Explain the formation of Ionic bond in NaCl.

Explanation

Sodium has elecrtronic configuration `, 1s^(2), 2s^(2) 2p^(6) 3s^(1)` when it loosed 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 ionic bond.
`(1) underset((1s^(2)2s^(2)2p^(6)3s^(1)))(Na) to underset((1s^(2)2s^(2)2p^(6)))(Na^(+)+e^(-))`
(2) `: overset( * *)underset(* *)Cl* +e^(-) to : overset( * *)underset(* *)Cl:^(-)`
`Na^(@)+: overset( * *)underset(* *)Cl* to Na^(+)[: overset( * *)underset(* *)Cl:^(-)]`