The dipole moment of H₂O is 1.85D.HOH angle is 105°. O-H bond length is 0.94A˚. The magnitude of charge-separated on the oxygen atom is (cos 52.5° − 0.61)

The correct option is B

1/3 of the magnitude of the charge on the electron

The explanation for the correct answer:-

Option (B)$\frac{1}{3}$of the magnitude of the charge on the electron.

Step 1: Given

Bond length =$0.94A˚$

Bond angle

= $\cos 52.5°$

=$0.61$

The dipole moment of water is $1.85D$

Step 2: Formula used

Dipole moment $\mathrm{\mu }=\mathrm{charge}\times \mathrm{displacement}\mathrm{distance}$

$\mathrm{e}=\frac{\mathrm{\mu }}{\mathrm{d}}$

Step 3: Calculating displacement distance

$\mathrm{d}=\mathrm{l}\cos 52.5^{\mathrm{o}}$

$$\begin{gathered} =0.94\times 0.61 \\ =0.572A˚ \\ =0.572\times {10}^{-8}cm \end{gathered}$$

Step 4: Evaluating the charge of electron

Putting all the values above we get,

$$\begin{gathered} \mathrm{e}=\frac{1.85\times {10}^{-18}\mathrm{esu}\mathrm{cm}}{0.572\times {10}^{-8}\mathrm{cm}} \\ =3.23\times {10}^{-10}\mathrm{esu} \end{gathered}$$

Which is $1/3$ of the charge on an electron.

Hence, option (B) is the correct answer.