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CHAPTER NINECovalent Bonding: Orbitals
Hybridization Some atoms will form
hybridized orbitals when bonding with other atoms. There are three that are included on the AP exam. sp, sp2 and sp3
To form the molecule methane, carbon must bond with four hydrogens. If the electron configuration for carbon is 1s22s22p2 then that only leaves 2 electrons to bond with 2 hydrogens but its been shown by experiment that carbon bonds with 4 hydrogens. How? Carbon forms a hybridized sp3 orbital
sp hybrid
sp2 hybrids
sp2 hybridization occurs with ethylene. A double bond can act as one effective pair. The ethylene carbon forms a trigonal planar arrangement with bond angles of 120o. sp3 hybridization forms 109.5o angles so that won’t work. Combining one s and two p orbitals allows for this arrangement.
sp3 Hybridization
This is the most common type of hybridization and it is found in molecules whose central atom is surrounded by a stable octet. CH4 is a classic example.
Each hydrogen has an electron in the 1s while carbon has 2 in the s and 2 in the p lobe.
These s and p orbitals combine to make an sp3 hybrid orbital.
Hybridization
# bonds
# lone pairs
Molecular shape Bond angle
Example
sp 2 0 Linear 180 CO2
sp2 3 0 Trigonal planar 120 SO3
sp2 2 1 Angular <120 SO2
sp3 4 0 Tetrahedral 109.5 CH4
sp3 3 1 Trigonal pyramidal <109.5 NH3
sp3 2 2 Bent (angular) <109.5 H2O
sp3d 5 0 Trigonal pyramidal 120, 90 PCl5
sp3d 4 1 Seesaw <120 SF4
sp3d 3 2 T-Shaped <90 CF3
sp3d 2 3 Linear 180 XeF2
sp3d2 6 0 Octahedron 90 SF8
sp3d2 5 1 Square pyramidal <90 IF5
sp3d2 4 2 Square planar 90 XeF4
Deducing molecular shape and hybridization for water
1. Determine Lewis structure
2. Determine the number of electron domains around the central oxygen atom.
The 4 electron domains are arranged in a tetrahedron with 2 lone pairs.
Four sp3 hybrid orbitals are oriented in a tetrahedral array. Two of the sp3 orbitals are occupied by bonding electron pairs and two are occupied by non-bonding pairs leading to an angular or bent arrangement.
Four electron domains will be most likely to exist as four sp3 hybrid orbitals.
Due to the large interelectron repulsion of the lone-paired electrons, the usual 109.5o bond angle found in a tetrahedron will be reduced and closer to 104.5o between hydrogen atoms.
Molecular Orbitals and Sigma, Pi Bonds Explained
Pi bonds are weaker than sigma bonds because the side-on overlap of p orbitals is less effective than end-on overlap.
In a σ bond, the electrons are in orbitals between the nuclei of the bonding atoms. Electron density is greatest between the nuclei. The electrons attract the nuclei and form a σ bond — the strongest type of covalent bond.
In a π bond, the p orbitals overlap side-on. The overlap is less efficient, because the electron density is off to the sides of the σ bond. The electrons are not as effective in attracting the two nuclei. Thus, a π bond is weaker than a σ bond.
ChemisNate 8:05
Steric number (9:47)
The steric number of a molecule is the number of atoms bonded to the central atom of a molecule plus the number of lone pairs on the central atom. It is often used in VSEPR theory (valence shell electron-pair repulsion theory) in order to determine the particular shape, or molecular geometry, that will be formed.
Formal Charge
Sometimes atoms will have extra or not enough electrons. This imbalance of electrons is denoted with a formal charge. A negative formal charge means there are too many electrons on atom. (Remember electrons are negatively charged.) A positive formal charge means there are not enough electrons on an atom. One confusing thing about formal charges is that we do not simply count up all of the electrons around an atom. Different types of electrons are counted differently. Non-bonding electrons are counted individually. However, electrons in bonds are counted as being shared and so each pair of electrons in a bond counts as only one electron.
Formal Charge (12:26)
Formal charge is the difference between the number of valence electrons on the free element and the number of electrons assigned to the atom in the molecule.
