Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The iodine monobromide molecule, IBr, has a bond length of \(249 \mathrm{pm}\) and a dipole moment of \(1.21 \mathrm{D} .(\mathbf{a})\) Which atom of the molecule is expected to have a negative charge? (b) Calculate the effective charges on the I and Br atoms in IBr in units of the electronic charge, \(e\).

Short Answer

Expert verified
(a) The Bromine atom is expected to have a negative charge. (b) The effective charges on the I and Br atoms in IBr are approximately 10.1e and -10.1e, respectively.

Step by step solution

01

(Step 1: Identifying Electronegativity of Atoms in IBr)

Determine the electronegativity of both Iodine (I) and Bromine (Br) atoms. From the periodic table, we have the following approximate electronegativity values: Iodine (I): 2.66 Bromine (Br): 2.96 Since Bromine is more electronegative than Iodine, the electrons in the I-Br bond will be closer to Bromine, making it slightly negatively charged.
02

(Step 2: Finding the Direction of the Dipole Moment)

As we established in the previous step, Bromine is more electronegative than Iodine, and thus has a partial negative charge. This means that the dipole moment vector points towards the Bromine atom. Answer to part (a): Bromine atom is expected to have a negative charge.
03

(Step 3: Converting Dipole Moment to SI Units)

The given dipole moment is in Debye (D), and we need to convert it to SI units (Coulomb meter, C m) before proceeding to the next step. The conversion factor is 1 D = \(3.336 \times 10^{-30}\) C m. Dipole moment in SI units: \(1.21 \, \mathrm{D} \times \frac{3.336 \times 10^{-30} C m}{1 \, \mathrm{D}} \approx 4.03 \times 10^{-30}\) C m
04

(Step 4: Calculating Effective Charges in IBr)

To calculate the effective charges on Iodine and Bromine atoms in IBr, we will use the following equation for the dipole moment: \( \mathrm{Dipole\, moment} = \mathrm{Effective\, charge\, difference} \times \mathrm{Bond\, length} \) Let the effective charge on Iodine be \(q_I\) and on Bromine be \(q_{Br}\). Since the total charge of the molecule is neutral, we have: \(q_I + q_{Br} = 0 \) \( q_{Br} = -q_I \) Now, we plug the values for the dipole moment and bond length into the equation: \(4.03 \times 10^{-30}\, \mathrm{C\, m} = (q_{Br} - q_I) \times 249 \times 10^{-12}\, \mathrm{m} \) Rearrange and solve for \(q_I\): \(q_I = \frac{4.03 \times 10^{-30}\, \mathrm{C\, m}}{249 \times 10^{-12}\, \mathrm{m}} = 1.62 \times 10^{-18}\) C With \(q_{Br} = -q_I\), we have: \(q_{Br} = -1.62 \times 10^{-18}\) C Now, divide the effective charges by the electronic charge \(e (1.6 \times 10^{-19}\, \mathrm{C})\) to get the effective charges in units of e: \(q_I (e) = \frac{1.62 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 10.1\, e \) \(q_{Br} (e) = \frac{-1.62 \times 10^{-18}}{1.6 \times 10^{-19}} \approx -10.1\, e \) Answer to part (b): The effective charges on the I and Br atoms in IBr are approximately 10.1e and -10.1e, respectively.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Iodine Monobromide
Iodine monobromide, abbreviated as IBr, is a diatomic molecule comprising one iodine atom and one bromine atom bonded together. It is noteworthy due to its intermediate properties between iodine and bromine elements.
  • Appearance: It appears as a red-brown solid under normal conditions.
  • Bond Characteristics: The bond between iodine and bromine results from the attraction between their nuclei and shared electrons.
The molecular structure of IBr plays a significant role in determining its chemical behavior. It is characterized by a polar bond due to differences in electronegativity between iodine and bromine, affecting its physical properties such as its dipole moment.
Electronegativity
Electronegativity is an essential concept for understanding why atoms attract shared electrons differently in a chemical bond. It refers to the tendency of an atom to attract electrons towards itself when chemically bonded to another atom.
  • Bromine Electronegativity: 2.96
  • Iodine Electronegativity: 2.66
Because bromine has a higher electronegativity compared to iodine, it means that in the IBr molecule, electrons are more strongly attracted to the bromine atom. This results in bromine obtaining a partial negative charge while iodine gains a partial positive charge. The difference in electronegativity is what creates the molecule's dipole moment, pointing from iodine to bromine.
Bond Length
Bond length is a measure of the distance between the nuclei of two bonded atoms. For iodine monobromide, this length is critical in understanding the molecule's dipole moment.
  • IBr Bond Length: 249 picometers (pm)
The bond length is influenced by the sizes of the iodine and bromine atoms. In general, longer bond lengths are weaker and typically found in molecules with larger atoms. In IBr, the bond length also plays a role in determining the dipole moment. The dipole moment is directly proportional to both the effective charge and the bond length. This means, for a given dipole moment, larger bond lengths may indicate smaller effective charges.
Effective Charge
Effective charge quantifies the net charge resulting from the imbalance of electron distribution in a molecule. In the context of the iodine monobromide molecule, this concept helps in determining the charge each atom appears to carry as a result of their differing electronegativities.Using the dipole moment equation:\[\text{Dipole Moment} = \text{Effective Charge Difference} \times \text{Bond Length} \]we calculated the effective charges:
  • Effective Charge on Iodine: 10.1e
  • Effective Charge on Bromine: -10.1e
These values are noted in terms of electronic charge units, where one electronic charge (e) equals \(1.6 \times 10^{-19}\) coulombs. The opposite signs indicate that the charges are balanced out in the molecule, with electrons being drawn more towards bromine, leaving iodine with an effective positive charge.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For Group 13-17 elements in the third row of the periodic table and beyond, the octet rule is often not obeyed. A friend of yours says this is because these heavier elements are more likely to make double or triple bonds. Another friend of yours says that this is because the heavier elements are larger and can make bonds to more than four atoms at a time. Which friend is more correct?

