chemistry

EAMCET PROBLEMS WITH SOLUTIONS FROM SOLIDSTATE CHAPTER PART-2

Example 21. Why is co-ordination number of 12 not found in ionic crystals?

Solution: 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.

Example 22. Iron changes its crystal structure from body-centred to cubic close-packed structure when heated to 916oC. Calculate the ratio of the density of the bcc crystal to that of ccp crystal, assuming that the metallic radius of the atom does not change.

Solution: n the bcc packing, the space occupies is 68% of the total volume available while in ccp, the space occupied is 74%. This means that for the same volume masses of bcc and ccp are in the ratio of 68 : 74. As the volume is same, ratio of density is also same viz 68 : 74

i.e. d(bcc) / d(ccp) = 68/74 = 0.919

Example 23. In BeO (Zinc Blende structure), is introduced in available tetrahedral voids. Then ions are removed from a single body diagonal of the unit cell. What will be the molecular formula of the unit cell?

Solution: In BeO (Zinc Blende structure), alternative ‘Td’ voids of FCC of ions are occupied by So can be introduced in four available alternative ‘Td’ voids and unit cell formula will be. If now ions are removed from a single body diagonal then unit cell formula will be.

Example 24. In a crystal oxide ions are arranged in fcc and A+2 ions are at 1/8th of the tetrahedral voids, and ions B+3 occupied ½ of the octahedral voids. Calculate the packing fraction of the crystal if O-2 of the removed from alternate corner and A+2 is being place at 2 of the corners.

Solution: Since oxide ions are fcc so 4O-2 ∴ unit cell

A+2 are at 1/8th of the tetrahedral so 1A+2|unitcell

B+3 occupies ½ of the octahedral voids ∴ 2B+2/unit cell

After removing O-2 ions

Oxide ion / unit cell = 4/8 + 3 = 3.5

A+2 ions/ unit cell = 2/8 + 1 = 10/8 = 1.25

B+3 ions/unit cell = 2

∴ P.F = 3.5 × 4/3 πr3 + 1.25 × 43πr3A+2 + 2 × 4/3 πr3B+3

We know that a = 4r / √2

rA+2 = 0.225, ∴rA, = 0.225r–

rB+3 / r– =0.414 ∴ rB+3 = 0.414r-

Putting all the values

P.F = 0.676

Example 25. A binary solid has a rock salt structure. If the edge length is 400 pm, and radius of cation is 75 pm the radius of anion is

(A) 100 mm (B) 125 pm

Solution: (B) Edge = 2(r+ + r–)

=> 400 – 2(75 + r–)

∴r– = 125 pm

Example 26. In closest packing of atoms

(A) The size of tetrahedral void is greater than that of the octahedral void.

The size of the tetrahedral void is smaller than that of the octahedral void.

The size of tetrahedral void is equal to that of the octahedral void.

(D) The size of tetrahedral void may be larger or smaller or equal to that of the octahedral void depending upon the size of atoms.

Solution: (B)For tetrahedral voids

r+ / r– = 0.225, r+ = 0.225 r– …(i)

Similarly for octahedral voids

r+ = 0.414 r– …(ii)

From equation (i) and (ii) it is clear that size of octahedral void is larger than that of tetrahedral voids.

Example 27. An element having atomic mass 60 has face centred cubic unit cell. The edge length of the unit cell is 400 pm. Find out the density of the element?

Solution: Unit cell edge length = 400 pm

= 400 × 10–10 cm

Volume of unit cell = (400 × 10-10)3 = 64 × 10-24 cm3

Mass of the unit cell = No. of atoms in the unit cell × mass of each atom

No. of atoms in fcc unit cell = 8 × 1/8 + 6 × 1/2 = 4

∴ Mass of unit cell = 4 × 60 / 6.023 × 1023

Density of unit cell = mass of unit cell / volume of unit cell = 4 × 60 / 6.023 × 1023× 64 × 10–24 = 6.2 g/cm3

Example 28. An element has a body centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2 g/cm3. How many atoms are present in 208 g of the element?

