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Explain the formation of Ionic bond in NaCl. |
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Answer» Solution :Sodium has elecrtronic configuration `, 1s^(2), 2s^(2) 2p^(6) 3s^(1)` whenit 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:^(-)]` |
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