Question

The iodine bromide molecule (IBr) has a
bond length of 2.49 A° and a dipole moment
of 1.21 D. Calculate the effective charges
on the I and Br atoms in IBr in units of
electronic charge, e. (1D = 3.34 x 10 -3° C m,
electronic charge, e= 1.602 x 10 -19 C)

Answer 1

Answer on Question #48814 - Chemistry - Inorganic Chemistry

Question:

The iodine bromide molecule (IBr) has a bond length of $2.49\,\mathrm{A}^{\circ}$ and a dipole moment of 1.21 D. Calculate the effective charges on the I and Br atoms in IBr in units of electronic charge, e. ($1\,\mathrm{D} = 3.34 \times 10^{-3}\,\mathrm{C}\,\mathrm{m}$, electronic charge, $e = 1.602 \times 10^{-19}\,\mathrm{C}$)

Solution:

The dipole moment is given by $\mu = qr$, where $q$ is the charge separated (in C) and $r$ is the distance separating the charge. Since we're given the dipole moment (1.21 D) and the bond length (2.49 Angstrom), we solve for the charge $q$. Dipole moments are given in D units - recall that $1\,\mathrm{D} = 3.34 \times 10^{-30}\,\mathrm{C}\,\mathrm{m}$. Convert the bond length into meters:

$$r = 2.49\,\mathrm{Angstrom} \left(\frac{1 \times 10^{-10}\,\mathrm{m}}{1\,\mathrm{Angstrom}}\right)$$ $$r = 2.49 \times 10^{-10}\,\mathrm{m}$$

Solve the definition of the dipole moment for the charge

$$q = \frac{\mu}{r}$$ $$q = 1.623 \times 10^{-20}\,\mathrm{C}$$

So, the amount of charge separated by the difference in EN is $1.623 \times 10^{-20}\,\mathrm{Coulombs}$. If you need the answer in units of electronic charge e:

$$e = 1.60 \times 10^{-19}\,\mathrm{C}$$ $$q = 1.623 \times 10^{-20}\,\mathrm{C} \left(\frac{1 \times e}{1.60 \times 10^{-19}\,\mathrm{C}}\right)$$ $$q = 0.101\,\mathrm{e}$$

Answer: Br has a charge of -0.101 and I has a charge of +0.101.

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