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Question

Chemistry Question on Bond Parameters

The dipole moment of LiHLiH is 1.964×10291.964 \times 10^{-29} cm and the interatomic distance between LiLi and HH in this molecule is 1.596?1.596\, ?. The percentage of ionic character in LiHLiH is

A

75.075.0

B

76.876.8

C

79.879.8

D

100100

Answer

76.876.8

Explanation

Solution

The dipole moment of 100%100 \% ionic molecule =(1electronic charge)×(interatomic distance)= \left(1 {\text{electronic charge}}\right) \times\left({\text{interatomic distance}}\right) =1.602×1019×1.596×1010= 1.602 \times10^{-19 }\times 1.596\times 10^{-10} =2.557×1029cm= 2.557 \times 10^{-29} cm Fractional ionic character =Exp. value of dipole momentTheoretical value of dipole moment=1.964×10292.557×1029= \frac{{\text{Exp. value of dipole moment}} }{{\text{Theoretical value of dipole moment}} } =\frac{ 1.964\times10^{-29}}{2.557\times10^{-29}} =0.768 =0.768 The bond in LiHLiH is 76.8%76.8\% ionic.