Unit 4-Covalent Compounds Ch. 5 Barrons Book

Unit 4-Covalent Compounds Ch. 5 Barrons Book

Warm Up Name the following: 1. CoCO3 2. NH4I 3. MgCl2 4. Ca3(PO4)2
Write the formula for the following: 5. Cobalt (I) selenide Silver cyanide 7. Iron (II) oxide Copper (II) carbonate 9. Ammonium sulfate Lead (IV) sulfite

Naming Covalent Compounds
In order to name molecular compounds, a different naming system is required. WHY? How else would we differentiate between CO & CO2? Still follow the same naming rules, but add PREFIXES to distinguish how many atoms of each element are in the compound.

Prefixes – need to memorize
1 = mono 2 = di 3 = tri 4 = tetra 5 = penta 6 = hexa 7 = hepta 8 = octa 9 = nona 10 = deca

Naming Covalent Compounds
Write the name of the first element. Add a prefix if there is more than one of that first atom. Write the name of the second element and change the ending to –ide Add a prefix to the second atom even if there is only one! If “ao” or “oo” appears in the name, remove the first “a” or “o”

H2O – dihydrogen monoxide N2O3 – dinitrogen trioxide
Examples: H2O – dihydrogen monoxide N2O3 – dinitrogen trioxide BI3 – boron triiodide

N2O NO P3N2 P2S5 SO3 Practice: Dinitrogen monoxide Nitrogen monoxide
Triphosphorus dinitride Diphosphorus pentasulfide Sulfur trioxide

Naming Diatomic Molecules
7 diatomic elements: H2, O2, F2, Br2, I2, N2, Cl2 You will name the molecule the same as the element! H2 = hydrogen or hydrogen gas

Warm Up Write the following formulas: Barium sulfide
Lead (II) phosphate Ammonium hydroxide Aluminum cyanide Diphosphorus pentoxide Name the following: FeO Ba(OH)2 KNO3 C4H2

Chapter 8 Covalent bonding

Molecular Compounds… Are held together by covalent bonds
Are compounds composed of molecules Are also known as Covalent Compounds Are held together by covalent bonds Covalent Bond : atoms held together by the SHARING of electrons

Diatomic Molecules Molecules that contain 2 of the same atom
There are 7 diatomic molecules: H2, O2, F2, Br2, I2, N2, Cl2 Ways to remember: Make the shape of a seven on the periodic table Excluding HYDROGEN! Name: H-O-F Br-I-N-Cl

Comparing Ionic & Covalent Compounds
Ionic Compounds Covalent Compounds Complete transfer of electrons Cation + Anion Metal present Have to look at charges Electrons shared Nonmetal + Nonmetal No metals! No ions! Have to look at valence electrons Example: Example: K1+Cl1- O H

Warm Up Give the name for the following: 1.Fe2(SO4)3 2. Na2S
3. P3Br O3H NaOH 5. Ca(OH)2 Write the formula for the following: Lead (II) nitrate dicarbon tetrachloride magnesium carbonate Tin(IV) sulfate tetrafluorine diselenide calcium hydroxide 7. heptabromine trioxide

Warm Up Thursday, October 8
Draw the lewis structures for the following: Cl Br Ca C What would the following look like: Ca + Br K + Cl

Properties of Covalent Bonds
Electrons are shared so that the atoms can attain the electron configuration of noble gases. Weak inter-particle forces in contrast to ionic compounds. Many are liquids or gases at room temperature, but some are solids (i.e. sugar). Have low melting and boiling points compared to ionic compounds. Do not conduct electricity. Less soluble in water than ionic compounds, in general.

MUST OBEY THE OCTET RULE
Sharing of Electrons H H Single Covalent Bond One shared pair of electrons Represented by two dots or one line Longest bond - weakest H - H MUST OBEY THE OCTET RULE Double Covalent Bond Two shared pairs of electrons Represented by four dots or two lines O = O O O Triple Covalent Bond Three shared pairs of electrons Represented by six dots or three lines Shortest bond - strongest N N N Ξ N

How to show a covalent bond…
Example: HCN Draw Lewis dot structure for each atom based on how many valence electrons the atom has. 2. Share only between electrons that are not paired 3. Make sure each atom obeys the octet rule after sharing. N H C C H N Triple Covalent Bond Single Covalent Bond C H N 2 e- 8 e- 8 e-

H Cl Cl Cl H N H H PRACTICE:
Did you check to make sure they obey the octet rule? Cl Cl H N H H

MORE PRACTICE: C + 2 O  N + N  Double Covalent Bonds
Carbon Dioxide: CO2 Double Covalent Bonds Nitrogen Gas: N2 Triple Covalent Bond

Exceptions to the Octet Rule
The octet rule cannot be satisfied in molecules whose total number of valence electrons is an odd number. The octet rule cannot be satisfied in molecules in which an atom has fewer, or more than a complete octet of valence electrons. What does this mean?! Let’s take a closer look…

1. Odd number of TOTAL valence electrons…
NO2 – 17 total electrons N – 5 valence electrons O – 6 valence electrons (x 2) = 12 valence electrons Count the total electrons around each atom… Resonance Structure: structure that occurs when it is possible to draw two or more valid electron dot structures that have the same number of electron pairs for a molecule or ion.

2. less than a complete octet…
BF3 Count the valence electrons around each atom… Boron is the exception It is missing 2 electrons Why? Because it only has 3 electrons to share!

