chapter 3 bonding: general conceptsboth nonmetals attain noble gas electron configurations in a...

Chapter 3

Bonding: General Concepts

Chapter 3Table of Contents

(3.1) Types of chemical bonds

(3.2) Electronegativity

(3.3) Ions: Electron configurations and sizes

(3.4) Energy effects in binary ionic compounds

(3.5) Partial ionic character of covalent bonds

(3.6) The covalent chemical bond: A model

(3.7) Covalent bond energies and chemical reactions

(3.8) The localized electron bonding model

Chapter 3Table of Contents

(3.9) Lewis structures

(3.10) Exceptions to the octet rule

(3.11) Resonance

(3.12) Naming simple compounds

Chapter 3

Questions to Consider

What is meant by the term “chemical bond”?

Why do atoms bond with each other to form compounds?

How do atoms bond with each other to form compounds?

Section 3.1Types of Chemical Bonds

Representation of Molecular Structures

The force that holds atoms together is called a chemical bond

A covalent bond is formed when atoms share electrons

A collection of atoms makes a molecule

Molecules can be represented using:

Chemical formula

Structural formula

H—O—H

Representation of Molecular Structures

Space-filling model: Indicates the relative sizes of the atoms as well as their relative orientation in the molecule

Ball and stick method: A three-dimensional model using spheres and rods

Attributes of Molecules

Physical properties

Melting point

Hardness

Electrical and thermal conductivity

Solubility

Electric charge

Bond energy

Amount of energy needed to break a bond

Types of Chemical Bonding

Ionic bonding: Occurs between an atom that easily loses electrons and an atom that has a high affinity for electrons

Ionic compounds are formed when metals react with non-metals

Coulomb’s Law

It is used to calculate the energy of interaction between ions

E has units of joules

r is the distance between ions in nanometers

Q1 and Q2 are numerical ion charges

–19 1 2 = (2.31 × 10 J . nm)Q Q

Figure 3.2 - Interaction Between Two Identical Atoms

Types of Chemical Bonding

Bond length is defined as the distance between two atoms where the energy is lowest

Based on diagram 3.2 (b):

Energy terms involved are net potential energy and kinetic energy

Zero point of energy is defined with the atoms at infinite separation

Energy rises steeply due to the repulsive forces when atoms are in close proximity

Stability of Atoms in Bonding

Consider the H2 molecule

Electrons reside in the space between the two nuclei

Simultaneous attraction leads to increased stability compared to single H2 atoms

Increased attractive forces cause a decrease in potential energy of electrons

Figure 3.3 - The Effect of an Electric Field on Hydrogen Fluoride Molecules

Polar Covalent Bond

Unequal sharing of electrons between atoms in a molecule

Results in a charge separation in the bond

Partial positive and partial negative charge

Concept Check

What is meant by the term “chemical bond?”

Why do atoms bond with each other to form molecules?

How do atoms bond with each other to form molecules?

Section 3.2Electronegativity

What is Electronegativity?

The ability of an atom in a molecule to attract shared electrons to itself

Linus Pauling’s method of determining values of electronegativity

For a molecule HX, the relative electronegativities of the H and X atoms are determined by comparing the measured H—X bond energy with the “expected” H—X bond energy

H—H bond energy + X—X bond energyExpected H—X bond energy =

Determining Values of Electronegativity

Δ between actual and expected bond energies can be calculated by the formula

Δ = 0 if H and X have identical electronegativities

Shared electrons are closer to the X atom if the electronegativity of X is greater than X

The charge distribution is:

act exp = (H X) (H X)

+ –δ δH — X

Figure 3.4 - The Periodic Table

Table 3.1 - The Relationship Between Electronegativity and Bond Type

Interactive Example 3.1 - Relative Bond Polarities

Order the following bonds according to polarity: H—H, O—H, Cl—H, S—H, and F—H

Solution

The polarity of the bond increases as the difference in electronegativity increases. From the electronegativity values in Figure 3-4, the following variation in bond polarity is expected (the electronegativity value appears in parentheses below each element)

H—H < S—H < Cl—H < O—H < F—H(2.1) (2.1) (2.5) (2.1) (2.1) (2.1) (2.1)(3.0) (3.5) (4.0)

Concept Check

If lithium and fluorine react, which has more attraction for an electron? Why?

