Answer
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Hint: For calculation of ionic character in a molecule, we need to determine the dipole moment within the molecule. The dipole moment is the physical property which is a measure of the polarity of a chemical bond between two atoms in a molecule. The higher the electronegativity difference between the two atoms, the more will be the ionic character to the bond and vice versa.
Complete answer:
- The dipole moment is the mathematical product of the total amount of positive charge or negative charge and the distance between the center of the charge distribution.
-For calculating the percent ionic character in a molecule, we need to follow the steps below-
(a) We start by calculating the dipole moment by taking the product of the magnitude of charge and separation between charges, i.e
\[\mu =q\times r\]
where $\mu $ is the dipole moment
q is the separated charge
r is the distance between them
Here, q= $1.6\times {{10}^{-19}}C$ ; r= \[1.275A{}^\circ \times {{10}^{-10}}m\]
So, dipole moment ($\mu $ ) $=1.6\times {{10}^{-19}}\times 1.275\times {{10}^{-10}}m$
$=2.04\times {{10}^{-29}}cm$
(b) Now, we have to convert the centimeter into the standard unit of dipole moment, i.e Debye.
We know that, $1D=3.335\times {{10}^{-30}}cm$
\[=\dfrac{2.04\times {{10}^{-29}}}{(3.335}\] \[=\dfrac{2.04\times {{10}^{-29}}}{\begin{align}
& 3.335\times {{10}^{-30}} \\
& =\dfrac{20.4}{3.335} \\
& =6.1169D \\
\end{align}}\]
(c) Now, percentage ionic character \[=\dfrac{\exp value}{theoretical value}\times 100\]
\[=\dfrac{1.03D}{6.1169D}\times 100\]
$=16.83%$
Note: For calculating the dipole moment you need to remember the charge of an electron i.e,$1.6\times {{10}^{-19}}$ . For the calculation of percentage ionic character, remember the conversion units i.e,$1D=3.335\times {{10}^{-30}}cm$ .
Complete answer:
- The dipole moment is the mathematical product of the total amount of positive charge or negative charge and the distance between the center of the charge distribution.
-For calculating the percent ionic character in a molecule, we need to follow the steps below-
(a) We start by calculating the dipole moment by taking the product of the magnitude of charge and separation between charges, i.e
\[\mu =q\times r\]
where $\mu $ is the dipole moment
q is the separated charge
r is the distance between them
Here, q= $1.6\times {{10}^{-19}}C$ ; r= \[1.275A{}^\circ \times {{10}^{-10}}m\]
So, dipole moment ($\mu $ ) $=1.6\times {{10}^{-19}}\times 1.275\times {{10}^{-10}}m$
$=2.04\times {{10}^{-29}}cm$
(b) Now, we have to convert the centimeter into the standard unit of dipole moment, i.e Debye.
We know that, $1D=3.335\times {{10}^{-30}}cm$
\[=\dfrac{2.04\times {{10}^{-29}}}{(3.335}\] \[=\dfrac{2.04\times {{10}^{-29}}}{\begin{align}
& 3.335\times {{10}^{-30}} \\
& =\dfrac{20.4}{3.335} \\
& =6.1169D \\
\end{align}}\]
(c) Now, percentage ionic character \[=\dfrac{\exp value}{theoretical value}\times 100\]
\[=\dfrac{1.03D}{6.1169D}\times 100\]
$=16.83%$
Note: For calculating the dipole moment you need to remember the charge of an electron i.e,$1.6\times {{10}^{-19}}$ . For the calculation of percentage ionic character, remember the conversion units i.e,$1D=3.335\times {{10}^{-30}}cm$ .
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