Ch. 1 - A Review of General ChemistryWorksheetSee all chapters
All Chapters
Ch. 1 - A Review of General Chemistry
Ch. 2 - Molecular Representations
Ch. 3 - Acids and Bases
Ch. 4 - Alkanes and Cycloalkanes
Ch. 5 - Chirality
Ch. 6 - Thermodynamics and Kinetics
Ch. 7 - Substitution Reactions
Ch. 8 - Elimination Reactions
Ch. 9 - Alkenes and Alkynes
Ch. 10 - Addition Reactions
Ch. 11 - Radical Reactions
Ch. 12 - Alcohols, Ethers, Epoxides and Thiols
Ch. 13 - Alcohols and Carbonyl Compounds
Ch. 14 - Synthetic Techniques
Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect
Ch. 16 - Conjugated Systems
Ch. 17 - Aromaticity
Ch. 18 - Reactions of Aromatics: EAS and Beyond
Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition
Ch. 20 - Carboxylic Acid Derivatives: NAS
Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon
Ch. 22 - Condensation Chemistry
Ch. 23 - Amines
Ch. 24 - Carbohydrates
Ch. 25 - Phenols
Ch. 26 - Amino Acids, Peptides, and Proteins
Johnny Betancourt

Molecular geometry is dictated by VSEPR theory. A central atom’s bond angles can be predicted number of substituents and lone pairs directly attached. 

VSEPR theory:

Valence Shell Electron Pair Repulsion theory is based on the idea that groups of atoms and electrons will repel each other as much as possible. This repulsion results in particular optimal bond angles, which in turn result in molecular geometries. 

Using bonding preferences and hybridization of a central atom, we can accurately predict the molecular geometry (fancy way of saying molecular shape). Hybridization can be discerned by the number of groups (atoms and lone pairs) attached to an atom. 

Hybridization:

The hybridization of an atom can help us quickly determine the shape of a molecule. Hybridization of a central atom is determined by how many bonds and/or lone pairs that atom has. Collectively, lone pairs and bonds are called bond sites. Empty orbitals do not count as bond sites!

4 bond sites = sp3 hybridized

3 bond sites = sp2 hybridized

2 bond sites = sp hybridized 

Bonding preferences

Atoms of different groups (columns of the periodic table) will actually have different numbers of bonds and lone pairs in their neutral states. Here’s a pretty helpful chart to show what I mean. Each atom is drawn in its neutral state—that is, with no charge.

Bonding preferencesBonding preferences

Bonding preferences are extremely helpful when determining the hybridization of atoms because your professor might not draw lone pairs or hydrogens. Generally, bondline structure omits hydrogens attached to carbon atoms and lone pairs attached to heteroatoms (any atom that isn’t a carbon or hydrogen). If no charge is indicated, it’s assumed that the atom meets its bonding preference through “implied” hydrogens or lone pairs. 

Molecular Geometry

Molecular Geometry is determined by the number of groups attached to a central atom. Think about it this way: what’s the farthest two bond sites can be separated if they’re attached to the same central atom? 180º apart. That’s what we would refer to as linear geometry. Remember that atoms and lone pairs will repel each other as far away as possible. They just want equal elbow room, basically. 

Of course, things get a bit more complicated when there are more bond sites to consider. Let’s take a look at neutral boron. Boron likes to have three atoms attached and no lone pairs, so what’s the maximum angle between the atoms attached?

Boron bond anglesBoron bond angles

Using X to represent any atom, let’s think about this. If boron has three groups attached, the absolute maximum angle at which the three groups can be distributed is 120º. Hang on a second… that means that the molecule will actually be flat! This is what we refer to as trigonal planar geometry, and it occurs with three bond sites and zero lone pairs. 

Instead of going through each iteration, let’s try to understand the repulsive effects of lone pairs and atoms attached to a central atom. 

C, N, O hybridization and geometriesC, N, O hybridization and geometries

Notice that the nitrogen has two different categories of sp3. A neutral nitrogen generally has three bonds and one lone pair (4 groups total) and a cationic nitrogen generally has four bonds and no lone pairs. The presence or absence of that lone pair makes a difference on the geometry of the molecule. Let’s use that chart to find the hybridizations and geometries of the central atoms in the following molecules:

Geometries blankGeometries blank

Try these out on your own before looking below for the answers. Hint: remember that lone pairs can make geometry a little bit different. 

Geometries answersGeometries answers

To do this accurately, we need to do three things: 

1)    Find how many bonds and/or lone pairs the central atom has

2)    Use that information to determine the hybridization of the central  atom

3)    Use the number of bonds and lone pairs to determine the geometry 

We talked earlier about how lone pairs can affect geometry. Notice that carbon (top left) and nitrogen (top middle) each have for bonding sites but different geometry. Basically, you can imagine the bonds as legs or sides of the shape and the lone pair as a little cap. 