2. More than a complete octet…
SF6 Count the valence electrons around each atom… S Original Lewis Structure Sulfur has more than 8 valence This is an expanded octet Each of sulfur’s unshared pairs were split apart

Warm Up Draw the shape and name the shape for the following: HF NH3
CH4 BF3 NH3 SeCl2

Bonding Theories Chapter 8

O = O O = O VSEPR Theory Valence Shell Electron Pair Repulsion
Predicts the shapes of covalent molecules The repulsion between electron pairs causes molecular shapes to adjust so that the valence-electron pairs stay as far apart as possible. O = O O = O

- - Molecular Shapes 180o 120o 104.5o 109.5o 107o Trigonal Planar
SHAPE DESCRIPTION ILLUSTRATION Linear Bond Angle = 180o One or two bonding pairs, no lone pairs X A 180o X A 104.5o Bent Bond Angle = 104.5o Two bonding pairs, two lone pairs X A 107o X A 120o Trigonal Planar Bond Angle = 120o 3 bonding pairs, no lone pairs X A 109.5o Trigonal Pyramidal Bond Angle = 107o 3 bonding pairs, one lone pair Tetrahedral Bond Angle = 109.5o Four bonding pairs, no lone pairs

Lone Pairs Take up MORE space SQUEEZE the bond angle X A X A

HF CH4 Practice: LINEAR TETRAHEDRAL
Determine the shape of each molecule: HF CH4 H F LINEAR H C TETRAHEDRAL

More Practice: BF3 NH3 SeCl2 TRIGONAL PLANAR TRIGONAL PYRAMIDAL BENT F

Octet Rule Atoms bond with other atoms by sharing or transferring electrons in order to achieve a stable octet (8 valence electrons). When bonds are formed energy is ___________. When bonds are broken energy is ___________. released absorbed

Ionic Bonds Transfer e- Metals and nonmetals High mp and bp Soluble
Conduct as liquid Mostly solid

Ionic Bonds have a crystalline structure.

Metallic Bonds Sea of e- Metals only High mp and bp Insoluble
Conduct always All other metallic properties

Covalent Bonds (Molecular)
Share e- Nonmetals only Low mp and bp Insoluble unless polar Never conduct Creates molecules

Ionic, Metallic,Covalent or both I & C?
NaCl Mg H2S MgCl2 HCl CH4 AlCl3 NH4Cl PBr3 CuI2 SiO2 Au O2 RbF I I/C M C C I I C I/C M C C I I

Bond Polarity A B Due to differences in ELECTRONEGATIVITY,
Equal & Unequal Sharing of Electrons Shared Electron Pair A B  Due to differences in ELECTRONEGATIVITY, the sharing of electrons is not always equal between the atoms.

Electronegativity Ability of an atom to attract electrons.
Across a period, electronegativity increases due to stronger nuclear charge and needing to fill the octet. Down a group, electronegativity decreases due to higher shielding from the nucleus by inner electrons. These values range from 0-4 and are not energy values. Notice the trend and explanations are similar to IE. Example: X + e-  X- or X+ + e-  X

Electronegativity Electronegativity (EN) is the ability of an atom to attract electrons F=largest electronegativity Atoms with the largest electronegativity attract electrons more strongly than those with a small electronegativity.

Electronegativity Differences
Another way to determine the polarity of a bond is by determining the difference in electronegativity (ΔEN) of two atoms. Electronegativity Differences & Bond Types: IONIC: ΔEN > or = 1.7 Polar COVALENT: ΔEN <1.7 Nonpolar Covalent: ΔEN = O ΔEN = Ι EN atom 1 – EN atom 2 Ι

Periodic Table of Electronegativities
Why are the Noble Gases not listed?

PRACTICE: H – F N – O C – H Si – P B – Cl Ι 2.1-4.0Ι = 1.9 IONIC
Determine if the bond is IONIC or COVALENT: H – F N – O C – H Si – P B – Cl Ι Ι = 1.9 IONIC Ι 3.0 – 3.5 Ι = 0.5 Covalent Ι 2.5 – 2.1 Ι = 0.4 Covalent Ι 2.1 – 2.8 Ι = 0.7 Covalent Ι 2.0 – 3.0 Ι = 1.0 Covalent

Periodic Table of Electronegativities
Why are the Noble Gases not listed?

So What Does It Mean To Be Polar?
EXPLAINATION: EXAMPLE: In a polar bond, the more electronegative atom attracts electrons more strongly & gains a slightly negative charge. The less electronegative atom has a slightly positive charge. Chlorine is the more electronegative atom POSITIVE NEGATIVE H Cl

Two poles have been created
A molecule that has two poles is said to be a dipolar molecule or DIPOLE. Two poles have been created positive pole & negative pole When placed between oppositely charged plates, they tend to orientate with respect to the positive & negative plates. POSITIVE NEGATIVE H Cl DIPOLES

Polar Covalent Bonds GUIDELINES FOR IDENTIFYING A POLAR BOND:
Bonding electrons are shared unequally. Examples: HCl, H2O, HCN, PBr3, CH3F GUIDELINES FOR IDENTIFYING A POLAR BOND: If only two atoms involved, both atoms are different. Lone pairs are on the central atom. At least one atom bonded to the central atom is different from the rest. There is more than one bond type around the central atom.

THERE ARE MORE POLAR COMPOUNDS THAN THERE ARE NONPOLAR!
Br H Cl ATOMS ARE DIFFERENT THERE ARE MORE POLAR COMPOUNDS THAN THERE ARE NONPOLAR! LONE PAIRS ON CENTRAL ATOM O H H C F LONE PAIRS ON CENTRAL ATOM ATOMS BOUND TO CENTRAL ATOM ARE NOT THE SAME DIFFERENT BOND TYPES AROUND CENTRAL ATOM C H N WHAT MAKES THIS A NONPOLAR COMPOUND?

Nonpolar Covalent Bonds
Bonding electrons are shared equally. Examples: H2, CO2, BF3, CH4 GUIDELINES FOR IDENTIFYING A NONPOLAR BOND: If only two atoms involved, both atoms are identical. NO lone pairs are on the central atom. All atoms bonded to the central atom are the same. The bond types are identical around the central atom.

= = NONPOLAR COMPOUNDS F H B H C BOTH ATOMS ARE IDENTICAL
ALL ATOMS BOUND TO CENTRAL ATOM ARE THE SAME BOTH ATOMS ARE IDENTICAL BOND TYPES ARE IDENTICAL AROUND CENTRAL ATOM NO LONE PAIRS ON CENTRAL ATOM H C = = O C BOND TYPES ARE IDENTICAL AROUND CENTRAL ATOM ALL ATOMS BOUND TO CENTRAL ATOM ARE THE SAME NO LONE PAIRS ON CENTRAL ATOM WHAT MAKES THIS A NONPOLAR COMPOUND?

