In the given compound Cu2O, the charge on Cu is +1. But the charge on copper in normal compounds is +2.
So Cu+2 will try to replace Cu+1 in atmospheric conditions.
This will lead to a generation of positive charge holes which are the cause of conductivity in p-type semiconductors.
So given compound can act as p-type semiconductors.

Semiconductors are the solids with have conductivities in the range from to . In the case of semiconductors, the gap between the valence band and the conduction band is small. Thus, some electrons may jump to the conduction band and may show some conductivity.
Based on the doping process we have two types of semiconductors.
(i) n-type semiconductors:- In the case of n-type semiconductor...

Given formula is Ni0.98O1.00.
Ratio of Ni to O is 0.98.
This means 98 Ni are required for 100 atoms of O.
Let us assume that Ni+2 ions are x.
So the number of Ni+3 ions will be 98 - x.
Using the charge neutrality principle, the net charge of Ni should be equal to a net charge of O.
So the equation becomes :
x(2) + (98-x)3 = 2(100)
or 2x + 294 - 3x = 200
or ...

The relation between density and edge lenght gives :
Since it crystallises in fcc lattice, thus z = 4.
Molar mass of copper = 63.546 u.
So,

From the figure, it is clear that we can use Pythagoras theorem and find the relation between R and r.
Using pythagoreas theorem :
or
or
or

Here we will use the relation between density and edge length.
It is given that Niobium crystallises in body-centred cubic structure so the value of z = 2.
So we will put the value of density, molar mass and z.
We get,
or
or ...

It is given that element Q is at corners of the cube.
So total atoms of element Q per unit cell = 1
And it is also given that element P is present at body centre.
So the total number of atoms of element P per unit cell = 1.
Therefore formula of the compound becomes: PQ
It is clear that it is a bcc lattice so coordination number of P and Q is 8.

The relation between density, molar mass and edge length is given by:-
It is given that silver crystallises in fcc lattice, so the value of z = 4. (since it is known that 4 atoms are completely efficient in fcc lattice)
...

In fcc, we know that we have a total of 4 efficient atoms present.
Also, radius and edge length of the sphere can be related by the following relation :
The above relation can be found out by equating diagonal of 1 face to 4r. (2r from the atom at face centre and r from each atom at a corner.)
Thus packing efficiency becomes:-
...

(ii) Body centred cubic:- In body centred cubic, we have atoms at all corners and at body centre.
Clearly, the atom at the centre will be in touch with the other two atoms diagonally arranged.
; and
Also the length of body diagonal is equal to 4r.
Thus,
...

(i) Simple cubic :- In a simple cubic lattice the atoms are located only on the corners of the cube.
Thus, the side of the cube ‘a’, and the radius of each particle, r are related as:
a = 2r
Volume of cubic unit cell = =
...

Similarities:- Both ionic and metallic crystals are hard in nature due to the good force of attraction between molecules. Both have fairly high melting points. In both ionic and metallic bond is non-directional.
Differences:- In ionic solids, attractive forces are coulombic or electrostatic whereas in case of metallic solids forces attractive forces are metallic bonding.
Moreover, ionic solids...

In body centred atoms are present at all 8 corners and 1 atom is present at body centre. So total number of lattice points are 9 in bcc.

In face centred tetragonal atoms are present at all 8 corners and at centre of each face (total 6 faces).
So total lattice points are 14.

In face centred cubic atoms are present at all 8 corners and at the centre of each face (6 faces in one unit cell). So total no. of lattice points are 14.

(iii) Tetrahedral void:- When a sphere (atom) of the second layer is above the void of the first layer (or vice versa) production of a tetrahedral void takes place. These voids are known as tetrahedral voids as, when the centres of these four spheres are joined a tetrahedron is formed.
(iv) Octahedral void:- The voids having a triangular in shape (or triangular voids) in the second layer are...

(i) Crystal lattice:- Crystal lattice is a three-dimensional array of points. The crystal structure is generated by using structural motifs with lattice points.Each point in a crystal lattice denotes one constituent particle which can be either an atom, a molecule (a group of atoms).
(ii) Unit cell:- It is the smallest unit of a crystal lattice which when repeated gives the crystal structure.

(i) Hexagonal close-packing:- In this, tetrahedral voids of the second layer are covered by the spheres of the third layer. In this case, the atoms (spheres) of the third layer are aligned with those of the first layer. This results in the pattern of spheres to be repeated in alternate layers. We get a pattern as ABAB ....... . This structure is known as the hexagonal close-packed (hcp)...

If a crystal has a high melting point implies that it will require high temperature or say energy to break the intermolecular bonds. Also if the intermolecular forces are strong then the crystal is more stable. Thus it can be said that higher the melting point high is the stability.
Melting points of compounds are given below:-
(i) Water:- 273 K
(ii) Ethyl alcohol:- 155.7 K
(iii) Diethyl...

We know that density is related to molar mass and edge length by the formula :
where, d=Density
a3 = The volume of the unit cell
M= Atomic mass
Z = No. of atoms in unit cell
Na= Avogadro’s number.

Exams

Articles

Questions