· web viewchapter nine models of chemical bonding the type of bonding within a substance explains...

CHAPTER NINE

MODELS OF CHEMICAL BONDING

The type of bonding within a substance explains why…

Ionic substances (like table salt) are hard, solids have high-melting points, and conduct a current only when molten or dissolved in water

Most covalent substances (like candle wax) are soft, have low-melting points and do not conduct electricityo Some covalent substances such as diamond have high

melting points, can conduct a current, and are extremely hard

Atomic Properties and Chemical BondsWhy do atoms bond?

Bonding lowers the potential energy between positive and negative particles

Types of Bonding: Three Ways Metals and Nonmetals CombineRemember that elements become less metallic across a period and more metallic down a group…

Metal with Nonmetal

Electron transfer and ionic bondingo Observed between atoms with large differences in

their tendencies to lose atoms This difference occurs between reactive metals

[Groups 1A(1) and 2A(2)] and nonmetals [Group 7A(17) and the top of Group 6A(16)]

Metals lose electrons and nonmetals gain electrons (form a noble gas configuration when doing so)o The electrostatic attractions between

the positive and negative ions draw them into a three dimensional arrangement to form an ionic solid

The chemical formula of an ionic compound is the empirical formula (cation-to-anion ratio)

Nonmetal with nonmetal

Electron sharing and covalent bonding

o Observed between atoms that have very little or no difference in their tendencies to gain or lose electrons

Occurs most commonly with nonmetals The nucleus of each atom attracts the

valence electrons of the other, which draws the atoms togethero The shared electron pair is typically

localized between the two atoms The chemical formula is the

molecular formula (it gives the actual number of atoms in each molecule)

Metal with Metal

Electron pooling and metallic bondingo Metals atoms lose their outer electrons easily

Metal atoms share their electrons, but do not form covalent bonds

The enormous number of atoms in a sample of metal pool their valence electrons into a “sea” of electrons that “flow” between and around each metal-ion core (nucleus plus inner electrons) attracting them togethero Unlike the localized electrons in

covalent bonding, electrons in metallic bonding are delocalized (they can move freely throughout the entire piece of metal)

Exceptions to idealized models…

Can’t always predict the type of bond from where the elements are found on the periodic tableo The metal beryllium (Be) combines with nonmetal

chlorine and there is more sharing than there is transfer

The bonding is more covalent than it is ionic

Lewis Symbols and the Octet RuleLewis electron-dot symbol – the element symbol represents the nucleus and the inner electrons, and the dots represent the valence electrons.

Note that elements in the same group have the same number of dots

The specific placement of the dots is not important For a metal, the total number of dots is the number of

electrons an atom loses to form a cation. For a nonmetal, the number of unpaired dots equals either

the number of electrons an atom gains to form an anion or the number it shares to form covalent bonds.

Octet rule – when atoms bond, they lose, gain, or share electrons to attain a filled outer level of eight electrons (or two, for H and Li)

Holds for nearly all of the compounds of Period 2 elements and a large number of others as well.

The Ionic Bonding ModelThe central idea of ionic bonding model…

There is transfer of electrons from metal atoms to nonmetal atoms to form ions that attract each other into a solid compound.

Three ways to represent (model) the reaction between lithium and fluorine (note that the total number of electrons lost by the metal atom(s) equals the total number of electrons gained by the nonmetal atom(s)…

Sample Problem 9.1 p. 332Depicting Ion Formation

Follow-Up Problem 9.1 Work on the whiteboard…

Why Ionic Compounds Form: The Importance of Lattice EnergyEnergy is absorbed during electron transfer…

Energy is released after electron transfer…

The electron-transfer process

Example… formation of lithium fluorideo Involves a gaseous lithium atom losing an electron and

a gaseous fluorine atom gaining it Li (g) Li+ (g) + e- IE1 = 520 kJ (absorbed) F(g) + e- F- (g) EA1 = -328 kJ (released) Li (g) + F(g) Li+(g) + F-(g)

IE1 + EA1 = 192 kJo This is a net 192 kJ absorbed in the

electron transfer

Other steps that absorb energy

Metallic lithium must be made into gaseous atoms (161 kJ/mol)

Fluorine molecules must be broken into separate atoms (79.5 kJ/mol)

Steps that release energy

Despite the endothermic steps already mentioned, the standard enthalpy of formation (∆Hof) of solid LiF is -617 kJ/molo That is 617 kJ released when 1 mol of LiF(s) forms

from its elements This is typical of reactions between active metals

and nonmetals Li+(g) + F-(g) LiF(g) ∆Ho = -755 kJ Li+(g) + F-(g) LiF(s) ∆Ho = -1050 kJ

o The negative of this enthalpy change is 1050 kJ, the lattice energy for LiF

∆Holattice is the change that accompanies the reverse of this reaction – 1 mol of ionic solid separating into gaseous ions

Determining Lattice Energy with a Born-Haber Cycle

The magnitude of the lattice energy is a measure of the strength of the ionic attractions and influences macroscopic properties, such as melting point, hardness, and solubility.

