Question

The experimental bond moment (approximate, in Debye) of the X-Y bond, if electronegativity of element X and Y are 1 and 2.5, respectively, and bond length is 3.92\[\overset{\text{o}}{\mathop{\text{A}}}\,\] is:

Answer 1

Hint: Bond moment in a molecule is defined as the dipole moment of each individual bond in that molecule. The bond formed between two atoms in a molecule with different electronegativities will be polar in nature and have some dipole moment.

Complete step by step solution:
In the question it is given that the electronegativity of elements X and Y are 1 and 2.5, and the bond length is 3.92\[\overset{\text{o}}{\mathop{\text{A}}}\,\].
Initially we have to calculate the electronegativity difference in the given molecule.
\[\Delta \]X = electronegativity of Y – electronegativity of X
        = 2.5-1.0 = 1.5
Now from the electronegativity difference we can calculate the percentage of ionic character of the molecule.
Formula to calculate Percentage ionic character = \[16\Delta X+3.5{{(\Delta X)}^{2}}\]
There is a relation between Percentage ionic character and the dipole moment of the molecule.

Percentage ionic character = \[\dfrac{{{\mu }_{\exp er}}}{{{\mu }_{theory}}}\times 100\]
\[\begin{align}
  & 16\Delta X+3.5{{(\Delta X)}^{2}}=\dfrac{{{\mu }_{\exp er}}}{{{\mu }_{theory}}}\times 100 \\
 & 16(1.5)+3.5{{(1.5)}^{2}}=\dfrac{{{\mu }_{\exp er}}\times \dfrac{1}{3}\times {{10}^{29}}}{1.6\times {{10}^{-19}}\times 3.92\times {{10}^{-10}}}\times 100 \\
 & {{\mu }_{\exp er}}=5.99=6 \\
\end{align}\]
Therefore, the experimental bond moment (approximate, in Debye) of the X - Y bond is 6.

Note: Don’t be confused with percentage ionic character and bond moment of the molecule.
Percentage ionic character decides the bond moment of the molecule. If the percentage ionic character for a particular molecule is zero then the bond moment of the molecule is also zero. Both are directly proportional to each other for the given molecule.