Question

These species do not obey the octet rule. Draw a Lewis structure for each, and state the type of octet-rule exception: (a) \(\mathrm{BrF}_{3}\) (b) \(\mathrm{ICl}_{2}^{-}\) (c) \(\mathrm{BeF}_{2}\)

Short answer

(a) 10 electrons on Br - Expanded octet. (b) 10 electrons on I - Expanded octet. (c) 4 electrons on Be - Incomplete octet.

Step-by-step solution

- Identify the central atom

For each species, determine the central atom. Generally, the least electronegative element is the central atom. In these cases: (a) Bromine (Br) is central in \(\text{BrF}_3\), (b) Iodine (I) is central in \(\text{ICl}_2^{-}\), and (c) Beryllium (Be) is central in \(\text{BeF}_2\).

- Count the total valence electrons

Calculate the total number of valence electrons for each molecule or ion. Remember to add or subtract electrons for charges on ions:(a) \(\text{BrF}_3\): Br (7) + 3*F (3*7) = 28 valence electrons.(b) \(\text{ICl}_2^{-}\): I (7) + 2*Cl (2*7) + 1(extra electron for the negative charge) = 22 valence electrons.(c) \(\text{BeF}_2\): Be (2) + 2*F (2*7) = 16 valence electrons.

- Draw the basic skeleton structure

Connect each peripheral atom to the central atom with single bonds:(a) \(\text{Br-F-F-F}\)(b) \(\text{Cl-I-Cl}\)(c) \(\text{F-Be-F}\)

- Distribute remaining electrons

Distribute the remaining valence electrons as lone pairs around the peripheral atoms first, and then place any leftover electrons on the central atom:(a) \(\text{Br-F-F-F}\): Each F has 6 more electrons as lone pairs (18 electrons on F) leaving 28-18=10 electrons placed on Br.(b) \(\text{Cl-I-Cl}\): Each Cl has 6 more electrons as lone pairs (12 electrons on Cl) leaving 22-12=10 electrons placed on I.(c) \(\text{F-Be-F}\): Each F has 6 more electrons as lone pairs (12 electrons on F) leaving 16-12=4 electrons placed on Be.

- Check for octet-rule exceptions

Identify if the central atom follows the octet rule or has exceptions:(a) \(\text{BrF}_3\): Bromine has 10 electrons -> Expanded octet.(b) \(\text{ICl}_2^{-}\): Iodine has 10 electrons -> Expanded octet.(c) \(\text{BeF}_2\): Beryllium has 4 electrons -> Incomplete octet.

Key concepts

Lewis structure

Lewis structures are graphical representations of molecules that show all the valence electrons and bonds between atoms. These structures help visualize the arrangement of atoms and electron pairs in a molecule. To draw a Lewis structure:

• Identify the central atom (the least electronegative one).
• Count the total number of valence electrons.
• Arrange the atoms and connect them with single bonds.
• Distribute the remaining electrons to satisfy the octet rule or determine the kind of exception.

For example, in \(\text{BrF}_{3}\), Bromine (Br) is the central atom while Fluorine (F) atoms are bonded to it. You can place the remaining electrons around these atoms according to the rules.

Valence electrons

Valence electrons are the electrons in the outermost shell of an atom. These are the electrons that are involved in forming bonds in molecules. Knowing the number of valence electrons is crucial for drawing Lewis structures and understanding molecular behavior.

• For Br, it has 7 valence electrons.
• Each F in \(\text{BrF}_{3}\) also has 7 valence electrons.

When you sum these up for the molecule, you get the total number of valence electrons which guides how you distribute them in the Lewis structure. The same principle applies to the other molecules like \(\text{ICl}_{2}^{-}\) and \(\text{BeF}_{2}\).

Expanded octet

Some atoms can hold more than 8 electrons in their valence shell; this is called an expanded octet. These exceptions occur typically in elements from the third period or higher in the periodic table.

• In \(\text{BrF}_{3}\), Br holds 10 electrons.
• In \(\text{ICl}_{2}^{-}\), Iodine also holds 10 electrons.

This is allowed because these atoms have available d-orbitals which can accommodate extra electrons, spreading the electron density and stabilizing the molecule.

Incomplete octet

Unlike the expanded octet, some elements do not need to satisfy the octet rule and can be stable with fewer than 8 electrons around them. This is known as an incomplete octet.

• In \(\text{BeF}_{2}\), Beryllium has only 4 electrons around it.

Such exceptions often occur with elements like Beryllium and Boron, which are stable with 4 and 6 valence electrons, respectively. This happens because these atoms typically form highly covalent bonds and do not require a full octet to attain stability.