Warm Up 1. Write the name for the following:
MnO2 AlBr3 C2O6 CaSO4 N2C 2. Write the formula for the following: Tetraphosphorus decoxide Carbon dioxide Magnesium perchlorate Draw the shape, name, and if it is polar or nonpolar. NH3 H2Se CBr4 H2O

When finished, place quiz in bin and pick up sheet up front…
Help yourself to a dry erase board and work on the example problems. If your structure can show resonance, make sure you draw the other resonance structures. REMEMBER TO Ve B L your structure like we did yesterday 

Para/Diamagnetism Paramagnetism
attracted to a magnetic field due to unpaired electrons; the more unpaired electrons the stronger the field. Diamagnetism repelled by a magnetic field due to the paired electrons

7.8 Paramagnetic Diamagnetic unpaired electrons all electrons paired

Determining Para/Diamagnetism
Substance is placed between electromagnets. If the substance appears heavier, it is attracted in the magnetic field, and is paramagnetic. If the substance appears lighter, it is diamagnetic.

NaCl Na+ Cl- CaSO4 Ca2+ ZnSO4 Zn2+ CuSO4 Cu2+ MnSO4 Mn2+ MnO2 Mn?
Write electron configuration for each and predict para- or dia-magnetic NaCl Na+ Para or Dia? Cl- CaSO4 Ca2+ ZnSO4 Zn2+ CuSO4 Cu2+ MnSO4 Mn2+ MnO2 Mn? KMnO4 1s22s22p6 Dia Dia 1s22s22p63s23p6 1s22s22p63s23p6 Dia 1s22s22p63s23p63d10 Dia Para 1s22s22p63s23p63d9 1s22s22p63s23p63d5 Para +4 Para 1s22s22p63s23p63d3 +7 1s22s22p63s23p6 Dia

Covalent Bonding (Molecular Bonding)
If 2 atoms or more form a bond with the same electronegativity the bonds are nonpolar and they share e- equally.

Equal sharing

If there is an electronegativity difference between bonded atoms, the bonds are polar and e- are pulled toward the more electronegative atom.

If a molecule is polar, it will have a slightly negative and slightly positive side, like 2 poles of a magnet. This is called a dipole moment or a dipolar molecule.

Dipolar Molecules As you can see, normally polar molecules are unaligned. When a electric source comes by, the molecules quickly align themselves.

AP Video 3.4 Covalent Bonds

Lewis Structures Rules: CCl4
Add up all valence e- Draw a skeletal structure with bonds between elements. Subtract 2e- from total for each bond drawn. Draw in remaining e- to fill each atom’s octet. Evaluate: each atom should have 8 e- only. C: Cl: 7 = 32 valence e- Cl Cl—C—Cl 32-8=24

Geometry Molecules have a specific geometry:
Linear: The molecule is on one plane (flat) such as CO2 or H2. Bent: The molecule is bent at angle like H2O due to unshared electrons and two bonding pairs on the central atom. Pyramidal: The molecule has a triangular shape like NH3 due to a lone pair and three bonding pairs on the central atom. Tetrahedral: The molecule has four bonding pairs and no lone pairs on the central atom like CH4.

Dipolar Molecules Dipolar bonds can create polar or nonpolar molecules. A polar molecule will have polar bonds and be asymmetrical. A nonpolar molecule will either have nonpolar bonds or polar bonds with a symmetrical shape. Water is polar, and like dissolves like, so only polar molecules are soluble in water. Polar (dipolar) molecules are also attracted to an electric field.

Draw the following. Identify the geometry and polarity:
H2O SI2 NH3 F2 O2 PI3 CF4 SiH4 L, NP L, NP B, P B, P P, P L, NP L, NP P, P T, NP T, NP Notice, 4 elements is showing pyramidal, 5 elements is showing tetrahedral. Its not always that easy… stay tuned for more shapes! 

Warm Up-Work on your structure sheet and the question below…
Draw the electric field lines through the molecule , head of the arrow toward the more negative side: H H---Cl H--C--H Cl---Cl S---H nonpolar nonpolar

I. VSEPR = Valence Shell Electron-Pair Repulsion
Electron pairs repel each other This applies to bonding pairs and lone pairs alike Steps in Applying VSEPR Draw the Lewis Structure Count atoms and lone pairs and arrange them as far apart as possible Determine the name of the geometry based only on where atoms are

Trigonal pyramidal Tetrahedral B. Complications to VSEPR
Lone pairs count for arranging electrons, but not for naming geometry Example: NH3 (ammonia) Lone pairs are larger than bonding pairs, resulting in adjusted geometries Bond angles are “squeezed” to accommodate lone pairs Trigonal pyramidal Tetrahedral

b. Lone pairs must be as far from each other as possible. XeF4
4. Double and triple bonds are treated as only 1 pair of electrons

linear trigonal planar bent tetrahedral bent trigonal bipyramidal.
5. Names for and examples of complicated VSEPR Geometries linear trigonal planar bent trigonal pyramidal tetrahedral bent trigonal bipyramidal. see-saw T-shaped linear octahedral square pyramidal square planar

II. Hybridization Localized Electron Model sp3 Hybridization
Molecules = atoms bound together by sharing e- pairs between atomic orbitals Lewis structures show the arrangement of electron pairs VSEPR Theory predicts the geometry based on e- pair repulsion Hybridization describes formation and properties of the orbitals involved in bonding more specifically sp3 Hybridization Methane, CH4, will be our example Bonding only involves valence e- a. H = 1s b. C = 2s, 2p

If we use atomic orbitals directly, the predicted shape doesn’t match what we predict by VSEPR and what we observe experimentally One C2s--H1s bond would be shorter than others Three C2p--H1s bonds would be at right angles to each other We know CH4 is tetrahedral, symmetric Free elements (C) use pure atomic orbitals Elements involved in bonding will modify their orbitals to reach the minimum energy configuration

When C bonds to four other atoms, it hybridized its 2s and 2p atomic orbitals to form 4 new sp3 hybrid orbitals The sp3 orbital shape is between 2s/2p; one large lobe dominates When you hybridize atomic orbitals, you always get an equal number of new orbitals

