Equilibrium Separation of HCl with Covalent Bonding: Expert Assistance Needed

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The discussion revolves around calculating the equilibrium separation of HCl, given its electric dipole moment of 3.60 x 10^-30 C.m and a covalent bonding percentage of 82.5%. A formula involving the dipole moment and covalent character is mentioned, yielding a separation of approximately 27.3 picometers. Participants seek clarification on the formulas used, particularly the origin of the constant 6.25 x 10^18. The original poster plans to share the complete working out after consulting their lecturer.
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Hey guys, I am currently studying Engieering in Australia and there is one question which was given to us which i have no idea on how to complete and i was wondering if you guys could give me a hand with it;

What is the equilibrium separation of HCl? ITs measured electric dipole moment is 3.60 x 10^-30C.m and the percentage of covalent bonding in HCl is 82.5%

ANy help would be greatly appreciated. Thanx
 
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rushil01 said:
Hey guys, I am currently studying Engieering in Australia and there is one question which was given to us which i have no idea on how to complete and i was wondering if you guys could give me a hand with it;

What is the equilibrium separation of HCl? ITs measured electric dipole moment is 3.60 x 10^-30C.m and the percentage of covalent bonding in HCl is 82.5%

ANy help would be greatly appreciated. Thanx

r =\ \mu_D/q\cdot P\ =\ 6.25\cdot10^{18}\ \mu_D/P\ =\ <br /> <br /> 6.25\cdot10^{18}\cdot3.6\cdot10^{-30}/0.825\ =\ 2.73\cdot10^{-11}\ m\ =\ 27.3\ pm
 
What are those formulas you came up with and where did 6.25*10^18 come from.
Thanks
 
i got my lecturer to do it and he used the formula p=qd. I'll post the full working out when i have a little more time
 

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