Lattice energy cannot be measured directlyo One way to determine the lattice energy is through

Hess’s law (section 6.5) in a Born-Haber cycle A series of steps from elements to ionic solid for

which all the enthalpies are known except the lattice energy

Hess’s law tells us that…

∆Hof of LiF(s) = sum of ∆Ho values for multistep path

Hess’s law lets us choose hypothetical steps whose enthalpy changes we can measure, even though they are not the actual steps that occur when lithium reacts with fluorine

Step 1 – From solid Li to Li atoms

Li(s) Li(g) ∆Hostep 1 = ∆Ho

atom = 161 kJ

Step 2 – From F2 molecules to F atoms

½ F2(g) F(g) ∆Hostep 2 = ½ (BE of F2) = ½ (159 kJ) = 79.5 kJ

Step 3 – From Li to Li+

Li(g) Li+(g) + e- ∆Hostep 3 = IE1 = 520 kJ

Step 4 – From F to F-

F(g) + e- F- ∆Hostep 4 = EA1 = -328 kJ

Step 5 – From gaseous ions to ionic solid

Li+(g) + F-(g) LiF(s) ∆Hostep 5 = -∆Ho

lattice of LiF = ???

Li(s) + ½ F2(g) LiF(s) ∆Hooverall = ∆Ho

f = -617 kJ

To solve for ∆Holattice of LiF…

∆Hof =∆Ho

step 1 +∆Hostep 2 +∆Ho

step 3 +∆Hostep 4 + (-∆Ho

lattice of LiF)

Solving for -∆Holattice

-∆Holattice of LiF=∆Ho

f – (=∆Hostep 1+∆Ho

step 2+∆Hostep 3 +∆Ho

step 4)

= 617 kJ – [161 kJ + 79.5 kJ + 520 kJ + (-328 kJ)

= -1050 kJ

Changing the sign gives ∆Holattice of LiF = 1050 kJ

Key point…

The Born-Haber cycle shows that the energy required for elements to form ions is supplied by the attraction among the ions in the solid. Ionic solids exist only because the lattice energy far exceeds the total energy needed to form the ions.

Periodic Trends in Lattice EnergyThe lattice energy results from the electrostatic interactions among ions, so its magnitude depends on ionic size, ionic charge, and ionic arrangement in the solid. Because of this, we see periodic trends in lattice energy.

Explaining the Trends with Coulomb’s Law

Coulomb’s law states that the electrostatic energy between particles A and B is directly proportional to the product of their charges and inversely proportional to the distance between them.

Electrostatic energy = charge A x charge B / distance

Lattice energy is directly proportional to electrostatic energy. In an ionic solid, cations and anions lie as close to each other as possible, so the sum of the ionic radii.

Electrostatic energy = cation charge x anion charge = ∆Holattice

cation radius + anion radius

This relationship helps us explain the effects of ionic size and charge on trends in lattice energy…

Effect of ionic sizeo As we move down a group, ionic size increases, so the

electrostatic energy between cation and ions decreases Lattice energies decrease as well

Effect of ionic chargeo Across a period, ionic charge changes

Lithium fluoride and magnesium oxide have cations and anions of about equal radii

The major difference is between singly charged Li+ and F- and doubly charged Mg2+ and O2-

o ∆Holattice of LiF = 1050 kJ

o ∆Holattice of MgO = 3923 kJ

The nearly fourfold increase in ∆Ho

lattice reflects the fourfold increase in the product of the charges (1x1 vs. 2x2)

Why does MgO Exist?

