Answer
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Hint: The dipole moment is calculated by multiplying the charge of the electron with the distance separated by the ions. The percent ionic character is calculated by dividing observed dipole moment with the theoretical dipole moment and multiplying with 100.
Complete step by step answer:
The observed dipole moment of HBr is $2.60 \times {10^{ - 30}}$ Cm.
The interatomic spacing is 1.41 \[\mathop A\limits^o\].
The dipole moment in a molecule arises due the separation of charge by the atoms into anion and cation. The dipole moment is due to the electronegativity difference of the atoms present in the molecule.
The dipole moment is measured by multiplying the magnitude of the charge by the distance between the two ions cations and anions.
The formula for calculating the dipole moment is shown below.
\[D = Q \times r\]
Where,
D is the dipole moment.
Q is the charge
r is the distance between the ions.
The dipole moment is calculated in terms of Debye unit which is denoted by D.
The charge of the electron is \[1.6 \times {10^{ - 19}}C\].
To calculate the theoretical dipole moment of HBr, substitute the values in the above equation.
\[\Rightarrow D = (1.6 \times {10^{ - 19}}) \times (1.41 \times {10^{ - 10}})\]
\[\Rightarrow D = 2.256 \times {10^{ - 29}}\]
The percent ionic character is calculated by dividing observed dipole moment by the theoretical dipole moment multiplied by 100.
The formula to calculate the % ionic character is shown below.
\[\% ionic\,character = \dfrac{{Observed\;dipole\;moment}}{{Theoretical\;dipole\;moment}}\]
To calculate the %ionic character of the HBr, substitute the values in the above equation.
\[\Rightarrow \% ionic\,character = \dfrac{{2.6 \times {{10}^{ - 30}}}}{{2.256 \times {{10}^{ - 29}}}} \times 100\]
\[\Rightarrow \% ionic\,character = 11.5\%\]
Thus, the %ionic character of HBr is 11.5%.
So, the correct answer is Option C.
Note: The dipole moment exists in both ionic compound and covalent compound. It is a vector quantity since it contains both direction and magnitude. Make sure to convert the given distance value in Armstrong into meters.
Complete step by step answer:
The observed dipole moment of HBr is $2.60 \times {10^{ - 30}}$ Cm.
The interatomic spacing is 1.41 \[\mathop A\limits^o\].
The dipole moment in a molecule arises due the separation of charge by the atoms into anion and cation. The dipole moment is due to the electronegativity difference of the atoms present in the molecule.
The dipole moment is measured by multiplying the magnitude of the charge by the distance between the two ions cations and anions.
The formula for calculating the dipole moment is shown below.
\[D = Q \times r\]
Where,
D is the dipole moment.
Q is the charge
r is the distance between the ions.
The dipole moment is calculated in terms of Debye unit which is denoted by D.
The charge of the electron is \[1.6 \times {10^{ - 19}}C\].
To calculate the theoretical dipole moment of HBr, substitute the values in the above equation.
\[\Rightarrow D = (1.6 \times {10^{ - 19}}) \times (1.41 \times {10^{ - 10}})\]
\[\Rightarrow D = 2.256 \times {10^{ - 29}}\]
The percent ionic character is calculated by dividing observed dipole moment by the theoretical dipole moment multiplied by 100.
The formula to calculate the % ionic character is shown below.
\[\% ionic\,character = \dfrac{{Observed\;dipole\;moment}}{{Theoretical\;dipole\;moment}}\]
To calculate the %ionic character of the HBr, substitute the values in the above equation.
\[\Rightarrow \% ionic\,character = \dfrac{{2.6 \times {{10}^{ - 30}}}}{{2.256 \times {{10}^{ - 29}}}} \times 100\]
\[\Rightarrow \% ionic\,character = 11.5\%\]
Thus, the %ionic character of HBr is 11.5%.
So, the correct answer is Option C.
Note: The dipole moment exists in both ionic compound and covalent compound. It is a vector quantity since it contains both direction and magnitude. Make sure to convert the given distance value in Armstrong into meters.
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