In a bond between fluorine and iodine, which has more attraction for an electron? Why?

Concept Check

What is the general trend for electronegativity across rows and down columns on the periodic table?

Explain the trend

Concept Check

Which of the following bonds would be the least polar yet be considered polar covalent?

Mg–O C–O O–O Si–O N–O

Figure 3.4 - The Periodic Table

Concept Check

Which of the following bonds would be the most polar without being considered ionic?

Mg–O C–O O–O Si–O N–O

Section 3.3Ions: Electron Configurations and Sizes

Electron Configuration of Compounds

In a reaction between two nonmetals, electron sharing ensures complete valence electron configuration of both atoms Both nonmetals attain noble gas electron configurations

In a reaction between a nonmetal and a representative-group metal, a binary ionic compound is formed Ions are formed in such a way that both ions achieve noble

gas electron configuration

Concept Check

Choose an alkali metal, an alkaline earth metal, a noble gas, and a halogen so that they constitute an isoelectronic series when the metals and halogen are written as their most stable ions.

What is the electron configuration for each species?

Determine the number of electrons for each species

Determine the number of protons for each species

Rank the species according to increasing radius

Rank the species according to increasing ionization energy

Predicting Formulas of Ionic Compounds

Ionic compound - Chemists’ term for compounds in solid state

Attraction between positive and negative ions is maximum

Caused by close proximity of the positive and negative ions, unlike those of compounds in the gaseous phase

Large attractive forces tend to increase the stability of a compound

Consider the valence electron configurations of oxygen and calcium

Ca [Ar]4s2

O [He]2s22p4

Electronegativity of oxygen is higher than that of calcium

The prediction made can be based on the fact that noble gas configurations are generally stable

An important fact that applies is that chemical compounds are always electrically neutral

In the reaction between calcium and oxygen:

There are equal numbers of calcium and oxygen ions

The empirical formula of the compound formed is CaO

Sizes of Ions

Ion size influences the structure and stability of ionic solids

Size is determined by measuring the distance between ion centers

Factors that influence ionic size

Size of the parent ion

Position in the periodic table

Isoelectric ions are ions that contain the same number of electrons

Figure 3.5 - Size of Ions with Relation to the Periodic Table

Interactive Example 3.3 - Relative Ion Size II

Choose the largest ion in each of the following groups:

a. Li+, Na+, K+, Rb+, Cs+

b. Ba2+, Cs+, I–, Te2–

Solution

a. The ions are all from Group 1A elements. Since size increases down a group (the ion with the greatest number of electrons is largest), Cs+ is the largest ion.

b. This is an isoelectronic series of ions, all of which have the xenon electron configuration. The ion with the smallest nuclear charge is largest.

Te2– ˃ I– ˃ Cs+ ˃ Ba2+

Z = 52 Z = 53 Z = 55 Z = 56

Section 3.4Energy Effects in Binary Ionic Compounds

Lattice Energy

The change in energy that takes place when separated gaseous ions are packed together to form an ionic solid

It is mostly defined as the energy released when an ionic solid forms from its ions

It possesses a negative energy sign

Energy decreases

+ –M ( ) + X ( ) MX( )g g s

Energy Changes in the Formation of Lithium Fluoride

Consider the reaction

The reaction follows a certain sequence

Solid lithium undergoes sublimation

Energy of sublimation is 161 kJ/mol

Lithium atoms are ionized, forming lithium ions

520 kJ/mol of energy is released

1Li( ) + F ( ) LiF( )

