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It can be seen now that both fcc and Hexagonal Primitive Structure have the same packing fraction. Moreover this is also the highest packing fraction of all the possible unit cells with one type of atom with empty voids. Can you explain this?
HCP is of the ABAB… type and FCC is of ABCABC… type. In HCP and FCC both, the layers are formed of same atoms of the same type in the same arrangement (each sphere surrounded by six) but the placement of the layers on top of each other is different.
Both will have same packing fraction.
his is the highest packing fraction possible because in both types, maximum number of atoms of the same size are present around the atom (namely six).
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The radius of a calcium ion is 94 pm and of an oxide ion is 146 pm. Predict the crystal structure of calcium oxide.
The ratio r+/r- = 94/146 = 0.644 . The prediction is an octahedral arrangement of the oxide ions around the calcium. Because the ions have equal but opposite charges, there must also been an octahedral arrangement of calcium ions around oxide ions. Thus we would expect a rock salt (NaCl) structure.
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The unit cell of silver iodide (AgI) has 4 iodine atoms in it. How many silver atoms must be there in the unit cell.
The formula, AgI, tells us that the ratio of silver atoms to iodine atoms is 1:1. Hence, if there are four iodine atoms in the unit cell, there must be four silver atoms.
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The co-ordination number of the barium ions, Ba2+, in barium chloride (BaF2) is 8. What must be the co-ordination number of the fluoride ions?
The co-ordination number of barium ions tells us that it is surrounded by eight fluoride ions (charge 8 ´ (-1) = -8). In order to balance out the eight negative charges, we need four barium ions (charge 4 ´ (+2) = +8). Hence, the co-ordination number of the fluoride ions must be 4. You will find that the co-ordination number and the charge on ions always balance out to neutrality.
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A solid between A and B has the following arrangement of atoms
Atoms A are arranged in ccp array
Atoms B occupy all the octahedral voids and half the tetrahedral voids.
What is the formula of the compound?
In a close packing, the number of octahedral voids is equal to the number of atoms and the number of tetrahedral voids is twice the number of atoms Since all the octahedral voids and half the tetrahedral voids are filled there will be one atom of B in tetrahedral void and one atom in octahedral void corresponding to each A. Thus, there will be two atoms of B corresponding to each A.
Hence, formula of the solid is AB_{2}
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In corundum, oxide ions are arranged in hcp array and the aluminium ions occupy two thirds of octahedral voids. What is the formula of corundum?
In ccp or hcp packing there is one octahedral void corresponding to each atom constituting the close packing. In corundum only 2/3^{rd} of the octahedral voids are occupied. It means corresponding to each oxide are 2/3 aluminium ions. The whole number ratio of oxide and aluminium ion in corundum is therefore 3:2 Hence formula of corundum is Al_{2}O_{3}
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Calculate the ratio of the alkali metal bromides on the basis of the data given below and predict the form of the crystal structure in each case. Ionic radii (in pm) are given below
Li^{+} = 74, Na^{+} = 102, K^{+} = 138
Rb^{+} = 148, Cs^{+} = 170, Br^{-} = 195
The ratio of cation to that of anion i.e. r^{+}/r^{–} gives the clue for crystal structure
Li^{+} / Br^{–}
74 / 195 = 0.379
Tetrahedral
Na^{+} / Br^{–}
102 / 195 = 0.523
Octahedral
K^{+} / Br^{–}
138 / 195 = 0.708
Rb^{+} / Br^{–}
148 / 195 = 0.759
Body centered
Cs^{+} / Br^{–}
170 / 195 = 0.872
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In the close packed cation in an AB type solid have a radius of 75 pm, what would be the maximum and minimum sizes of the anions filling the voids?
For close packed AB type solid
r^{+} / r^{–} = 0.414 – 0.7342
∴ Minimum value of r^{–} = r^{+} / 0.732 = 75 / 0.732 = 102.5 pm
Maximum value r^{–} = r^{+} / 0.414 = 75 / 0.414 = 181.2 pm
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NH_{4}Cl crystallizes in a body centered cubic lattice, with a unit cell distance of 387 pm. Calculate (a) the distance between the oppositely charged ions in the lattice, and (b) the radius of the NH_{4}^{+} ion if the radius of the Cl^{-} ion is 181 pm.
(a) In a body centered cubic lattice oppositely charged ions touch each other along the cross - diagonal of the cube. Hence, we can write,
2r^{+} 2r^{–} = √3a, r^{+} + r^{–} = √3a / 2 = √3 / 2 (387 pm) = 335.15 pm
(b) Now, since
r^{–} = 181 pm
wehaver^{+} = (335.15 – 181) pm = 154.15 pm.
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Copper has the fcc crystal structure. Assuming an atomic radius of 130pm for copper atom (Cu = 63.54):
(a) What is the length of unit cell of Cu?
(b) What is the volume of the unit cell?
(c) How many atoms belong to the unit cell?
(d) Find the density of Cu.
As we know
p = n × M_{m} / N_{A} × a^{3}
(a) for fcc structure
4r = √2 a
a = 2√2r =2√2 × 130 pm = 367.64 pm
(b) volume of unit cell = a^{3} = (367.64 × 10^{–10} cm)^{3 }= 4.968 × 10^{–23} cm^{3}
(c) n = 4
(d) p = 4 × 63.54 / 6.023 × 10^{23} × (3.67 × 10^{–8} cm^{3})^{3} = 8.54 gm / cm^{3}
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The density of CaO is 3.35 gm/cm^{3}. The oxide crystallises in one of the cubic systems with an edge length of 4.80 Å. How many Ca^{++} ions and O^{–2} ions belong to each unit cell, and which type of cubic system is present?
From equation
r(density) = 3.35 gm/cm^{3}
a = 4.80 Å
M_{m} of CaO = (40 + 16) gm = 56 gm CaO
Q r = where n = no. of molecules per unit cell
p = n × M_{m} / a^{3 }× N_{A}
∴ n = 3.35 × (4.8 ×10^{–8})^{3} × 6.023 × 10^{23} / 56 = 3.98
or n ≈ 4
So, 4-molecules of CaO are present in 1 unit cell
So, no. of Ca^{+ +} ion = 4
No. of O^{– –} ion = 4
So, cubic system is fcc type.
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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.
fcc unit cell length = 3.5Å
bcc unit cell length = 3.0Å
Density in fcc = n_{1} × stomic weight / V_{1} × Avogardo number
Density in bcc = n_{2} × stomic weight / V_{2} × Avogardo number
D_{fcc} / D_{bcc} = n_{1} / n_{2} × V_{2} / V_{1}
n_{1} for fcc = 4; Also V_{1} = a^{3} = (3.5 × 10^{–8})^{3}
n_{2} for bcc = 2; Also V_{2} = a^{3} = (3.0 × 10^{–8})^{3}
= 1.259
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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.
In face centred cubic arrangement face diagonal is four times the radius of atoms face diagonal = 4×128 = 512 pm
Face diagonal = √2 × edge length
Edge length = 512 / √2 = 362 × 10^{–10} cm
Volume of the unit cell = (362 × 10^{-10})^{3} cm^{3} = 47.4 × 10^{-24} cm^{3}
In a face centred cubic unit cell, there are four atoms per unit cell
Mass of unit cell = 4 × 63.5 / 6.023 × 10^{23} g = 4.22 × 10^{–22} g
Density = mass of unit cell / volume of unit cell = 4.22 × 10^{–22} / 47.4 × 10^{–24} = 8.9 g cm^{–3}
The first order reflections of a beam of X – rays of wavelength of 1.54A° from the (100) face of a crystal of the simple cubic type occurs at an angle 11.29°. Calculate the length of the unit cell.
Applying Bragg’s equation
2 d Sinθ = nλ
Given θ = 11.29°, λ = 1.54A° = 1.54×10^{-8} cm
n = 1
d = 1.54 × 10^{–8} / 2 × Sin 11.29^{o} = 1.54 × 10^{–8} / 2 × 0.1957 = 3.93 × 10^{–8} cm
dhkl = a √h^{2} + k^{2} + l^{2} = a
a = 3.93 × 10^{–8} cm (length of the unit cell)
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X- rays of wavelength equal to 0.134 nm give a first order diffraction from the surface of a crystal when the value of θ is 10.5°. Calculate the distance between the planes in the crystal parallel to the surface examined.
Given λ = 0.134 nm, θ = 10.5°
2d Sinθ = nλ
d = nλ / 2Sinθ = 1 × 0.134 / 2 × Sin 10.5^{o} = 0.134 / 2 × 0.1822 = 3.68 A
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What is the difference in the semiconductors obtained by doping silicon with Al or with P?
Silicon doped with Al produces P – type semiconductors i.e. flow is due to creation of positive holes whereas silicon doped with P produces n – type semi conductors i.e. flow is due to extra electrons carrying negative charge.
Non stoichiometric cuprous oxide, Cu_{2}O can be prepared in the laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1 can you account of the fact that this substance is a p – type semiconductor
The ratio less that 2:1 in Cu_{2}O show that some cuprous (Cu^{+}) ions have been replaced by cupric (Cu^{+2}) ions. To maintain electrical neutrality, every two Cu^{+} ions will be replaced by one Cu^{+2} ion. Thereby creating a hole. As conduction will be due to presence of these positive holes, hence it is a p – type semi conductor.
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Classify each of the following as being either a p – type or n – type semiconductor.
Ge doped with In
B doped with Si
Ge is group 14 element and In is group 13 element. Hence an electron deficient hole is created and therefore, it is p – type.
B is group 13 elements and Si is group 14 elements, there will be a free electron. Henc, it is n – type
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If NaCl is doped with 10^{-3} mole% SrCl_{2} what is the concentration of cation vacancies?
Doping of NaCl with 10^{-3} mol% SrCl_{2} means that 100 moles of NaCl are doped with 10^{-3} mol of SrCl_{2}
∴1 mole of NaCl is doped with SrCl_{2} = 10^{–3} / 100 mole = 10^{–5} mole
As each Sr^{+2} ion creates one cation vacancy, therefore concentration of cation vacancies = 10^{-5} mol / mol of NaCl =10^{-5}×6.023×10^{23} = 6.023×10^{18} mol^{–1}
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In LiI Crystal, I– ions form a cubical closest packed arrangement and Li+ ions occupy octahedral holes. What is the relationship between the edge-length of the unit cells and the radii of the I– ions? Calculate the limiting ionic radii of Li+ and I–- if a= 600pm.
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