Explain the formation of sodium chloride

CBSE Class 10 CBSE Class 10 Science
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1

Explain the formation of sodium chloride and calcium oxide on the basis of the concept of electron transfer from one atom to another atom.

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  1. When sodium (Na) atom loses one electron to get octet electron configuration it forms a cation (${{Na}^{+}}$)

${ }_{ 11 }{ Na }$ ------> ${ }{ _{ 11 }{ Na } }^{ + }$+${{e}^{-}}$

  1. Chlorine has shortage of one electron to get octet in its valence shell. So it gains electron that was lost by Na and form anion.

${ }_{ 17 }{ Cl }$ + ${{e}^{-}}$ ----------> ${{Cl}^{-}}$

  1. Transfer of electrons between Na and Cl atoms they form ${{Na}^{+}}$ and ${{Cl}^{-}}$ ions.
  2. These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound sodium chloride (NaCl).

${{Na}^{+}}$ + ${{Cl}^{-}}$ --------> ${{Na}^{+}}$ ${{Cl}^{-}}$ or NaCl

II. Formation of Calcium Oxide :

  1. Calcium (Ca) atom is ready to lose two electrons to get octet electron configuration it forms a cation (${{Ca}^{2+}}$).

Ca -------> ${{Ca}^{2+}}$ + ${{2e}^{-}}$

  1. Oxygen (O) has shortage of two electrons to get octet in its valence shell. So it gains electron that was lost by Ca and form anion.

${ }_{ 8 }{ O }$ + ${{2e}^{-}}$ -------> ${{O}^{2-}}$

  1. Transfer of 2 electrons between ‘Ca’ and ‘O’ atoms form ${{Ca}^{2+}}$ and ${{O}^{2-}}$ ions.
  2. These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound calcium oxide (CaO).
    ${{Ca}^{2+}}$ + ${{O}^{2-}}$ --------> CaO