General Chemistry 202-SN1-RE • CEGEP, Montreal

Revision sheet: chemical bonding and molecular geometry (202-SN1-RE)

Q: What is the difference between electronic geometry and molecular geometry?

Electronic geometry describes the arrangement of all the domains around the central atom, lone pairs included, and gives the starting angles. Molecular geometry describes only the atoms, so it ignores the lone pairs while still being shaped by their bulk. Water is tetrahedral from the first point of view and bent from the second.

Q: Why is carbon dioxide nonpolar when its bonds are polar?

Because the dipole moment of a molecule is a vector sum. The two carbon-oxygen bonds are strongly polar, but they point in exactly opposite directions and have the same magnitude: they cancel. The molecule is therefore nonpolar overall, despite very polar bonds.

Q: Where does a lone pair go in a trigonal bipyramid?

Always in an equatorial position. There, the lone pair has only two neighbours at ninety degrees, against three if it were axial. Since a lone pair is bulkier than a bond, it chooses the least crowded position. This is what gives a molecule with four bonds and one lone pair the seesaw shape.

Q: Which atoms can exceed the octet?

Only those of the third period and beyond, such as phosphorus, sulfur, chlorine or xenon, because they have accessible d orbitals. Carbon, nitrogen, oxygen and fluorine never exceed eight electrons. This is why phosphorus pentachloride exists and nitrogen pentachloride does not.

Q: When does a molecule form hydrogen bonds?

Only when a hydrogen atom is bonded directly to a nitrogen, an oxygen or a fluorine. You need both a hydrogen strongly depleted of electrons and a small, very electronegative acceptor carrying a lone pair. A hydrogen carried by a carbon meets neither condition.

Q: Why are the angles of ammonia not equal to one hundred and nine point five degrees?

Because nitrogen carries a lone pair in addition to its three bonds. A lone pair is held by only one nucleus, so it takes up more room than a bonding pair and pushes the bonds harder. The angles close to about one hundred and seven degrees, and to one hundred and four and a half in water, which has two.

In 202-SN1-RE (formerly 202-NYA), this chapter strings together four moves that are always made in the same order: count the valence electrons, write the Lewis structure, count the domains around the central atom, then deduce the shape and the polarity. Each step depends entirely on the previous one, which is why a counting error on the first line costs the whole question.

This sheet gives that sequence in order, with the two tables to know by heart and the list of traps that come back from one exam to the next, in particular on polarity and on geometries with five and six domains.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

Two geometries coexist in every molecule: that of the electron DOMAINS, which decides the angles, and that of the ATOMS, which decides the shape and the polarity. Confusing them costs half the questions of the chapter.

This chapter is part of 202-SN1-RE, General Chemistry

The essentials

The cutoffs and formulas to know by heart

• Bond type from $\Delta EN$ : nonpolar covalent if $\Delta EN<0.4$ ; polar covalent if $0.4\le\Delta EN<1.7$ ; ionic if $\Delta EN\ge 1.7$ .
• Formal charge: $FC=N_{v}-N_{nb}-\dfrac{N_{b}}{2}$ , where $N_{b}$ is the number of BONDING electrons, counted for half. The sum of the $FC$ equals the charge of the species.
• Lattice energy proportional to $\dfrac{Q_{1}Q_{2}}{r}$ : the CHARGE weighs far more than the radius, which explains why MgO melts at $2850$ °C and NaCl at $801$ °C.
• Dipole moment: $\mu=Qd$ , in debyes, with $1$ D $=3.336\times 10^{-30}$ C·m. Ionic character $=\dfrac{\mu_{measured}}{ed}$ .
• Intermolecular forces in increasing order: London, present everywhere and growing with molar mass; dipole-dipole; hydrogen bonding, reserved for H bonded to N, O or F.

The cutoffs

0.4

and

1.7

slice up a continuous axis: H−Cl is polar, Na−Cl ionic, and a bond between two identical atoms has no polarity at all.

The $\Delta EN$ cutoffs are conventions, not laws: a $\Delta EN$ of $1.6$ is commented on (“strongly polar, at the limit of ionic”) rather than settled to the hundredth.

The VSEPR table, in the only usable form

• $AX_{2}$ linear $180°$ ; $AX_{3}$ trigonal planar $120°$ ; $AX_{4}$ tetrahedral $109.5°$ ; $AX_{5}$ trigonal bipyramid; $AX_{6}$ octahedral $90°$ .
• Each lone pair replaces an $X$ by an $E$ and TIGHTENS the angles: $AX_{3}E$ trigonal pyramidal ( $107°$ ), $AX_{2}E_{2}$ bent ( $104.5°$ ).
• Five domains: the lone pair ALWAYS goes equatorial. $AX_{4}E$ seesaw, $AX_{3}E_{2}$ T-shaped, $AX_{2}E_{3}$ linear.
• Six domains: $AX_{5}E$ square pyramid, $AX_{4}E_{2}$ square planar with the two lone pairs in opposite positions.
• A double or triple bond counts as ONE domain only: $CO_{2}$ has two domains, so it is linear.

