Question

The experimental value of the dipole moment of $HCl$ is $1.03D$ . The length of the $H-Cl$ bond is $1.275Å$ . The percentage of ionic character in $HCl$ is:

[Given: $1\text{D}=10^{-18}\text{esu}\text{cm}$ ]

• A. 43
• B. 21
• C. 17
• D. 7

Answer 1

Answer: (C)

$\%$ Ionic character $=\frac{\text{ Experimental value of dipole moment }}{\text{ Theoritical value of dipolemoment }}\times 100$

$=\frac{{\mu }_{\mathrm{cxp}}}{{\mu }_{th}}\times 100$

Where:

${\mu }_{\exp }=1.030$ and ${\mu }_{\text{theo }}=|q|\times d$

$d=1.275AA=1.275\times 10^{-8}cm$

$|q|=4.8\times 10^{-10}$ esu

${\mu }_{\text{theo }}=4.8\times 10^{-10}\times 1.275\times 10^{-8}$ esu $-cm$

$=4.8\times 1.275\times 10^{-18}esu-cm$

$=4.8\times 1.275D$

$1D=10^{-18}esucm$

$\%$ Ionic character $=\frac{1.03}{4.8\times 1.275}\times 100=17$