The new sp3 orbitals are degenerate due to the mixing
The H atoms of methane can only use their 1s orbitals for bonding. The shared electron pair can be found in the overlap area of H1s—Csp3

NH3 and H2O also use sp3 hybridization, even though lone pairs occupy some of the hybrid orbitals.
C2H4, ethylene is our example Lewis and VSEPR structures tell us what to expect H atoms still can only use 1s orbitals C atom hybridizes 2s and two 2p orbitals into 3 sp2 hybrid orbitals

The new sp2 orbitals are degenerate and in the same plane
One 2p orbital is left unchanged and is perpendicular to that plane One C—C bond of ethylene forms by overlap of an sp2 orbital from each of the two sp2 hybridized C atoms. This is a sigma (s) bond because the overlap is along the internuclear axis. The second C—C bond forms by overlap of the remaining single 2p orbital on each of the carbon atoms. This is a pi (p) bond because the overlap is perpendicular to the internuclear axis.

9. Pictorial views of the orbitals in ethylene

sp Hybridization CO2, carbon dioxide is our example
Lewis and VSEPR predict linear structure C atom uses 2s and one 2p orbital to make two sp hybrid orbitals that are 180 degrees apart We get 2 degenerate sp orbitals and two unaltered 2p orbitals

Oxygen uses sp2 orbitals to overlap and form sigma bonds with C
Free p orbitals on the O and C atoms form pi bonds to complete bonding

d-orbitals can also be involved in hybridization
dsp3 hybridization in PCl5 d2sp3 hybridization in SF6

F. The Localized Electron Model
1. Draw the Lewis structure(s) 2. Determine the arrangement of electron pairs (VSEPR model). 3. Specify the necessary hybrid orbitals.

Polar Bonds A. Any bond between atoms of different electronegativities is polar Electrons concentrate on one side of the bond One end of the molecule is (+) and one end is (-) B. Complex molecules: vector addition of all bond dipoles

Polarity greatly effects molecular properties
Examples: Polarity greatly effects molecular properties Oppositely charged ends of polar molecules Attract each other like opposite poles of magnets

Covalent Bonding Forces
Electron – electron repulsive forces Proton – proton repulsive forces Electron – proton attractive forces

Bond Length and Energy Bond Bond type Bond length (pm) Bond Energy (kJ/mol) C - C Single 154 347 C = C Double 134 614 C  C Triple 120 839 C - O 143 358 C = O 123 745 C - N 305 C = N 138 615 C  N 116 891 Bonds between elements become shorter and stronger as multiplicity increases.

Bond Energy and Enthalpy
Energy required Energy released D = Bond energy per mole of bonds Breaking bonds always requires energy Breaking = endothermic Forming bonds always releases energy Forming = exothermic

Bond Energy Using bond energies in your textbook calculate the heat of formation. (It is recommended that you draw the compounds) CH4 + 2Cl2+ 2F2 CF2Cl2 + 2HF + 2HCl (all gases) [4(413) + 2(242) + 2(155)] – [2(485) + 2(328) + 2(567) + 2(431) = kJ

Basics of Bonding A chemical bond occurs when atoms or ions are strongly attached to each other. Ionic bonds involve the transfer of e– and the subsequent electrostatic attractions. -- “metal/nonmetal” (“cation/anion”) Covalent bonds involve the sharing of e– between two atoms. -- “nonmetal/nonmetal” metallic bonds: each metal atom is bonded to several neighboring atoms -- bonding e– (i.e., valence e–) are free to move throughout the material

C N S Lewis symbols show ONLY the valence e– (i.e., the ones
involved in bonding). C N S octet rule: atoms “want” 8 v.e– -- several exceptions (a topic for later) Gilbert Lewis (1875–1946)

[ Cl ] Na Cl + Na+ + Ionic Bonding Na(s) + ½ Cl2(g) NaCl(s) –
“Salts” are brittle solids with high melting points. Na Cl [ Cl ] – + Na+ + lattice energy: the energy required to separate 1 mole of solid ionic compound into gaseous ions DHfo = –410.9 kJ/mol enthalpy (i.e., heat) of formation -- a measure of stability NaCl(s) Na+(g) + Cl–(g) DHlatt = +788 kJ/mol In general, ionically bonded substances have... …big, (+) lattice energies.

Because lattice energies are electrostatic in nature,
two variables are involved in how big they are: 1. the magnitude of the charges 2. the separation between the ions (“wears the pants”) Charge can easily be twice as large (i.e., 2+ and 2– vs. 1+ and 1–), but the separation never varies by that much. Put the following in order of increasing lattice energy: LiBr, FeN, CdO. Now these: MgS, MgCl2, MgO. LiBr < CdO < FeN MgCl2 < MgS < MgO

With transition-metal ions, the e– lost first come
from the subshell with the largest value of n. e.g., Ni = [Ar] 4s2 3d8 Ni2+ = [Ar] 3d8 Ni3+ = [Ar] 3d7 Recall that polyatomic ions are groups of atoms that stay together and have a net charge. -- e.g., NO3– CH3COO– -- their atoms are held to each other by… covalent bonds

Cations are smaller than the neutral atoms from
which they are derived. e.g., Li 1s2 2s1 Li s2 -- orbitals might be completely vacated -- e–/e– repulsion always decreases Fe Fe2+ Fe3+

Anions are larger than the neutral atoms from
which they are derived. -- more e–/e– repulsion Cl Cl– 17 p+, 17 e– 17 p+, 18 e– An isoelectronic series is a list of species having an identical electron configuration. e.g., N3–, O2–, F–, Ne, Na+, Mg2+, Al3+ (big radius) (small radius) EX. Which has the Rb Sr Y3+ largest radius? (smallest Zeff) (largest Zeff)

+ + Covalent Bonding -- atoms share e– -- covalent (molecular)
compounds tend to be solids with low melting points, or liquids or gases (–) charge density -- one shared pair = a single covalent bond of e– (i.e., 2 e–) -- two shared pairs = a double covalent bond of e– (i.e., 4 e–) -- three shared pairs = a triple covalent bond of e– (i.e., 6 e–)

bond polarity: describes the sharing of e– between atoms nonpolar covalent bond: e– shared equally polar covalent bond: e– NOT shared equally electronegativity (EN): the ability of an atom in a molecule to attract e– to itself -- A bonded atom w/a large EN has a great ability to attract e–. max. EN = 4.0 (F) -- A bonded atom w/a small EN does not attract e– very well. min. EN = 0.7 (Cs) -- EN values have been tabulated. (see p. 308)

The DEN between bonded atoms approximates
the type of bond between them. DEN < 0.5 nonpolar covalent 0.5 < DEN < 2.0 polar covalent DEN > 2.0 ionic bond As DEN increases, bond polarity... increases. DEN for a C–H bond = 0.4, so all C–H bonds (such as those in candle wax) are nonpolar.