2s g s

Li(s) Li(g)

+ –Li(g) Li (g) + e

Fluorine molecules undergo dissociation

F—F bonds in a half mole of F2 molecules are broken, forming a mole of fluorine atoms

154 kJ/mol of energy is required

Formation of F– ions occurs with an energy change of –328 kJ/mol

1 F (g) F(g)

– –F(g) + e F (g)

Gaseous ions of lithium and fluoride combine to form solid lithium fluoride

The overall process is shown in the table below

–Li (g) + F (g) LiF(s)

Figure 3.6 - Energy Changes Involved in the Formation of Lithium Fluoride

Lattice Energy Calculations

A modification of Coulomb's law is used

k is a proportionality constant that depends on the structure of the solid and the electronic configurations of the ions

Q1 and Q2 are the charges on the ions

r is the shortest distance between the centers of the anions and the cations

1 2Lattice energy = Q Q

Energies Involved in the Formation of NaF(s) and MgO(s)

Energy released by Mg2+ and O2– ions is four times greater than the energy released by the combination of Na+ and F– ions, forming NaF

The energy needed to remove two electrons from the magnesium atom (2180 kJ/mol) is greater than the energy required to remove one electron from the sodium atom (495 kJ/mol)

737 kJ/mol of energy is required to add two electrons to the oxygen atom in the gas phase

Figure 3.8 - Formation of NaF(s) Vs MgO(s)

Section 3.5Partial Ionic Character of Covalent Bonds

Formula for Percent Ionic Character of a Bond

There are probably no totally ionic bonds between discrete pairs of atoms

Calculations of the percent ionic character for bonds of various binary compounds in the gas phase

Percent ionic character of a bond can be calculated using the formula

Measured dipole moment of X—Y 100%

Calculated dipole moment of X Y

Figure 3.10 - Relationship Between Ionic Character of a Covalent Bond and Electronegativity Difference of Bonded Atoms

Ionic Compounds

Problems faced in the identification of ionic compounds

No individual bonds are totally ionic

Many substances contain polyatomic ions

Operational definition of ionic compounds

Any compound that conducts an electric current when melted will be classified as ionic

Section 3.6The Covalent Chemical Bond: A Model

Models

They are used to explain how nature operates on the microscopic level based on experiences in the macroscopic world

They are created by observing the properties of nature

The bonding model provides a framework to systematize chemical behavior

Molecules are viewed as collections of common fundamental components

Section 3.6The Covalent Chemical Bond: A Model

Fundamental Properties of Models

A model does not equal reality

Models are oversimplifications and are therefore often wrong

Models become more complicated and are modified as they age

The underlying assumptions of a model must be properly understood

More is learnt when a model is wrong than when it is right

Section 3.7Covalent Bond Energies and Chemical Reactions

Establishing the Sensitivity of Bonds to their Molecular Environment

Consider the stepwise decomposition of methane

Process Energy Required

CH ( ) CH ( ) + H( ) 435

CH ( ) CH ( ) + H( ) 453

CH ( ) CH( ) + H( )

CH( ) C( ) + H( ) 339

Total = 1652

1652 Average = = 413

Establishing the Sensitivity of Bonds to their Molecular Environment

Establishing the sensitivity of bonds to their molecular environment

Measured C—H

bond energy

Molecule (kJ/mol)

HCCl 380

HCF 430

C H 410

Types of Bonds

Single Bond: A bond in which a single pair of electrons is shared

Double bond: A bond in which two pairs of electrons are shared

Triple bond: A bond in which three pairs of electrons are shared

Table 3.3 - Average Bond Energies

Bond Energies

Used to calculate approximate values for reactions

Consider the reaction

In the reaction above, two H—F bonds are formed by breaking one H—H bond and one F—F bond

Bonds can be broken when positive energy is added to the system

2 2H ( ) + F ( ) 2HF( )g g g

Bond Energies

The formula for energy change

Σ = The sum of terms

D = Bond dissociation energy per mole of bonds

D is always positive

n = Moles of a particular type of bond

Energy required Energy released

Δ = (bonds broken) (bonds formed)E n D – n D

Table 3.4 - Bond Lengths for Selected Bonds

Interactive Example 3.4 - ΔE from Bond Energies

Using the bond energies listed in Table 3-3, calculate ΔE for the reaction of methane with chlorine and fluorine to give Freon-12 (CF2Cl2).

