The water molecule's dipole moment is \(6.2 \times 10^{-18} \mathrm{C} \cdot \mathrm{m}\). What would be the separation distance if the molecule consisted of charges \(\pm e\) ? (The effective charge is actually less because \(\mathrm{H}\) and O atoms share the electrons.)

Short Answer

Expert verified
The separation distance \(d\) is approximately 0.387 \(\mathrm{nm}\)

Step by step solution

01

Understand the given parameters

The dipole moment of the water molecule is given as \(6.2 \times 10^{-18} \, \mathrm{C} \cdot \mathrm{m}\) and the charges represented in the molecule is the charge of an electron, denoted as \(\pm e\), where \(e = 1.60217662 \times 10^{-19}\, \mathrm{C}\). These parameters will be used in the formula of dipole moment.
02

Use the formula of dipole moment

The dipole moment formula is \(p = q \cdot d\), where \(p\) is the dipole moment, \(q\) is the charge and \(d\) is the separation distance.
03

Solve for the separation distance

Substitute the values of \(p\) and \(q\) in the formula and solve for \(d\): \(d = \frac{p}{q} = \frac{6.2 \times 10^{-18} \, \mathrm{C} \cdot \mathrm{m}}{1.60217662 \times 10^{-19}\, \mathrm{C}}\)
04

Calculation

Calculate the above expression to obtain the separation distance.

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