Question

0. The observed dipole moment of HCl is 1.03 D.
Bond length is 1.275 A then the percentage of
ionic character is
(1) 16.83%
(2) 21%
(3) 30.72%
(4) 14.21%​

Answer 1

answer : option (1) 16.83 %

explanation : dipole moment is the product of magnitude of charge and seperation between charges .

i.e., P = q × d

here compound is HCl,you know as well, HCl is ionic compound in which one electron transfers from hydrogen to chlorine.

so, magnitude of charge on dipole , q = 1.6 × 10^-19C

and seperation between charges = bond length = 1.275 A° = 1.275 × 10^-10 m

so, dipole moment , P = 1.6 × 10^-19 × 1.275 × 10^-10 Cm

= 2.04 × 10^-29 Cm

we know, 1D = 3.335 × 10^-30 Cm [ Debay, D is the unit of dipole moment ]

= 2.04 × 10^-29/(3.335 × 10^-30)

= 20.4/3.335

= 6.1169D

now, percentage ionic character = experimental value/ theoretical value × 100

= 1.03D/6.1169D × 100

= 16.83 %

Answer 2

Hii Students,

◆ Answer - (1)

Inoic character = 16.83 %

● Explanation -

# Given -

μo = 1.03 D

d = 1.275×10^-10 m

# Solution -

Expected dipole moment of HCl is calculated as -

μe = d × e

μe = 1.275×10^-10 × 1.6×10^-19

μe = 2.04×10^-29 Cm

μe = 2.04×10^-29 × 3×10^29 D

μe = 6.12 D

Thus, percentage of ionic character is given as -

Ionic character = μo / μe × 100

Ionic character = 1.03 / 6.12 × 100

Inoic character = 16.83 %

Therefore, ioinic character of HCl is 16.83 %.

Thanks dear..