4 2 2 2 2CH ( ) + 2Cl ( ) + 2F ( ) CF Cl ( ) + 2HF( ) + 2HCl( )g g g g g g

Solution

The idea here is to break the bonds in the gaseous reactants to give individual atoms and then assemble these atoms into the gaseous products by forming new bonds

The energy changes are combined to calculate ΔE

ΔE = energy required to break bonds – energy released when bonds form

The minus sign gives the correct sign to the energy terms for the exothermic processes

Energy Energy

required releasedReactants atoms products

Solution

Reactant bonds broken

4CH : 4 mol C—H 4 mol413 kJ

× mol

= 1652 kJ

2Cl : 2 mol Cl—Cl 2 mol239 kJ

= 478 kJ

2F : 2 mol F—F 2 mol154 kJ

× mol

= 308 kJ

Total energy requied = 2438 kJ

Solution

Product bonds formed

2 2CF Cl : 2 mol C—F 2 mol485 kJ

× mol

= 970 kJ

2 mol C—Cl 2 mol339 kJ

= 678 kJ

HF: 2 mol H—F 2 mol565 kJ

× mol

= 1130 kJ

HCl: 2 mol H—Cl 2 mol427 kJ

× mol

= 854 kJ

Total energy released = 3632 kJ

Solution

Calculating ΔE

Since the sign of the value for the energy change is negative, this means that 1194 kJ of energy is released per mole of CF2Cl2 formed

ΔE = energy required to break bonds – energy released when bonds form

= 2438 kJ – 3632 kJ

= –1194 kJ

Section 3.8The Localized Electron Bonding Model

Localized Electron (LE) Model

A molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms

Pairs of electrons localized on a specific atom are called lone pairs

Electron pairs found in the space between the atoms are called bonding pairs

Parts of the LE Model

Description of the valence electron arrangement in the molecule using Lewis structures

Prediction of the geometry of the molecule using the valence shell electron-pair repulsion (VSEPR) model

Description of the type of atomic orbitals used by the atoms to share electrons or hold lone pairs

Lewis Structure

Shows how valence electrons are arranged among atoms in a molecule

Named after G.N. Lewis

Atoms achieving noble gas configurations are central to the formation of a stable compound

Writing Lewis Structures

Only valence electrons are included

Dots represent electrons

Lewis structure for KBr

K has no valence electrons

Br– has a filled valence shell

Writing Lewis Structures

The duet rule

A stable H2 molecule is formed by the combination of two hydrogen molecules

Sharing of electrons results in a filled valence shell

Rules in Writing Lewis Structures

The octet rule

Elements form stable molecules when surrounded by eight electrons

Each fluorine atom has bonding pairs as well as lone pairs of electrons

Section 3.9Lewis Structures

Problem Solving Strategy - Steps for Writing Lewis Structures

Sum the valence electrons from all the atoms

Use a pair of electrons to form a bond between each pair of bound atoms

Arrange the remaining electrons to satisfy the duet rule for hydrogen and the octet rule for the second-row elements

Lewis Structure of Water

Stepwise approach

Sum the valence electrons for H2O

1 + 1 + 6 = 8 valence electrons

Using a pair of electrons per bond, draw the two O—H single bonds

H—O—H

A line replaces a pair of dots to indicate each pair of bonding electrons

Lewis Structure of Water

Distribute the remaining electrons to achieve a noble gas configuration for each atom

