Dipole moment of HCl = 1.03 D, HI-0.38 D. Bond length of `HCI-1.3 A^(@)` and `HI= 1.6A^(@)` The ratio of fraction of electric charge, delta, existing on each atom in HCI and HI is

Options

A. `12:1`
B. `27:1`
C. 3.3:1
D. `1:3.3`

Answer

( c)

Explanation

Dipole moment, `mu_(HCI)=1.03D`
`mu_(HI)=0.38D`
Bond length, `d_(HCI)=1.3A^(@)`
`d_(HI)=1.6A^(@)`
Dipole moment `(mu)` = Electric charge `(delta) xx` bond length (d)
`Rightarrow delta=(mu)/(d)`
`therefore (delta_(HCI))/(delta_(HI))=(mu_(HCI))/(d_(HCI))xx(d_(HI))/(mu_(HCI))=(1.03)/(1.3)xx(1.6)/(0.38)`
`therefore delta_(HCI)=3.33:1`