Chemical Bonding And Molecular Structure Question 94

Question: The dipole moment of HBr is $ 1.6\times {10^{-30}} $ cm and interatomic spacing is 1Å. The % ionic character of HBr is

Options:

A) 7

B) 10

C) 15

D) 27

Show Answer

Answer:

Correct Answer: B

Solution:

Charge of $ {e^{-}}=1.6\times {10^{-19}} $ Dipole moment of $ HBr=1.6\times {10^{-30}} $ Inter atomic spacing $ =1\AA=1\times {10^{-10}}m $ % of ionic character in $ HBr=\frac{dipole,moment,of,HBr\times 100}{inter,spacing,distance,\times q} $

$ =\frac{1.6\times {10^{-30}}}{1.6\times {10^{-19}}\times {10^{-10}}}\times 100 $

$ ={10^{-30}}\times 10^{29}\times 100={10^{-1}}\times 100 $

$ =0.1\times 100=10% $



NCERT Chapter Video Solution

Dual Pane