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Question: The observed dipole moment of HI is \(0.38\times {{10}^{-18}}\operatorname{D}\). Calculate the perce...

The observed dipole moment of HI is 0.38×10−18D⁡0.38\times {{10}^{-18}}\operatorname{D}. Calculate the percentage ionic character of the bonding HI if bond distance is 1.61 Ao \overset{\text{o}}{\mathop{\text{A}}}\,.
(A) 16.4%
(B) 12%
(C) 5.0%
(D) 20%

Explanation

Solution

The formula of percentage ionic character is-
%\text{ionic character = }\dfrac{\text{Observed dipole moment}}{\text{Calculated dipole moment}}\text{ x 100}
To calculate the “Calculated dipole moment” use the formula:
Dipole moment = (Charge) x (Distance of separation)\text{Dipole moment = (Charge) x (Distance of separation)}

Complete answer:
You already know that an electrovalent bond is also known as an ionic bond. Unlike a covalent bond, the ionic bond is born from complete give and take of electrons between two or more atoms. This is because there is a significant difference in the electronegativity between these atoms. But it is a fact that no bond is completely covalent or electrovalent.
In case of ionic bonds, Fajan’s rule states that there is a covalent character to them which varies from bond to bond. Therefore, we can calculate the percent ionic character which shows how much of the bond is actually electrovalent or ionic in nature.
First let us calculate the dipole moment of HIHI from the given data. Hydrogen iodide is formed when one electron is donated by hydrogen atom to iodine atom, which then takes its anion form- iodide. So, only one electron transfer is involved and therefore the charge involved is the charge of one electron, which has a constant value of 1.6 x 10−191.6\text{ x 1}{{\text{0}}^{-19}} C, which when converted to stat coulomb is4.7952×10−10statC⁡4.7952\times {{10}^{-10}}\operatorname{statC}. The distance of separation is given as 1.61 Ao \overset{\text{o}}{\mathop{\text{A}}}\, which when converted to centimetres becomes1.61×10−8cm⁡1.61\times {{10}^{-8}}\operatorname{cm}. We know that,
Dipole moment = (Charge) x (Distance of separation)\text{Dipole moment = (Charge) x (Distance of separation)}
Putting the respective values in the above equation we get-
DP = (4.7952 x 10−10) x (1.61 x 10−8) D\text{DP = (4}\text{.7952 x 1}{{\text{0}}^{-10}}\text{) x (1}\text{.61 x 1}{{\text{0}}^{-8}})\text{ D}
DP = 7.72 x 10−18 D\text{DP = 7}\text{.72 x 1}{{\text{0}}^{-18}}\text{ D}
We can now calculate the percentage ionic character as follows:
%\text{ionic character = }\dfrac{\text{Observed dipole moment}}{\text{Calculated dipole moment}}\text{ x 100}
We have been provided with the observed dipole moment in the question already. Therefore, putting the respective values in the above equation we get:
%\text{ionic character = }\dfrac{0.38\text{ x 1}{{\text{0}}^{-18}}}{\text{7}\text{.72 x 1}{{\text{0}}^{-18}}}\text{ x 100}
%\text{ionic character = }4.9\text{ }\\!\\!%\\!\\!\text{ }

The option which is the nearest to our answer is option (C) 5.0%.

Note:
Debye, which is represented as ‘D’ is actually the CGS unit of dipole moment. Here coulomb is in stat coulomb and the distance between the ionic species in the bond is in centimetres. Make sure you observe these things and add suitable changes to the calculations in order to get the perfect answer.
Also mark, that percent ionic character does not have a unit, as it is the ratio between similar quantities.