Question

The percentage ionic character of a single bond may be estimated from ratio of the observed dipole moment to the calculated moment, assuming oppositely charged ions located at a distance equal to the bond length. The observed dipole moments of $HCl, HBr$ and $HI$ are $1.03, 0.79$ and $0.38 D$ , respectively. Calculate the percentage ionic character of the bond in each of these compounds. Do the results obtained are parallel to the magnitudes of the extra ionic resonance energies, $δ$ , for these molecules?

(Given, then $d_{H - Cl} = 1 \cdot 27 \overset{\circ}{A}, d_{H - Br} = 1 \cdot 42 \overset{\circ}{A}, d_{H - I} = 1.61 \overset{\circ}{A}$ ).

Accepted Answer

Given: $d_{H - Cl} = 1.27 \overset{\circ}{A}$
$d_{H - Br} = 1.42 \overset{\circ}{A}$
$d_{H - I} = 1.61 \overset{\circ}{A}$

Dipole moment, $μ = δ$ (charge) $\times d$ (distance).

For $HCl$ :

$δ = \frac{μ}{d} = \frac{1.03 \times 10^{- 18} esu \cdot cm}{1.27 \times 10^{- 8} cm} = 8.11 \times 10^{- 11} esu$ .

If $HCl$ is $100 %$ ionic, each end would carry charge equal to one unit, $4.8 \times 10^{- 10} esu$

$= \frac{q}{e} \times 100 = \frac{8.11 \times 10^{- 11}}{4.8 \times 10^{- 10}} \times 100 = 16.9 %$ ionic for $HCl$ .

For $HBr$ :

$δ = \frac{0.79 \times 10^{- 18} esu \cdot cm}{1.42 \times 10^{- 8} cm} = 5.6 \times 10^{- 11} esu$ .

If $HBr$ is $100 %$ ionic, each end would carry charge equal to one unit $4.8 \times 10^{- 10} esu$

$= \frac{q}{e} \times 100 = \frac{5.6 \times 10^{- 11}}{4.8 \times 10^{- 10}} \times 100 = 12 %$ ionic for $HBr$ .

For $HI$ :

$δ = \frac{0.38 \times 10^{- 18} esu \cdot cm}{1.61 \times 10^{- 8} cm} = 2.4 \times 10^{- 11} esu$ .

$= \frac{q}{e} \times 100 = \frac{2.4 \times 10^{- 11}}{4.8 \times 10^{- 10}} \times 100 = 5.0 %$ ionic for $HI$ .

Using pauling equation: Pauling scale is based on the empirical relation between the energy of the bond
and the electronegativities of the bonded atoms in molecules.

$Δ = {[\frac{EN (X) - EN (H)}{0.208}]}^{2} = {(\frac{0.9}{0.208})}^{2} = 18$ for $HCl$

$= {(\frac{0.7}{0.208})}^{2} = 11$ for $HBr = {(\frac{0.4}{0.208})}^{2} = 4$ for $HI$ .

The result parallels the value of $δ$ rather well in these cases.

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