C]+

Total 3 ions

Example 19. The two ions A+ and B- have radii 88 and 200 pm respectively. In the close packed crystal of compound AB, predict the coordination number of A+.

Solution: r+ / r– = 88/200 = 0.44

It lies in the range of 0.414 – 0.732

Hence, the coordination number of A+ = 6

Example 20. 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: (i) Radius of the cations just filling into the tetrahedral hole

= Radius of the tetrahedral hole = 0.225´195

= 43.875 pm

(ii) For cation A+ with radius = 82 pm

Radius ratio r+ / r– = 82/195 = 0.4205

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-.

hari

Sunday, April 20, 2014

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