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The observed dipole moment of HCl is 1.0...

The observed dipole moment of `HCl` is `1.03D`. If the bond length of HCL is `1.3Å`, then the percent ionic character of `H-Cl` bond is

A

`17%`

B

`34%`

C

`40%`

D

`10%`

Text Solution

Verified by Experts

The correct Answer is:
A
We can use dipole moment to calculate the percent ionic character. For this purpose, we calculate the dipole moment of `HCl` assuming it to be 100% ionic. If `HCl` is 100% ionic, then the bonding `e^(-)` pair will be completely in control of `Cl`. As a result, the H atom will carry one unit of positive charge while the Cl atom will carry one unit of negative charge. Assuming that during this polarization the bond length of the molecule does not change, we have
`mu_("ionic")=Qxxr`
`=(4.8xx10^(-10) "esu")(1.3xx10^(-8)cm)`
`=6.24xx10^(-18)"esu" cm`
`=6.24D`
Thus, `6.24D` implies 100% ionic character. Then `1.3D` will imply
% ionic character `=(mu_("observed"))/(mu_("ionic"))xx100%`
`=(1.03D)/(6.24D)xx100%=16.5%=17%`
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Knowledge Check

  • The dipole moment of HCl is 1.03D, if H-Cl bond distance is 1.26Å, what is the percentage of ionic character in the H-Cl bond?

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