Forming 1 mol of Mg2+ involves the sum of the first and second ionization energies…o ∆Ho = IE1 + IE2 = 738 kJ + 1450 kJ = 2188 kJ

Forming 1 mol of O2- involves the sum of the first and second electron affinities…o ∆Ho = EA1 + EA2 = (-141 kJ) + 878 kJ = 737 kJ

There is also an endothermic step for converting Mg(s) to Mg(g)…o 148 kJ/mol

The step for converting O2 molecules to O atoms is endothermic…o 498 kJ/mol

The 2+ and 2- ionic charges make the lattice energy so large (∆Ho

lattice = 3923 kJ/mol) that solid MgO forms readily whenever Mg burns in air (∆Ho

f = -601 kJ/mol)

How the Model Explains the Properties of Ionic CompoundsThe central role of a model is to explain the facts…

Physical behavior of ionic compounds

Hard (does not dent) Rigid (does not bend) Brittle (cracks without deforming)

o These properties arise from the strong attractive forces that hold the ions in specific positions

If enough force is applied (picture below) ions of like charge are brought next to each other, and the repulsions between them crack the sample.

Electrical conductivity of ionic compounds

Ionic compounds typically do not conduct electricity in the solid state, but do conduct when melted or dissolvedo The model explains that the solid consists of fixed

ions, but when it melts or dissolves, the ions can move and carry a current.

Thermal conductivity of ionic compounds

Because of the strong attraction between ions, ionic solids have high melting points and much higher boiling points

The interionic attraction is so strong that the vapor consists of ion pairs (gaseous ionic molecules) rather than individual ions

The Covalent Bonding ModelSharing electrons is the main way that atoms interact…

The Formation of a Covalent BondLooking at the figure below…

At point 1o The atoms are far apart, and each acts as though the

other were not present At point 2

o The distance between the atoms has decreased enough for each nucleus to start attracting the other atoms electrons, which lowers the potential energy

As atoms get closer, these attractions increase, but so do the repulsions between the nuclei and between the electrons

At point 3 (bottom of the energy “well”)o The maximum attraction is achieved in the face of the

increasing repulsion, and the system has its minimum energy

At point 4o If this point were reached, the atoms would be too

close, and the rise in potential energy from the increased repulsions would push them toward position 3 again

Covalent bondo Arises from the balance between nuclei attracting the

electrons and the nuclei and the electrons and nuclei repelling each other

o Formation of a covalent bond always results in greater electron density between the nuclei

Bonding Pairs and Lone PairsTo achieve a full outer (valence) level of electrons, each atom in a covalent bond “counts” the shared electrons as belonging entirely to itself

The two shared electrons in H2 simultaneously fill the outer level of both H atomso The shared pair, or bonding pair is represented by a

pair of dots or a line See the example on whiteboard for H2

An outer-level electron pair that is not involved in bonding is called a lone pair, or unshared pairo See the example on the whiteboard for HF and F2

o This text generally shows bonding pairs as lines and lone pairs as dots

Properties of a Covalent Bond: Order, Energy, and LengthThree important properties of covalent bonds…

Bond order

Bond order is the number of electrons being shared by a given pair of atomso A single bond (examples H2, HF, F2) is the most

common bond and consists of one bonding pair of electrons

A single bond has a bond order of 1o A double bond (usually involve C, O, and/or N)

consists of two bonding electron pairs (four electrons shared between two atoms)

A double bond has a bond order of 2 Example C2H4 on whiteboard…

o A triple bond consists of three shared pairs (two atoms share six electrons)

A triple bond has a bond order of 3 Example N2 on the whiteboard…

Bond Energy

The strength of a covalent bond depends on the magnitude of the attraction between the nuclei and the shared electronso The bond energy (BE) also called bond enthalpy or

bond strength is the energy needed to overcome the attraction between nuclei and shared electrons (breaking the bond in 1 mol of gaseous molecules)

Bond breakage is an endothermic process (energy is absorbed) so bond energy is always positive

Bond formation is an exothermic process (energy is released) so bond energy is always negative

Stronger bonds are lower in energy (have a deeper energy well)

Weaker bonds are higher in energy (have a shallower energy well)

Bond length

The bond length is the distance between the nuclei of two bonded atomso Bond lengths for a series of similar bonds, as in the

halogens [Group 7A(17)], increase with atomic size

Key relationships between order, energy, and bond length…

Two nuclei are more strongly attracted to two shared pairs than to oneo Double-bonded atoms are drawn closer together and

are more difficult to pull apart than single-bonded atoms

A higher bond order results in a shorter bond length and a higher bond energy

A shorter bond is a stronger bond Variations within a group

o The trend in carbon-halogen single bond lengths parallels the trend in atomic size

