Name: Hybridisation in SF₆ Molecule
Uploaded: 2019-07-19
Duration: 2 min 31 s
Description: Understand the bonding present in the case of sulphur hexafluoride. This video illustrates that the hybridisation present in the case of sulphur hexafluoride...

Question

What do you understand by partial ionic character of covalent bond? How it is calculated? Explain taking example of

HCl

, given that its observed dipole moment is

1.03 D

and bond length is

1.275 Å

?

Accepted Answer

If two atoms linked together have different electronegativities, the bond formed is polar. In other words, the bond is said to possess partial ionic character.
The extent of partial ionic character is determined by the difference in electronegativities of the combining atoms. More is the difference in electronegativities, greater will be the ionic character.

If the electronegativity difference between two atoms is $1.9$ , the bond is said to have $50 %$ ionic character and $50 %$ covalent character.
If the electronegativity difference between the two atoms is more than $1.9$ , the partial ionic character of the bond is more than $50 %$ and the bond is taken as ionic.
If the electronegativity difference between two atoms is less than $1.9$ , the bond· is predominantly covalent.

Several empirical equations have been proposed to calculate the percentage ionic character from the
electronegativity difference,( $χ_{A} - χ_{B}$ ), of the bonded atoms. Two such equations are :

Pauling equation: $% Ionic character = [1 - e^{- \frac{1}{4}} (χ_{A} - χ_{B}) \times 100 or % Ionic character = 18 (χ_{A} - χ_{B})^{1.4}$

Henry and Smith equation:

$% Ionic character = 16 (χ_{A} - χ_{B}) + 3.5 (χ_{A} - χ_{B})^{2}$ .

If $HCl$ were $100 %$ ionic, each end would carry charge equal to one unit, viz, $4 \cdot 8 x 10^{- 10} esu$ .

As bond length of $HCl$ is $1.275 A^{°}$ , its dipole moment for $100 %$ ionic character would be

$\begin{array}{rcl} µ_{ionic} & = & q \times d = 4 \cdot 8 \times 10^{- 10} esu \times 1 \cdot 275 x 10^{- 8} cm \\ = & 6 \cdot 12 x 10^{- 18} esu cm \\ = & 6 \cdot 12 D \\ µ_{observed} & = & 1 \cdot 03 D (Given) \\ % ionic character & = & \frac{µ_{observed}}{µ_{ionic}} \times 100 = \frac{1.03}{6.12} \times 100 = 16 \cdot 83 % \end{array}$

What do you understand by partial ionic character of covalent bond? How it is calculated? Explain taking example of $HCl$ , given that its observed dipole moment is $1.03 D$ and bond length is $1.275 Å$ ?

Important Questions on Chemical Bonding and Molecular Structure

Learn and Prepare for any exam you want !

What do you understand by partial ionic character of covalent bond? How it is calculated? Explain taking example of HCl, given that its observed dipole moment is 1.03D and bond length is 1.275Å ?

Important Questions on Chemical Bonding and Molecular Structure

Learn and Prepare for any exam you want !

What do you understand by partial ionic character of covalent bond? How it is calculated? Explain taking example of $HCl$ , given that its observed dipole moment is $1.03 D$ and bond length is $1.275 Å$ ?