Solution: Volume of unit cell = (288×10–10)3 cm3 = 2.39 × 10-23cm3

Volume of 208 g of the element = mass / volume = 208 / 7.2 = 28.88cm3

No of unit cells in this volume = 28.88 / 2.39 × 10–23 = 12.08 × 1023

Since each bcc unit cell contains 2 atoms

∴ no of atom in 208 g = 2 × 12.08 × 1023 = 24.16 × 1023 atom

Example 29. A compound formed by elements X & Y, Crystallizes in the cubic structure, where X is at the corners of the cube and Y is at the six face centers. What is the formula of the compound? If side length is 5A°, estimate the density of the solid assuming atomic weight of X and Y as 60 and 90 respectively.

Solution: From eight corner atoms one atoms (X) contributes to one unit cell.

From six face centres, three atoms (Y) contributes to one unit cell.

So, the formula of the compound is XY3.

As we know that,

p = n × Mm / NA × a3 =, here n = 1

Molar mass of XY3

Mm = 60 + 3 × 90 = 330 gm

p = 1×330 / 6.023 ×1023 × (5×10–8)3 gm/cm3

a = 5Å = 5 × 10–8 cm

= 330 / 6.023 ×1023 × 125 × 10–24 gm/cm3 = 4.38 gm / cm3

Example 30. Lithium borohydrides, LiBH4, crystallizes in an orthorhombic system with
4 molecules per unit cell. The unit cell dimensions are: a = 6.81 Å, b = 4.43 Å and c = 7.17 Å. Calculate the density of the crystal. Take atomic mass of Li = 7,
B = 11 and H = 1 a.m.u.

Solution: Molar mass of LiBH4 = 7 + 11 + 4 = 22 g mol-1

Mass of the unit cell = 4 × 22 gmol–1 / 6.02 × 1023 mol–1 = 14.62 × 10–23 g

Volume of the unit cell = a × b × c

= (6.81 × 10-8 cm) (4.43 × 10-8 cm) (7.17 × 10-8 cm)

= 21.63 × 10-23 cm3

∴ Density of the unit cell = Mass / Volume = 14.62 × 10–23 g / 21.63 × 10–23cm3 = 0.676 gcm–3

Example 31. A binary solid (A+B–) has a rocksalt structure. If the edge length is 400 pm and the radius of cation is 75 pm, the radius of anion is

(A) 100 pm (B) 125 pm

Solution: (B)

Example 32. The vacant space in bcc lattice unit cell is about

(A) 32% (B) 10%

Solution: (A)

Example 33. A substance has density of 2 kg dm-3 & it crystallizes to fcc lattice with
edge-length equal to 700pm, then the molar mass of the substance is

(A) 74.50gm mol-1 (B) 103.30gm mol-1

Solution: (B)p = n × Mm / NA × a3

2 = 4 × Mm / 6.023 × 1023 × (7 × 10–8)3

(since, effective number of atoms in unit cell = 4)

On solving we get Mm = 103.03 gm / mol

Example 34. X – Rays of wavelength 1.54A° strike a crystal and are observed to be deflected at an angel 22.5°. Assuming that n = 1, calculate the spacing between the planes of atom that are responsible for this reflection.

Solution: Applying Bragg’s equation

nγ = 2d Sinθ

Giving n = 1, γ = 1.54A°, θ = 22.5°

Using relation nγ = 2 d Sinθ

d = 1.54 / 2 Sin 22.5o = 1.54 / 2 × 0.383 = 2.01 Ao

Example 35. Calculate the angle at which second order reflection will occur in an X- ray spectrometer. When X – ray of wavelength 1.54 A° are diffracted by atoms of a crystal. The interplanar distance is 4.04A°.

Solution: According to Bragg’s equation

nγ = 2 d Sinθ

n = 2

γ = 1.54A°; d = 4.04 A°

γ = Sin-1 (2 × 1.54 / 2 × 4.04) => Sin–1 (1.54 / 4.04)

= Sin-1 (0.381)

= 22° 24'

hari

Sunday, April 20, 2014

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