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Question: The dipole moment of \(HBr\) is \(2.6 \times {10^{ - 30}}C - m\) and the inter atomic spacing is \(1...

The dipole moment of HBrHBr is 2.6×10−30C−m2.6 \times {10^{ - 30}}C - m and the inter atomic spacing is 1.41\buildrel∘A1.41{{\buildrel _{\circ} \over {\mathrm{A}}}}. The percentage of ionic character in HBrHBr is:
A. 8.7%
B. 11.5%
C. 12.5%
D. 13.5%

Explanation

Solution

Ionic character percentage 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. Bond length is the length of interatomic spacing.

Complete step by step answer:
We are given that the dipole moment of HBrHBr is 2.6×10−30C−m2.6 \times {10^{ - 30}}C - m and the inter atomic spacing is 1.41\buildrel∘A1.41 {{\buildrel _{\circ} \over {\mathrm{A}}}}.
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
Inter atomic spacing can also be called as the bond length between two atoms.
Bond length is 1.41\buildrel∘A1.41 {{\buildrel _{\circ} \over {\mathrm{A}}}} and charge of an electron is 1.6×10−19C1.6 \times {10^{ - 19}}C
⇒μ=BL×q ⇒1\buildrel∘A=10−10m ⇒BL=1.41\buildrel∘A,q=1.6×10−19C ⇒μ=1.41×10−10m×1.6×10−19C=2.256×10−29C−m  \Rightarrow \mu = BL \times q \\\ \Rightarrow 1{{\buildrel _{\circ} \over {\mathrm{A}}}} = {10^{ - 10}}m \\\ \Rightarrow BL = 1.41{{\buildrel _{\circ} \over {\mathrm{A}}}},q = 1.6 \times {10^{ - 19}}C \\\ \Rightarrow \mu = 1.41 \times {10^{ - 10}}m \times 1.6 \times {10^{ - 19}}C = 2.256 \times {10^{ - 29}}C - m \\\
Therefore, the calculated dipole moment of HBrHBr is 2.256×10−29C–m2.256 \times {10^{ - 29}}C – m.
The actual dipole moment of HBrHBr is 2.6×10−30C−m2.6 \times {10^{ - 30}}C - m.
Therefore, the ionic character percent of HBrHBr will be
⇒IC%=Actual  dipole  momentCalculated  dipole  moment×100 ⇒Actual  dipole  moment=2.6×10−30C−m ⇒Calculated dipole  moment=2.256×10−29C−m ⇒IC%=2.6×10−302.256×10−29×100 ⇒IC%=1.15×10−1×100 ∴IC%=11.5  \Rightarrow IC\% = \dfrac{{Actual\,\,dipole\,\,moment}}{{Calculated\,\,dipole\,\,moment}} \times 100 \\\ \Rightarrow Actual\,\,dipole\,\,moment = 2.6 \times {10^{ - 30}}C - m \\\ \Rightarrow Calculated\,dipole\,\,moment = 2.256 \times {10^{ - 29}}C - m \\\ \Rightarrow IC\% = \dfrac{{2.6 \times {{10}^{ - 30}}}}{{2.256 \times {{10}^{ - 29}}}} \times 100 \\\ \Rightarrow IC\% = 1.15 \times {10^{ - 1}} \times 100 \\\ \therefore IC\% = 11.5 \\\
The ionic character percent of Hydrobromic acid (HBrHBr) with bond length 1.41\buildrel∘A1.41 {{\buildrel _{\circ} \over {\mathrm{A}}}} and dipole moment 2.6×10−30C−m2.6 \times {10^{ - 30}}C - m is 11.5%.

Hence we can conclude that option B 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 Coulomb meters and interatomic spacing is given in \buildrel∘A{\buildrel _{\circ} \over {\mathrm{A}}}. So we have converted \buildrel∘A{\buildrel _{\circ} \over {\mathrm{A}}} into meters and the charge of the electron is taken in Coulombs. So be careful with the units of the values.