Q.Assertion (A): Sodium chloride formed by the action of chlorine gas on sodium metal is a stable compound.
Reason (R): This is because sodium and chloride ions acquire octet in sodium chloride formation.

Concept: Ionic bonding and the octet rule.

Sodium metal ($\text{Na}$, electronic configuration $2,8,1$) loses one electron to form $\text{Na}^+$ ($2,8$), achieving a stable noble-gas configuration. Chlorine gas ($\text{Cl}_2$, configuration $2,8,7$) accepts one electron per atom to form $\text{Cl}^-$ ($2,8,8$), also completing an octet. The electrostatic attraction between $\text{Na}^+$ and $\text{Cl}^-$ produces sodium chloride, $\text{NaCl}$, which is indeed a stable ionic compound.

The reason directly explains the stability: both ions attain the octet (or duplet for inner shells), minimizing their potential energy. The formation reaction …

Why ionic stability depends on electronic configuration

When sodium metal reacts with chlorine gas, an electron transfer occurs: sodium loses one electron to become $\text{Na}^+$, and chlorine gains that electron to become $\text{Cl}^-$. The resulting compound, $\text{NaCl}$, is remarkably stable—it doesn't spontaneously decompose, it has a high melting point (801 °C), and it forms robust crystals. The question is whether the octet rule genuinely accounts for this stability.

The octet rule captures a deep truth: atoms with eight valence electrons (or two, for the first shell) possess filled outer shells, which correspond to particularly low-energy, stable configurations. Noble gases are chemically inert precisely because they already have this arrangement. When sodium and chlorine form ions, they mimic the electronic structure of the nearest noble gases.

Let's verify both the assertion and the reason step by step.

Examining the assertion and reason

1. Sodium's electron configuration before and after ionization

   Neutral sodium has the configuration $1s^2 \, 2s^2 \, 2p^6 \, 3s^1$—eleven electrons total, with one lonely electron in the third shell. When it loses that $3s^1$ electron, $\text{Na}^+$ becomes $1s^2 \, 2s^2 \, 2p^6$, which is the neon configuration: eight electrons in the outermost (now second) shell. The ion has achieved an octet.

2. Chlorine's electron configuration before and after gaining an electron

   Neutral chlorine is $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^5$—seventeen electrons, with seven in the valence (third) shell. By accepting one electron, $\text{Cl}^-$ becomes $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6$, the argon configuration: eight electrons in the third shell. This ion, too, has an octet.

3. Why the octet confers stability

   Filled shells are energetically favorable because they correspond to closed quantum states with no unpaired electrons and minimal electron–electron repulsion in the valence region. The large energy cost of removing an electron from a filled shell (high ionization energy) and the negligible tendency to accept another (electron affinity near zero for noble gases) mean these configurations resist further change. When $\text{Na}^+$ and $\text{Cl}^-$ form, each ion "locks in" this stable arrangement.

4. Electrostatic attraction and lattice stability …