Chemistry books

 Q1. Define the term ‘amorphous’. Give a few examples of amorphous solids.

Ans:

Amorphous solids are solids without a regular/definitive arrangement of its constituent particles (ions, atoms or molecules) and they possess something called the short-range order i.e., a regular and periodically repeating arrangement is seen only over short distances, e.g., rubber, glass.

Q2. What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?

Ans:

The arrangement of the constituent particles differentiates glass from quartz. The constituent particles in glass have a short-range order, but the constituent particles of quartz possess long range orders.
Quartz is converted into glass by heating it and then rapidly cooling it.

Q3. Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
(a) Tetra phosphorus decaoxide (P4O10)
(b) Ammonium phosphate (NH4)3PO4
(c) SiC
(d) I2
(e) P4
(f) Plastic
(g) Graphite
(h) Brass
(i) Rb
(j) LiBr
(k) Si

Ans:

Metallic : ( h ) Brass, ( i ) Rb

Molecular : (a) Tetra phosphorus decaoxide (P4O10), (d) I2, (e) P4.

Ionic : ( b ) Ammonium phosphate (NH4)3PO4, ( j ) LiBr

Amorphous : ( f ) Plastic

Covalent : ( c ) SiC, ( g ) Graphite, ( k ) Si

Q4. (i) What is meant by the term ‘coordination number’?
(ii) What is the coordination number of atoms:
(a) in a cubic close-packed structure?
(b) in a body-centred cubic structure?

Ans:

(i) Coordination number is the number of nearest neighbours of a particle.
(ii) 

(a) coordination number =12

(b) coordination number = 8

Q5. How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.

Ans:

Given,
We know the dimension and density of its unit cell.
Let,
The edge length of a unit cell = a
The volume of the cell = a3
Density = d
Atomic mass = M
Mass of unit cell = No. of atoms in unit cell x Mass of each atom = Z × m
Mass of an atom present in the unit cell, m = M/ Na
where Na is the Avogadro’s number.
We know,
d = Mass of unit cell / Volume of unit cell = Zm/a3 = Z.M / a3Na
Therefore, Atomic mass, M =( da3 Na) / Z

Q6. ‘Stability of a crystal is reflected in the magnitude of its melting points’. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?

Ans:

A substance with a high melting point is more stable than a substance with a low melting point, this is because higher the melting point, stronger the intermolecular forces of attraction in that substance/ crystal. Thus, implying greater stability.
The melting points of the above substances are:
Methane → 89.34 K
Diethyl ether → 156.85 K
Solid water → 273 K
Ethyl alcohol → 158.8 K
Observing the data we can say that solid water has the strongest intermolecular force and methane has the weakest.

Q7. How will you distinguish between the following pairs of terms:
(i) Hexagonal close-packing and cubic close-packing?
(ii) Crystal lattice and unit cell?
(iii) Tetrahedral void and octahedral void?

Ans:

(a) Cubic close packing: When a third layer is placed over the second layer in a manner that the octahedral voids are covered by the spheres, a layer different from the first (A) and second (B) is obtained. If we continue packing in this manner we get the cubic close packing.
Hexagonal close packing: When the third layer is placed over the second layer in a way that the tetrahedral voids are covered by the spheres, a 3D close packing is produced where spheres in each third or alternate layers are vertically aligned. If we continue packing in this order we get hexagonal close packing.
(b) Unit cell: It is the smallest 3D dimensional portion of a complete space lattice, which when repeated over and over again in different directions from the crystal lattice.
Crystal lattice – it is a regular orientation of particles of a crystal in a 3D space.

(c) Octahedral void – it is a void surrounded by 6 spheres.

Tetrahedral void – it is a void surrounded by 4 spheres.

Q8. How many lattice points are there in one unit cell of each of the following lattice?
(i) Face-centred cubic
(ii) Face-centred tetragonal
(iii) Body-centred

Ans:

(i) There are 14 lattice points in face-centred cubic, 8 from the corners and 6 from the faces.

(ii) There are 14 lattice points in face-centred tetragonal, 8 from the corners and 6 from the faces.

(iii) There are 9 lattice points in body-centred cubic, 1 from the centre and 8 from the corners.

Simple cubic:
In a simple cubic lattice, particles are present only at the corners and they touch each other along the edge.

