Question

Dipole moments of HCI = 1.03 D, HI= 0.38 D. Bond length of $HCl = 1.3 \mathring{A}$ and $HI = 1.6 \mathring{A}$. The ratio of fraction of an electric charge, e existing on each atom in HCl and HI is

• A. 1.2 : 1
• B. 2.7 : 1
• C. 3.3 : 1
• D. 01:03.3

Answer 1

Answer: (C)

3.3 : 1

Answer 2

$\mu=e\times d$
where, e = magnitude of electric charge
d = distance between particles (here bond length)
$\therefore\, \, \, \, \, \, \, \, \, \, \, \, \, \, e=\frac{\mu}{d}$
$or, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{e_{HCl}}{e_{HCI}}=\frac{\mu_{HCl}}{d_{HCl}}\times\frac{d_{HCl}}{\mu_{HCl}}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{1.3 \times 1.6}{1.3\times 0.38}=3.3 : 1$