Dipole Moments Polar covalent molecules have a partial (–) and a partial (+) charge and are said to have a dipole moment. d+ d– d+ d+ partial charge H–F H–F O H d– big big big DEN = _____ polarity = _____ dipole moment

Polar molecules tend to align themselves with each
other and with ions. H–F NO3– NH4+ H–F H–F H–F NH4+ NO3– H–F H–F ** Nomenclature tip: For binary compounds, the less electronegative element comes first. -- Compounds of metals w/high ox. #’s (e.g., 4+ or higher) tend to be molecular rather than ionic. e.g., TiO2, ZrCl4, Mn2O7

The Octet Rule—Ions The octet rule signifies the general observation that atoms of main-group (or representative) elements tend to gain, lose or share electrons in order to have the same electronic configuration as a noble gas. With the formation of monatomic ions, the octet rule is very useful. Element Electron Configuration of Element Ion Electron Configuration of Ion Na 1s2 2s2 2p6 3s1 Na+ 1s2 2s2 2p6 = [Ne] Ca 1s2 2s2 2p6 3s2 3p6 4s2 Ca2+ 1s2 2s2 2p6 3s2 3p6 = [Ar] O 1s2 2s2 2p4 O2- P 1s2 2s2 2p6 3s2 3p3 P3-

The Octet Rule—Covalent Bonds
For molecular compounds and polyatomic ions, the octet rule is also useful tool when building the skeleton of Lewis structures. However, a more useful rule to keep in mind is the total number of electrons that can occupy a given shell (i.e. all the orbitals with the same principal quantum number, n). Principal Quantum Number (Shell) Angular Momentum Quantum Numbers (Subshells) Total Electrons Total Bonds (= electrons/2) 1 0 (s) 2 0, and 1 (s and p) 2 + 6 = 8 4 3 0, 1, and 2 (s, p and d) = 18 9

The Octet Rule—Covalent Bonds
F Hydrogen is a period 1 element and can have a maximum of 2 electrons in its valence shell. Oxygen and fluorine are period 2 elements and can have a maximum of 8 electrons in their valence shells. Sulfur is a period 3 element and can have a maximum of 18 electrons in its valence shell, but in sulfur hexafluoride there are only 12.

Single Bond Single bonds between atoms is the sharing of a single pair of electrons. For example, in water the central oxygen is bonded to two hydrogen atoms each by a single pair of electrons creating a single bond with each Single covalent bonds Single covalent bonds

Double Bond Double bonds
Double bond – two atoms share two pairs of electrons. For example, carbon dioxide has two sets of double bonds, one each with each oxygen. Double bonds

:N:::N: :N N: Triple Bond Triple bond
Triple bond – two atoms share three pairs of electrons. For example, the bond in nitrogen gas (N2) is a triple bond. :N:::N: Triple bond :N N:

Electronegativity Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond. Electron Affinity - measurable, Cl is highest X (g) + e X-(g) Electronegativity - relative, F is highest

Electronegativity decreases from top to bottom of the periodic table
Electronegativity increases from left to right across the periodic table Electronegativity decreases from top to bottom of the periodic table

Electronegativity EN Difference = 2.1 – 2.1 = 0 Equal sharing
Unequal sharing EN Difference = 4.0 – 0.9 = 3.1 Electron transfer

Classification of Bonds by Electronegativity Difference
Difference in electronegativity (EN) = Larger EN – Smaller EN Increasing difference in electronegativity Covalent share e- Polar Covalent partial transfer of e- Ionic transfer e- Difference Bond Type ≤ 0.4 Covalent 0.4 < and ≤1.8 Polar Covalent 1.8 < and ≤ 2.2 Partially Ionic > 2.2 Ionic

Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H2S; the NN bond in H2NNH2; and the bond in HF. Atom 1 (Greatest EN) EN1 Atom 2 (Least EN) EN2 ∆EN= EN1 – EN2 Bond Cl 3.0 Cs 0.7 2.3 Ionic S 2.5 H 2.1 0.4 Covalent N 0.0 F 4.0 1.9 Partially Ionic

Dipole Moments and Polar Bonding
H F electron rich region electron poor d+ d- The arrow points to the electron rich region which has a greater than expected electron density and, therefore, a partial negative charge. The partial negative charge is designated by the lower case Greek letter delta with a negative sign (δ-). At the beginning of the arrow is the electron poor region which has a less than expected electron density and, therefore, a partial positive charge. The partial positive charge is designated by the lower case Greek letter delta with a positive sign (δ+).

6.3 Drawing Lewis Structures 6.4 Lewis Structures & Formal Charge

Drawing Lewis Structures 1
Draw skeletal structure of compound showing what atoms are bonded to each other and set up valence electron accounting table (see the generic valence electron accounting table below). Hydrogen and fluorine atoms are always in a terminal (outside) position. Put least electronegative element in the center. Atom Central Atom Peripheral Atom 1 Peripheral Atom 2 Peripheral Atom 3 Overall Charge Total* Valence electrons Bonds Lone electrons Formal Charge** *Total = Sum of Atomic Electrons – Charge **Formal Charge = Valence Electrons – Bonds – Lone Electrons

Lewis Structure for CO2 (1)
Carbon is the least electronegative atom and is placed in the center. The two oxygen atoms are attached to the central carbon. Atom C O Overall Charge Total Valence Electrons Bonds Lone Electrons Formal Charge

