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Question: Compare the magnitude of the partial charge on the hydrogen atom in \({\text{HCl}}\) with that of a ...

Compare the magnitude of the partial charge on the hydrogen atom in HCl{\text{HCl}} with that of a singly charged positive ion which is 4.80×10−10esu4.80 \times {10^{ - 10}}{\text{esu}}
A. The charge on the hydrogen atom of HCl{\text{HCl}} is about one-third the charge on a monopositive ion.
B. The charge on the hydrogen atom of HCl{\text{HCl}} is about one-fourth the charge on a monopositive ion.
C. The charge on the hydrogen atom of HCl{\text{HCl}} is about one-sixth the charge on a monopositive ion.
D. The charge on the hydrogen atom of HCl{\text{HCl}} is about one-fifth the charge on a monopositive ion.

Explanation

Solution

A system that contains two separate charges of opposite sign is called an electric dipole. It is described quantitatively by the dipole moment, μ→\overrightarrow \mu . The dipole moment is a vector quantity since it has both direction and magnitude.

Complete step by step answer:
Dipole moment refers to the quality of a system to behave like a dipole. It is defined as the product of the magnitude of charge and the distance between two bonded atoms. Its formula is given below:
Dipole moment, μ→\overrightarrow \mu ==Charge, Q{\text{Q}} ×\times distance between the atoms, R{\text{R}} →(1) \to \left( 1 \right)
It is given that the charge on monopositive ion =4.8×10−10esu = 4.8 \times {10^{ - 10}}{\text{esu}}
Dipole moment of HCl{\text{HCl}} molecule, μ=1.03D\mu = 1.03{\text{D}}. Unit of dipole moment is Debye.
Since the unit given in the question is in esu{\text{esu}}, we have to convert Debye to esu{\text{esu}}
1D=10−18esu1{\text{D}} = {10^{ - 18}}{\text{esu}}
Hence dipole moment, μ=1.03×10−18esu\mu = 1.03 \times {10^{ - 18}}{\text{esu}}
And the distance between the atoms, R=127pm = 1.27×10−8cm{\text{R}} = 127{\text{pm = 1}}{\text{.27}} \times {\text{1}}{{\text{0}}^{ - 8}}{\text{cm}}
Substituting these values in (1)\left( 1 \right), we get
μ=Q×R\mu = {\text{Q}} \times {\text{R}}
1.03×10−18esu = Q×1.27×10−8cm1.03 \times {10^{ - 18}}{\text{esu = Q}} \times {\text{1}}{\text{.27}} \times {\text{1}}{{\text{0}}^{ - 8}}{\text{cm}}
Hence the charge can be determined.
Q=1.03×10−181.27×10−8=8.11×10−11esu{\text{Q}} = \dfrac{{1.03 \times {{10}^{ - 18}}}}{{1.27 \times {{10}^{ - 8}}}} = 8.11 \times {10^{ - 11}}{\text{esu}}
Thus charge on the hydrogen atom on HCl{\text{HCl}} =8.11×10−11esu = 8.11 \times {10^{ - 11}}{\text{esu}}
Now we have to compare the hydrogen atom on HCl{\text{HCl}} with the charge on a monopositive ion, thereby finding the ratio between them.
It is given that charge on monopositive ion =4.8×10−10esu = 4.8 \times {10^{ - 10}}{\text{esu}}
Ratio of charge of hydrogen atom on HCl{\text{HCl}} to the charge on monopositive ion=8.11×10−11esu4.8×10−10esu=16 = \dfrac{{8.11 \times {{10}^{ - 11}}{\text{esu}}}}{{4.8 \times {{10}^{ - 10}}{\text{esu}}}} = \dfrac{1}{6}

Hence option C is correct .

Additional Information:
Dipole moment has many applications:
In determining the polarity of bonds
In determining the symmetry of the molecules
To distinguish between cis and trans isomer.
To distinguish between ortho, meta and para isomers.

Note:
Bigger the electronegativity difference more the polar bond. The centers of positive and negative charge do not coincide. Such molecules behave as permanent dipoles. Molecular dipole depends upon the geometry of the molecule. HCl{\text{HCl}} has a polar covalent bond.