C-I > C-Br > C-Cl (I > Br > Cl) Opposite to the trend in bond energy

o C-Cl > C-Br > C-I

Variations within a periodo Trends in bond lengths

C-N > C-O > C-Fo Trends in bond energies

C-F > C-O > C-N

Sample Problem 9.2 p. 341Comparing Bond Length and Bond Strength(a)

Bond length S-Br > S-Cl > S-F

Bond strength S-F > S-Cl > S-Br

(b)

Bond length C-O > C=O > C=O

Bond strength C=O > C=O > C-O

Follow-Up Problem 9.2 p. 341(a)

Bond length Si-F < Si-O < S-C

Bond strength Si-C < Si-O < S-F

(b)

Bond length N=N < N=N < N-N

Bond strength N-N < N=N < N=N

How the Model Explains the Properties of Covalent Substances

Physical properties of molecular covalent substances

Most are… o gases (examples… methane and ammonia)

o liquids (examples… benzene and water)o low melting point solids (examples… sulfur and

paraffin wax) If covalent bonds are strong, why do covalent bonds have

low melting and boiling points?o There are two different forces at work…

Strong bonding forces that hold the atoms together within the molecule

Weak intermolecular forces that that hold separate molecules close to each other

It is the weak intermolecular forces that account for the observed physical properties of molecular covalent substanceso In the picture above weak forces

between pentane (C5H12) molecules are overcome, not the strong C-C and C-H bonds within the molecule

Physical properties of network covalent solids

Some covalent substances do not consist of separate molecules

Network covalent solids are held together by covalent bonds between atoms throughout the sample and their properties do

reflect the strength of covalent bonds (examples… quartz and diamond to the left)

Quartz (SiO2) has silicon-oxygen bonds in three-dimensions without any SiO2 molecules.

It is very hard and melts at 1550oC

Diamond has covalent bonds connecting carbon atoms to four other carbons

It is the hardest natural substance and melts at around 3550oC

Key point…

Covalent bonds are strong, but most covalent substances consist of separate molecules with weak forces between them

Electrical conductivity of covalent substances

Most covalent substances are poor electrical conductors, whether melted or dissolved, because their electrons are localized as either shared or unshared pairs and no ions are present.

Bond Energy and Chemical ChangeThe relative strengths of bonds in reactants and products determine whether heat is released or absorbed in a chemical reaction.

Changes in Bond Energy: Where does ∆H orxn Come From?

When 1 mol of H2 and 1 mol of F2 react to form 2 mol of HF and 546 kJ is released, where does that enthalpy of reaction (∆Ho

rxn) come from?

The heat released or absorbed during a chemical change is due to differences between reactant bond energies and product bond energies

Using Bond Energies to Calculate ∆H orxn

Hess’s law allows us to think of any reaction as a two-step process, whether or not it actually occurs that way…

1. A quantity of heat is absorbed (∆Ho > 0) to break the reactant bonds and form separate atoms.

2. A different quantity of heat is then released (∆Ho < 0) when atoms form product bonds.

∆Horxn = ∑∆Ho

reactant bonds broken + ∑∆Horeactant bonds formed

In an exothermic reaction, ∆Horxn is negative (heat is released)

In an endothermic reaction, ∆Horxn is positive (heat is absorbed)

∆Horxn = ∑BEreactant bonds broken - ∑BEreactant bonds formed

The negative sign is needed because all bond energies are positive…

Using Hess’s law, this is the simplest way to calculate ∆Horxn…

1. Break all reactant bonds to obtain individual atoms.2. Use the atoms to form all the product bonds.3. Add the bond energies, with appropriate signs, to obtain the

enthalpy of reaction.

2 example problems…Formation of HFH2 + F2 2HF

H-H + F-F 2 H-F

Bonds Broken

1 x H-H = (1 mol)(432 kJ/mol) = 432 kJ

1 x F-F = (1 mol)(159 kJ/mol) = 159 kJ

∑∆Horeactant bonds broken = 591 kJ

Bonds Formed

2 x H-F = (2 mol)(-565 kJ/mol) = -1130 kJ

∑∆Horeactant bonds broken = -1130 kJ

Substituting into the following equation…

∆Horxn = ∑∆Ho


∆Horxn = 591 kJ + (-1130 kJ)

∆Horxn = - 539 kJ (exothermic reaction… energy is released)

