Question

Question:

(A)The$H-Cl$ bond length is $136pm$. What would the dipole moment of $H-Cl$ be if the molecule were $100\%$ionic, $H^{+}C{l}^{-}$?

(B) The actual dipole moment of HCl is $1.08D$. What is the percent

ionic character of the$H-Cl$bond?

Short answer

Answer

The dipole moment, $\mu =6.523D$.

$Percentioniccharacter=16.5\%$.

Step-by-step solution

Dipole moment

The dipole momentis defined as the magnitude of the charge $\left(Q\right)$ at both ends of the molecular dipole multiplied by the distance r between the charges which is the bond length.

$\mu =Q\times r$

Dipole moments are expressed in debyes(D), where $1D=3.336\times {10}^{-30}\left(C.m\right)$, coulomb meters in SI units. Bond length is usually in pm which is ${10}^{-12}$m.

Step 2: Percent ionic character

Ionic character refers to the percentage of the difference between the electronegativity of two covalently bonded atoms.

It is calculated as:

$Percentioniccharacter=\frac{{\mu }_{Experimental}}{{\mu }_{Theoretical}}$

Calculation of dipole moment

Formula to calculate dipole moment is:

$\mu =Q\times r$

$\mu =\left(1.6\times {10}^{-19}C\right)\left(136\times {10}^{-12}m\right)\left(\frac{1D}{3.336\times {10}^{-30}C.m}\right)$

$\mu =6.523D$

Calculation of percent ionic character

Formula to calculate percent ionic character:

$Percentageioniccharacter=\frac{{\mu }_{Experimental}}{{\mu }_{Theoretical}}$

$=\frac{1.08}{6.523}\times 100$

$=16.56\%$