Question

_____.Na (atomic no. 11) reacts with cl (atomic no. 17) to become stable. In the reaction, na will ____________, while cl will ____________.

Answer 1

Answer:

             Na (atomic no. 11) reacts with cl (atomic no. 17) to become stable. In the reaction, Na will give up one electron, while cl will accept on electron.

Explanation:

                   NaCl is formed by Na⁺ cation and Cl⁻ anion as follow,

Oxidation of Na;

                                     Na  →  Na⁺  +  1 e⁻

In this step sodium looses one electron due to less ionization energy as the resulting Na⁺ ion has attained a stable noble gas configuration of Neon (i.e. 1s², 2s², 2p⁶).

Reduction of Cl₂;

                                   Cl₂  +  2 e⁻  →  2 Cl⁻

In this step each Chlorine atom accepts one electron and forms Chloride ion (Cl⁻). Again the driving force for this step is attaining the noble gas configuration of Argon (i.e. 1s², 2s², 2p⁶, 3s², 3p⁶).

Crystal Lattice formation is as follow,

                                 Na⁺  +  Cl⁻   →  NaCl