Question

The $H-OH$ bond angle in water is ${105}^o$. The $H-OH$ bond distance being 0.94 $\stackrel{\circ }{\text{A}}$. The dipole moment for the molecule is 1.85 D. Calculate the charge on oxygen atom.
Given: cos ${105}^o$ = $-0.25$

• A. $1.617\times {10}^{-10}$ esu
• B. $1.617\times {10}^{-10}$ C
• C. $1.167\times {10}^{-12}$ esu
• D. $1.167\times {10}^{-15}$ esu

Answer 1

The correct option is A $1.617\times {10}^{-10}$ esu

We know, resulatnt dipole moment can be calculated using
R = $\sqrt{P^2+Q^2+2PQcos\theta }$
where R = ${\mu }_{H_{2}O}$ = Resultant dipole moment
P = ${\mu }_{OH}$ = dipole moment of $OH$ bond
So, ${\mu }_{H_{2}O}$ = $\sqrt{{\mu }_{OH}^2+{\mu }_{OH}^2+2{\mu }_{OH}^{2}cos\theta }$
Put up the values,
1.85 D = $\sqrt{2{\mu }_{OH}^2(1+cos{105}^o)}$
On solving, we get ${\mu }_{OH}$ = 1.52 D
Now 1 D = ${10}^{-18}$ esu cm
So, ${\mu }_{OH}$ = $1.52\times {10}^{-18}$ esu cm
Also, ${\mu }_{OH}$ = $\delta \times d$
where $\delta$ = partial charge on oxygen atom
d = bond distance of $O-H$ bond = $0.94\times {10}^{-8}$ cm
$\delta$ = $\frac{1.52\times {10}^{-18}}{0.94\times {10}^{-8}}$ = $1.617\times {10}^{-10}$ esu