Revision sheet: hybridization and the quantum model of bonding (202-SN1-RE)
VSEPR theory gives the shapes, valence bond theory gives the reasons. A bond is an overlap of orbitals, and the central atom mixes its valence orbitals so that its bonds point in the right directions. The count is exactly the same as in VSEPR, which makes the chapter much shorter than it looks.
This sheet fixes the count, separates σ bonds from π bonds once and for all, and gathers the traps of molecular orbital theory, in particular bond order and the paramagnetism of oxygen gas. It is written for CEGEP students in General Chemistry 202-SN1-RE (formerly 202-NYA). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.
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The thread of the chapter
Hybridization is not read from the number of bonds, it is read from the number of DIRECTIONS: σ bonds plus lone pairs. Counting π bonds in that tally is the error that runs through the whole chapter.
•Count the DIRECTIONS around the atom: σ bonds plus lone pairs. It is the same count as the electron domains in VSEPR.
•4 directions give sp3 (109.5°); 3 give sp2 (120°); 2 give sp (180°).
•5 directions give sp3d and 6 directions give sp3d2, reserved for the third period and beyond.
•The number of hybrid orbitals obtained ALWAYS equals the number of atomic orbitals mixed: s+3p give four sp3.
•The UNhybridized p orbitals that remain are used only for π bonds: one for sp2, two for sp.
Four, three or two directions around carbon: sp3 at 109.5°, sp2 at 120°, sp at 180°. A double or triple bond counts only once.
Hybridization is read from the DOMAINS, geometry from the bonds alone. The oxygen of water is sp3 (four domains) although the molecule is bent (two bonds): both statements are true at once.
Sigma and pi, two overlaps and two behaviours
•σ: HEAD-ON overlap, along the axis of the nuclei. It allows FREE ROTATION about the bond.
•π: SIDEWAYS overlap of two parallel p orbitals, above and below the axis. It PREVENTS rotation.
•Single =1σ; double =1σ+1π; triple =1σ+2π. Every bond, whatever it is, contains exactly ONE σ bond.
•Counting the σ bonds of a molecule therefore amounts to counting its bond LINES, and counting the π bonds amounts to counting the EXTRA lines of the multiple bonds.
•The rotation blocked by a π bond is what makes cis-trans isomerism possible: two distinct compounds, with distinct properties.
On the left, the overlap lies ALONG the axis of the nuclei: turning one of the two atoms changes nothing. On the right, it lies above and below: any rotation would destroy it, so the bond blocks it.
A π bond never adds a direction: it is superimposed on the σ bond already counted. That is why the carbon of an alkene, with three lines leaving it of which one is double, stays sp2.
Molecular orbitals, bond order and magnetism
•n atomic orbitals combined give n molecular orbitals: half BONDING, half ANTIBONDING, marked with an asterisk.
•BO=2nbonding−nantibonding. A bond order of zero means the species does not exist: that is the case for He2.
•A high order gives a SHORT, STRONG bond. Populating an antibonding orbital lengthens and weakens it, hence the series O2+>O2>O2−>O22−.
•A species with at least one UNPAIRED electron is PARAMAGNETIC, hence attracted by a magnet; with all electrons paired, it is diamagnetic.
•s character: 25% for sp3, 33% for sp2, 50% for sp. The higher it is, the shorter the bond and the closer to the nucleus the lone pair is held.
Two bonding electrons give the order 1 of H2; in He2 the two antibonding electrons cancel the two bonding ones, the order is zero and the molecule does not exist.
The paramagnetism of oxygen gas is THE result that established molecular orbital theory: the Lewis structure, with its double bond and paired lone pairs, predicts it diamagnetic, which experiment denies.
The mistakes that cost marks
These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.
1.Counting π bonds as directions in the hybridization count
the whole question, 5 marks
What not to write
“The carbon of ethylene has four bonds in all, two to the hydrogens and a double bond to the other carbon: so it is sp3.”
What to write
“It has THREE directions: two σ bonds to the hydrogens and one σ bond to the carbon. The π bond does not count. It is sp2, at 120°.”
Four lines do leave the carbon, but two of them are the same double bond: three directions, hence sp2 and angles of 120°.
