PERCENTAGE OF IONIC CHARACTER
Every ionic compound having some percentage of covalent character according to Fajan’s rule. The percentage of ionic character in a compound having some covalent character can be calculated by the following equation.
The percent ionic character = Observed dipole moment/Calculated dipole moment assuming 100% ionic bond × 100
Example: Dipole moment of KCl is 3.336 × 10–29 coulomb metre which indicates that it is highly polar molecule. The interatomic distance between k+ and Cl– is 2.6 ×10–10 m. Calculate the dipole moment of KCl molecule if there were opposite charges of one fundamental unit located at each nucleus. Calculate the percentage ionic character of KCl.
Solution: Dipole moment μ = e × d coulomb metre
For KCl d = 2.6 × 10–10 m
For complete separation of unit charge
e = 1.602 × 10–19 C
Hence μ = 1.602 × 10–19 × 2.6 × 10–10 = 4.1652 × 10–29 Cm
μKCl = 3.336 × 10–29 Cm
∴ % ionic character of KCl = 3.336×10–29/4.165×10–29 = 80.09%
Example. Calculate the % of ionic character of a bond having length = 0.83 Å and 1.82 D as it’s observed dipole moment.
Solution: Tocalculate μ considering 100% ionic bond
= 4.8 × 10–10 × 0.83 × 10–8esu cm
= 4.8 × 0.83 × 10–18 esu cm = 3.984 D
∴ % ionic character = 1.82/3.984 × 100 = 45.68
The example given above is of a very familiar compound called HF. The % ionic character is nearly 43.25%, so the % covalent character is (100 – 43.25) = 56.75%. But from the octet rule HF should have been a purely covalent compound but actually it has some amount of ionic character in it, which is due to the electronegativity difference of H and F. Similarly knowing the bond length and observed dipole moment of HCl, the % ionic character can be known. It was found that HCl has 17% ionic character. Thus it can be clearly seen that although we call HCl and HF as covalent compounds but it has got appreciable amount of ionic character. So from now onwards we should call a compound having more of ionic less of covalent and vice versa rather than fully ionic or covalent.