(a) The nitrate ion, \(\mathrm{NO}_{3}^{-}\), has a trigonal planar structure with the \(\mathrm{N}\) atom as the central atom. Draw the Lewis structure(s) for the nitrate ion. (b) Given \(S=\mathrm{O}\) and \(\mathrm{S}-\mathrm{O}\) bond lengths are \(158 \mathrm{pm}\) and \(143 \mathrm{pm}\) respectively, estimate the sulphuroxygen bond distances in the ion.

A metal oxide has a lattice energy in the range of \(3300 \mathrm{~kJ} /\) mol. Is the charge on the metal ion M likely to be \(1^{+}, 2^{+},\) or \(3^{+} ?\) Explain.

Ammonia reacts with boron trifluoride to form a stable compound, as we saw in Section 8.7. (a) Draw the Lewis structure of the ammonia-boron trifluoride reaction product. \((\mathbf{b})\) The \(\mathrm{B}-\mathrm{N}\) bond is obviously more polar than the C-C bond. Draw the charge distribution you expect on the \(\mathrm{B}-\mathrm{N}\) bond within the molecule (using the delta plus and delta minus symbols mentioned in Section 8.4 ). (c) Boron trichloride also reacts with ammonia in a similar way to the trifluoride. Predict whether the \(\mathrm{B}-\mathrm{N}\) bond in the trichloride reaction product would be more or less polar than the \(\mathrm{B}-\mathrm{N}\) bond in the trifluoride product, and justify your reasoning.

Write the electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Fe}^{2+}\), (b) \(\mathrm{V}^{3+}\), (c) \(\mathrm{Ni}^{2+}\), (d) \(\mathrm{Pt}^{2+}\), (e) \(\mathrm{Ge}^{2-}\), (f) \(\mathrm{Ba}^{2+}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free