Equilibrium Separation of HCl with Covalent Bonding: Expert Assistance Needed

  • Thread starter Thread starter rushil01
  • Start date Start date
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 2K views
rushil01
Messages
2
Reaction score
0
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
 
Physics news on Phys.org
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

[tex]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[/tex]
 
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