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Question

HBr has dipole moment 2.6×1030 Cm. If the ionic character of the bond is 11.5 %, the interatomic spacing(in oA) is (write the value to the nearest integer) :

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Solution

Let the inter atomic spacing be x cm.
Its dipole moment assuming 100 % ionic character would be the product of its charge and the inter atomic distance.
It would be 4.8×1010esu×xcm=4.8x×1010esu.cm
But 1esu.cm=3.356×1012C.m

Hence, the dipole moment assuming 100% ionic character is
4.8x×1010×3.3356×1012=1.601x×1021Cm
But actual dipole moment is 2.6×1030Cm
Percent ionic character is the ratio of the observed dipole moment to the dipole moment assuming 100% ionic character. This ratio is multiplied by 100.
11.5=2.6×1030Cm1.601x×1021Cm×100
x=1.4×108cm=1.4Ao

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