Let’s try finding the hybridization and geometry of three atoms circled on this next molecule: 

Hybridizations and geometries of three atoms blankHybridizations and geometries of three atoms blank

Try this one on your own before looking at the answer below. All lone pairs and hydrogens attached to carbon atoms have been implied. Heads up! Notice that there is a carbon with a positive charge. What does that mean? How many bonds does it have? How many lone pairs? 

Hybridizations and geometries of three atoms answeredHybridizations and geometries of three atoms answered

Both oxygen molecules have two bonds and two lone pairs, but they have different hybridization. Why? Because they have a different number of atoms attached; the oxygen on the left has single bonds to a carbon and to a hydrogen while the oxygen on the right has both bonds to a single carbon. 

The carbon is a bit more complicated, though. Check it out. It’s got two bonds to carbon that are drawn explicitly and it’s got a bond to an implied hydrogen attached. It has zero lone pairs. That means it has a total of three bond sites; remember that empty orbitals like that in the carbocation do not count as bond sites. 

That wasn’t so bad, right? So, let’s try another—potentially weirder—example. Let’s take a look at the different hybridizations and geometries of NH3, BF3, and AlCl3. Remember that boron is in group 3 (to the left of carbon) and aluminum is right underneath boron. 

Ammonia, boron triflouoride, and aluminum trichloride geometryAmmonia, boron triflouoride, and aluminum trichloride geometry

NH3 is sp3, BF3 is sp2, and AlCl3 is sp2. Even though they all have three bonds, the geometries are different. The nitrogen’s geometry is trigonal pyramidal while the geometry of both boron and aluminum is trigonal planar. Again, that’s because of that lone pair on the nitrogen. Imagine we somehow added a lone pair to the boron. What would happen? The lone pair would repulse (push away) the fluorines and achieve a trigonal pyramidal structure. If we added another fluorine instead of the lone pair, a tetrahedral geometry would develop.

Let’s try to figure out the molecular geometries of a few more. How about the central atoms of NH4+, CH4, H2O, CO2, and H2CO? Try drawing the Lewis structures of each on your own.

A few more examplesA few more examples

Both the sp3 nitrogen and sp3 carbon have tetrahedral geometries since they have four bonds and no lone pairs; the oxygen in water has two bonds and two lone pairs, so it’s sp3 and bent; the carbon in formaldehyde has three atoms attached and no lone pairs, so it’s sp2 and trigonal planar; and the carbon in carbon dioxide has only two atoms attached and no lone pairs, so it has a linear geometry. 



Johnny Betancourt

Johnny got his start tutoring Organic in 2006 when he was a Teaching Assistant. He graduated in Chemistry from FIU and finished up his UF Doctor of Pharmacy last year. He now enjoys helping thousands of students crush mechanisms, while moonlighting as a clinical pharmacist on weekends.


