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The most common type of exception to the octet rule are compounds or ions with central atoms having more than eight electrons around them. PF_. \(\mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{ClF}_{3}\) and \(\mathrm{Br}_{3}^{-}\) are examples of this type of exception. Draw the Lewis structure for these compounds or ions. Which elements, when they have to, can have more than eight electrons around them? How is this rationalized?

Short Answer

Expert verified
The Lewis structures for the given compounds or ions are: PF5: F - P - F | | F - P - F SF4: F - S - F | | F F ClF3: F - Cl - F | F Br3-: Br - Br - Br | | 2e 2e Elements in period 3 or below, like P, S, and Cl, can have more than eight electrons around them due to their empty d-orbitals, which allow for an expanded octet or hypervalency.

Step by step solution

01

Count Valence Electrons

For each compound or ion, calculate the total number of valence electrons. PF5: P: 5 valence electrons (group 15) F: 7 valence electrons each (group 17) Total valence electrons = 5 + (5 × 7) = 40 SF4: S: 6 valence electrons (group 16) F: 7 valence electrons each (group 17) Total valence electrons = 6 + (4 × 7) = 34 ClF3: Cl: 7 valence electrons (group 17) F: 7 valence electrons each (group 17) Total valence electrons = 7 + (3 × 7) = 28 Br3-: Br: 7 valence electrons each (group 17) Extra electron due to negative charge: 1 electron Total valence electrons = 3 × 7 + 1 = 22
02

Draw Skeleton Structures

Now, draw the skeleton structure for each compound or ion with the central atom bonded to the surrounding atoms. PF5: F | P - F | F SF4: F | S - F | F ClF3: F | Cl | F - F Br3-: Br - Br - Br
03

Place Remaining Electrons

Next, place the remaining valence electrons around each central atom to fill their octets and complete the Lewis structures: PF5: F - P - F | | F - P - F SF4: F - S - F | | F F ClF3: F - Cl - F | F Br3-: Br - Br - Br | | 2e 2e
04

Determine Elements with More Than Eight Electrons and Rationalize

In the Lewis structures for PF5, SF4, and ClF3, the central atoms have more than eight electrons around them: P has 10 electrons, S has 10 electrons, and Cl has 10 electrons. These elements can accommodate more than eight electrons because they are in period 3 or below and have empty d-orbitals to accommodate additional electrons. This phenomenon is known as the "expanded octet" or "hypervalency." In conclusion, elements in period 3 or below, like P, S, and Cl, can have more than eight electrons around them due to their empty d-orbitals, which allow for an expanded octet or hypervalency. The Lewis structures for the given compounds or ions are as follows: PF5: F - P - F | | F - P - F SF4: F - S - F | | F F ClF3: F - Cl - F | F Br3-: Br - Br - Br | | 2e 2e

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Key Concepts

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

Lewis Structures
Lewis structures are diagrams that depict the bonding between atoms within a molecule as well as the lone pairs of electrons that may exist. The main purpose of a Lewis structure is to provide a visual representation of the arrangement of valence electrons in a molecule. These structures help determine the number of bonds and the geometry of the compound by placing electrons around the atoms involved.

Here's a simple way to build a Lewis structure:
  • Identify the total number of valence electrons available by adding up the outermost electrons of each atom within the molecule.
  • Create a skeleton of the molecule by connecting atoms with single bonds to determine the structure’s framework.
  • Distribute the remaining electrons to satisfy the octet rule for each atom, filling outer shells with pairs of electrons.
  • If necessary, form double or triple bonds to ensure each atom achieves a full octet.
Understanding Lewis structures is a fundamental skill in chemistry, helping students predict molecular shape, bond angles, and overall stability.
Valence Electrons
Valence electrons are the electrons in the outermost shell of an atom. They play a crucial role in chemical bonding, as they are responsible for the formation of bonds between atoms. The number of valence electrons largely determines how reactive an element is, with elements in the same group of the periodic table typically having the same number of valence electrons.

