Question

The $H-O-H$ bond angle in the water molecule is ${105}^o$, the $H-O$ bond distance being $0.94\mathring{A}$. The dipole moment for the molecule is $1.85D$.The charge on the oxygen atom is $X\times {10}^{-10}esu$ $cm$. Find the value of $X$. Write the answer to the nearest interger.

Answer 1

$\therefore {\mu }_{H_{2}O}=\sqrt{{\mu }_{OH}^2+{\mu }_{OH}^2+2{\mu }^{2}cos\left(105\right)}$

Since $H_{2}O$ has two vectors of $O-H$ bond acting at ${105}^{\circ }$ . Let dipole moment of $O-H$ bond be ${}^{\prime }{a}^{\prime }$

$\therefore$ $1.85=\sqrt{2a^2(1+cos105)}$

or $a$ , i.e., ${\mu }_{O-H}=1.52$ debye = $1.52\times {10}^{-18}esucm$

Now $M_{O-H}=\delta \times d$ where $\delta$ is charge on either end

$\therefore$ $1.52\times {10}^{-18}=\delta \times 0.94\times {10}^{-8}$

$\therefore$ $\delta =1.617\times {10}^{-10}esu$

Since $O$ acquire $2\delta$ charge , one $\delta$ charge from each bond and thus

Charge on $O$ atom = $2\delta =2\times 1.617\times {10}^{-10}=3.23\times {10}^{-10}esucm$

$X\simeq 3$