Question

The iodine monobromide molecule, IBr, has a bond length of \(2.49 \mathrm{~A}\) and a dipole moment of \(1.21 \mathrm{D}\). (a) Which atom of the molecule is expected to have a negative charge? Explain. (b) Calculate the effective charges on the \(\mathrm{I}\) and \(\mathrm{Br}\) atoms in IBr, in units of the electronic charge, \(e\).

Short answer

The Bromine (Br) atom is expected to have a negative charge because it is more electronegative than the Iodine (I) atom. In the IBr molecule, the effective charges on the Iodine (I) and Bromine (Br) atoms are approximately +0.1011e and -0.1011e, respectively.

Step-by-step solution

Determine the electronegativity difference between Iodine (I) and Bromine (Br)

From the periodic table, we know the electronegativities of Iodine and Bromine are 2.5 and 2.8 respectively. Now, calculate the difference in electronegativities: Electronegativity difference = \(|2.8 - 2.5|\) = \(0.3\) Since the electronegativity difference is positive, it means the Bromine (Br) atom is more electronegative than Iodine (I), and will have a partial negative charge.

Answer the first part of the question

The Bromine (Br) atom is expected to have a negative charge because it is more electronegative than the Iodine (I) atom.

Use the dipole moment and bond length to compute the effective charges

We are given the bond length as \(2.49 \mathrm{~A}\) and the dipole moment as \(1.21 \mathrm{D}\). In order to calculate the effective charges, we need to convert these units into SI units. Convert the bond length into meters: \(2.49 \mathrm{~A} * (1 \times 10^{-10} \mathrm{m/A}) = 2.49 \times 10^{-10} \mathrm{m}\) Convert the dipole moment into Coulomb meters: \(1.21 \mathrm{D} * (3.336 \times 10^{-30} \mathrm{C\cdot m/D}) = 4.03616 \times 10^{-30} \mathrm{C\cdot m}\) Now, we can relate the dipole moment to the effective charge and bond length using the following formula: Dipole moment (µ) = Effective charge (Q) × Bond length (r) So, we can calculate the effective charge (Q) as follows: Effective charge (Q) = Dipole moment (µ) / Bond length (r) = \(4.03616 \times 10^{-30} \mathrm{C\cdot m} / 2.49 \times 10^{-10} \mathrm{m} = 1.62094 \times 10^{-20} \mathrm{C}\).

Express the effective charge in terms of electronic charge

Since the electronic charge (e) is equal to \(1.602 \times 10^{-19} \mathrm{C}\), we can express the effective charge in terms of the electronic charge by dividing the effective charge by the electronic charge: Effective charge (Q) in units of electronic charge (e) = \(1.62094 \times 10^{-20} \mathrm{C} / 1.602 \times 10^{-19} \mathrm{C} = 0.1011\)

Answer the second part of the question

In the IBr molecule, the effective charges on the Iodine (I) and Bromine (Br) atoms are approximately +0.1011e and -0.1011e, respectively.

Key concepts

Electronegativity

Electronegativity is a fundamental concept in chemistry that describes the tendency of an atom to attract a bonding pair of electrons. In the IBr molecule, the atoms involved are iodine (I) and bromine (Br). Each element has its own electronegativity value, which helps predict the behavior of electrons in a chemical bond.

• Iodine has an electronegativity value of 2.5.
• Bromine has a higher electronegativity value of 2.8.

The difference in electronegativity between bromine and iodine causes the electrons in the bond to be slightly more attracted towards bromine.
This uneven sharing of electrons is the reason bromine becomes partially negative, while iodine becomes partially positive. It is important to note that a small difference in electronegativity, like 0.3 in this case, indicates a polar covalent bond rather than an ionic one.

Dipole Moment

The dipole moment is a measure of the polarity of a chemical bond within a molecule. It occurs because of the difference in electronegativity between atoms creating a partial charge separation. The dipole moment is the product of the magnitude of the charge difference and the distance between the two charges.
In the case of the iodine monobromide (IBr) molecule, the dipole moment is given as 1.21 Debye (D). A dipole moment in this magnitude indicates that the molecule has a significant degree of charge separation, marking it as polar.
When calculating dipole moments in units of Coulombs and meters, remember to use the following conversion:

• 1 Debye (D) equals approximately 3.336 × 10^-30 Coulomb meters (C·m).

Thus, the dipole moment in IBr plays a critical role in understanding the effective charge distribution between the bonded atoms.

Bond Length

Bond length is an important factor that influences the chemical properties of a molecule. It refers to the average distance between the nuclei of two bonded atoms. For iodine monobromide (IBr), the bond length is found to be 2.49 Å. This value needs to be converted for calculations involving dipole moments, typically to meters (1 Å = 1 × 10^-10 meters).
A shorter bond length often indicates a stronger bond, as the overlapping electron clouds between the atoms are more extensive. In cases of polar molecules such as IBr, the bond length helps in quantifying the dipole moment, as it directly affects how far apart the partial charges are.
The bond length reflects the size and electron distribution between the iodine and bromine atoms and is crucial for calculating other molecular properties such as effective charges.

Effective Charge

Effective charge is a key concept for understanding the distribution of electron density in a polar molecule. It quantifies the net charge appearing on each atom as a result of bond polarity. In IBr, bromine is more electronegative than iodine, contributing to effective charges on each atom.
The effective charge can be calculated using the relation to dipole moment and bond length:

• \[Q = \frac{\mu}{r}\]where \(\mu\) is the dipole moment and \(r\) is the bond length.

For IBr, the effective charge is found to be approximately 0.1011e, indicating the partial positive and negative charges on iodine and bromine, respectively. The effective charge measured in units of electronic charge, \(e\), provides a convenient means to understand the behavior and interactions of the molecule at the electronic level, helping predict chemical reactivity and bond nature.