Solveeit Logo

Question

Question: A slightly polar molecule \(AB\) has a dipole moment of \(0.24D\) . If bond length is \[1{A^ \circ }...

A slightly polar molecule ABAB has a dipole moment of 0.24D0.24D . If bond length is 1A1{A^ \circ } , ionic character is (%)\left( \% \right) :

Explanation

Solution

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 =μoμa×100%= \dfrac{{{\mu _o}}}{{{\mu _a}}} \times 100\%
Here μo{\mu _o} is the observed dipole moment and μa{\mu _a} is the dipole moment calculated under the assumption that the compound is 100%100\% ionic character.

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

Now in the given question if we assume 100%100\% Ionic character the magnitude of charge will be 1.602×1019C1.602 \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 1A1{A^ \circ }or 1×1010m1 \times {10^{ - 10}}m
We can calculate μa{\mu _a} as
μa=1.602×1019×1010Cm{\mu _a} = 1.602 \times {10^{ - 19}} \times {10^{ - 10}}Cm
μa=1.602×1029Cm{\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 (D)\left( D \right) .
1D=3.336×1030Cm1D = 3.336 \times {10^{ - 30}}Cm
So μa{\mu _a} in Debye is given by 1.602×10293.336×1030D\dfrac{{1.602 \times {{10}^{ - 29}}}}{{3.336 \times {{10}^{ - 30}}}}D
μa=4.8021D{\mu _a} = 4.8021D
Also μo=0.24D{\mu _o} = 0.24D (given in the question)
Now Percent Ionic Character =μoμa×100%= \dfrac{{{\mu _o}}}{{{\mu _a}}} \times 100\%
Here μo{\mu _o} is the observed dipole moment and μa{\mu _a} is the dipole moment calculated under the assumption that the compound is 100%100\% ionic character.
Hence Percent Ionic Character =0.244.8021×100%= \dfrac{{0.24}}{{4.8021}} \times 100\%
Answer =5%= 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