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Determine the partial negative charge on the fluorine atom in a C-F bond. The bond length is 1.39 1.39A* , and the bond dipole moment is 1.06D. The charge on an electron is 4.80⨯10-10esu.

Short Answer

Expert verified

The partial negative charge over fluorine in the bond is 0.24 D.

Step by step solution

01

Definition of Dipole moment

Dipole moment is the product of charge and the distance between the two charged positive and negative centers.

02

Determining the dipole moment

1 Angstrom = 10-8cm.

If fluorine was fully negatively charged then,

Dipole moment = charge on electron ⨯ bond length

=(4.80 ⨯10-10esu) ⨯(1.39 ⨯10-8cm)

= 6.67 ⨯10-18esu cm

= 6.67 D

03

Calculating the partial negative charge over fluorine

To find out the partial negative, we should divide the given dipole moment with the calculated dipole moment if fluorine would have been completely negatively charged.

Partial negative charge over fluorine = 1.60D/6.67D =0.24D

Thus, the partial negative charge over fluorine in the bond is 0.24 D.

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