Q.2-2-31b - Organic Chemistry (Mcmurry)

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?

Step-by-Step Solution

Answer

Answer

The dipole moment, $\mu =6.523 D$ .

$Percent ionic character = 16.5\%$.

Step 1: 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:

$Percent ionic character = \frac{{\mu }_{Experimental}}{{\mu }_{Theoretical}}$

 

Step 3: Calculation of dipole moment

Formula to calculate dipole moment is:

 $\mu =Q\times r$

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

 $\mu =6.523D$

Step 4: Calculation of percent ionic character

Formula to calculate percent ionic character:

 $Percentage ionic character =\frac{ {\mu }_{Experimental}}{{\mu }_{Theoretical}}$

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

$=16.56\%$