NCERT Solution (part 1)
Question – 1.1 - Define the term 'amorphous'. Give a few examples of amorphous solids.
Answer: Solids having constituent particles with irregular shapes and short range order are called amorphous solids. Amorphous solids are isotropic in nature and melt over a range of temperature. Thus, amorphous solids are also referred as pseudo solids or super cooled liquids. Amorphous solids do not have definite heat of fusion. Amorphous solids give irregular surfaces, when cut with sharp tool. Glass, rubber, plastic, etc. are some examples of amorphous solid.
Question – 1.2 - What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
Answer: It is the arrangement of constituent particles of glass which makes it different from quartz. The constituent particles of glass have short range order while quartz has constituent particles in long range order and short range order both. By heating and cooling rapidly quartz can be converted into glass.
Question – 1.3 - Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
- Tetra phosphorous decoxide (P4O10)
- Ammonium phosphate (NH4)3PO4
|Tetra phosphorous decoxide (P4O10)||Molecular|
|Ammonium phosphate (NH4)3PO4||Ionic|
Question – 1.4 - (i) What is meant by the term 'coordination number'?
(ii) What is the coordination number of atoms:
Answer: Coordination number is the number of nearest neighbours of any constituent particle present in the crystal lattice.
(a) in a cubic close-packed structure?
Answer: In a cubic close-packed structure is 12
(b) in a body-centred cubic structure?
Answer: In a body-centered cubic structure is 8
Question – 1.5 - How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
Answer: The atomic mass of an unknown metal can be determined by knowing its density and the dimension of unit cell.
Let ‘a’ be the edge length of a unit cell of a crystal.
'd’ is the density of the metal
‘m’ is the atomic mass of the metal
‘z’ is the number of atoms in the unit cell
Now, the density of the unit cell
`=text(Mass of unit cell)/text(Volume of unit cell)`
As we know that, mass of the unit cell = Number of atoms in the unit cell X Atomic mass
And Volume of the unit cell = (Edge length of the cubic unit cell) 3
Hence, `d=(zm)a^3` …………..(i)
Now, since mass of the metal (m)
`=(text(Atomic mass) (M))/(text(Avogadro number) (N_A))`
From equation (ii) we have
Or, `M=(da^3N_A)/z` …………….(iii)
If the edge lengths are different (say a, b and c), therefore, equation (iii) can be written as
Thus, using equation (iii) and the atomic mass of an unknown metal can be determined.
Question – 1.6 - '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?
Answer: Stability of a crystal is reflected in the magnitude of its melting points because higher the melting point, greater is the intermolecular force and greater the intermolecular force greater is the stability. And hence, a substance with higher melting point would be more stable.
The melting points of the given substances are as follows:
|Substance||Melting Point (K)|
As we can see the melting point of solid water is highest and melting point of methane is lowest among the given substance. This says that intermolecular force in solid water is strongest and the intermolecular force in methane is weakest.