Solid State
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1.43 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.
1.43 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.
1.43 There is one octahedral hole for each atom in hexagonal close packed arrangement. If the number of oxide ions (O2−) per unit cell is 1, then the number of
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1.42 Non-stoichiometric cuprous oxide, Cu2O 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?
1.42 Non-stoichiometric cuprous oxide, Cu2O 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?
1.42 The ratio less than 2:1 in Cu2O shows that some cuprous (Cu+) ions have been replaced by cupric (Cu+2) ions.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 the presence of these positive holes, hence
Upvote 1.41 What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.
1.41 What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.
1.41 These solids have conductive in the intermediate range from 10−6 to 104ohm−1m−1. As there is rise in
the temperature, conductivity also increases because electrons from the valence band jump to
conduction band.
Types of semiconductors
(a) n - type semiconductor when silicon or germanium
Upvote Analysis shows that nickel oxide has the formula Ni0.98 O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?
1.40 It is given that nickel oxide has the formula as Ni0.98 O1.00.
As per the formula, there are 98 Ni ions for 100 oxide ions.
Out of 98 Ni ions, let x ions be in +2 oxidation s
Upvote Analysis shows that nickel oxide has the formula Ni0.98 O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?
1.41 It is given that nickel oxide has the formula as Ni0.98 O1.00.
As per the formula, there are 98 Ni ions for 100 oxide ions.
Out of 98 Ni ions, let x ions be in +2 oxidation s
Upvote 1.38 If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R.
1.38 If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R.
1.38 We apply pythagoras theorem AC2= AB2+ BC2
(2R)2= (R+r)2 +(R+r)2 = 2(R+r)2
4R2 = 2(R+r)2
(2R)2= (R+r)2
√2(R)2= √(R+r)2 = √2r = R+r
r = √2 R – R
r = (√2-1) R
r = (1.4114-1)R
r= 0.414 R
Upvote 1.37 Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm^–3, calculate atomic radius of niobium using its atomic mass 93 u.
1.37 Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm^–3, calculate atomic radius of niobium using its atomic mass 93 u.
1.37 Calculation of edge length of unit cell(a)
Atomic mass of the element (M)= 93g mol−1
Number of particles in bcc type unit cell (Z) = 2
Mass of the unit cell = Z × MNA = 2 × (93 g mol−1) (6.022×1023mol−1)
=30.89×10−23g
Density of unit cell (d) =8.55 g cm−3
Volume of unit cell (a3)=Mass of unit cell
Den
Upvote 1.36 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?
1.36 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?
1.36 It is given that the atoms of Q are present at the corners of the cube.
Therefore, number of atoms of Q in one unit cell = 8 x 1/8 = 1
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
This means that the ratio of the number of
Upvote 1.35 Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10^–8 cm and density is 10.5 g cm^-3, calculate the atomic mass of silver.
1.35 Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10^–8 cm and density is 10.5 g cm^-3, calculate the atomic mass of silver.
1.35 P= density
A= edge length of the cell
NA= Avogadro Number
Z= no. of atoms in F.C.C unit cell
M= mass of the metal
Edge length of the cell = d = 4.07*10-8 cm
Density = P =10.5g/cm3
No. of unit cell of face centered cubic (F.C.C) lattice is 4, Z=4
Avogadro Number (NA) = 6.022*1023
Mass of silver = M=
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