BONDING

Methods for assigning Geometries to Molecular Structures

It does not matter what method you use, but you must be able to assign geometries to molecular structures.

Method 1 Memorize what each shape looks like, draw your structures accurately by using electron pair repulsion, VSEPR, in three dimensions and then after drawing each structure assign a geometry that matches.

Method 2 Use hybridization to shorten the list and then combine with method-1.

Method 3 Use LAX and the number of lone pairs on the central atom. Use the if-then statements and combine that with the LAX charts (listed below within the if-then statements).

To do LAX – “L” is the number of lone electron pairs on the central atom (0, 1,2,or 3),

“A” is the central atom

“X” is the number of atoms attached (just the atom count, not the number of bonds)

Pairs on central atom / Hybridization / LAX / GEOMETRY / Bond Angle
4 / N/A / AX / Linear / 180º
4 / sp, sp2, sp3 / LAX, L2AX, L3AX / Linear / 180º
4 / sp / AX2 / Linear / 180º
4 / sp2 / AX3 / Trigonal Planar / 120º
4 / sp2, sp3 / LAX2, L2AX2 / Bent / 104.5º (~109.5º)
4 / sp3 / LAX3 / Trigonal Pyramidal / 107º (~109.5º)
4 / sp3 / AX4 / Tetrahedral / 109.5º
5 / sp3d / AX5 / Trigonal Bipyramidal / Axially – 180 º
Equatorially – 120 º
Ax. to Eq. – 90 º
5 / sp3d / LAX4 / See-Saw
5 / sp3d / L2AX3 / T-shape
5 / sp3d / L3AX2 / Linear
6 / sp3d2 / AX6 / Octahedral / Axially – 180 º
Equatorially – 90 º
Ax. to Eq. – 90 º
6 / sp3d2 / LAX5 / Square Pyramid
6 / sp3d2 / L2AX4 / Square Planar

To draw a Lewis Structure:

1.  Determine the number of electron pairs by adding the total number of valence e’ (account for charge) and then divide by 2.

2.  Select the least electronegative atom (not H) as the central atom.

3.  Bond all the atoms to the central atom using single bonds. Subtract the number of single electron pairs used from the total in step 1.

4.  Complete the octets (most electronegative atoms first) using the remaining e pairs as lone pairs (unshared pairs). Subtract the added electron pairs from the total calculated in step 3.

5.  If electron pairs are still remaining, then the extra electron pairs go on the central atom to form 5 or 6 electron pair systems (see table)

Or!

If there are no electron pairs remaining and all the octets are not completed, then the atom that needs to complete an octet will share an electron pair with an adjoining atom that has lone (unshared) electron pairs thus forming double or triple bonds.

BONDING

Electronegativity Difference and Bond Character

Covalent bond - a shared pair of electrons between two atoms

Electronegativity - the ability of an atom to attract electrons in a covalent bond.

Electronegativity Difference, DE, DE = E1-E2,

For example, each H—–O bond in H2O has: DE = EO - EH

DE = 3.44 - 2.10

DE = 1.44

As the electronegativity difference grows from 0.00 to >3.00, the sharing of electrons becomes more skewed. First one atom becomes partially negative, d-, while the other becomes partially positive, d+, and the bond is polarized, a polar covalent bond. Then as the sharing becomes more and more unequal, the electrons reside only on one atom. The atoms are now ionized (+ or -), and the bond is called ionic. This is shown in Table 1.

Table 1 nonpolar covalent ↔ polar covalent ↔ ionic

Electronegativity Difference, DE / 0.00 / 0.65 / 0.94 / 1.19 / 1.43 / 1.67 / 1.91 / 2.19 / 2.54 / 3.03
Percent Ionic Character / 0% / 10 % / 20 % / 30 % / 40 % / 50 % / 60 % / 70 % / 80 % / 90 %
Percent Covalent Character / 100 % / 90 % / 80 % / 70 % / 60 % / 50 % / 40 % / 30 % / 20 % / 10 %

For a shared pair of electrons, if one atom is able to attract the electrons to itself (more electronegative) that atom will begin to become negatively charged, d-, (a negative pole) while the other atom (least electronegative) begins to become positively charge, d+, (a positive pole). The two atoms become a dipole (meaning 2 poles), and the bond will become a polar covalent bond. [see Table 2]

If the difference in attracting the electrons, DE, is so great, then one atom may just take the electrons for itself. This stops any sharing of electrons, and the bond is an ionic bond. The atom that took the electrons is the anion (negative), and the atom that lost the electrons is the cation (positive). [see Table 2]

If the difference in attracting the electrons, DE, is very small, then the sharing remains relatively equal and no charges develop. No developed charges means there are no pole, which makes the bond a nonpolar covalent bond. In the special case that the electronegativity difference, DE, is zero, then no atom attracts the electrons to itself and the sharing is perfectly equal. Such a bond is called a pure covalent bond and is nonpolar also. [see Table 2]

Table 2

Electronegativity Difference, DE / Polarity / Bond Type
DE = 0.00 / nonpolar / pure covalent
0.00 < DE < 0.65 / nonpolar / nonpolar covalent
0.65 < DE < 1.67 / polar / polar covalent
1.67 < DE / ionic

More on Dipoles

·  An electric dipole consists of two opposite charges that are the same magnitude and separated in space.

·  Unequal sharing of electrons (DE > 0) results in one end of the bond being negative and one end being positive. This is a dipole (2 poles, one negative, one positive)

·  The atom with the greater electronegativity becomes partially negative, d-,

·  The atom with the smaller electronegativity becomes partially positive, d+.

·  Drawing an arrow from d+ to d- shows the electric field between the charges.

Practice Problem: Hydrogen fluoride, HF

o  Draw the Lewis Structure

o  Determine the polarity and bond type for each bond.

o  Draw the dipole for each bond.

HF, would be written as H—– F

Since DE = 1.10, the bond is polar covalent. {DE = 3.98 - 2.10 = 1.10}

F has the greater electronegative so it is partially negative, d-, and H with the smaller electronegativity is partially positive, d+.

Final Answer: H —– F

d+ → d- polar covalent

Bond energy and the Enthalpy of a Reactions:

Bonds are potential energy and every bond therefore has a value of energy called bond energy.

The enthalpy or heat of reaction may be calculated by determining the difference in bond energy of bonds broken (reactant side) and the bonds made (products side). These values are supplied in a table and are not expected to be memorized.

For example

Consider the burning of natural gas. Calculate the enthalpy of reaction.

CH4(g) + 3O2(g) ® CO2(g) + 2H2O

solution:

Problem Set on Bonding

Name: AP Chemistry

Period: Date: R.F. Mandes, PhD, NBCT

Directions For each formula, • draw the Lewis Structure

• determine the hybridization of the central atom

• determine the geometry

• determine the formal charge on each atom

• determine the bond type for each different bond

1. CBr4

2. HCN

3. CO2

4. XeF4

5. BrF2

6. IF5

7. NbBr5 (Nb has 5 val e’s)

8. KrF2