Question

Explain the formation of Ionic bond in NaCl.

Answer 1

Sodium has electronic configuration $1s^{2}2s^{2}2p^{6}3s^1$, when it looses one electron, attains stability where as chlorine $1s^{2}2s^{2}2p^{6}3s^{2}3p^5$, it gain one electron to get stability.

The electron lost by Na is gained by the chlorine forming an ionic bond.

(1) $Na\to$ $N{a}^{+}+e^{-}$
(2) $Cl+e^{-}$ $\to$ $C{l}^{-}$
(3) $Na+Cl$ $\to$ $N{a}^{+}\left[C{l}^{-}\right]$