The dipole moments of diatomic molecules $\mathrm{AB}$ and $\mathrm{CD}$ are $10.41 \mathrm{D}$ and $10.27…

The dipole moments of diatomic molecules $\mathrm{AB}$ and $\mathrm{CD}$ are $10.41 \mathrm{D}$ and $10.27 \mathrm{D}$, respectively while their bond distances are $2.82$ and $2.67 Å$, respectively. This indicates that
  1. bonding is $100 \%$ ionic in both the molecules
  2. $\mathrm{AB}$ has more ionic bond character than $\mathrm{CD}$
  3. $\mathrm{AB}$ has lesser ionic bond character than $\mathrm{CD}$
  4. bonding is nearly covalent in both the molecules

Solution

As dipole moment $=$ electric charge $\times$ bond length
D. M. of AB molecule
$=4.8 \times 10^{-10} \times 2.82 \times 10^{-8}=13.53 \mathrm{D}$
D.M. of CD molecule
$=4.8 \times 10^{-10} \times 2.67 \times 10^{-8}=12.81 \mathrm{D}$
Now \% ionic character
$=\frac{\text { Actual dipole moment of the bond }}{\text { Dipole moment of pure ionic compound }} \times 100$ then $\%$ ionic character in $\mathrm{AB}$ $=\frac{10.41}{13.53} \times 100=76.94 \%$
$\%$ ionic character in $\mathrm{CD}$
$=\frac{10.27}{12.81} \times 100=80.17 \%$ /

Asked in: JEE-TOPICTESTS-CHEMISTRY

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