CBSE Class 12 Chemistry Chapter 2: Solutions NCERT Solutions

Amorphous solids are characterized by the absence of a long-range, ordered arrangement of their constituent particles. These particles are arranged in a disordered manner, leading to a short-range order only. Consequently, amorphous solids lack a definite geometric shape and exhibit isotropic properties, meaning their physical properties are the same in all directions. They also do not have a sharp melting point and melt over a range of temperatures. Due to these properties, they are sometimes referred to as pseudo solids or super cooled liquids. They do not possess a definite heat of fusion. When cut with a sharp tool, they produce irregular surfaces.

Examples of amorphous solids include:

• Glass
• Rubber
• Plastic

The fundamental difference between glass and quartz lies in the arrangement of their constituent particles. Quartz is a crystalline solid, meaning its atoms or molecules are arranged in a highly ordered, repeating three-dimensional pattern, exhibiting both long-range and short-range order. In contrast, glass is an amorphous solid; its particles are arranged in a disordered manner, possessing only short-range order. This structural difference impacts their physical properties.

Quartz can be transformed into glass through a process of heating and rapid cooling. When quartz is heated to a sufficiently high temperature and then cooled quickly, the ordered crystalline structure is disrupted, and the particles get locked into a disordered arrangement characteristic of glass.

The classification of the given solids is as follows:

• Ionic solids: These are formed by electrostatic attraction between cations and anions. Examples are (ii) Ammonium phosphate ((NH₄)₃PO₄) and (x) LiBr.
• Metallic solids: These consist of metal atoms held together by metallic bonds, forming a lattice of positive ions surrounded by a sea of mobile electrons. Examples are (viii) Brass (an alloy) and (ix) Rb (Rubidium).
• Molecular solids: These are composed of discrete molecules held together by weaker intermolecular forces (van der Waals forces or dipole-dipole interactions). Examples are (i) Tetra phosphorus decoxide (P₄O₁₀), (iv) I₂ (Iodine), and (v) P₄ (Phosphorus).
• Covalent (network) solids: In these solids, atoms are linked by a network of covalent bonds, forming a large molecule or crystal. Examples are (iii) SiC (Silicon carbide), (vii) Graphite, and (xi) Si (Silicon).
• Amorphous solids: These solids lack a regular, ordered internal structure. An example is (vi) Plastic.

1. The coordination number of an atom in a crystal lattice is defined as the number of nearest neighbouring atoms or particles that are in direct contact with it. It essentially indicates how closely packed the atoms are around a central atom.
2. The coordination numbers for the specified structures are:
3. a) In a cubic close-packed (CCP) structure, each atom is in contact with 12 nearest neighbours (6 in its own plane, 3 above, and 3 below). Therefore, the coordination number is 12.
4. b) In a body-centred cubic (BCC) structure, each atom is surrounded by 8 nearest neighbours. The central atom touches the 8 corner atoms, and each corner atom touches the central atom. Therefore, the coordination number is 8.

The atomic mass of an unknown metal can be determined if its density and the dimensions of its unit cell are known. This is achieved by relating these macroscopic properties to the microscopic structure of the crystal lattice using the following formula derived from the definition of density:

Density (d) = $$\backslash frac\{\backslash text\{Mass\; of\; the\; unit\; cell\}\}\{\backslash text\{Volume\; of\; the\; unit\; cell\}\}$$

The mass of the unit cell can be expressed as the product of the number of atoms in the unit cell ($z$) and the mass of a single atom ($m$). The volume of a cubic unit cell is $a^3$, where $a$ is the edge length of the unit cell.

Thus, the formula becomes:

$$d\; =\; \backslash frac\{z\; \backslash times\; m\}\{a^3\}$$

To find the atomic mass ($m$), we can rearrange this equation:

$$m\; =\; \backslash frac\{d\; \backslash times\; a^3\}\{z\}$$

By substituting the known values of density ($d$), edge length ($a$), and the number of atoms per unit cell ($z$) for the specific crystal structure (e.g., $z=4$ for FCC, $z=2$ for BCC, $z=1$ for simple cubic), the atomic mass ($m$) of the unknown metal can be calculated.