Why: A π bond is superimposed on a σ bond already counted, it does not point in a new direction. It is exactly the same rule as in VSEPR, where a multiple bond counts as a single domain.
2.Forgetting the lone pairs in the count of directions
the whole question, 5 marks
What not to write
“The oxygen of water carries two bonds, so two directions, so it is sp and water is linear.”
What to write
“Two bonds PLUS two lone pairs make FOUR directions: oxygen is sp3, and it is the crowding of the lone pairs that closes the angle to 104.5°.”
Without its two lone pairs, water looks linear; with them, oxygen has four directions, it is sp3 and the angle closes to 104.5°.
Why: A lone pair occupies a hybrid orbital, exactly as a bond does. Ignoring it amounts to predicting a linear water molecule, which all of water's chemistry denies, starting with its polarity.
3.Taking a double bond for two sigma bonds
3 marks
What not to write
“Ethylene contains six σ bonds: four C-H and two between the carbons.”
What to write
“It contains FIVE: four C-H and a single one between the carbons. The double bond is 1σ+1π, never 2σ.”
Each line carries a σ bond, the second line of the double bond is the π bond: five σ and one π, never six σ.
Why: Two orbitals can overlap head-on only once. Any further overlap is necessarily sideways, hence π.
4.Believing that rotation stays free about a double bond
4 marks, and the whole isomerism question
What not to write
“cis-dichloroethylene and trans-dichloroethylene are the same molecule seen from two angles.”
What to write
“The π bond PREVENTS rotation: they are two DIFFERENT compounds, with different boiling points and different dipole moments.”
The two drawings have exactly the same formula and the same bonds, but you cannot go from one to the other without breaking the π bond: they are two distinct compounds, one polar, the other not.
Why: Turning one carbon would break the sideways overlap of the p orbitals, which costs the full energy of the π bond. At room temperature, this barrier cannot be crossed.
5.Calculating a bond order without dividing by two
2 marks, but also the whole comparison of lengths that follows
What not to write
“O2 has 8 bonding and 4 antibonding electrons, so the bond order is 4.”
What to write
“BO=28−4=2, which does match the double bond of the Lewis structure.”
Why: Each bond takes TWO electrons: the division by two expresses exactly that. A bond order of 4 for oxygen gas would exceed that of nitrogen, which is absurd.
6.Predicting oxygen gas diamagnetic from its Lewis structure
4 marks, and the most classic reasoning question of the chapter
What not to write
“The Lewis structure of O2 shows a double bond and four lone pairs, all paired: the molecule is diamagnetic.”
What to write
“The molecular orbital diagram puts TWO unpaired electrons in the π∗ orbitals: O2 is PARAMAGNETIC, and it really does stick to a magnet.”
The Lewis structure pairs everything; the molecular orbital diagram leaves two electrons alone in the two π∗ orbitals: O2 is paramagnetic.
Why: It is the most famous failure of the Lewis model, and the very reason for molecular orbital theory. Experiment decides: liquid oxygen is attracted by a magnet, which Lewis cannot explain.
7.Believing that sp3 gives the shortest bond and the most acidic atom
3 marks on comparison questions
What not to write
“The sp3 carbon has more orbitals mixed, so its bonds are shorter and its hydrogen more acidic.”
What to write
“It is the opposite: the s character is 25% for sp3 and 50% for sp. The higher it is, the closer the electrons are to the nucleus, so the bond is shorter and the hydrogen more acidic.”
As s character increases, the pKa FALLS: the hydrogen of an alkyne is twenty-five orders of magnitude more acidic than that of an alkane, for the sole reason that its electrons are held closer to the nucleus.
Why: An s orbital is spherical and concentrated around the nucleus, a p orbital is elongated. The more s a hybrid contains, the closer to the nucleus it holds its electrons.
8.Confusing the name of the hybridization with the name of the geometry
3 marks
What not to write
“The nitrogen of ammonia is sp3, so ammonia is tetrahedral.”
What to write
“Nitrogen is indeed sp3, since it has four directions, but one of them is a lone pair: the molecule is TRIGONAL PYRAMIDAL.”
Why: Hybridization describes the orbitals of the central atom, geometry describes the position of the atoms. Both are deduced from the same count but they do not concern the same objects.
