1.

HBr molecule has internuclear distance of 1.27xx10^(-10)m. The electronic charge is 4.8xx10^(10) esu. Observed dipole moment is 1.03 D. find % ionic character of the bond.

Answer»

Solution :% lonic CHARACTER `=(mu_(("observed")))/(mu_(("THEO")))xx100`
`=(1.3xx100)/(1.27xx10^(-8)xx4.8xx10^(-18))=(1.03)/(6.096)xx100=16.80`


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