Additional Problems
For each molecule below, fill in the table to determine the electron geometry & molecular geometry about the CENTRAL ATOM. Then sketch a perspective drawing of the substance and LABEL ALL BOND ANGLES.  1. HCN      2. NO2-      3. NF3
VSEPR theory predicts an identical shape for all of the following, except: (hint: draw the Lewis structures before deciding ) a. OCl3+ b. Br2CO c. CH3● d. Et2O e. All have the same geometry f. none of the above
Predict the geometry of each atom except hydrogen in the compounds below: 
Predict the geometry of each atom except hydrogen in the compounds below: 
Predict the geometry of each atom except hydrogen in the compounds below: 
Predict the geometry of each atom except hydrogen in the compounds below: 
Give the hybridization, VSEPR, geometry, charge and angle of the indicated atoms.
What is the hybridization of each carbon in CH 3CH=CHC≡CH?  What are the CCC bond angles?   
Predict the bond angles for all bonds in the following compounds:  (b)CH2O 
Predict the bond angles for all bonds in the following compounds:  (c)C2H4 
Predict the bond angles for all bonds in the following compounds:  (d)C2H2 
Predict the bond angles for all bonds in the following compounds:  (e) CH3OCH3 
Predict the bond angles for all bonds in the following compounds:  (f) CH3NH2 
Predict the bond angles for all bonds in the following compounds:  (h)CH3CN 
Identify the expected hybridization state and geometry for the central atom in each of the following compounds: 
Of the overlaps between an  s and a p orbital as shown in the illustration, one is bonding, one is antibonding, and the third is nonbonding (neither bonding nor antibonding). Which orbital overlap corresponds to which interaction? Why? 
Identify the hybridization state and geometry of each carbon atom in the following compounds: 
Identify the hybridization state and geometry of each carbon atom in the following compounds: 
Consider the three compounds shown below and then answer the questions that follow:  (b) Which compound contains a nitrogen atom with trigonal pyramidal geometry? 
Consider the three compounds shown below and then answer the questions that follow:  (f) Which compound contains an sp2-hybridized carbon atom?
Consider the three compounds shown below and then answer the questions that follow:  (g) Which compound contains only sp3-hybridized atoms (in addition to hydrogen atoms)?   
Propose at least two different structures for a compound with six carbon atoms that exhibits the following features: (a) All six carbon atoms are sp2 hybridized.  
Propose at least two different structures for a compound with six carbon atoms that exhibits the following features: (c) There is a ring, and all of the carbon atoms are sp3 hybridized.
Propose at least two different structures for a compound with six carbon atoms that exhibits the following features: (d) All six carbon atoms are sp hybridized, and the compound contains no hydrogen atoms (remember that a triple bond is linear and therefore cannot be incorporated into a ring of six carbon atoms). 
A compound with molecular formula C5H11N has no π bonds. Every carbon atom is connected to exactly two hydrogen atoms. Determine the structure of the compound.   
Isonitriles (A) are an important class of compounds because of the versatile reactivity of the functional group, enabling the preparation of numerous new compounds and natural products. Isonitriles can be converted to isonitrile dihalides (B), which represents a useful procedure for temporarily hiding the reactivity of an isonitrile (Tetrahedron Lett. 2012, 53, 4536–4537).  (a) Identify the hybridization state for each highlighted atom in A.   
Isonitriles (A) are an important class of compounds because of the versatile reactivity of the functional group, enabling the preparation of numerous new compounds and natural products. Isonitriles can be converted to isonitrile dihalides (B), which represents a useful procedure for temporarily hiding the reactivity of an isonitrile (Tetrahedron Lett. 2012, 53, 4536–4537).  (d) Identify the hybridization state for each highlighted atom in B.
In Section 1.12 we discussed the effect that branching can have on the boiling point of a compound. In certain instances, branching may also affect how a molecule can react with different molecules. We use the term “steric hindrance” to describe how branching can influence reactivity. For example, greater “steric crowding” may decrease the reactivity of a C=C π bond (Org. Lett. 1999, 1, 1123–1125). In the following molecule, identify each π bond and determine which has a greater degree of steric crowding. (Note: This is not the same as “steric number”.)   
Ramelteon is a hypnotic agent used in the treatment of insomnia:  (g)  Identify the geometry of each atom (except for hydrogen atoms). 
Cycloserine is an antibiotic isolated from the microbe Streptomyces orchidaceous. It is used in conjunction with other drugs for the treatment of tuberculosis.  (g)  Identify the geometry of each atom (except for hydrogen atoms). 
Melatonin is an animal hormone believed to have a role in regulating the sleep cycle:  The structure of melatonin incorporates two nitrogen atoms. What are the hybridization state and geometry of each nitrogen atom? Explain your answer. 
What can be said about the stability of the molecule cyclopentyne? Would it be stable? Draw it and explain why or why not.
What is the hybridization and geometry of the indicated atom in the following molecule?
Use VSEPR to deduce the hybridization and molecular geometry of the ionized atoms in the following structures:
Consider a chemical species like the one in the previous problem in which a carbon atom forms three single bonds to three hydrogen atoms, but in which the carbon atom possesses an unshared electron pair.(a) What formal charge would the carbon atom have? (b) What total charge would the species have?(c) What total charge would the species have?(d) What would you expect the hybridization state of the carbon atom to be?
Use VSEPR theory to deduce the hybridization state of the negatively charged carbon in the following cumulene derived carbanion. In which orbital does the electron pair reside in? What is its geometry?
Label the molecular shape around each of the central atoms in the amino acid glycine.
What is the molecular geometry about NA the molecule below? a. Linear b. Bent c. Trigonal planar d. Trigonal pyramidal e. Tetrahedral
The F - •S - F bond angles in SF6 are a. 90° and 120°b. 90° and 180°c. 109.5°d. 180°e. 120°
If the analysis of carbon in a molecule indicates sp hybridization, what must be the shape around the carbon? a. Trigonal planar b. Linear c. Tetrahedral d. Cannot be determined
Why the following compound is not stable? (i) because the bond angles at the triple-bonded carbons, when the bonding orbitals overlap maximally, are 120° and therefore the sp-hybridized carbons can not simultaneously bind atoms other than hydrogen on both sides of triple-bond.(ii) because the bond angles at the triple-bonded carbons, when the bonding orbitals overlap maximally, are 180° and 180 degree angle cannot fit into the ring structure of such a size. (iii) because the bond angles at the triple-bonded carbons, when the bonding orbitals overlap maximally, are 180° and therefore the sp-hybridized carbons can not simultaneously bind atoms other than hydrogen on both sides of triple-bond.(iv) because the bond angles at the triple-bonded carbons, when the bonding orbitals overlap maximally, are 120° and 120° angle cannot fit into the ring structure of such a size.