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The dipole moment of HF molecule is 1.91...

The dipole moment of HF molecule is 1.91D and the bond distance is 0.92 Å. What is the fractional charge on H and F in the HF bond? (Electronic charge = `4.8 xx 10^(-10)` e.s.u.).

Text Solution

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Dipole moment=charge `xx` distance
`1.91xx10^(-18)` e.s.u. cm= `"Charge" xx 0.92 xx10^(-8)` cm
Charge = `1.91xx10^(-18)` e.s.u. `cm//0.92xx10^(-8) cm=2.08xx10^(-10)` e.s.u.
Therefore, fraction of electron charge, d= `("Charge")/("Electronic charge")=(2.08xx10^(-10) e.s.u.)/(4.8xx10^(-10) e.s.u.) = 0.43`
Thus `d_(H)^(+) = +0.43` and `d_(Cl)^(-)=0.43`
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