Count total number of electrons for the structure by: Summing the valence electrons for each atom in the molecule, Then add 1 for each negative charge or subtract 1 for each positive charge. Atom Central Atom Peripheral Atom 1 Peripheral Atom 2 Peripheral Atom 3 Overall Charge Total* Valence electrons A B C D E A+B+C+D-E Bonds Lone electrons Formal Charge** *Total = Sum of Atomic Electrons – Charge **Formal Charge = Valence Electrons – Bonds – Lone Electrons

Carbon has 4 valence electrons and each oxygen has 6. Since this is a neutral molecule, there is no overall charge. Atom C O Overall Charge Total Valence Electrons 4 6 16 Bonds Lone Electrons Formal Charge

Subtract bonded electrons (i.e., 2 electrons for each bond) from the total count in step 2; then, distribute remaining electrons by first completing octets on terminal atoms (except hydrogen which gets a duet). Since each terminal atom already has 2 electrons from the bond, 3 sets of lone pairs will be added to each terminal atom (6 electrons total). Then, place remaining electrons on the central atom.

Carbon has 2 bond and each oxygen has 1 for a total of 4. This leaves 12 electrons to be distributed on the structure Atom C O Overall Charge Total Valence Electrons 4 6 16 Bonds 2 1 Lone Electrons 12 Formal Charge

The 12 remaining valence electrons are place on the oxygen atoms with each receiving 6. Atom C O Overall Charge Total Valence Electrons 4 6 16 Bonds 2 1 Lone Electrons 12 Formal Charge

Calculate the formal charge on each atom by the following formula: Formal Charge = Valence e- - # of bonds - lone electrons The sum of the formal charges should equal the overall charge. The Total column should have a Formal Charge of 0 (i.e. all valence electrons should be accounted for.) Atom Central Atom Peripheral Atom 1 Peripheral Atom 2 Peripheral Atom 3 Overall Charge Total* Valence electrons Bonds Lone electrons Formal Charge** a) b) *Total = Sum of Atomic Electrons – Charge **Formal Charge = Valence Electrons – Bonds – Lone Electrons

Atom C O Overall Charge Total Valence Electrons 4 6 16 Bonds 2 1 Lone Electrons 12 Formal Charge +2 -1

Reduce formal charges on the central atom by creating double or triple bonds. Reduce a positive formal charge by moving a lone pair from a neighboring atom with a negative formal charge. Each movement of a pair of electrons increases the formal charge on the donating atom by one and reduces the formal charge on the receiving atom by one. Rework the table to recalculate the formal charges. Continue to reduce formal charges until either: All formal charges are reduced to zero. Further movement of electron pairs is not necessary except to obey the octet rule for certain 2nd period elements. Further movement of electrons would violate the octet rule for 2nd period elements. Note the following special cases regarding 2nd period elements Beryllium (Be) usually only forms 2 bonds. Boron (B) usually only forms 3 bonds, but occasionally will form 4.

Since each oxygen a negative formal charge, the best movement of electron pairs is from the oxygen to form double bonds with the carbon, which has a +2 formal charge

Reworked table after moving electrons and making double bonds. Atom C O Overall Charge Total Valence Electrons 4 6 16 Bonds 2 8 Lone Electrons Formal Charge

If more than one possible structure can be drawn (especially in the cases that double bonds can be formed in different locations in the molecule), then draw the “resonance” structures. Molecules with no formal charges do not need to go beyond step 5.

Since the final structure in step 5 has no formal charges it is the best possible structure. Nonetheless, resonance structures could be formed by moving two pairs of electrons from one oxygen to form a triple bond. However, these structures result in formal charges on each oxygen.

Choose structures with : The fewest number of formal charges Negative formal charges on the more electronegative atoms Positive formal charges on the less electronegative atoms

Choose the top structure because it has no formal charges.

Drawing the Lewis Structure of the Carbonate ion (CO32-)

Lewis Structure for CO32- (1)
Carbon is the least electronegative atom and is placed in the center. The three oxygen atoms are attached to the central carbon. Atom C O Overall Charge Total Valence Electrons Bonds Lone Electrons Formal Charge

Carbon has 4 valence electrons, each oxygen has 6 and there is an overall charge of -2. Atom C O Overall Charge Total Valence Electrons 4 6 -2 24 Bonds Lone Electrons Formal Charge

Carbon has 3 bonds and each oxygen has 1. The total bonding electrons is 6. The remaining 18 electrons are distributed to the 3 oxygen atoms to complete their octets. Atom C O Overall Charge Total Valence Electrons 4 6 -2 24 Bonds 3 1 Lone Electrons 18 Formal Charge

Carbon has 3 bonds and each oxygen has 1. The total bonding electrons is 6. The remaining 18 electrons are distributed to the 3 oxygen atoms to complete their octets. Atom C O Overall Charge Total Valence Electrons 4 6 -2 24 Bonds 3 1 Lone Electrons 18 Formal Charge -1

Since carbon has only a single + formal charge, only one pair of electrons needs to be moved to create a double bond. Note the changes that occur in the formal charges. Atom C O Overall Charge Total Valence Electrons 4 6 -2 24 Bonds 2 1 Lone Electrons 18 Formal Charge -1

The double bond could have been formed to any one of the three oxygen atoms creating three resonance structures.

6.5 Resonance

Resonance Structure A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure. O + - What are the resonance structures of the carbonate (CO32-) ion? O C - O C - O C -

Ben Mills, 29 Aug 2008, www.Wikipedia.com
Carbonate Bonding Ben Mills, 29 Aug 2008,

6.6 Exceptions to the Octet Rule

The Incomplete Octet Be – 2e- 2H – 2x1e- 4e- H Be BeH2 3 single bonds (3x2) = 6 9 lone pairs (9x2) = 18 Total = 24 B – 3e- 3F – 3x7e- 24e- BF3 F B

Odd-Electron Molecules NO N – 5e- O – 6e- 11e- N O The Expanded Octet (or Hypervalent central atom) (central atom with principal quantum number n > 2) S – 6e- 6F – 42e- 48e- SF6 S F 6 single bonds (6x2) = 12 18 lone pairs (18x2) = 36 Total = 48

Lewis Structures (or “electron-dot structures”) 1. Sum the valence e– for all atoms. If the species is an ion, add one e– for every (–); subtract one e– for every (+). 2. Write the element symbols and connect the symbols with single bonds. 3. Complete octets for the atoms on the exterior of the structure, but NOT for H. 4. Count up the valence e– on your L.S. and compare that to the # from Step 1. -- If your LS doesn’t have enough e–, place as many e– as needed on central atom. -- If LS has too many e– OR if central atom doesn’t have an octet, use multiple bonds.

