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Topic: [CHMA] Question about Dipole Moments  (Read 1347 times)

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Offline GreenAssailant

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[CHMA] Question about Dipole Moments
« on: December 10, 2011, 09:12:17 PM »
The dipole moment [μ], is given by
μ = (q)(r)

The dipole moment is measured in Debyes, which is 3.34 x 10^-39 C*m.

Where q is the charge for one electron/proton, and r is the distance between the nuclei of the bonding atoms.

We have been using 1.6 x 10^-19C for compounds like ClF, HF and LiF. These compounds all have a 1+ charge, and 1- charge. So, how would this work for a compound like MgO, where we have 2+ charge and 2- charge? Would it be

3.2 x 10^-19 C instead? This is double the regular charge.

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