Question

Dipole moments: ${\text{HCl}}\left( {{\text{1}}{\text{.03D}}} \right){\text{ and HI}}\left( {{\text{0}}{\text{.38D}}} \right)$. Compare the magnitude of the partial charge on the hydrogen atom in ${\text{HCl}}$ with that on the hydrogen atom of ${\text{HI}}$. If it is assumed that the partial positive and negative charges are central on the hydrogen and halogen atoms, respectively. the distance between charge centers are calculated to be ${\text{1}}{\text{.27 and 1}}{\text{.61 }}^\circ {\text{A for HCl and HI}}$ respectively;
A ) The charge on the hydrogen atom of ${\text{HCl}}$ is over 3 times that on the hydrogen atom of ${\text{HI}}$.
B ) The charge on the hydrogen atom of ${\text{HCl}}$ is over 2 times that on the hydrogen atom of ${\text{HI}}$.
C ) The charge on the hydrogen atom of ${\text{HCl}}$ is over 4 times that on the hydrogen atom of ${\text{HI}}$.
D ) The charge on the hydrogen atom of ${\text{HCl}}$ is over 5 times that on the hydrogen atom of ${\text{HI}}$.

Answer 1

Dipole moment gives the measure for bond polarity. Electric dipole moment arises due to charge separation. Dipole moment is the product of the charge and the distance between two charges.

Complete step by step answer:
When a chemical bond is present between two atoms of elements having different electronegativities, a bond dipole is formed. One end of bond gets partial positive charge and the other end of bond gets partial positive charge. Such a bond dipole is a result of charge separation and is associated with dipole moment. Dipole moment is a vector quantity as it has magnitude and direction. The dipole moment of the molecule is the vector addition of individual bond dipoles.
When a molecule has charge separation, it has a dipole moment.
Write the formula for the dipole moment:
$\mu = \delta \times d$
Write the expression for the ${\text{HCl}}$ molecule.
$\mu \left( {{\text{HCl}}} \right) = \delta \left( {{\text{HCl}}} \right) \times d\left( {{\text{HCl}}} \right)$
Substitute values in the above expression:

\delta \left( {{\text{HCl}}} \right) = \dfrac{{1.03}}{{1.27}} = 0.811 \\\\$$ Write the expression for $${\text{HI}}$$ molecule. $$\mu \left( {{\text{HI}}} \right) = \delta \left( {{\text{HI}}} \right) \times d\left( {{\text{HI}}} \right)$$ Substitute values in the above expression: $$0.38 = \delta \left( {{\text{HCl}}} \right) \times 1.61 \\\ \delta \left( {{\text{HCl}}} \right) = \dfrac{{0.38}}{{1.61}} = 0.236 \\\\$$ Calculate the ratio $$\dfrac{{\delta \left( {{\text{HCl}}} \right)}}{{\delta \left( {{\text{HI}}} \right)}}$$ : $$\dfrac{{\delta \left( {{\text{HCl}}} \right)}}{{\delta \left( {{\text{HI}}} \right)}} = \dfrac{{0.811}}{{0.236}} \\\ = 3.4 \\\\$$ The charge on the hydrogen atom of $${\text{HCl}}$$ is over 3 times that on the hydrogen atom of $${\text{HI}}$$. _**Hence, the A ) is the correct option.**_ **Note:** Do not arrive at the answer by considering only the value of dipole moments as it will lead to wrong answers. Also consider the charge separation in the calculations.