Interactive Example 3.5 - Writing Lewis Structures

Give the Lewis structure for each of the following

Solution

Apply the three steps for writing Lewis structures

Lines indicate shared electron pairs

Dots indicate nonbonding pairs

Solution

Concept Check

Draw a Lewis structure for each of the following molecules:

Concept Check

Section 3.10Exceptions to the Octet Rule

Boron tends to form compounds in which the boron atom has fewer than eight electrons around it

It does not have a complete octet

The Lewis structure of boron fluoride

Boron has only 6 electrons around it

Adding a double bond satisfies the octet rule

In stable compounds such as H3NBF4, boron has an octet of electrons

Exceeding the Octet Rule

Observed only in elements in Period 3 of the periodic table and beyond

Consider the Lewis structure of sulfur hexafluoride

6 + 6 (7) = 48 electrons

Interactive Example 3.6 - Lewis Structures that Violate Octet Rule I

Write the Lewis Structure for PCl5 Solution

Sum the valence electrons

5 + 5(7) = 40 electrons

Indicate single bonds between bound atomsP Cl

Solution

Distribute the remaining electrons

30 electrons remain

Phosphorus, a third-row element, has exceeded the octet rule by two electrons

Concept Check

PCl5 SF6

Exceeding the Octet Rule in Certain Elements

When it is necessary to exceed the octet rule for one of several third-row (or higher) elements, assume that the extra electrons should be placed on the central atom

Calculating the valence electrons present in the triiodide ion (I3

3(7) + 1 = 22 valence electrons

–1 chargeI

Interactive Example 3.7 - Lewis Structures for Molecules that Violate the Octet Rule II

Write the Lewis structure for each molecule or ion

RnCl2 BeCl2 ICl4

Solution

The chlorine atom accepts the extra electrons

All atoms obey the octet rule

Solution

Radon, a noble gas in Period 6, accepts the extra electrons

Beryllium is electron-deficient

Iodine exceeds the octet rule

Section 3.11Resonance

Resonance

More than one valid Lewis structure can be written for a particular molecule

Consider the nitrate ion

NO3– = 24 valence electrons

The correct description is given only by the superposition of all three structures

Resonance

It is the condition in which more than one valid Lewis structure can be written for a specific molecule

Atoms need to be connected in the same order

The resulting electron structure is derived from the average of the resonance structures

Represented by double-headed arrows

Interactive Example 3.8 - Resonance Structures

Describe the electron arrangement in the nitrite anion (NO2

–), using the localized electron model

Solution

Step 1 - Determine the valence electrons

5 + 2 (6) + 1 = 18

Step 2 - Use single bonds to indicate the structure

O—N—O

Step 3 - Distribute the remaining 14 electrons to produce Lewis structures

Odd-Electron Molecules

There are few molecules from non-metals that possess electrons in odd numbers

Nitric oxide (NO), emitted by automobiles, reacts with oxygen in the air to from nitrogen dioxide

A more sophisticated model than the localized electron model is required

The localized electron model is based on electron pairs

Concept Check

OCN–

Formal Charge

It is used to estimate the charge on all possible nonequivalent Lewis structures

It is defined as the difference between the number of valence electrons on the free atom and the number of electrons assigned to the atom in the molecule

Formal charge = (number of valence electrons on a free atom) – (number of valence electrons assigned to the atom in the molecule)

Computing the Formal Charge of an Atom

Valence electrons are assigned to various atoms based on the following assumptions:

Lone pair electrons belong entirely to the atom in question

Shared electrons are divided equally between the two sharing atoms

The number of valence electrons is calculated

(Valence electrons)assigned = (number of lone electrons) + 1/2(number of shared electrons)

Formal Charge

Fundamental assumptions about formal charges in evaluating Lewis structures

Atoms in molecules try to achieve charges as close to zero as possible

Any negative formal charges are expected to reside on the most electronegative atoms

