Section E: Answer the case-based question given below.
37.a The melting points and boiling points of some ionic compounds are given below:
| Compound | Melting Point (K) | Boiling Point (K) |
| \(NaCl\) | 1074 | 1686 |
| \(LiCl\) | 887 | 1600 |
| \(CaCl_2\) | 1045 | 1900 |
| \(CaO\) | 2850 | 3120 |
| \(MgCl_2\) | 981 | 1685 |
iii. A. While forming an ionic compound say sodium chloride how does sodium atom attain its stable configuration?
iii. (A) Stable configuration of sodium atom in sodium chloride:
When forming sodium chloride (NaCl), the sodium atom (Na) attains its stable configuration by losing electron to achieve the electron configuration of neon:
Sodium (Na) has the electronic configuration: \(1s^2\), \(2s^2\), \(2p^6\), \(3s^1\)
By losing one electron, sodium becomes \(Na^+\) with the configuration: \(1s^2\), \(2s^2\), \(2p^6\) (same as neon).
Chlorine (Cl) gains this electron to complete its octet:
Chlorine (Cl) has the electronic configuration: \(1s^2\), \(2s^2\), \(2p^6\), \(3s^2\), \(3p^5\).
By gaining electron, chlorine becomes \(Cl^−\) with the configuration: \(1s^2\), \(2s^2\), \(2p^6\), \(3s^2\), \(3p^6\) (same as argon).
The reaction:
\(Na\) → \(Na^+\) + \(e^−\)
\(Cl\) + \(e^−\) → \(Cl^−\)
\(Na\) + \(Cl\) → \(NaCl\)