# Calculate the percentage of ionic and covalent character of HF molecule having Bond distance = 0.92 angstrom and dipole moment = 1.78 D. But my question is why do we take q = 4.8 x (10 ^ -10) esu, Since electronic charge e = 4.8 x (10 ^ -10) esu. Explain this question in detail and please don't send any link

Dipole moment of HF observed = 1.78 D

Dipole moment of HF calculated = Charge x bond distance

= $=1.6\times {10}^{-19}\times 0.92\times {10}^{-10}\phantom{\rule{0ex}{0ex}}=1.472x{10}^{-29}Cm$

As we know 1 Debye = 3.33 x 10

^{-30}Cm

Dipole moment of HF calculated = $\frac{14.72\times {10}^{-30}}{3.33\times {10}^{-30}}=4.42D$

So, the percentage ionic character = $\frac{{\mu}_{obs}}{{\mu}_{calculated}}\times 100$

$=\frac{1.78}{4.42}\times 100\%\phantom{\rule{0ex}{0ex}}=40.27\%$

So, the percentage covalent character of HF= 100 - 40.27 = 59.73 %

Above calculation can also be done by taking the charge = 4.8 x 10

^{-10}e.s.u

As 1 debye = 10

^{-10 }e.s.u A

^{o}, gives the similar result as dipole moment of HF calculated will be equal = 4.418 D

So, the percentage ionic character and covalent character of HF = 40.27% and 59.73% respectively.

Regards.

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