Draw Lewis structures for the following species.
.. .. .. .. .. PCl3 26 e– Cl– .. P –Cl .. Cl .. .. .. .. HCN 10 e– H–C–N .. .. .. H–C=N .. H–C–N .. .. [ ] .. .. O 3– .. .. PO43– 32 e– .. .. .. O– P –O .. O .. .. ..

.. .. .. .. .. .. .. .. .. .. .. .. H2 2 e– H–H H H–C–C–O–H CH3CH2OH
This CO bond is longer (and weaker) than this CO bond. CO2 16 e– .. .. .. .. .. O–C–O .. .. .. O=C=O .. .. As the # of bonds between two atoms increases, the distance between the atoms... decreases.

formal charge: the charge a bonded atom would have
if all the atoms had the same electronegativity; to find it, you must first draw the Lewis structure formal charge # of v.e– in = the isolated – atom # of e– assigned to the atom in the Lewis structure all unshared e– plus half of the bonding e– When several Lewis structures are possible, the most stable is the one in which: (1) the atoms have the smallest formal charges, and (2) the (–) charges reside on the most electronegative atoms

Find the formal charge on each
atom in the following species. [ ] .. .. .. 2– S O SO32– .. .. .. O– S –O .. v.e– 6 6 O .. .. .. assigned e– 5 7 26 e– +1 –1 S: +1 O: –1 (N is more EN than S.) – .. .. – “Mr. B, tell me what the electronegativity value is.” “All right… FO NCl.” – .. .. [ ] .. N–C–S .. [ ] .. [ ] NCS– N=C=S .. N–C–S .. .. .. .. 5 4 6 5 4 6 5 4 6 16 e– 7 4 5 6 4 6 5 4 7 2– 1+ 1– 1–

[ ] [ ] [ ] .. .. .. resonance structures: the two or more Lewis
structures that are equally correct for a species e.g., For NO3–... 24 e– [ ] N =O O– .. O – [ ] N –O O= .. O – [ ] N –O O– .. O – Resonance structures are a blending of two or more Lewis structures. In drawing resonance structures, e–s move. Atoms don’t move.

Aromatic compounds are based on
resonance structures of the benzene molecule. C H C H In the interest of efficiency, benzene is often represented like this…

.. .. Exceptions to the Octet Rule
There are a few cases (other than for H) in which the octet rule is violated. These are: 1. particles with an odd number of valence e– -- e.g., ClO2, NO, NO2 2. atoms with less than an octet B .. F F– –F -- e.g., B, Be (24 e–) BF3 B –F F= .. F -- would require resonance -- F is WAY more EN than B and won’t share extra e– -- inconsistent w/chemical behavior

.. 3. atoms with more than an octet
-- this occurs when an atom gains an expanded valence shell P .. Cl Cl– –Cl PCl5 (40 e–) -- other e.g., AsF6–, ICl4–, SF4 Expanded valence shells occur only for atoms in periods > 3. -- unfilled d orbitals are involved -- large central atom = more space to accept e– pairs -- small exterior atoms = fit more easily around central atom

.. .. . The strengths of a molecule’s covalent bonds are
related to the molecule’s stability and the amount of energy required to break the bonds. bond enthalpy: the DH req’d to break 1 mol of a particular bond in a gaseous substance -- always + At room temp., I2 is a silver-purple solid. -- big DH = strong bond (g) I–I .. I .. (g) 2 . -- e.g., DH = +149 kJ/mol

.. .. . atomization: the process of breaking a molecule into
its individual atoms H H–C–Cl .. C (g) H (g) Cl (g) .. . (g) methylchloride (chloromethane) -- DH for a given bond (e.g., the C–H bond) varies little between compounds. CH3Cl was once used as a refrigerant, but it is toxic and so is now used primarily in the production of silicone polymers. It is also used as an organic solvent. e.g., C–H bonds in CH4 vs. those in CH3CH2CH3 have about the same DH

-- Typical values of bond enthalpies for specific
bonds have been tabulated. (p. 326) -- To find bond enthalpy for atomization, add up bond enthalpies for each bond broken. Calculate the bond enthalpy for the atomization of dichloromethane. H H–C–Cl Cl (i.e., break 2 C–H bonds and 2 C–Cl bonds) DHatm = 2 (413 kJ/mol) + 2 (328 kJ/mol) = kJ/mol

You can approximate reaction enthalpy using
Hess’s law and tabulated bond enthalpies. DHrxn S(DHbroken bonds) – S(DHformed bonds) Approximate the reaction enthalpy for the combustion of propane. C3H O2 4 H2O CO2 H–C–C–C–H H O=O .. H–O H .. O=C=O .. Broken: 2 C–C: 2 (348) –2023 kJ 8 C–H: 8 (413) 6475 kJ 5 O=O: 5 (495) (Actual = –2220 kJ) Formed: 8 O–H: 8 (463) 8498 kJ 6 C=O: 6 (799)

bond length: the center-to-center distance
between two bonded atoms -- fairly constant for a given bond (e.g., the C–H bond), no matter the compound e.g., C–H bonds in CH4 are about the same length as those in CH3CH2CH3 -- Average bond lengths have been tabulated for many bonds. (p. 329) -- As the number of bonds between two atoms increases, bond length… and bond enthalpy… e.g., C–C C=C C–C 1.54 A 348 kJ/mol 1.34 A 614 kJ/mol 1.20 A 839 kJ/mol

Molecular Geometry and Bonding Theories

The properties of a molecule depend on its shape and
and the nature of its bonds. In this unit, we will discuss three models. (1) a model for the geometry of molecules -- valence-shell electron-pair repulsion (VSEPR) theory (2) a model about WHY molecules form bonds and WHY they have the shape they do -- valence-bond theory (3) a model of chemical bonding that deals with the electronic structure of molecules -- molecular orbital (MO) theory

bond angles: the angles made by the lines joining
the nuclei of a molecule’s atoms carbon dioxide methane formaldehyde CO2 CH4 CH2O 180o 109.5o 120o

.. .. VSEPR electron domain: a region in which at least two
electrons are found -- they repel each other because… they are all (–) bonding domain: 2-to-6 e– that are shared by two atoms; they form a… covalent bond nonbonding domain: 2 e– that are located on a single atom; also called a… lone pair For ammonia, there are three bonding domains and one nonbonding domain. NH3 N –H H– H .. 4 e– domains N H .. Domains arrange themselves so as to minimize their repulsions.

