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

we know ,
HBr is an ionic compound in which H has + 1 and Br has -1 ion
e.g charge one each have equal to charge of one electron or charge of proton .
dipole moment =charge x distance between them
=1.6 x 10^-19 x 141 x 10^-12 Cm
=225.6 x 10^-31 Cm
=22.56 x 10^-30 Cm
now,
% of ionic characters
=dipole moment by experimental/dipole moment by theoretical x100
=(2.6 x 10^-30/22.56 x 10^-30 ) x 100
=260/22.56 %
=11.52 %

Answer 2

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

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