Question

if dipole moment of Hcl is 3.2*10-³¹cm.Fimd out percent ionic charecter if d=1A
note:1A=10¹0
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Answer 1

Thus the percentage of ionic character is 66 %

Explanation:

We are given that:

• As bond length of HCl = 1 A
• Dipole moment of HCl = 4.8×10−10
• To find: Percentage ionic character = ?

Solution:

μ ionic = q x d = 4.8×10−10  x 1 A = 4.8×10−10

μobserved = 3.2 x 10^-31  cm

% ionic character = μobserved  / μ ionic  

% ionic character = 3.2  / 4.8  x 100

% ionic character = 0.66 x 100

% ionic character = 66 %

Thus the percentage of ionic character is 66 %