.. .. N The electron-domain geometry is one H
of five basic arrangements of domains. -- it depends only on the total # of e– domains, NOT the kind of each domain The molecular geometry describes the orientation of the atoms in space. .. -- it depends on how many of each kind of e– domain

Electron-Domain Geometry Possible Molecular Geometries
Total # of Domains Electron-Domain Geometry Possible Molecular Geometries 2 3 4 5 6 linear linear (CO2) trigonal planar trigonal planar (BF3), bent (NO2) tetrahedral “atoms – axial” tetrahedral (CH4), trigonal pyramidal (NH3), bent (H2O) trig. bipyramidal (PCl5), linear (XeF2) seesaw (SF4), T-shaped (ClF3) trigonal bipyramidal octahedral octahedral (SF6), sq. pyr. (BrF5), square planar (XeF4)

[ ] .. .. .. .. To find the electron-domain geometry (EDG) and/or
molecular geometry (MG), draw the Lewis structure. Multiple bonds count as a single domain. Predict the EDG and MG of each of the following. [ ] Sn –Cl Cl– .. Cl – SnCl3– EDG: tetrahedral MG: trig. pyramidal 26 e– .. O –O O= .. O O– O= O3 EDG: trig. planar 18 e– MG: bent Cl–Se–Cl .. SeCl2 EDG: tetrahedral 20 e– MG: bent

[ ] .. .. .. .. .. .. .. C= O O– 2– CO32– EDG: trig. planar (res.)
[ ] C= O O– .. 2– CO32– EDG: trig. planar (res.) MG: trig. planar 24 e– .. S –F F– F SF4 EDG: trig. bipyr. .. MG: seesaw 34 e– I .. F –F F– IF5 EDG: octahedral .. MG: sq. pyramidal 42 e– Cl –F F– .. F ClF3 EDG: trig. bipyr. .. MG: T-shaped 28 e–

[ ] .. .. .. .. I –Cl Cl– Cl – ICl4– EDG: octahedral MG: sq. planar
[ ] .. I –Cl Cl– Cl – ICl4– EDG: octahedral MG: sq. planar 36 e– For molecules with more than one central atom, simply apply the VSEPR model to each part. Predict the EDG and MG around the three interior atoms of ethanoic (acetic) acid. CH3COOH PORTION EDG MG H–C–C–O–H .. H O –CH3 tetra. tetra. C=O – .. trig. plan. trig. plan. –OH .. tetra. bent

.. .. .. Nonbonding domains are attracted to only one nucleus;
therefore, they are more spread out than are bonding domains. The effect is that nonbonding domains (i.e., “lone pairs”) compress bond angles. Domains for multiple bonds have a similar effect. e.g., CH4 NH H2O the ideal bond angle for the tetrahedral EDG is 109.5o COCl2 N H .. O H .. .. Cl O=C C H 124.3o 104.5o 111.4o 107.0o 109.5o 124.3o EDG = trig. plan. ideal = 120o

Polarity of Molecules A molecule’s polarity is determined by its overall dipole, which is the vector sum of the dipoles of each of the molecule’s bonds. Consider CO2 v. H2S... bond dipoles bond dipoles S H .. O=C=O .. overall dipole = zero overall dipole = (NONPOLAR) (POLAR)

.. .. Classify as polar or nonpolar: PCl3 Cl– P –Cl polar BCl3 26 e–
Boron can be extracted by the electrolysis of molten boron trichloride. Boron is an essential nutrient for plants, and is also a primary component of control rods in nuclear reactors.

Valence-Bond Theory: merges Lewis structures
w/the idea of atomic orbitals (2s, 3p, etc.) Lewis theory says… covalent bonding occurs when atoms share electrons V-B theory says… covalent bonding occurs when valence orbitals of adjacent atoms overlap; then, two e–s of opposite spin (one from each atom) combine to form a bond V-B theory is like Velcro: “No overlap, no bond.”

Consider H2, Cl2, and HCl... H (unpaired e– is in 1s orb.) Cl [Ne] (unpaired e– is in 3p orb.) H2 Cl2 HCl = orbital overlap region (responsible for bond)

There is always an optimum distance between two
bonded nuclei. At this optimum distance, attractive (+/–) and repulsive (+/+ and –/–) forces balance. H–H distance Energy H2 molecule optimum dist. (min. PE)

Intermolecular Forces
Are weaker than either ionic or covalent bonds. Types of Intermolecular Forces: Hydrogen Bonds Van der Waals Forces Dipole Interactions Dispersion Forces

Hydrogen Bonds Attractive forces in which a hydrogen covalently bonded to a very electronegative atom is also weakly bonded to an unshared electron pair of another electronegative atom.

Properties of Hydrogen Bonds
Hydrogen bonds are very easy to break 5% the strength of a covalent bond Strongest of the intermolecular forces Water molecules are held together by hydrogen bonds. Hydrogen bonding accounts for many properties of water. Liquid at room temperature Cohesive properties

Van der Waals Forces Dipole Interactions Dispersion Forces
Occur when polar molecules are attracted to one another Like Hydrogen bonds, but with other atoms Dispersion Forces Weakest of all molecular interactions Caused by the motion of electrons of a molecule when close to a neighboring molecule

Unit 4-Covalent Compounds Ch. 5 Barrons Book

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Unit 4-Covalent Compounds Ch. 5 Barrons Book

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