Formation of ionic Bonds in NaCL

by • 22/08/2013 • GeneralComments (0)342

Formation of ionic Bonds in NaCL

The formation of ionic bond in NaCL consists of following steps.

 

1. Formation Of Cations:
 

When we observed the electronic configuration of ‘Na’ we found that it has only one valence electron, therefore it has to lose that electron in order to gain stability and have a complete valence shell. When 495 Ks/mole energy provided to it, ‘Na‘. Lose an electron and become cation.

 

Na — Na + 1 e- ∆H = + 495 KJ/mol

 

2. Formation of Anion:

In case of chlorine, it has 7 electrons in its valence shell therefore, in order to gain stability it accept the electron of sodium and become anion. This process is an exothermic process and 348l<j/mole energy is released during the process.

Cl + e- —> Cl- ∆H = – 348 KJ/mol

 

3. Combinations Of Cation and Anion:

Now electrostatic force of attraction is present between cation and anion and crystal lattices is formed of NaCl in which each ion is surrounded by six opposite ions, this process is exothermic and a large amount of heat is released during the process and ionic bond is formed between sodium and chlorine ions as:

 

Na + Cl- —-> NaCl ∆H = – 788 KJ/mol

This energy released in the formation of crystal lattice “Lattice energy” which can be defined as:

Energy evolved by 1 mole of gaseous ion when they form crystal lattice is called “Lattice Energy”.

Over all Energy Change During The Process

Na — Na + 1 e- ∆H = + 495 KJ/mol

Cl + e- —> Cl- ∆H = – 348 KJ/mol

Na + Cl- —-> NaCl ∆H = – 788 KJ/mol

Na + Cl- —-> NaCl ∆H = – 644 KJ/mol

 

The above equation shows that the formation of ionic bond in NaCl is an exothermic process.

 

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