Question

The dipole moment of XY is 1.6×10⁻³⁰ Cm. if the internuclear distance is 2×10⁻¹⁰ m then % ionic character of XY is (charge of electron =1.6×10⁻¹⁹)

Answer 1

Step 1: Given data

Dipole moment of xy is $1.6\times {10}^{-30}$ Cm

Inter nuclear distance is $2\times {10}^{-10}\mathrm{m}$

Step 2: Formula used

For a $100\%$ ionic character, the dipole moment will be the product of charge and distance between two ions :
${\mathrm{\mu }}_{\mathrm{ionic}}=\mathrm{q}\times \mathrm{d}$

Step 3: Calculation of dipole moment

${\mathrm{\mu }}_{\mathrm{ionic}}$= $1.6\times {10}^{-19}\mathrm{C}\mathrm{x}2.0\mathrm{x}{10}^{-10}\mathrm{m}$
${\mathrm{\mu }}_{\mathrm{ionic}}$$=3.2\times {10}^{-29}$ Cm
Step 4: Calculating The percentage ionic character

= $$\begin{gathered} \frac{\mathrm{Observed}\mathrm{dipole}\mathrm{moment}}{\mathrm{Dipole}\mathrm{moment}\mathrm{for}\mathrm{complete}\mathrm{ionic}\mathrm{character})}\times 100 \\ =\left[\frac{(1.6\times {10}^{-30})}{(3.2\times {10}^{-29})}\right]\mathrm{x}100 \\ =5\% \end{gathered}$$
Hence, the ionic character proportion is $5\%.$