What is an Ionic bond? Explain with an example.
Ionic Bond Definition:
“The chemical bond which is formed due to complete transfer of electrons from one atom to another atom is called Ionic bond or electrovalent bond”.
Conditions for Ionic Bonding:
Following are the conditions which are necessary for the formation of an Ionic bond.
- One element should be metal while other elements should be non-metal.
- The electronegativity difference between two bonded atoms should be 1.7 or above.
- The ionization energy of metal should be low.
- The electronegativity of non-metal should be high.
- The transfer of electrons between atoms completes the octet or duplet.
- This type of bonding is always between a metal and non-metal.
- Metals always lose electrons to form cations and non-metals always gain electrons to form Anion.
- In ionic bond formation, one atom loses an electron and the other gain it.
- The atom that losses the electron acquires a positive charge and the other atom, which gains an electron becomes a negatively charged particle.
- Due to opposite charges, the electrostatic force of attraction is set up between them.
- This force holds these ions together.
- This force of attraction is referred to as an ionic bond.
- An atom having low ionization energy will lose one or more electrons to form a cation.
- An atom having high electronegativity will gain one or more electrons to form an anion.
Formation Of Sodium Chloride (NaCl):
- Sodium (Na) and chloride (Cl) combine to form sodium chloride (NaCl).
- The atomic number of sodium is 11.
The electronic configuration of sodium based On Shell:
K = 2
L = 8
M = 1
The electronic configuration of sodium based On Sub-Shell:
1s2, 2s2, 2p6, 3s1
- Sodium belongs to group 1 A
- Chlorine belongs to group 7 A
- The electronegativity of sodium is 0.9 and chlorine is 3.0
- The electronegativity difference between sodium and chlorine is 2.1, hence, an ionic bond is formed.
- Sodium has one electron in its outermost shell. Sodium loses one electron to form sodium ion (Na+) and attain the electronic configuration of neon (Ne).
Na(s) ——— > Na+ + 1e–
- The atomic number of chlorine is 7.
- The electronic configuration of chlorine is:
The electronic configuration of chlorine based On Shell:
The electronic configuration of chlorine based On Sub-Shell:
1s2, 2s2, 2p6, 3s2, 3p5
- Chlorine has 7 electrons in its outermost shell.
- Cl gains one electron to form a chloride ion (Cl–) and to attain the electronic configuration of Argon (Ar).
Cl + 1e– ——- > Cl–
- The electrostatic force f attraction is established between sodium and chlorine
- Therefore, a crystal of NaCl is formed.
Formation Of Calcium Chloride (CaCl2):
- Ca belongs to group II A
- Cl belongs to group VII A
- The electronegativity of calcium is 1.0 while chlorine is 3.0
- The electronegativity difference between calcium and chlorine is 2.0
- Thus, an ionic bond is formed
- Calcium has 2 electrons in its valence shell
- It has the tendency to lose electrons to attain the inert gas electronic configuration of Neon (Ne)
- Thus, the calcium ion (Ca+) is formed
Ca ——- > Ca+2 + 2 e-
- Chlorine needs one electron to complete its valence shell
- It has the tendency to gain one electron to attain the inert gas electronic configuration of Argon (Ar)
- So, it forms chlorine ion
2Cl +2e– ——- > Cl–
- The electrostatic force of attraction appears between Ca+2 and two chloride ions (Cl- )
- Thus, white-coloured crystalline CaCl2 is formed
Ca+2 + 2Cl– —— > CaCl2
Formation Of AlCl3:
- Al belongs to group III A
- Cl belongs to group VII A
- Al has three electrons in its valence shell while Cl has 7 electrons in its valence shell
- Al loses three electrons to form Aluminum ion (Al+3)
Al —> Al +3 + 3e–
- Each Cl atom gains one electron to form a chloride ion (Cl–)
3Cl – 3e– —> 3Cl–
- The electrostatic force that appears between (Al+3) and (Cl–) ions
- Thus, AlCl3shows ionic character