Energy, Bonds & Chemical Structure Chapter 7, 8 & 9

Determination of Atomic Radius: Half of the distance between nuclei in covalently bonded diatomic molecule "covalent atomic radii"

Determination of Atomic Radius: Periodic Trends in Atomic Radius Radius decreases from left to right across a period Increased effective nuclear charge due to decreased shielding Radius increases down a group Addition of principal quantum levels

Table of Atomic Radii

Ionic Radii Cations Positively charged ions Lose electrons Smaller than the corresponding atom Anions Negatively charged ions Gains electrons Larger than the corresponding atom

Table of Ion Sizes

Ionization Energy The energy required to remove one mole electrons from a mole of gaseous atoms to produce one mole of gaseous ions M(g)  M+(g) + e- Second ionization energy – energy change accompanying M+(g)  M2+(g) + e- Measured in kJ/mol Positive energy values – requires energy to remove electrons (endothermic)

Ionization Energy Magnitude is determined by the attraction of the positive nucleus for the negative electrons that are being removed Attraction is dependant on Nuclear energy Shielding effect of inner electrons

Ionization of Magnesium Mg + 738 kJ  Mg+ + e- Mg+ + 1451 kJ  Mg2+ + e- Mg2+ + 7733 kJ  Mg3+ + e-

Electron Affinity The energy change when one mole of gaseous atoms gains one mole of electrons to form one mole of gaseous ions X(g) + e-  X- Measured in kJ/mol

Electron Affinity Affinity tends to increase across a period Affinity tends to decrease as you go down in a period Electrons farther from the nucleus experience less nuclear attraction Some irregularities due to repulsive forces in the relatively small p orbitals

Lattice Energy Atomization of Lithium +155.2 kJ/mol Atomization of fluorine +150.6 kJ/mol Ionization of lithium +520 kJ/mol Electron affinity of fluorine -328 kJ/mol Lattice Energy for LiF -1016 kJ/mol

Electronegativity A measure of the ability of an atom in a chemical compound to attract electrons Electronegativities tend to increase across a period Electronegativities tend to decrease down a group or remain the same

Periodic Tables of Electronegativities

Example Which bond is more polar? H—F Na—Cl C—H

Ionic bonds Electrons are transferred Electronegativity differences are generally greater than 2.0 The formation of ionic bonds is always exothermic!

General Rule In most chemical compounds, atoms are reacting so that they have the same electronic structure as a noble gas (they are “isoelectronic” with a noble gas). General Rule For any set of isoelectronic atoms, the highest nuclear charge (Z) will be smallest, because that nucleus can pull on the electron clouds the hardest. Ex Put these species in order of increasing size: Se2-, Br-, Rb+ and Sr2+

Covalent Bonds Covalent bonds – sharing of electrons to form discrete molecules Polar-Covalent bonds Electrons are unequally shared Electronegativity difference between 0 and 2.0 Nonpolar-Covalent bonds Electrons are equally shared Electronegativity difference of 0

Calculating Bond Energy Values come from a table (Pg 372 in your book) Energy (∆H) is the sum of the energy required to break old bonds plus the sum of the energy required to make new bonds Old bonds have a positive sign (require energy to break) New bonds have a negative sign (take in energy to make)

Bond Energy Bond energy is given in kJ/mol More bonds (double or triple) contain more energy and are harder to break If the overall value of ∆H is positive, energy must be added into the reaction (endothermic) If the overall value of ∆H is negative, energy is released into the environment (exothermic)

Example Methane is burned in oxygen gas creating water and carbon dioxide. Calculate the ∆H for the reaction. Write the reaction. Figure out what bonds are in each molecule. Figure out how many of each bond there are and what the energy for each is (Use Pg. 372) ∆H = Σ(reactants) – Σ(products)

Lewis Structures Shows how valence electrons are arranged among atoms in a molecule. Reflects central idea that stability of a compound relates to noble gas electron configuration.

The Octet Rule Combinations of elements tend to form so that each atom, by gaining, losing, or sharing electrons, has an octet of electrons in its highest occupied energy level. Bonding for fluorine

Drawing Lewis Structures 1. Calculate the total number of valence electrons for the compound - don’t forget to take charge into account 2. Decide which atom is the central atom - if this is not obvious, use the least electronegative atom available 3. Use a line to indicate covalent bonds to the outer atoms 4. Arrange the rest of the electrons around the atoms to form octets 5. Form multiple bonds (double or triple bonds/lines) if necessary to complete the octets

Completing a Lewis Structure Draw Lewis structures for the following Fluorine gas Hydrochloric acid Carbon tetrabromide PCl6- NH4+

Comments on the Octet Rule 2nd row elements C, N, O, F observe the octet rule (HONC rule as well). 2nd row elements B and Be often have fewer than 8 electrons around themselves - they are very reactive. 3rd row and heavier elements CAN exceed the octet rule using empty valence d orbitals. When writing Lewis structures, satisfy octets first, then place electrons around elements having available d orbitals.

Multiple Covalent Bonds: Double Bonds Two pairs of shared electrons

Multiple Covalent Bonds: Triple Bonds Three pairs of shared electrons

Resonance Resonance is invoked when more than one valid Lewis structure can be written for a particular molecule. The actual structure is an average of the resonance structures. Benzene, C6H6 The bond lengths in the ring are identical, and between those of single and double bonds.

Resonance, Bond Length and Bond Energy Resonance bonds are shorter and stronger than single bonds. Resonance bonds are longer and weaker than double bonds.

