Question

• The fractional charge on each atom in HBr molecule
is [Given that dipole moment of HBr = 0.78 D, Bond
length of HBr = 1.41 Å and 1 D= 1 * 10-18 esu cm]
(1) 0.251
(2) 0.472
(3) 0.115
(4) 0.362​

Answer 1

answer : option (3) 0.115

explanation : practical value of dipole moment of HBr, μ = 0.78D = 0.78 × 10^-18 esu.cm

let's find theoretical value of dipole moment.

dipole moment of HBr = magnitude of charge × seperation between charges.

as HBr dissociates into H+ and Br-

then, charge on each ion = |±e| = 1.6 × 10^-19 C

as we know, 1 esu = 3.33 × 10^-10 C

so, charge = (1.6/3.33) × 10^-9 esu

and seperation between charges = bond length of HBr = 1.41A° = 1.41 × 10^-8cm

so, theoretical value of dipole moment, μ' = (1.6/3.33) × 10^-9 × 1.41 × 10^-8 esu.cm

= 0.67 × 10^-17 esu.cm

= 6.77 × 10^-18 esu.cm

now fractional charge = μ/μ'

= (0.78 × 10^-18)/(6.77 × 10^-18)

≈ 0.115

Answer 2

Answer:

0.115 is correct answer