Formal charge = valence e- - nonbinding e- - bonding e- ÷ 2.
OR: Formal Charge = (valence number) - (number of bonds) - (non-bonding e's)
Calculate the formal charges for the practice problems in the NMSI handout.
Formal charge for XeO3 explained.
Bond Order
Bond order indicates bond strength. A high bond order indicates more attraction between electrons. A higher bond order also means that the atoms are held together more tightly. With a lower bond order, there is less attraction between electrons and this causes the atoms to be held together more loosely. Bond order also indicates the stability of the bond. The higher the bond order, the more electrons holding the atoms together, and therefore the greater the stability.
Bond order = (number of electrons – number of antibonding electrons) ÷ 2
Bond order is the number of chemical bonds between a pair of atoms. For example, in diatomic nitrogen N≡N the bond order is 3, in acetylene H−C≡C−H the bond order between the two carbon atoms is also 3, and the C−H bond order is 1. Bond order gives an indication of the stability of a bond.
Bond order and length are inversely proportional to one another. ie: the longer the bond length the lower the bond order (less attraction/strength)
Polyatomic molecules
Draw the Lewis structure. Count the total number of
bonds. 4 Count the number of bond
groups between individual atoms.
3 Divide the number of
bonds between individual atoms by the total number of bonds.
4 ÷3 = 1.33 Bond order is 1.33
Determine the bond order for a nitronium ion- NO2 +.
Bond total = 4 Bond groups = 2 Divide = 4 ÷ 2 = 2 Bond order is 2
Molecular Orbital Theory (12:01)
Write electron configuration.Homo vs heteronuclear. If homonuclear the diagram is symmetrical. If not, the more electronegative atom goes lower in the diagram (extra electrons= more stability)Fill in orbitals from bottom to top.Edvantage pages 440, 442
More examples (9:00)
Practice
Explain the following in terms of the electronic structure and/or bonding of the compounds involved.
At ordinary conditions, HF (normal boiling point = 20oC is a liquid, whereas HCl normal boiling point = -114oC) is a gas.
1. The hydrogen is bonded to the highly electronegative F atom and exhibits a special type of dipole-dipole IMF called hydrogen bonding.
2. H-Cl molecules are held together with only dipole-dipole forces (not FON) therefore they do not have hydrogen bonding. It takes more energy to separate the HF molecules than the HCl molecules due to the added H-bonding of the HF molecule.
1 pt IMF, 1 pt energy
Molecules of ASF3 are polar, whereas molecules of AsF5 are nonpolar, why?
1. First draw out the Lewis structure to see what you’re dealing with.
2. AsF3 is trigonal pyramidal and AsF5 is trigonal bipyamidal.
3. AsF3 is polar because it has a net dipole moment due to the presence of a lone pair of e- with no other lone pair (of e-) oriented to cancel out the dipole.
4. AsF5 has all of its dipole moments cancelling each other.
1 pt for AsF3 having a lone pair of electrons making it polar, and 1 pt for AsF5 being nonpolar because its dipole moments cancel.
The N-O bonds in the NO2- ion are equal
in length, whereas they are unequal in HNO2.
Draw structures first. When drawing acids, the hydrogen will always bond to the oxygen.
NO2- has resonance structures with a bond order
of 1.5 for each N-O bond. This is because there are two single and one double bond for 2 structures. HNO2 does not have resonance and has one single bond which is longer than the other double bond which has a shorter length.
1 pt resonance structure (you don’t have to have bond order).
1 pt for HNO2 having two distinctly different bonds-you can even go into sigma, pi if you want to.
For sulfur, SF2, SF4 and SF6 are known to exist whereas for oxygen only OF2 is known to exist.
Sulfur has valence electrons in the 3rd principle energy level (n=3). But since its on the 3rd level, those electrons can expand its octet into the d sublevel and form 6 bonds around the central atom (SF6). Oxygen has valence electrons in the 2nd energy level (n=2) and cannot expand into the d orbitals because they are not available. Therefore, oxygen cannot expand its octet.