9.Counting the core electrons in a molecular orbital diagram
the whole question, 5 marks
What not to write
“N2 has 14 electrons: I place all of them in the valence diagram, 12 bonding and 2 antibonding, so a bond order of 5.”
What to write
“Only the 10 VALENCE electrons enter the molecular orbital diagram: BO=28−2=3, the expected triple bond.”
Why: The 1s electrons stay localized on each nucleus and do not make the bond: counting them among the valence electrons wrongly adds four bonding electrons. In their own orbitals they cancel exactly as bonding and antibonding and would change nothing in the result; mixed into the valence diagram, they spoil the count.
10.Believing that delocalization gives whole-number bonds
3 marks on the reasoning question
What not to write
“In the nitrate ion there is one double bond and two single bonds, so the three N-O bonds have different lengths.”
What to write
“The π bond is DELOCALIZED over the three bonds: they are IDENTICAL, of order 34, with an intermediate length.”
A single Lewis structure would suggest different bonds; the π bond is spread over all three, all identical, of order 34.
Why: The four parallel p orbitals combine into molecular orbitals extended over the whole ion. Diffraction confirms three rigorously equal bonds, which no single Lewis structure can represent.
Which method to choose
From the formula to the hybridization, atom by atom
The number of directions around the atom considered: distinct bond lines plus lone pairs
If 4 directions → sp3, angles close to 109.5°, no free p orbital so no π bond
Example: the carbon of methane, the oxygen of water, the nitrogen of ammonia
If 3 directions → sp2, angles close to 120°, one free p orbital so ONE π bond
Example: the carbons of ethylene, the carbon of a carbonyl
If 2 directions → sp, angle of 180°, two free p orbitals so TWO π bonds
Example: the carbons of acetylene, the nitrogen of a nitrile
If 5 directions, atom of the third period or beyond → sp3d, trigonal bipyramid
Example: the phosphorus of PCl5, the sulfur of SF4
If 6 directions, atom of the third period or beyond → sp3d2, octahedron
Example: the sulfur of SF6, the xenon of XeF4
The count is made ATOM BY ATOM: within a single molecule, one carbon can be sp3 and its neighbour sp. The question “what is the hybridization of this molecule?” makes no sense, only that of an atom does.
Counting the sigma and pi bonds of a molecule
The structural formula, line by line
One line, one σ bond; each extra line is a π bond. Counting the σ bonds means counting the lines. In the drawing, the σ bond is on the axis, in blue, and the extra π bond in orange.
If a single line → 1σ, free rotation
Example: every C-H, every C-C of an alkane
If a double line → 1σ+1π, rotation blocked, cis-trans isomerism possible
Example: the C=C of an alkene, the C=O of a carbonyl
If a triple line → 1σ+2π, linear geometry imposed
Example: the C≡C of an alkyne, the C≡N of a nitrile
If you need the total of σ bonds → simply count the bond LINES, without worrying about their multiplicity
every bond, single or multiple, contains exactly one σ bond
If you need the total of π bonds → count the EXTRA lines: one per double bond, two per triple bond
an aromatic molecule counts its π bonds as if the double bonds were localized, then you state that they are delocalized
This count also answers “how many atoms lie in the same plane?”, since each sp2 carbon forces its three neighbours to be coplanar, and each sp carbon forces its two neighbours into alignment.
How the answer is expected to be written
A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.
Analyzing an organic molecule atom by atom
When to use it: The question gives a structural formula and asks for the hybridizations, the angles, the number of σ and π bonds, or the geometry.
1Redraw the structural formula showing ALL the bond lines and ALL the lone pairs, especially on oxygen and nitrogen.
2Number the non-hydrogen atoms so that you can refer to them without ambiguity afterwards.
3For each numbered atom, count the directions, writing the count down, then deduce the hybridization and the expected angle.
4Count the σ bonds by counting the lines, then the π bonds by counting the extra lines.
5Conclude on the local geometry of each atom and, if asked, on the planarity of all or part of the molecule.
Concluding sentence
“Carbon 1 carries two bonds to hydrogens and one bond to carbon 2, that is, three directions: it is sp2, with angles close to 120°; the molecule has 6σ bonds and 3π bonds.”
The trap: Forgetting the lone pairs of nitrogen or oxygen. They are almost never drawn in an organic formula, and they are what moves a nitrogen from sp to sp2 or sp3.
Marking: Typically 1 mark per hybridization correctly justified by a count, 2 marks for the totals of σ and π bonds, 1 mark for the angles.
Building a molecular orbital diagram
When to use it: The question asks for the bond order, the magnetism, or the comparison of lengths between a molecule and its ions.
1Count the VALENCE electrons only, adjusting for the charge of the ion.
2Fill the diagram from the lowest level to the highest, applying Pauli and then Hund in degenerate orbitals.
3Count the electrons in the bonding orbitals and in the antibonding ones, then calculate BO=2nbonding−nantibonding.
4Conclude on the existence of the species, on the relative length of the bond, and on the magnetism according to whether or not there are unpaired electrons.
Concluding sentence
“With 12 valence electrons, O2 has 8 bonding and 4 antibonding electrons, that is, BO=2; the last two electrons occupy the two π∗ orbitals separately, so the molecule is PARAMAGNETIC.”
The trap: Pairing the last two electrons in the same π∗ orbital. Hund's rule also holds in degenerate molecular orbitals, and it is precisely what gives oxygen gas its paramagnetism.
Check before you hand in
Five minutes of checking recover more marks than one more problem started in a hurry.
The hybridization checked against the VSEPR geometry
The hybridization count and the VSEPR domain count are the SAME count. An sp hybridization for a molecule announced as bent is therefore contradictory.
A water oxygen announced as sp when the angle is 104.5°: the lone pair has been forgotten.
The bond order compared with the Lewis structure
For simple molecules, the order calculated from molecular orbitals must come back to the Lewis multiplicity: 1 for a single bond, 2 for a double, 3 for a triple.
An order of 4 for oxygen gas: the division by two was skipped.
The total number of bonds recounted on the drawing
Add up the σ and π bonds found: the total must equal the number of lines, counting multiple bonds for their multiplicity.
6 sigma and 3 pi make 9, which must match exactly the count of lines in the structural formula.
The magnetism checked against the total number of electrons
An ODD number of valence electrons guarantees at least one unpaired electron, hence paramagnetism. An even number guarantees nothing, you have to look at the diagram.
An ion announced as diamagnetic with an odd number of electrons: that is impossible.
The consistency of the angles announced
An angle of 120° requires sp2 hybridization, an angle of 180° requires sp. Check that the angle announced and the hybridization announced correspond.
A carbon declared sp3 with angles of 120°: one of the two is wrong.
The typical problem, taken apart
Acrylonitrile, three carbons and three hybridizations
Acrylonitrile has the structural formula CH2=CH−C≡N. Number the carbons 1, 2 and 3 from left to right.
Determine the hybridization of each carbon and of the nitrogen, the expected angles, the total number of σ and π bonds, then say which atoms are necessarily coplanar.
The structural formula from the problem, carbons numbered from left to right: at each atom, count the directions.
Step 1
Carbon 1: two bonds to hydrogens and one (double) bond to carbon 2, that is, THREE directions. It is sp2, with angles close to 120°.
Why
The double bond counts as only one direction. This is the point on which the whole question turns, and it is worth writing down the count rather than announcing the result.
Step 2
Carbon 2: one bond to a hydrogen, one (double) bond to carbon 1, one single bond to carbon 3, that is, THREE directions. It is also sp2, at 120°.
Why
Two neighbouring atoms can have the same hybridization for different reasons. Here one carries two hydrogens, the other only one plus a carbon-carbon bond: the count, for its part, does not change.
Step 3
Carbon 3: one single bond to carbon 2 and one (triple) bond to nitrogen, that is, TWO directions. It is sp, at 180°.
Why
The triple bond, like the double bond, is worth only one direction. The result is a rigorously linear chain between carbon 2, carbon 3 and nitrogen.
Step 4
Nitrogen: one (triple) bond to carbon 3 and ONE lone pair, that is, TWO directions. It is sp.
Why
This is the atom that students miss most often, because the lone pair is not drawn in the condensed formula. Without it, nitrogen would seem to have only one direction.
Step 5
σ bonds: two C-H on carbon 1, one C-H on carbon 2, one C-C between 1 and 2, one C-C between 2 and 3, one C-N between 3 and nitrogen, that is, SIX σ bonds.
