Question

If dipole moment of A-B molecule is 4.8 D and the bond length is 167 pm then the percentage ionic character of A-B bond is approximately :
40%
80%
70%
60%​

​

Answer 1

Answer:

The percentage of ionic character of the A-B bond is $60$ %·

Explanation:

Given that,

The observed dipole moment,  $\mu_{(obs)}$ $=$ $4.8$ D

The bond length, l  $=$  $167$ pm  $=\ 167\ *\ 10^{-10}$ cm

The formula to calculate the percentage of ionic character follows,

   % Ionic character $=$  $\frac{\mu _{(obs.)} }{\mu_{(theor.)} }$  ×  $100$ %  $----(1)$

where,

$\mu_{(obs)}$  $=$  The observed dipole moment

$\mu_{(theor.)}$ $=$ The theoretical dipole moment

To calculate the percentage of ionic character first, we need to find out the value of the theoretical dipole moment·

We know that dipole moment is the product of the charge and the bond length·

That is,

  $\mu\ =\ q\ *\ l$   $----(2)$

where,

q $=$ The charge  

l $=$  The distance between the A-B molecule

On substituting the values in equation No· $2$ we get

⇒ $\mu_{(theor.)} \ =\ q\ *\ l$  

⇒ $\mu_{(theor.)} \ =\ 4.8\ *\ 10^{-10} \ *\ 167\ *\ 10^{-10}$ esu cm

⇒ $\mu_{(theor.)} \ =\ 8.0\ *\ 10^{-18}$ esu cm

We know that,

   1 Debye  $=$  $1\ *\ 10^{-18}$ esu cm

∴  $8.0\ *\ 10^{-18}$ esu cm is $8.0$ D

Then substitute these values in equation No· $1$ we get,

⇒ % Ionic character $=$  $\frac{\mu _{(obs.)} }{\mu_{(theor.)} }$  ×  $100$ %

⇒ % Ionic character $=$  $\frac{4.8}{8.0}$   ×  $100$ %

⇒ % Ionic character $=$  $0.6$  ×  $100$ %

⇒ % Ionic character $=$  $60$ %

Therefore,

The percentage of ionic character of the A-B bond is $60$ %·