Rules Governing Formal Charge

To calculate the formal charge on an atom:

Take the sum of the lone pair electrons and one-half the shared electrons

Subtract the number of assigned electrons from the number of valence electrons on the free, neutral atom

Rules Governing Formal Charge

The sum of the formal charges of all atoms in a given molecule or ion must equal the overall charge on that species

If nonequivalent Lewis structures exist for a species, those with formal charges closest to zero and with any negative formal charges on the most electronegative atoms are considered to best describe the bonding in the molecule or ion

Interactive Example 3.9 - Formal Charges

Give possible Lewis structures for XeO3, an explosive compound of xenon. Which Lewis structure or structures are most appropriate according to the formal charges?

Solution

For XeO3 (26 valence electrons) we can draw the following possible Lewis structures:

Formal charges are indicated in parentheses

Solution

Based on the ideas of formal charge, it can be predicted that the Lewis structures with the lower values of formal charge would be most appropriate for describing the bonding in XeO3

Concept Check

Consider the Lewis structure for POCl3. Assign the formal charge for each atom in the molecule

Section 3.12Naming Simple Compounds

Binary Ionic Compounds (Type I)

They possess a cation in the formula followed by an anion

Rules in naming type I binary compounds

The cation is always named first and the anion second

A monatomic cation takes its name from the name of the element

A monatomic anion is named by taking the root of the element name and adding -ide

Table 3.5 - Common Monatomic Cations and Anions

Interactive Example 3.10 - Naming Type I Binary Compounds

Name each binary compound

AlCl3 LiH

Solution

CsF is cesium fluoride

AlCl3 is aluminum chloride

LiH is lithium hydride

Formulas from Names

In naming chemical compounds, the chemical formula is considered

The reverse process is also important

The formula of calcium chloride is written as CaCl2 Calcium forms two Ca2+ ions

Chloride is Cl–

So, two of these anions will be required to produce a neutral compound

Interactive Example 3.11 - Formulas from Names for Type I Binary Compounds

Given the following systematic names, write the formula for each compound:

Potassium iodide

Calcium oxide

Gallium bromide

Solution

Name Formula Comments

Potassium iodide KI Contains K+ and I–

Calcium oxide CaO Contains Ca2+ and O2–

Gallium bromide GaBr3 Contains Ga3+ and Br –

Must have 3Br– to balance charge of Ga3+

Binary Ionic Compounds (Type II)

Some metals form more than one type of positive ion

Results in the formation of more than one type of ionic compound with a given anion

The charge on the metal ion is specified using Roman numerals

Alternative for metals that form only two ions

Metals possessing ions with higher charge end in -ic

Metals possessing ions with lower charge end in -ous

Interactive example 3.12 - Naming Type II Binary Compounds

Given the following systematic names, write the formula for each compound:

Manganese (IV) oxide

Lead (II) chloride

Solution

Name Formula Comments

Manganese (IV) oxide MnO2 Two O2– ions (total charge 4–) are required by the Mn4+ ion [manganese (IV)]

Lead (II) chloride PbCl2 Two Cl– ions are required by the Pb2+ ion [lead (II)] for charge balance

Binary Ionic Compounds (Type II)

The Roman numeral naming system is used only in cases where more than one ionic compound forms between a given pair of elements

Elements that form only one cation are not identified by a Roman numeral

Group 1(A)

Group 2(A)

Aluminum

Silver

Figure 3.12 - Flowchart for Naming Binary Ionic Compounds

Interactive Example 3.13 - Naming Binary Compounds

Give the systematic name for each of the following compounds:

CaCl2 Al2O3

Solution

Formula

CoBr2 Cobalt (II) bromide Cobalt is a transition metal; the compoundname must have a Roman numeral. The two Br– ions must be balanced by a Co2+ ion.

CaCl2 Calcium chloride Calcium, an alkaline earth metal, forms only the Ca2+ ion. A Roman numeral is not necessary.