For instance:
  • Fluorine, chlorine, bromine, and iodine possess seven valence electrons, placing them in group 17 of the periodic table.
  • Phosphorus has five valence electrons, found in group 15, while sulfur has six, residing in group 16.
By knowing the count of valence electrons, chemists can predict how atoms will bond and which structures are possible when drawing Lewis structures. It's essential to correctly count these electrons for constructing accurate molecular diagrams and understanding interactions in compounds.
Expanded Octet
The expanded octet refers to a phenomenon where certain atoms can hold more than eight electrons in their valence shell. This is an exception to the standard octet rule that typically limits atoms to eight valence electrons for a stable configuration. The expanded octet is notably present in elements from period 3 or below on the periodic table.

These elements have accessible d-orbitals that can accommodate additional electrons beyond their standard octet. For example:
  • Phosphorus in the molecule e.g., PF5, holds 10 electrons in its valence shell.
  • Sulfur in SF4 similarly accommodates more than eight electrons.
The ability to host an expanded octet is crucial for the formation of more complex molecules, allowing atoms like phosphorus and sulfur to participate in varied chemical reactions. This concept extends the versatility of elements in forming stable structures by accommodating more electrons than typically possible.
Hypervalency
Hypervalency is closely related to the concept of an expanded octet. It occurs when atoms have more than eight electrons in their valence shell, using available space in d-orbitals to house additional electron pairs. Hypervalency is a term often used interchangeably with expanded octet.

Hypervalent molecules are possible primarily with nonmetals found in period 3 or beyond. This phenomenon is evident in molecules like ClF3, where chlorine as the central atom has 10 electrons, displaying hypervalency.

Some key points about hypervalency include:
  • It allows elements to exceed the traditional eight-electron arrangement, providing the flexibility needed for forming complex compounds.
  • Hypervalency enhances the understanding of bond types and strengths, showing how atoms with more than eight electrons can still form stable structures.
This concept greatly expands the diversity of molecular structures that can occur, contributing to the variety and complexity seen in chemical substances and reactions.

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Most popular questions from this chapter

The standard enthalpies of formation for \(\mathrm{S}(g), \mathrm{F}(g), \mathrm{SF}_{4}(g),\) and \(\mathrm{SF}_{6}(g)\) are \(+278.8,+79.0,-775,\) and \(-1209 \mathrm{kJ} / \mathrm{mol}\) respectively. a. Use these data to estimate the energy of an \(\mathrm{S}-\) F bond. b. Compare your calculated value to the value given in Table \(8.5 .\) What conclusions can you draw? c. Why are the \(\Delta H_{f}^{\circ}\) values for \(\mathrm{S}(g)\) and \(\mathrm{F}(g)\) not equal to zero, since sulfur and fluorine are elements?

Write Lewis structures and predict the molecular structures of the following. (See Exercises 115 and 116.) a. \(\mathrm{OCl}_{2}, \mathrm{KrF}_{2}, \mathrm{BeH}_{2}, \mathrm{SO}_{2}\) b. \(\mathrm{SO}_{3}, \mathrm{NF}_{3}, \mathrm{IF}_{3}\) c. \(\mathrm{CF}_{4}, \mathrm{SeF}_{4}, \mathrm{KrF}_{4}\) d. \(\mathrm{IF}_{5}, \mathrm{AsF}_{5}\)

The lattice energies of \(\mathrm{FeCl}_{3}, \mathrm{FeCl}_{2},\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) are (in no particular order) \(-2631,-5359,\) and \(-14,774 \mathrm{kJ} / \mathrm{mol}\) . Match the appropriate formula to each lattice energy. Explain.

Lewis structures can be used to understand why some molecules react in certain ways. Write the Lewis structures for the reactants and products in the reactions described below. a. Nitrogen dioxide dimerizes to produce dinitrogen tetroxide. b. Boron trihydride accepts a pair of electrons from ammonia, forming \(\mathrm{BH}_{3} \mathrm{NH}_{3} .\) Give a possible explanation for why these two reactions occur.

Use bond energies to estimate \(\Delta H\) for the combustion of one mole of acetylene: $$\mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$

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