Let the edge length = a

Radius of each particle = r.
Thus, a = 2r
Volume of spheres = πr3(4/3)
Volume of a cubic unit cell = a 3 = (2r) 3 = 8r 3
We know that the number of particles per unit cell is 1.
Therefore,
Packing efficiency = Volume of one particle/Volume of cubic unit cell = [ πr3(4/3) ] / 8r 3 = 0.524 or 52.4 %

Body-centred cubic:

From ∆FED, we have:
b2 = 2a2
b = ( 2a )1/2
Again, from ∆AFD, we have :
c2 = a2 + b2
=> c2 = a2 + 2a2
c2 = 3a2
=> c = (3a)1/2 
Let the radius of the atom = r.
Length of the body diagonal, c = 4r
=> (3a)1/2 = 4r
=> a = 4r/ (3)1/2
or , r = [ a (3)1/2 ]/ 4
Volume of the cube, a3 = [4r/ (3)1/2 ] 3
A BCC lattice has 2 atoms.
So, volume of the occupied cubic lattice = 2πr3(4/3) = πr3( 8/3)
Therefore, packing efficiency = [ πr3( 8/3) ]/ [ { 4r/(3)1/2 }3 ] = 0.68 or 68%

Face-centred cubic:

Let the edge length of the unit cell = a
let the radius of each sphere = r
Thus, AC = 4r
From the right angled triangle ABC, we have :
AC = ( a2 + a2 )1/2 = a(2)1/2

Therefore, 4r = a(2)1/2
=> a = 4r/( 2)1/2
Thus, Volume of unit cell =a3 = { 4r/( 2)1/2 } 3
a3 = 64r3/2(2)1/2 = 32r3/ (2)1/2
No. of unit cell in FCC = 4
Volume of four spheres = 4 × πr3(4/3)

Thus, packing efficiency = [πr3(16/3) ] / [32r3/ (2)1/2 ] = 0.74 or 74 %

Q9. Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10–8cm and density is 10.5 g cm–3, calculate the atomic mass of silver.

Ans:

Given:
Edge length, a = 4.077 × 10−8 cm 
Density, d = 10.5 g cm−3
The given lattice is of fcc type,
Thus the number of atoms per unit cell, z = 4
We also know that NA = 6.022 × 1023 / mol
let M be the atomic mass of silver.
We know, d = zM/a3NA
=> M = da3Na / z

= (10.5 x 4.077 × 10−8 x 6.022 × 1023 ) / 4 = 107.13 g /mol

Q10. 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?

Ans:

Given:

Atoms of Q occupy the corners of the cube.
=> Number of Q atoms in a unit cell = 8 x 1/8 = 1
The atom of P occupies the body centre.
=> Number of Q atoms in a unit cell = 1
Therefore, the ratio of the number of P atoms to the number of Q atoms;
X : Y = 1 : 1
Thus, the formula of the compound is PQ
And the coordination number of both the elements is 8.

Q11. Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?

Ans:

The given formula of nickel oxide is Ni0.98 O1.00.
Thus, the ratio of the number of Ni atoms to the number of O atoms is,
Ni : O = 0.98 : 1.00 = 98 : 100
Now,
Total charge on 100 O2− ions = 100 × (−2) = −200
Let the number of Ni2+ ions = x.
So, the number of Ni3+ ions is 98 − x.
Now,
Total charge on Ni2+ ions = x(+2) = +2x
Similarly, total charge on Ni3+ ions = (98 − x)(+3) = 294 − 3x
As, the compound is neutral, we can write:
2x + (294 − 3x) + (−200) = 0
⇒ −x + 94 = 0
⇒ x = 94
Therefore, number of Ni2+ ions = 94
And, number of Ni3+ ions = 98 − 94 = 4
Thus, the fraction of nickel that exists as Ni2+ = 94/98 = 0.0959
And, the fraction of nickel that exists as Ni3+ = 4/98 = 0.041

Q12. Explain the following with suitable examples:
(a) Ferromagnetism
(b) Paramagnetism
(c) Ferrimagnetism
(d) Antiferromagnetism
(e) 12-16 and 13-15 group compounds.

Ans:

(a) Ferromagnetic:
These substances (ferromagnetic substances) are strongly attracted by magnetic fields. They could be permanently magnetized even when a magnetic field is absent. Few examples of ferromagnetic substances include cobalt, iron, nickel, CrO2 and gadolinium.
In a solid state, their metal ions come together to form small regions termed domains and each domain behaves like a tiny magnet.
In a magnetized piece of a ferromagnetic substance, these domains are randomly arranged thus, their net magnetic moment becomes zero. However, when it is kept in a magnetic field, the domains orient themselves in the direction of the field. This results in a powerful magnetic effect being produced. This orientation of domains persists even after the field is removed. Hence, the ferromagnetic substance is transformed into a permanent magnet.
Schematic alignment of magnetic moments in ferromagnetic substances: