Shape & structure
Hybridization and molecular geometry
Almost every argument to be made later in this course, about acidity, about which face a nucleophile attacks, about whether two molecules are the same, starts with knowing the shape of each atom. Assess the number of sigma bonds and lone pairs, name the hybridization pattern, and the geometry follows.
Basic shapes
Three geometries at carbon
There are three important geometries displayed by carbon, and then nitrogen and oxygen with small deviations. When C is connected to other atoms by only single bonds it is said to be sp3 hybridized with a tetrahedral geometry. This is exemplified by the molecule ethane below, in which all H–C–C–H bond angles will be close to 109°. For nitrogen, which has a lone pair in place of one bond, the angle deviates away from 109° since the lone pair repels more; likewise for oxygen with two lone pairs. The shape around those atoms may be described as distorted tetrahedral — trigonal pyramidal for N, bent for O.
When C (or N or O) is involved in one pi bond, the hybridization is now sp2 and the geometry changes to trigonal planar. Bond angles are approximately 120°, since the three sigma bonds dictate the shape and the pi bond plays no role in that. This may be seen in the alkene ethylene below. With two pi bonds the atom is sp hybridized, only two sigma bonds remain, and the geometry is linear at 180°, as in acetylene.
Nitrogen and oxygen
How lone pairs distort the ideal angles
Moving to nitrogen and oxygen we recognize the introduction of lone pair(s), which are known to take up more volume than bond pairs and so change these shapes slightly. For ammonia (N bonded to three hydrogens) and water (O bonded to two hydrogens) the tetrahedral parent shape still applies, but with slightly different bond angles — 107° and 105° rather than 109°.
Similarly, for N and O in double bonds the shape is roughly trigonal planar, but the associated lone pairs distort the angles away from 120°. The molecules will, however, still be flat overall. For the linear molecules, the N equivalent (nitriles) has the lone pair opposite the alkyl group attached, with the molecule still being linear overall.
sp3 → tetrahedral (~109°), sp2 → trigonal planar (~120°), sp → linear (180°). Lone pairs on N and O distort the ideal angles slightly downward.
Reference
Geometry summary
Count sigma bonds and lone pairs first. The two together give the parent geometry; the lone pairs then tell you how far the real angle sits below the ideal.
Self-check
Six questions before you move on
Work out an answer on paper, then reveal to check. If your reason is right but the answer is wrong, you are closer than you think.
A carbon belong to a C=C. Why does the pi bond not appear in the geometry you assess for it?
Geometry is set by the sigma framework. Three sigma bonds give trigonal planar at ~120°; the pi bond sits perpendicular to that plane and does not affect the bond angles.
Ammonia is 107° and water is 105°, both below the ideal 109.5°. What explains the trend?
Lone pairs occupy more volume than bond pairs, so they compress the bonded angles. Two lone pairs on oxygen compress more than one on nitrogen.
Where does the lone pair sit on the nitrogen of a nitrile, and is the molecule still linear?
The lone pair sits in the remaining sp orbital, pointing directly away from the C–N bond, so R–C≡N stays linear overall.
An imine nitrogen carries a lone pair and is part of a double bond. Is the molecule flat, and what is the C=N–R angle?
Flat, yes; the nitrogen is sp2, so all attached atoms lie in one plane. The angle is somewhat below 120° because the lone pair pushes the two sigma bonds closer together.
Carbon monoxide is drawn C≡O with a positive oxygen. Why is that oxygen sp hybridized rather than bent?
It is part of two pi bonds, so only one sigma bond and one lone pair remain. Two groups around the atom means sp and a linear arrangement, exactly as for the nitrile nitrogen.
You are shown a structure and asked for the shape at one atom. What do you count, and in what order?
Count sigma bonds plus lone pairs. Four gives sp3, three gives sp2, two gives sp. Then subtract a few degrees for each lone pair present.