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. 1 : 3.3

Answer 1

Answer: (C)

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