Q34P (II) The dipole moment, consider... [FREE SOLUTION]

91影视

Chapter 17: Q34P (page 473)

(II) The dipole moment, considered as a vector, points from the negative to the positive charge. The water molecule, Fig. 17鈥�42, has a dipole moment \({\bf{\vec p}}\) which can be considered as the vector sum of the two dipole moments, \({{\bf{\vec p}}_{\bf{1}}}\) and \({{\bf{\vec p}}_{\bf{2}}}\) as shown. The distance between each H and the O is about \({\bf{0}}{\bf{.96 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}\;{\bf{m}}\). The lines joining the centre of the O atom with each H atom make an angle of 104掳, as shown, and the net dipole moment has been measured to be \({\bf{p = 6}}{\bf{.1 \times 1}}{{\bf{0}}^{{\bf{ - 30}}}}\;{\bf{C}} \cdot {\bf{m}}\). Determine the charge q on each H atom.
FIGURE 17鈥�42 Problem 34

Short Answer

Expert verified

The charge on each H atom is\(5.16 \times {10^{ - 20}}\;{\rm{C}}\).

Step by step solution

Understanding of Dipole moment

The product of magnitude of either charge and the distance between the two charges is termed as dipole moment.

The dipole moment is given as:

\(p = Ql\)

Here, p is the dipole moment, Q is the charge and l is distance between the two equal and opposite charges.

Since it is considered that the magnitude of both dipole moment is same.So, the angle made by each dipole will be half of the angle made by H atoms.

Given Data

The dipole moment is,\(p = 6.1 \times {10^{ - 30}}\;{\rm{C}} \cdot {\rm{m}}\).

The separation distance is,\(l = 0.96 \times {10^{ - 10}}\;{\rm{m}}\).

The angle is,\(\theta = 104^\circ \)

Evaluation of the charge on the H atom

The relation of net dipole momentis given by,

\(\begin{aligned}p &= 2{p_1}\cos \left( {\frac{\theta }{2}} \right)\\p &= 2\left( {Q \times l} \right)\cos \left( {\frac{\theta }{2}} \right)\\Q &= \frac{p}{{2l\cos \left( {\frac{\theta }{2}} \right)}}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}Q &= \frac{{\left( {6.1 \times {{10}^{ - 30}}\;C \cdot m} \right)}}{{2\left( {0.96 \times {{10}^{ - 10}}\;m} \right)\cos \left( {\frac{{104^\circ }}{2}} \right)}}\\ &= \frac{{\left( {6.1 \times {{10}^{ - 30}}} \right)}}{{2\left( {0.96 \times {{10}^{ - 10}}} \right)\cos \left( {\frac{{104^\circ }}{2}} \right)}}\\Q &= 5.16 \times {10^{ - 20}}\;C\end{aligned}\)

Thus, the charge on each H atom is \(5.16 \times {10^{ - 20}}\;{\rm{C}}\).