Question

The dipole measured for HI is 0.380 D. The bond distance is 161 pm. What is the percent ionic character of the HI bond?

Answer 1

Answer:

Percent ionic character of HI bond is 4.91%.

Explanation:

Given Data:

Measured Dipole = 0.380D

bond distance = d = 161pm = 1.61*10^-8 cm

Calculation:

% ionic character is determined by following equation:

% ionic= (dipole measured/dipole calculated)*100

Now,

$Dipole(calc)=qd$

$Dipole(calc)= (1.6*10^{-19}*3*10^{9})esu *1.61*10^{-8}cm$

(In above step 3*10^8 is multiplied to convert coulomb into esu)

$Dipole(calc)=7.728*10^{-18} esu*cm$

As,

$10^{-18}esu*cm= 1D$

So,

$Dipole(calc)=7.728D$

Now we can % ionic character using above equation:

%ionic=(0.380D/7.728D)*100

% ionic character=4.91%