Question

A slightly polar molecule $AB$ has a dipole moment of $0.24D$ . If bond length is \[1{A^ \circ }\] , ionic character is $\left( \% \right)$ :

Answer 1

Hint:
Whether a compound is polar, covalent or ionic that can be determined by calculating the percentage ionic character, which is the ratio of a bond's measured dipole moment to the dipole moment assuming a complete electron transfer.
Formula Used: Percent Ionic Character \[ = \dfrac{{{\mu _o}}}{{{\mu _a}}} \times 100\% \]
Here ${\mu _o}$ is the observed dipole moment and ${\mu _a}$ is the dipole moment calculated under the assumption that the compound is $100\% $ ionic character.

Complete step by step answer:
As we all know that dipole moment $\mu $ of any compound is given by the formula
$\mu = q \times r$
Where $q$ is the magnitude of charge and $r$ is the distance between the centers of positive and negative charge

Now in the given question if we assume $100\% $ Ionic character the magnitude of charge will be $1.602 \times {10^{ - 19}}C$ that is the charge on a single electron since we are assuming transfer as complete ionic so magnitude of charge will be equal to the charge on one electron
Given the distance between both the centers as $1{A^ \circ }$or \[1 \times {10^{ - 10}}m\]
We can calculate ${\mu _a}$ as
\[{\mu _a} = 1.602 \times {10^{ - 19}} \times {10^{ - 10}}Cm\]
\[{\mu _a} = 1.602 \times {10^{ - 29}}Cm\]
Now to find Ionic Character $\left( \% \right)$ we need to convert the above dipole moment in a similar unit as that of the observed dipole moment that is Debye $\left( D \right)$ .
$1D = 3.336 \times {10^{ - 30}}Cm$
So ${\mu _a}$ in Debye is given by $\dfrac{{1.602 \times {{10}^{ - 29}}}}{{3.336 \times {{10}^{ - 30}}}}D$
${\mu _a} = 4.8021D$
Also ${\mu _o} = 0.24D$ (given in the question)
Now Percent Ionic Character \[ = \dfrac{{{\mu _o}}}{{{\mu _a}}} \times 100\% \]
Here ${\mu _o}$ is the observed dipole moment and ${\mu _a}$ is the dipole moment calculated under the assumption that the compound is $100\% $ ionic character.
Hence Percent Ionic Character $ = \dfrac{{0.24}}{{4.8021}} \times 100\% $
Answer $ = 5\% $

Note:Whenever taking the ratio for the calculation of Percent Ionic Character always remember taking proper units into consideration such that the ratio is always dimensionless