Resonance in Ozone, O3 Neither structure is correct. Oxygen bond lengths are identical, and intermediate to single and double bonds

Resonance in Polyatomic Ions Resonance in a carbonate ion: Resonance in an acetate ion:

Formal Charge Formal Charges a method of determining a. which resonance structure is most likely b. where the electrons are most concentrated in a structure Formal charge of an atom is (Family # ) – ( # of lone pair e-) - (# of bonds on that atom)

Formal Charge Structures that have formal charges closest to zero and any negative formal charges on the most electronegative atoms are most likely to be stable

Example Ex Which is the most likely Lewis structure for CO2 ? Ex For the SCN- ion?

Localized Electron Model Lewis structures are an application of the “Localized Electron Model” L.E.M. says: Electron pairs can be thought of as “belonging” to pairs of atoms when bonding Resonance points out a weakness in the Localized Electron Model.

VSEPR – Valence Electron Pair Repulsion X + E Overall Structure Forms 2 Linear AX2 3 Trigonal Planar AX3 , AX2E 4 Tetrahedral AX4 , AX3E , AX2E2 5 Trigonal bipyramidal AX5 , AX4E , AX3E2 , AX2E3 6 Octahedral AX6 , AX5E , AX4E2 A = central atom X = atoms bonded to A E = nonbonding electron pairs on A

VSEPR - Linear AX2 Example: CO2

VSEPR – Trigonal Planar AX3 BF3 AX2E SnCl2

VSEPR - Tetrahedral AX4 CCl4 AX3E PCl3 AX2E2 Cl2O

VSEPR – Trigonal Bipyramidal AX5 PCl5 AX4E SF4 AX3E2 ClF3 AX2E3 I3-

VSEPR - Octahedral AX6 SF6 AX5E BrF5 AX4E2 ICl4-

Sigma and Pi Bonds Sigma () bonds exist in the region directly between two bonded atoms. Pi () bonds exist in the region above and below a line drawn between two bonded atoms. Single Bond 1 sigma bond Double Bond 1 sigma, 1 pi bond Triple Bond 1 sigma, 2 pi bonds

Sigma and Pi Bonds Single Bonds 1 s bond

Sigma and Pi Bonds Double Bonds 1 p bond H H C C H H 1 s bond

Sigma and Pi Bonds Triple Bonds 1 s bond 1 p bond 1 p bond

The De-Localized Electron Model Pi bonds () contribute to the delocalized model of electrons in bonding, and help explain resonance Electron density from  bonds can be distributed symmetrically all around the ring, above and below the plane.

Hybridization The blending of orbitals poodle + cocker spaniel = cockapoo S orbital + p orbital = sp orbital

What proof exists for hybridization? We have studied electron configuration notation and the sharing of electrons in the formation of covalent bonds. Lets look at a molecule of methane, CH4. Methane is a simple natural gas. Its molecule has a carbon atom at the center with four hydrogen atoms covalently bonded around it.

Carbon Ground State Configuration What is the expected orbital notation of carbon in its ground state? Can you see a problem with this? (Hint: How many unpaired electrons does this carbon atom have available for bonding?)

Carbon’s Bonding Problem You should conclude that carbon only has TWO electrons available for bonding. That is not enough! How does carbon overcome this problem so that it may form four bonds?

Carbon’s Empty Orbital The first thought that chemists had was that carbon promotes one of its 2s electrons…

A Problem Arises… However, they quickly recognized a problem with such an arrangement… Three of the carbon-hydrogen bonds would involve an electron pair in which the carbon electron was a 2p, matched with the lone 1s electron from a hydrogen atom. 1s 1s 1s 1s 1s 2s 2p 2p 2p

Unequal Bond Energy This would mean that three of the bonds in a methane molecule would be identical, because they would involve electron pairs of equal energy. But what about the fourth bond…?

Unequal Bond Energy The fourth bond is between a 2s electron from the carbon and the lone 1s hydrogen electron. Such a bond would have slightly less energy than the other bonds in a methane molecule. 1s 1s 1s 1s 1s 2s 2p 2p 2p

Unequal Bond Energy This bond would be slightly different in character than the other three bonds in methane. This difference would be measurable to a chemist by determining the bond length and bond energy. But is this what they observe?

Enter Hybridization The simple answer is, “No”. Measurements show that all four bonds in methane are equal. Thus, we need a new explanation for the bonding in methane. Chemists have proposed an explanation – they call it Hybridization. Hybridization is the combining of two or more orbitals of nearly equal energy within the same atom into orbitals of equal energy.

In the case of methane, they call the hybridization sp3, meaning that an s orbital is combined with three p orbitals to create four equal hybrid orbitals. 1s 2sp3 These new orbitals have slightly MORE energy than the 2s orbital… … and slightly LESS energy than the 2p orbitals. sp3 Hybrid Orbitals

sp3 hybrid orbitals Here is another way to look at the sp3 hybridization and energy profile…

sp hybrid orbitals While sp3 is the hybridization observed in methane, there are other types of hybridization that atoms undergo. These include sp hybridization, in which one s orbital combines with a single p orbital. This produces two hybrid orbitals, while leaving two normal p orbitals

sp2 hybrid orbitals Another hybrid is the sp2, which combines two orbitals from a p sublevel with one orbital from an s sublevel. One p orbital remains unchanged

Hybridization Involving “d” Orbitals Beginning with elements in the third row, “d” orbitals may also hybridize dsp3 = five hybrid orbitals of equal energy d 2sp3 = six hybrid orbitals of equal energy

Hybridization and Molecular Geometry Forms Overall Structure Hybridization of “A” AX2 Linear sp AX3, AX2E Trigonal planar sp2 AX4, AX3E, AX2E2 Tetrahedral sp3 AX5, AX4E, AX3E2, AX2E3 Trigonal bipyramidal dsp3 AX6, AX5E, AX4E2 Octahedral d2sp3 A = central atom X = atoms bonded to A E = nonbonding electron pairs on A