Question

Q. Calculate percentage ionic character of HF molecule having bond distance 0.92Å and dipole moment = 1.78 D.
Can anyone please answer this question!​

Answer 1

Answer:

Q. Calculate percentage ionic character of HF molecule having bond distance 0.92Å and dipole moment = 1.78 D.

Can anyone please answer this question

Answer 2

Given - Bond distance - 0.92 angstrom

Dipole moment - 1.78 D or debye.

Find - Percentage ionic character

Solution - Ionic character is calculated by the formula - (Observed dipole moment/Calculated dipole moment)*100

We know, 1 D = 10-¹⁸ esu cm (in c.g.s system)

Calculated dipole moment is further calculated by the formula - charge on electron*distance

Charge on electron = 4.8*10-¹⁰ e.s.u (in c.g.s system)

Now, dipole moment = 4.8*10-¹⁰*0.92*10-⁸ e.s.u. cm.

Dipole moment = 4.4*10-¹⁸ e.s.u. cm

= 4.4 D

Keeping the values in equation-

Percentage Ionic character = 1.78/4.4*100

Percentage Ionic character = 40.45%

Hence, percentage ionic character = 40.45%