Combustion of CH4

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

4 x C-H + 2 x O=O 4 x C=O + 4 x O-H

Bonds Broken

4 x C-H = (4 mol)(413 kJ/mol) = 1652 kJ

2 x O=O = (2mol)(498 kJ/mol) = 996 kJ


Bonds formed

2 x C=O = (2 mol)(-799 kJ/mol) = -1598 kJ

4 x O-H = (4 mol)(-467 kJ/mol) = -1868 kJ



∆Horxn = ∑∆Ho


∆Horxn = 2648 kJ + (-3466 kJ)


Sample Problem 9.3 p. 347Using Bond Energies to Calculate ∆H orxn

Follow-Up Problem 9.3 p. 347Bonds broken

1 x N=N = (1 mol)(945 kJ/mol) = 945 kJ

3 x H-H = (3 mol)(432 kJ/mol) = 1296 kJ


Bonds formed

6 x N-H = (6mol)(-391 kJ/mol) = -2346 kJ



∆Horxn = ∑∆Ho


∆Horxn = 2241 kJ + (-2346 kJ)


Between the Extremes: Electronegativity and Bond Polarity

Scientific models are idealized descriptions of reality. The ionic and covalent boding models portray compounds as formed by either complete electron transfer (ionic) or complete electron sharing (covalent).

In real substances, most atoms are joined by polar covalent bonds – partly ionic and partly covalent.

ElectronegativityElectronegativity – is the relative ability of a bonded atom to attract shared electrons

American chemist Linus Pauling explained that the fluorine atom attracts the shared electrons in HF more than the hydrogen atom. Since the fluorine atom has the electrons (negative charges) closer to it than the hydrogen, the fluorine end of the molecule is more negative and the hydrogen end is more positive. It creates negative and positive “poles”.o From studies with many other compounds, Pauling

derived a scale of relative EN values based on fluorine having the highest EN value.

Trends in Electronegativity

In general, electronegativity is inversely related to atomic sizeo The nucleus of a smaller atom is closer to the shared

pair than the nucleus of the larger atom, so it attracts the electrons more strongly

Down a group… electronegativity decreases because atoms get bigger down a group

Across a period…electronegativity increases because atoms get smaller across a period (from left to right)

Nonmetals are more electronegative than metals Most electronegative element is fluorine Least electronegative element is francium (very

rare and radioactive, so for all practical purposes, cesium is the least electromagnetic)

Electronegativity and Oxidation Number

1. The more electronegative atom in a bond is assigned all the shared electrons; the less electronegative atom is assigned none.

2. Each atom in a bond is assigned all of its unshared electrons

3. The oxidation number is given by

O.N. = # of valence e- - (# of shared e- + # of unshared e-)

Example… (HCl)

Chlorine is more electronegative than hydrogeno Chlorine is has 7 valence electrons and is assigned 8

(2 shared + 6 unshared) So its O.N. is 7 – 8 = -1

o Hydrogen has 1 valence electron and is assigned none (0 shared and 0 unshared)

So its O.N. is 1 – 0 = +1

Bond Polarity and Partial Ionic CharacterWhenever atoms of different electronegativities form a bond, the bonding pair is shared unequally …

This unequal distribution of electrons density results in a polar covalent bondo It is depicted by a polar arrow ( ) pointing toward

the partially negative pole See example of HF on the whiteboard

In H-H and F-F (or other diatomic molecules) where the atoms are identical, the bonding pair is shared equallyo This equal sharing is called a nonpolar covalent bond

The Importance of Electronegativity Difference (∆EN)

Electronegativity difference (∆EN) – the difference between the EN values of the bonded atoms

∆EN is 0.0 in diatomic molecules (H2, O2, Cl2, etc.)

∆EN is as high as 3.3 between the most electronegative atom fluorine and the most electropositive element cesium in CsF

Partial ionic character – the greater the ∆EN, the greater the partial ionic character (the more it is like an ionic bond than a covalent bond)

Example of partial ionic character…

Consider LiCl, HCl, and Cl2

o For LiCl [Cl = 3.0, Li = 1.0] ∆EN = 2.0o For HCl [Cl = 3.0, H = 2.1] ∆EN = 0.9o For Cl2 ∆EN = 0.0

The bond in LiCl has more ionic character than the other two examples

Two approaches that quantify ionic character…

1. ∆EN range… this approach divides bonds into mostly ionic (1.7 – 3.3), polar covalent (0.4 – 1.7), mostly covalent (0.0 – 0.4), and nonpolar covalent (0.0) Figure 9.24 p. 351

2. Percent ionic character… based on the following relationship

Sample Problem 9.4 p. 352Determining Bond Polarity from EN Values(a)

