Question

03) Calculate the fractional charge on each atom in HBr molecule. Given that Dipole moment of HBr = 0.78
D, Bond distance of HBr = 1.41 A, Electronic charge, e = 4.8 * 10-esu​

Answer 1

Answer:74755

Explanation:

Answer 2

Answer:

0.115.

Explanation:

From the question we get that the values of dipole moment of HBr molecule is, μ = 0.78D which is 0.78 × 10^-18 esucm.

So, now we know that the dipole moment is having a formulae which is dipole moment of HBr is the charge magnitude x the charge separation .

Since, the dissociation of HBr occurs into H+ and Br- in which the charge on each ion is given as  1.6 × 10^-19 C since, 1 esu = 3.33 × 10^-10 C. Therefore the charge will be  (1.6/3.33) × 10^-9 esu.

While the charge separation is the HBr bond length which is = 1.41A° which on cm we will have 1.41 × 10^-8cm.

Hence, the value of the dipole moment after calculating μ1 = (1.6/3.33) × 10^-9 × 1.41 × 10^-8 esucm .

=0.67 × 10^-17 esucm.

=6.77 × 10^-18 esucm.

So, the fractional charge will be = μ/μ1.

Which on solving we will get that = (0.78 × 10^-18)/(6.77 × 10^-18) =0.115.