Covalent Bonding: Orbitals

 Some atoms will form hybridized orbitals when bonding with other atoms. There are three that are included on the AP exam. sp, sp 2 and sp 3  To form the molecule methane, carbon must bond with four hydrogens. If the electron configuration for carbon is 1s 2 2s 2 2p 2 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 sp 3 orbital

 sp 2 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 120 o. sp 3 hybridization forms o angles so that won’t work. Combining one s and two p orbitals allows for this arrangement.

 This is the most common type of hybridization and it is found in molecules whose central atom is surrounded by a stable octet. CH 4 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 sp 3 hybrid orbital.

Hybridization# bonds# lone pairsMolecular shapeBond angleExample sp20Linear180CO 2 sp 2 30Trigonal planar120SO 3 sp 2 21Angular<120SO 2 sp 3 40Tetrahedral109.5CH 4 sp 3 31Trigonal pyramidal<109.5NH 3 sp 3 22Bent (angular)<109.5H2OH2O sp 3 d50Trigonal pyramidal120, 90PCl 5 sp 3 d41Seesaw<120SF 4 sp 3 d32T-Shaped<90CF 3 sp 3 d23Linear180XeF 2 sp 3 d 2 60Octahedron90SF 8 sp 3 d 2 51Square pyramidal<90IF 5 sp 3 d 2 42Square planar90XeF 4

 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 sp 3 hybrid orbitals are oriented in a tetrahedral array. Two of the sp 3 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 sp 3 hybrid orbitals.  Due to the large interelectron repulsion of the lone-paired electrons, the usual o bond angle found in a tetrahedron will be reduced and closer to o between hydrogen atoms.

 Pi bonds are weaker than sigma bonds because the side-on overlap of p orbitals is less effective than end-on overlap. Pi bondssigma bonds  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. bondingdensity  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

 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 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)valence number  Calculate the formal charges for the practice problems in the NMSI handout.  Formal charge for XeO 3 explained. Formal charge for XeO 3

 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)

 Draw the Lewis structure.  Count the total number of bonds. 44  Count the number of bond groups between individual atoms. 3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- NO 2 +.  Bond total = 44  Bond groups = 22  Divide =  4 ÷ 2 = 2  Bond order is 2

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)

Crash Course summary (10:51)

 Explain the following in terms of the electronic structure and/or bonding of the compounds involved.  At ordinary conditions, HF (normal boiling point = 20 o C is a liquid, whereas HCl normal boiling point = o C) 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 ASF 3 are polar, whereas molecules of AsF 5 are nonpolar, why?  1. First draw out the Lewis structure to see what you’re dealing with.  2. AsF 3 is trigonal pyramidal and AsF 5 is trigonal bipyamidal.  3. AsF 3 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. AsF 5 has all of its dipole moments cancelling each other.  1 pt for AsF 3 having a lone pair of electrons making it polar, and 1 pt for AsF 5 being nonpolar because its dipole moments cancel.

 The N-O bonds in the NO 2 - ion are equal in length, whereas they are unequal in HNO 2.  Draw structures first. When drawing acids, the hydrogen will always bond to the oxygen.

 NO 2 - 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. HNO 2 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 HNO 2 having two distinctly different bonds-you can even go into sigma, pi if you want to.

 For sulfur, SF 2, SF 4 and SF 6 are known to exist whereas for oxygen only OF 2 is known to exist.  Sulfur has valence electrons in the 3 rd principle energy level (n=3). But since its on the 3 rd level, those electrons can expand its octet into the d sublevel and form 6 bonds around the central atom (SF 6 ). Oxygen has valence electrons in the 2 nd energy level (n=2) and cannot expand into the d orbitals because they are not available. Therefore, oxygen cannot expand its octet.