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

Dear Student,

Dipole moment of HF observed = 1.78 D
Dipole moment of HF calculated = Charge x bond distance 
                                                     = =1.6×10-19×0.92×10-10=1.472 x 10-29 Cm
As we know 1 Debye = 3.33 x 10-30 Cm
Dipole moment of HF calculated = 14.72 ×10-303.33 ×10-30= 4.42 D
So, the percentage ionic character = μobsμcalculated ×100
                                                    =1.784.42×100%= 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 Ao , 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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