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Question: The bond length of \(HCl\) molecule is \(1.275{A^\circ}\) and its dipole moment is 1.03D. The ionic ...

The bond length of HClHCl molecule is 1.275A∘1.275{A^\circ} and its dipole moment is 1.03D. The ionic character of the molecule (in percent) is? (Charge of the electron =4.8×10−10esu4.8 \times {10^{ - 10}}esu)
A. 100
B. 67.3
C. 33.66
D. 16.83

Explanation

Solution

Ionic character percent of a molecule can be obtained by using the formula Actual dipole momentCalculated dipole moment×100\dfrac{{Actual\,dipole\,moment}}{{Calculated\,dipole\,moment}} \times 100. The actual dipole moment is already given in the question, so obtain the calculated dipole moment using the formula μ=BL×q\mu = BL \times q, where BL is the bond length and q is the charge of the electron.

Complete step by step answer:
We are given that the bond length of HClHCl molecule is 1.275A∘1.275{A^\circ} and its dipole moment is 1.03D.
We have to calculate its ionic character percentage.
Ionic character percent is the amount of electron sharing between two atoms and if the electrons are shared limitedly then the ionic character percent of the molecule will be high.
The dipole moment can be obtained by the formula μ=BL×q\mu = BL \times q
Bond length is 1.275A∘1.275{A^\circ} and charge is 4.8×10−10esu4.8 \times {10^{ - 10}}esu
⇒μ=BL×q ⇒1A∘=10−8cm ⇒BL=1.275A∘,q=4.8×10−10esu ⇒μ=1.275×10−8cm×4.8×10−10esu=6.12×10−18esu  cm  \Rightarrow \mu = BL \times q \\\ \Rightarrow 1 {A^\circ} = {10^{ - 8}}cm \\\ \Rightarrow BL = 1.275 {A^\circ},q = 4.8 \times {10^{ - 10}}esu \\\ \Rightarrow \mu = 1.275 \times {10^{ - 8}}cm \times 4.8 \times {10^{ - 10}}esu = 6.12 \times {10^{ - 18}}esu\;cm \\\
Therefore, the calculated dipole moment of HClHCl is 6.12×10−18esu  cm6.12 \times {10^{ - 18}}esu\;cm

⇒1D=10−18esu  cm ⇒1.03D=1.03×10−18esu  cm  \Rightarrow 1D = {10^{ - 18}}esu\;cm \\\ \Rightarrow 1.03D = 1.03 \times {10^{ - 18}}esu\;cm \\\
Therefore, the actual dipole moment of HClHCl is 1.03×10−18esu  cm1.03 \times {10^{ - 18}}esu\;cm
Therefore, the ionic character percent of HClHCl will be
⇒IC%=Actual dipole momentCalculated dipole moment×100 ⇒Actual dipole moment=1.03×10−18esu  cm ⇒Calculated dipole moment=6.12×10−18esu  cm ⇒IC%=1.03×10−186.12×10−18×100 ⇒IC%=16.83  \Rightarrow IC\% = \dfrac{{Actual\,dipole\,moment}}{{Calculated\,dipole\,moment}} \times 100 \\\ \Rightarrow Actual\,dipole\,moment = 1.03 \times {10^{ - 18}}esu\;cm \\\ \Rightarrow Calculated\,dipole\,moment = 6.12 \times {10^{ - 18}}esu\;cm \\\ \Rightarrow IC\% = \dfrac{{1.03 \times {{10}^{ - 18}}}}{{6.12 \times {{10}^{ - 18}}}} \times 100 \\\ \Rightarrow IC\% = 16.83 \\\
The ionic character percent of Hydrochloric acid (HClHCl) with bond length 1.275A∘1.275{A^\circ} and dipole moment 1.03D is 16.83%.

Hence we can conclude that option D is correct.

Note: When calculating a percentage, the right-hand side values which are undergoing division must have the same units. If the values are given in different units then convert all of the values into having the same units. Here in the above question, the actual dipole moment is given in Debye and the charge of the electron is given in esu (electrostatic units). So we have converted the actual dipole moment in terms of esu.