Al2O3 Aluminum oxide Aluminum forms only the Al3+ ion. A Roman numeral is not necessary.

Figure 3.13 - The Common Cations and Anions

Ionic Compounds with Polyatomic Atoms

Polyatomic ions are assigned specific names that must be memorized to name the compounds containing them

Oxyanions: Anions that contain an atom of a given element and different numbers of oxygen atoms

In a series with two oxyanions:

The oxyanion with lesser number of oxygen atoms ends in -ite

The oxyanion with higher number of oxygen atoms ends in -ate

Ionic Compound with Polyatomic Atoms

In a series with more than two oxyanions:

Hypo- is used as a prefix in the names of members with the fewest oxygen atoms

Per- is used as a prefix in the names of members with the most oxygen atoms

Interactive Example 3.14 - Naming Compounds Containing Polyatomic Ions

Give the systematic name for each of the following compounds:

Na2SO4

KH2PO4

Fe(NO3)3

Mn(OH)2

Na2SO3

Na2CO3

Solution

Formula Name Comments

Na2SO4 Sodium sulfate

KH2PO4 Pottasium dihydrogen sulfate

Fe(NO3)3 Iron (III) nitrate Transition metal—name must contain aRoman numeral. The Fe3+ ion balancesthree NO3

– ions

Mn(OH)2 Manganese (II) hydroxide Transition metal—name must contain aRoman numeral. The Mn2+ ion balances three OH– ions

Na2SO3 Sodium sulfite

Na2CO3 Sodium carbonate

Binary Covalent Compounds (Type III)

They are formed between two non-metals

Their names are similar to that of binary ionic compounds

Compound Systematic Name Common Name

N2O Dinitrogen monoxide Nitrous oxide

NO Nitrogen monoxide Nitric oxide

NO2 Nitrogen dioxide

N2O3 Dinitrogen trioxide

N2O4 Dinitrogen tetroxide

N2O5 Dinitrogen pentoxide

Interactive Example 3.15 - Naming Type III Binary Compounds

Name each of the following compounds:

PCl5 PCl3 SO2

Solution

Name Formula

PCl5 Phosphorus pentachloride

PCl3 Phosphorus trichloride

SO2 Sulfur dioxide

Figure 3.14 - Naming Binary Compounds

Figure 3.13 - Naming Chemical Compounds

Interactive Example 3.16 - Naming Various Compounds

Given the following systematic names, write the formula for each compound

Vanadium (V) fluoride

Dioxygen difluoride

Rubidium peroxide

Gallium oxide

Solution

Name Chemical Formula Comment

Vanadium (V) fluoride

VF5 The compound contains V5+ ions and requires five F– ions for charge balance

Dioxygen difluoride

O2F2 The prefix di- indicates two of each atom

Rubidium peroxide

Rb2O2 Since rubidium is in Group 1A, it forms only 1+ ions. Thus two Rb+ ions are needed to balance the 2– charge on the peroxide ion (O2

Gallium oxide Ga2O3 Since gallium is in Group 3A, like aluminum, it forms only 3+ ions. Two Ga3+

ions are required to balance the charge on three O2– ions

Molecules in which one or more H+ ions are attached to an anion

If the name of the anion ends in -ide, the acid is named with the prefix hydro- and the suffix -ic

The names of oxygen-containing anions are based on the root name of the anion, along with a suffix of -ic or -ous

If the name of the anion ends in -ate, the suffix -ic is added to the root name

If the name of the anion ends in -ite, the -ite is replaced by -ous

AcidsAcid Anion Name

HClO4 Perchlorate Perchloric acid

HClO3 Chlorate Chloric acid

HClO2 Chlorite Chlorous acid

HClO Hypochlorite Hypochlorous acid

The 6 strong acids are: (All the others are “weak”)

Figure 3.16 - Naming Acids

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