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Question: Bond-length of HCl is 1.275 Angstrom (\[e = 4.8 \times {10^{ - 10}}esu\]). If \(\mu \)=1.02D, then H...

Bond-length of HCl is 1.275 Angstrom (e=4.8×10−10esue = 4.8 \times {10^{ - 10}}esu). If μ\mu =1.02D, then HCl is:
A: 100% ionic
B: 83% covalent
C: 50% covalent
D: 40% ionic

Explanation

Solution

Electric dipole moment takes place when separation of charge occurs. It can take place in electrically neutral molecules having ionic bonds or molecules possessing covalent bonds. Dipole moments are meant to measure the electric polarity of the system of charges.

Complete step by step solution: For each bond, the dipole moments can be estimated by using the electronegativity difference between the bonded atoms as a measure of dipole strength. The bond dipole is referred to as δ+ — δ– with a distance (denoted as d) between the partial charges being δ+ and δ–. The formula for dipole moment is mentioned below:
μ=δd\mu = \delta d
μ=dipole moment δ=charge d=distance between partial charges  \mu = dipole{\text{ }}moment \\\ \delta = charge \\\ d = distance{\text{ }}between{\text{ }}partial{\text{ }}charges \\\
In the question, we are given all these values i.e. dipole moment, bond length, and charge and we are asked to find out the percentage ionic character. The percentage ionic character can be found out by dividing the observed dipole moment with the calculated dipole moment times 100.
% ionic character=Observed dipole momentCalculated dipole moment×100\% {\text{ }}ionic{\text{ }}character = \dfrac{{Observed{\text{ }}dipole{\text{ }}moment}}{{Calculated{\text{ }}dipole{\text{ }}moment}} \times 100
The calculated dipole moment is actually the charge on the electron multiplied by the radius of the molecule (μ=δd\mu = \delta d).
Bond length = 1.275{\text{ }}{A^ \circ } = $$$$1.275 \times {10^{ - 8\;}}cm(Given)
charge on the electron, e=4.8×10−10esue = 4.8 \times {10^{ - 10}}esu(Given)
Observed dipole moment, μ=1.02D=1.02×10−18  esu cm\mu = 1.02D = 1.02 \times {10^{ - 18\;}}esu{\text{ }}cm(Given)
Calculated dipole moment μ=1.275×10−8×4.8×10−10=  6.12×10−18esu cm\mu = 1.275 \times {10^{ - 8}} \times 4.8 \times {10^{ - 10}} = \;6.12 \times {10^{ - 18}}esu{\text{ }}cm
Thus now, we can calculate the % ionic character as stated below:

∴% ionic character=1.02×10−18  esu cm  6.12×10−18esu cm×100=16.82%≃17%ionic ∴100−17=83%cov⁡alent  \therefore \% {\text{ }}ionic{\text{ }}character = \dfrac{{1.02 \times {{10}^{ - 18\;}}esu{\text{ }}cm}}{{\;6.12 \times {{10}^{ - 18}}esu{\text{ }}cm}} \times 100 = 16.82\% \simeq 17\% ionic \\\ \therefore 100 - 17 = 83\% \operatorname{cov} alent \\\

Hence, the correct answer is Option B: 83% covalent.

Note: More the electronegativity difference between the two atoms, more will be the ionic character of the bond. A bond can possess 100% ionic character only in case one of the bonded atoms takes the bonding electrons completely. This can occur only if one of the bonded atoms possess zero electronegativity. But as we know, none of the atoms possess zero electronegativity. Thus, no bond can possess 100% ionic character.