Question 8.5: The bond length in the HCl molecule is 127 pm. (a) Calculate...
The bond length in the HCl molecule is 127 pm. (a) Calculate the dipole moment, in debyes, that results if the charges on the H and Cl atoms were 1+ and 1- respectively. (b) The experimentally measured dipole moment of HCl(g) is 1.08 D. What magnitude of charge, in units of e, on the H and Cl atoms leads to this dipole moment?
Analyze and Plan We are asked in part (a) to calculate the dipole moment of HCl that would result if there were a full charge transferred from H to Cl. We can use Equation 8.10 to obtain this result.In part (b), we are given the actual dipole moment for the molecule and will use that value to calculate the actual partial charges on the H and Cl atoms.
\mu = Qr [8.10]
Solve
(a) The charge on each atom is the electronic charge,
e = 1.60 × 10^{-19} C. The separation is 127 pm. The dipole moment is therefore:
\mu =Qr=\left(1.60 × 10^{-19} \cancel{C})(127 \cancel{pm}\right)\left(\frac{10^{-12} \cancel{m}}{1 \cancel{pm}} \right)\left(\frac{1 D}{3.34 × 10^{-30} \cancel{C}-\cancel{m}} \right)= 6.08 D
(b) We know the value of μ, 1.08 D, and the value of r, 127 pm. We want to calculate the value of Q:
Q=\frac{\mu }{r}=\frac{\left(1.08 \cancel{D}\right)\left(\frac{3.34 × 10^{-30} C- \cancel{m}}{1 \cancel{D}} \right) }{\left(127 \cancel{pm}\right)\left(\frac{10^{-12} \cancel{m}}{1 \cancel{pm}} \right) }= 2.84 × 10^{-20} CWe can readily convert this charge to units of e:
Charge in e = \left(2.84 × 10^{-20} \cancel{C}\right)\left(\frac{1e}{1.60 × 10^{-19} \cancel{C}} \right) = 0.178 e
Thus, the experimental dipole moment indicates that the charge separation in the HCl molecule is:
\overset{0.178 +}{H}ـــــــــ\overset{0.178 -}{Cl}Because the experimental dipole moment is less than that calculated in part (a), the charges on the atoms are much less than a full electronic charge. We could have anticipated this because the H—Cl bond is polar covalent rather than ionic.