Why
Counting the bond lines, without worrying about their multiplicity, gives the number of σ bonds directly: each bond contains exactly one.
Step 6
π bonds: one for the double bond C1=C2 and two for the triple bond C3≡N, that is, THREE π bonds.
Why
The extra lines give the π bonds. The total 6+3=9 must match the count of lines in the structural formula, which serves as an immediate check.
Step 7
Coplanarity: carbons 1 and 2 being sp2, the two hydrogens of carbon 1, the hydrogen of carbon 2, the two carbons and carbon 3 lie in the same plane. Carbon 3 being sp, nitrogen is aligned with carbons 2 and 3: the whole molecule is PLANAR.
Why
Planarity is never given, it is a consequence of the hybridizations. It is the most frequent synthesis question on this type of problem.
Step 8
Check: each carbon does carry four bonds in all (1: two H and a double; 2: one H, a double and a single; 3: a single and a triple) and nitrogen carries three plus a lone pair.
Why
The tetravalence of carbon and the trivalence of nitrogen are checks independent of the count of directions: if they fail, the formula was misread before the reasoning even began.
The conclusion, written out
“Carbons 1 and 2 are sp2 at 120°, carbon 3 and nitrogen are sp at 180°; the molecule has 6σ bonds and 3π bonds, and all of its atoms are coplanar, the nitrile part being moreover linear.”
The classic mistake on this problem: Declaring carbon 3sp3 because it carries four bonds if the triple bond is counted as three. The count is of DIRECTIONS: it has two, so it is sp, and that is what forces the nitrile to be linear.
Learn by heart
•Hybridization = number of DIRECTIONS = σ bonds + lone pairs. The π bonds never count.
•4 directions give sp3, 3 give sp2, 2 give sp, 5 give sp3d, 6 give sp3d2.
•BO=2nbonding−nantibonding, calculated on VALENCE electrons only.
•O2 is PARAMAGNETIC: two unpaired electrons in the π∗ orbitals, which Lewis does not predict.
•s character: 25% (sp3), 33% (sp2), 50% (sp). The higher it is, the shorter the bond.
•Hybridization = count of DOMAINS; geometry = position of the ATOMS. Same count, two different answers.
Frequently asked questions
How do you find the hybridization of an atom quickly?
Count the directions around the atom: the number of sigma bonds, that is, the number of distinct bond lines, plus the number of lone pairs. Four directions give sp3 hybridization, three directions sp2, two directions sp. Pi bonds never count, because they are superimposed on a sigma bond already counted.
What is the difference between a sigma bond and a pi bond?
A sigma bond comes from an overlap along the axis joining the two nuclei and allows free rotation. A pi bond comes from the sideways overlap of two parallel p orbitals, above and below that axis, and it blocks rotation. Every bond contains exactly one sigma bond; multiple bonds add pi bonds.
Why is oxygen gas attracted by a magnet?
Because it has two unpaired electrons, placed separately in two antibonding molecular orbitals of equal energy, in accordance with Hund's rule. The Lewis structure, which pairs all the electrons, predicts the opposite. It is the most famous experiment in favour of molecular orbital theory.
Why does cis-trans isomerism exist only about a double bond?
Because the pi bond blocks rotation. About a single bond, the two halves of the molecule turn freely and all positions are equivalent. About a double bond, turning one carbon would break the sideways overlap, which costs too much energy: the two arrangements become two distinct compounds.
How do you calculate a bond order?
Take the number of electrons in the bonding orbitals, subtract the number of electrons in the antibonding ones, and divide the result by two. Only valence electrons enter the calculation. A bond order of zero means the species does not exist, and a high order signals a short, strong bond.
Does hybridization give the geometry of the molecule directly?
Not exactly. Hybridization describes the arrangement of all the directions around the atom, lone pairs included. Geometry describes only the atoms. A nitrogen with sp3 hybridization carries four directions, but since one of them is a lone pair, ammonia is pyramidal and not tetrahedral.
Practise it
Corrected exercises: Hybridization and the quantum model of bonding, 202-SN1-RE
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Get in touch for a first session. On hybridization, everything comes down to a three-second count per atom: we put it in place first, and the rest of the chapter follows by itself.
