Solid State
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New answer posted
9 months agoContributor-Level 10
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
New answer posted
9 months agoContributor-Level 10
1.42 The ratio less than 2:1 in Cu2O shows that some cuprous (Cu+) ions have been replaced by cupric (Cu+2) ions.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 it is a p -type semi conductor
New answer posted
9 months agoContributor-Level 10
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 crystal is doped with group 15 element like P or
As, the dopant atom forms four covalent bonds like a Si or Ge atom but the fifth electron no used in
bonding, becomes delocalised and contribute its share towards electrical conduction. Thus, silicon or
germanium doped with P or As is called n-type semiconductor (negative
New answer posted
9 months agoContributor-Level 10
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 state
98−x ions will be in +3 oxidation state.
Oxide ion has −2 charge.
To maintain electrical neutrality, total positive charge on cations = total negative charge on anions.
2x+3(98−x)+100(−2)=0
x=94
Fraction of Ni2+ ions = 94/98 = 0.96
Fraction of Ni2+ ions = 98-94/98 = 0.04
Hence, the fractions of nickel that exists as Ni2+ and Ni3+ are 0.
New answer posted
9 months agoContributor-Level 10
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 state
98? x ions will be in +3 oxidation state.
Oxide ion has ?2 charge.
To maintain electrical neutrality, total positive charge on cations = total negative charge on anions.
2x+3 (98? x)+100 (?2)=0
x=94
Fraction of Ni2+ ions = 94/98 = 0.96
Fraction of Ni2+ ions = 98-94/98 = 0.04
Hence, the fractions of nickel that exists as Ni2+ and Ni3+ are 0.96 a
New question posted
9 months agoNew answer posted
9 months agoContributor-Level 10
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
New answer posted
9 months agoContributor-Level 10
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
Density of unit cell=(30.89*10−23g)(8.55 g cm−3)
=36.16*10−24cm3
Edge length of unit cell (a) = (36.13*10−24cm3)13
=3.31 * 10−8cm
Step II: Calculation of radius of unit cell (r)
For bcc structure, r=√3a4
=√3*(3.31*10−8cm)4
=1.43*10−8cm
New answer posted
9 months agoContributor-Level 10
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 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.
New answer posted
9 months agoContributor-Level 10
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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