An element a crystallises in fcc structure. 200 g of this element has 4.12 x 1024atoms. The density of A is 7.2 g cm-3. Calculate the edge length of the unit cell?

solution:
we have given that
mass = 200g
density = 7.2 g cm-3
NA = 4.12 x 1024
Z = 4 
by using the following equation,
 
ρ = z× Ma3× NA
                     

               a3 = Z×Md×N
                 

         = 4×2007.2×4.12×1024= 80029.664×10-24


            a3 = 26.968×10-24a = 2.997×10-8a = 299.7 pm.
 
hence the edge length is 299.7pm
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How many lattice points are there in one unit cell of each of the following lattice?
(a) face centred cubic
(b) face centred tetragonal
(c) body centred cubic


A face-centred cubic (fcc) unit cell contains atoms at all the corners and at the centre of all the faces of the cube.each atom located at the face-centre is shared between two adjacent unit cells and only 1/2 of each atom belongs to a unit cell

Thus, in a face-centred cubic
(fcc) unit cell:
(i) 8 corners atoms × 1/8 atom per unit cell = 8×1/8 = 1 atom
(ii) 6 face-centred atoms ×1/2 atom per unit cell = 6 ×1/2 = 3 atoms
∴ Total number of atoms per unit cell = 4 atoms

(a) z = 4,
(b) z = 4,

body centre wholly belongs to the unit cell in which it is present. Thus
in a body-centered cubic (bcc) unit cell:
(i) 8 corners × 1/8 per corner atom= 8×1/8 = 1 atom
(ii) 1 body centre atom = 1 × 1 = 1 atom
∴ Total number of atoms per unit cell =1+1 = 2

(c) z = 2.

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What are paramagnetic and ferromagnetic substances. Account for the paramagnetic character of transition metal compounds. How does the paramagnetic character of the bivalent ions of the first transition metal series varies from Ti (Z = 22) to Cu (Z = 29)?

Paramagnetic substances are attracted by the magnetic field. This is due to the presence of unpaired electrons in the atoms/ions/molecules of the paramagnetic substance. Paramagnetism is a temporary effect.

Ferromagnetic substances are strongly attracted by the magnetic field. They show magnetism even when the magnetic field is removed. This is due to a spontaneous alignment of the magnetic dipole in the same direction.

Transition metal compounds contains one or more unpaired electrons. So, these compounds show para magnetism. The paramagnetism of the bivalent ions of the first transition metal series first increases in going from Ti2+ to Mn2+ (number of unpaired electrons increases) and then decreases upto Cu2+ (because the number of unpaired electrons decreases).
Transition metal compounds contains one or more unpaired electrons. So, these compounds show para magnetism. The paramagnetism of the bivalent ions of the first transition metal series first increases in going from Ti2+ to Mn2+ (number of unpaired electrons increases) and then decreases upto Cu2+ (because the number of unpaired electrons decreases).

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The ions of NaF and MgO all have the same number of electrons, and the internuclear distance are about the same (235 pm and 215 pm). Why then are the melting points of NaF and MgO so different. (992°C and 2642°C)?


The crystals of NaF and MgO are formed by Na+ and F- in NaF and Mg2+ and O2–ions in MgO respectively arranged in cubic closed structures. There are strong electrovalent bond forces (strong coulombic forces, attraction between ions) between Na+and F in NaF and Mg2+ and O2– in MgO.

But the magnitude of these coulombic forces of attraction is much higher in MgO as compared to that in NaF. (Mg
2+ is divalent while Na+is monovalent, similarly O2– and F).

The electrostatic forces of attraction between Mg
2+and O2– is almost 4 times as compared to Na+ and F ions. Melting point of ionic solids is almost the index of inter-ionic attraction in crystal lattice as lot of energy is required to break these forces/overcome these forces before the substance melts.

It changes into paramagnetic at hight temperature due to randomization of spins.

            (b)        The ions in MgO carry two unit charges. In NaCl only one unit charge. Hence electrostatic forces of attraction in MgO are stronger.

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An element occurs in BCC structure with cell edge of 300 pm. The density of the element is 5.2 g cm–3. How many atoms of the element does 200 g of the element contain?

z = 2, a = 300 pm, d = 5.2 gm cm–3
           
z = 2, a = 300 pm, d = 5.2 gm cm–3           
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