The EN of N = 3.0 and the EN of H = 2.1, so N-H, ∆EN = 0.9

The EN of F = 4.0 and the EN of N = 3.0, so F-N, ∆EN = 1.0

The EN of I = 2.5 and the EN of Cl = 3.0, so I-Cl, ∆EN = 0.5

(b)

The EN of H = 2.1 and the EN of N = 3.0, so ∆EN = 0.9

The EN of H = 2.1 and the EN of O = 3.5, so ∆EN = 1.4

The EN of H = 2.1 and the EN of C = 2.5, so ∆EN = 0.4

Increasing bond polarity H-C < H-N < H-O

Decreasing percent ionic character H-O > H-N > H-C

Follow-Up Problem 9.4 p. 352(a)

The EN of Cl = 3.0 and the EN of F = 4.0, so ∆EN = 1.0

(+)Cl-F(-)

The EN of Br = 2.8 and the EN of Cl = 3.0, so ∆EN = 0.2

(+)Br-Cl(-)

The ∆EN of Cl-Cl = 0.0

Cl-Cl

Increasing polarity Cl-Cl < Br-Cl < C-F

(b)

The EN of Si = 1.8 and the EN of Cl = 3.0, so ∆EN = 1.2

(+)Si-Cl(-)

The EN of P = 2.1 and the EN of Cl = 3.0, so ∆EN = 0.9

(+)P-Cl(-)

The EN of S = 2.5 and the EN of Cl = 3.0, so ∆EN = 0.5

(+)S-Cl(-)

The ∆EN for Si-Si = 0

Si-Si

Increasing polarity Si-Si < S-Cl < P-Cl < Si-Cl

The Graduation in Bonding Across a PeriodA metal and a nonmetal – elements from the left and right sides of the periodic table – have a relatively large ∆EN and typically form ionic compounds

Two nonmetals – both from the right side of the periodic table – have a small ∆EN and form a covalent compound

When we combine chlorine with each of the Period 3 elements, starting with sodium, we observe a steady decrease in ∆EN and a graduation in bond type from ionic through polar covalent to nonpolar covalent

General trends in bonding across Period 3…

As ∆EN decreases, the bond becomes more covalento ∆EN decreases as you go from left to righto Melting points decrease as you go from left to righto Electrical conductivity decreases as you go from left

to right in Period 3

An Introduction to Metallic BondingThe electron-sea model of metallic bonding proposes that all the metal atoms in a sample contribute their valence electrons to form a delocalized electron “sea” throughout the sample, with the metal ions (nuclei and core electrons)lying in an orderly array.

All the atoms in the sample share the electrons and the piece is held together by the mutual attraction of the metal cations for the mobile, valence electrons

Bonding in metals is different in two different ways…

o In contrast to ionic bonding, the metal electrons are not held in place as rigidly

o In contrast to covalent bonding, no particular pair of metals atoms is bonded through a localized pair of electrons

o Instead of forming compounds, two or more metals typically form an alloy

An alloy is a solid mixture of variable composition

Car and airplane bodies Bridges Coins and jewelry Dental fillings

How the Model Explains the Properties of MetalsMelting and boiling points…

Nearly all metals are solids with moderate to high melting points and much higher boiling points

Melting points are only moderately high because the cations can move without breaking the attraction to the surrounding electrons

Boiling points are much higher because each cation and its valence electron(s) must break away from the others

Example… gallium melts in your hand but boils at over 2400oC

Down a group… melting points decrease because the larger metal ions have weaker attraction to the electron sea

Across a period… melting points increaseGroup 2A(2) have higher melting points than Group 1A(1) because their +2 cations have stronger attractions to twice as many valence electrons

Mechanical properties…

When a piece of metal is deformed by a hammer, the metal ions do not repel each other, but rather slide past each other through the electron sea and end up in new positions

All the Group 1B(11) metals are soft enough to be machined into sheets (malleable) and wires (ductile)

Electrical conductivity…

Unlike ionic and covalent substances, metals are good conductors of electricity in both the solid and liquid states because of their mobile electrons.o Foreign atoms disrupt the metal atoms and reduce

conductivity

Copper used in electrical wiring is 99.99% pure because traces of other atoms drastically restrict electron flow.

Thermal conductivity…

Mobile electrons also make metals good conductors of heato Hold an aluminum can under hot water and see how

quickly the can heats up compared to holding a wooden spoon under the same hot water

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