Question

The dipole moment of HBr is 2.60 X10^-30 Cm and the interatomic distance is 141 pm. What is the percentage ionic character of HBr.

Answer 1

HBr is a ionic compund because , H + and Br- ions occur in a small distance, (e.g 141 pm) just like dipole .


hence,

dipole moment of HBr = charge × distance

= 1.6 × 10^-19 × 141 × 10^-12 Cm
= 1.6 × 141 × 10^-31 Cm
=225.6 × 10^-31 Cm
= 22.56 × 10^-30 Cm
now,
% ionic character = experimental D.P/theoretical D.P × 100

={2.6 × 10^-30/22.56 × 10^-30} × 100
=260/22.56
=11.54 %

Answer 2

Explanation:

As we known that

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\% \: of \: ionic \: character \:  =  \frac{observed \: dipole \: movement}{theoritical \: dipole \: movement}  \times 100 \\  

So

observed \: dipole \: movement \:  = 2.6 \times  {10}^{ - 30}  

Also

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theoritical \: dipole \: movement = q \times d \\  =  > 1.6 \times 10 {}^{ - 19}  \times 141 \times 10 {}^{ - 12}  \\  =  > 22.56 \times 10 {}^{ - 30}  

So

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ionic \: \% =  \frac{2.6 \times  {10}^{ - 30} }{22.56 \times  {10}^{ - 30} }  \times  100 \\  =  >  \frac{260}{22.56} \% \\  =  > 11.52