Question

Calculate % ionic character in HCl.

$Bondlength=1.275\stackrel{\circ }{A},{\mu }_{observed}=1.03D$

• A. $17\%$
• B. $20\%$
• C. $30\%$
• D. $50\%$

Answer 1

The correct option is A

$17\%$

$\%$ Ionic character = Observed dipole moment/ theoretical dipole moment $\times 100$ (for $100\%$ ionic bond)

Theoritical dipole moment $=4.8\times {10}^{-10}\times 1.275\times {10}^{-8}cm=6.12D=6.12\times {10}^{-18}esu.cm$

$\%IC=\left(\frac{1.03}{6.12}\right)\times 100$

$\%IC=0.168\times 100$

$\%IC=16.8\%$

From dipole moment standard scale,

When a proton (+1) and electron (-1) are separated by distance of $100pm$, the dipole moment in between is 4.8 D.