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Chapter 9: Problem 60

Write Lewis formulas for the following ions: a. \(\mathrm{IBr}_{2}^{+}\) b. \(\mathrm{ClF}_{2}^{+}\) c. \(\mathrm{CN}^{-}\)

Short Answer

Expert verified
IBr2鈦� and ClF2鈦� have 20 valence electrons each, forming linear structures. CN鈦� has 10 electrons, forming a triple bond.

Step by step solution

01

Count Valence Electrons for IBr2鈦�

Calculate the total number of valence electrons for the ion IBr2鈦�. Iodine (I) has 7 valence electrons and each bromine (Br) atom also has 7 valence electrons. Since the ion has a +1 charge, subtract 1 from the total. Thus, the total is 7 (I) + 7 (Br) + 7 (Br) - 1 = 20 electrons.
02

Draw the Skeleton Structure for IBr2鈦�

Place iodine (I) in the center because it is less electronegative than bromine (Br). Attach the two Br atoms with single bonds to the central iodine atom. Now distribute the electrons around the molecule to satisfy each atom's valence requirements.
03

Assign Remaining Electrons for IBr2鈦�

Each Br atom needs 8 electrons to fulfill the octet rule. Place three lone pairs on each Br. The iodine will have two lone pairs, using up all 20 electrons. This results in a linear shape for IBr2鈦�.
04

Count Valence Electrons for ClF2鈦�

Calculate the valence electrons for ClF2鈦�. Chlorine (Cl) has 7 valence electrons and each fluorine (F) has 7. Subtract 1 for the +1 charge, getting 7 (Cl) + 7 (F) + 7 (F) - 1 = 20 electrons.
05

Draw the Skeleton Structure for ClF2鈦�

Chlorine (Cl) is the central atom, with two fluorine (F) atoms bonded to it. Connect each fluorine atom with single bonds to chlorine.
06

Allocate Remaining Electrons for ClF2鈦�

Distribute the remaining electrons as lone pairs to fulfill the valence of each atom. Place three lone pairs around each F atom. This leaves chlorine with one lone pair and total valence electrons used is 20.
07

Count Valence Electrons for CN鈦�

Calculate the total valence electrons for CN鈦�. Carbon (C) has 4 valence electrons and nitrogen (N) has 5. Add 1 for the extra negative charge, for a total of 4 (C) + 5 (N) + 1 = 10 electrons.
08

Draw the CN鈦� Structure

Place carbon and nitrogen next to each other. Use a triple bond between them, which will use 6 electrons, leaving 4 electrons remaining.
09

Complete the Duo with Lone Pairs for CN鈦�

Distribute the remaining 4 electrons as lone pairs: two lone pairs on nitrogen and none on carbon. This fulfills the octet rule for both atoms.

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Key Concepts

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

Valence Electrons
Valence electrons are the outermost electrons of an atom. They play a key role in chemical reactions and bonding. When forming compounds, atoms tend to share, lose, or gain valence electrons to achieve more stable electron configurations.
  • For the ion IBr鈧傗伜, iodine and each bromine have 7 valence electrons.
  • In ClF鈧傗伜, chlorine and each fluorine also have 7 valence electrons.
  • For CN鈦�, carbon has 4 valence electrons and nitrogen has 5.
The ion's overall charge affects the total count. A positive charge reduces the total number of valence electrons, while a negative charge increases it. This must be considered when drawing Lewis structures, as the starting valence electron count impacts how atoms are bonded and how many electrons remain for lone pairs.
Octet Rule
The octet rule is a guiding principle of chemistry stating that atoms are generally most stable when surrounded by eight electrons. This principle helps explain the formation of chemical bonds.
  • Every bromine and chlorine should have 8 electrons around them after bonding, as seen in IBr鈧傗伜 and ClF鈧傗伜.
  • The structure of CN鈦� ensures each atom reaches its octet; carbon pairs with nitrogen using a triple bond.
Exceptions to the octet rule exist but rely on factors such as atomic size and electron availability. However, in these exercises, the rule is generally adhered to, supporting stable molecular structures.
Ions
Ions are atoms or molecules that carry an electric charge due to gaining or losing electrons. Cations are positively charged ions formed by losing electrons, while anions are negatively charged ions formed by gaining electrons.
  • IBr鈧傗伜 and ClF鈧傗伜 are examples of cations with a +1 charge, indicating a loss of one electron from the neutral species.
  • CN鈦�, on the other hand, is an anion with an extra electron added, giving it a -1 charge.
Understanding ions helps explain their role in reactions and bonds. These charges influence how the atoms within the ions share electrons, impacting the resulting Lewis structures.

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Most popular questions from this chapter

Solid phosphorus pentabromide, \(\mathrm{PBr}_{5}\), has been shown to have the ionic structure \(\left[\mathrm{PBr}_{4}^{+}\right]\left[\mathrm{Br}^{-}\right] .\) Write the electrondot formula of the \(\mathrm{PBr}_{4}{ }^{+}\) cation.

Assuming that the atoms form the normal number of covalent bonds, give the molecular formula of the simplest compound of arsenic and bromine atoms.

The \(2+\) ions of transition elements are common. Explain why this might be expected.

Iodic acid, \(\mathrm{HIO}_{3}\), is a colorless, crystalline compound. What is the electron-dot formula of iodic acid?

Decide which of the following bonds is least polar on the basis of electronegativities of atoms: \(\mathrm{H}-\mathrm{S}, \mathrm{Si}-\mathrm{Cl}, \mathrm{N}-\mathrm{Cl}\).
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