The five basic arrangements, without lone pairs: each domain moves as far as possible from the others, hence the angles to know by heart.

The number of domains gives the ELECTRONIC geometry and the starting angles; the number of $X$ gives the MOLECULAR geometry, the only one you draw and the only one that decides polarity.

Polarity: two conditions, not one

• A molecule is polar if it has polar bonds AND a geometry that does not make them cancel.
• Geometries that cancel everything, with identical substituents: linear $AX_{2}$ , trigonal planar $AX_{3}$ , tetrahedral $AX_{4}$ , bipyramid $AX_{5}$ , octahedral $AX_{6}$ , square planar $AX_{4}E_{2}$ , linear $AX_{2}E_{3}$ .
• Any geometry carrying an uncompensated lone pair is polar: $AX_{3}E$ , $AX_{2}E_{2}$ , $AX_{4}E$ , $AX_{3}E_{2}$ , $AX_{5}E$ .
• $CO_{2}$ , $CCl_{4}$ , $BF_{3}$ and $SF_{6}$ are NONPOLAR despite clearly polar bonds: symmetry decides.
• DIFFERENT substituents break the symmetry: $CH_{3}Cl$ is polar whereas $CCl_{4}$ is not.

Same polar bonds on both sides: it is the GEOMETRY that decides. Aligned, the two dipoles destroy each other; opened to

104.5°

, they add.

The question “is this molecule polar?” is never answered without first drawing the geometry. Answering from the molecular formula is the most frequent error of the chapter.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

The complete VSEPR table, one molecule per row

You count the domains around the central atom, bonding AND lone pairs, a double bond counting as ONE domain only. The number of domains gives the arrangement, the lone pairs give the observed shape.

Notation	Domains	Geometry and angles
$AX_{2}$	$2$	linear, $180°$
Example: $CO_{2}$ : two double bonds, so only two domains, and an angle of $180°$ . The molecule is NONPOLAR despite two clearly polar bonds.
$AX_{3}$	$3$	trigonal planar, $120°$
Example: $BF_{3}$ : boron has only six electrons, it does not follow the octet rule, and the molecule is nonpolar by symmetry.
$AX_{2}E$	$3$ , of which $1$ lone	bent, about $118°$
Example: $SO_{2}$ : same trigonal arrangement as $BF_{3}$ , but a lone pair replaces a substituent and squeezes the angle below $120°$ . The molecule is polar.
$AX_{4}$	$4$	tetrahedral, $109.5°$
Example: $CH_{4}$ and $CCl_{4}$ : both nonpolar, even though the C-Cl bonds are very polar. Symmetry cancels everything.
$AX_{3}E$	$4$ , of which $1$ lone	trigonal pyramidal, $107°$
Example: $NH_{3}$ : the angle falls from $109.5°$ to $107°$ because the lone pair pushes harder than the bonds. Polar molecule.
$AX_{2}E_{2}$	$4$ , of which $2$ lone	bent, $104.5°$
Example: $H_{2}O$ : two lone pairs, two pushes, and the angle drops to $104.5°$ . This geometry makes water polar and explains almost all its properties.
$AX_{5}$	$5$	trigonal bipyramid, $90°$ and $120°$
Example: $PCl_{5}$ : two different angles coexist, $120°$ in the equatorial plane and $90°$ toward the axial positions.
$AX_{4}E$	$5$ , of which $1$ lone	seesaw, about $102°$ and $173°$
Example: $SF_{4}$ : the lone pair goes EQUATORIAL, never axial, because it suffers fewer repulsions there. Polar molecule.
$AX_{3}E_{2}$	$5$ , of which $2$ lone	T-shaped, about $90°$
Example: $ClF_{3}$ : the two lone pairs occupy two of the three equatorial positions.
$AX_{2}E_{3}$	$5$ , of which $3$ lone	linear, $180°$
Example: $XeF_{2}$ : the three lone pairs take the whole equatorial plane, and only the two axial positions remain, hence $180°$ .
$AX_{6}$	$6$	octahedral, $90°$
Example: $SF_{6}$ : six identical bonds, a nonpolar molecule, and one of the most inert gases known.
$AX_{5}E$	$6$ , of which $1$ lone	square pyramid, about $85°$
Example: $BrF_{5}$ : the lone pair pushes the five bonds downward, so the angle goes below $90°$ . Polar molecule.
$AX_{4}E_{2}$	$6$ , of which $2$ lone	square planar, $90°$
Example: $XeF_{4}$ : the two lone pairs sit OPPOSITE each other, so their effects compensate and the molecule is nonpolar.

Two conditions make a molecule polar, not one: polar bonds AND a geometry that does not cancel them. The five geometries without a lone pair and with identical substituents cancel everything, as do the square planar $AX_{4}E_{2}$ and the linear $AX_{2}E_{3}$ ; all the others are polar.

The bond type is read from the electronegativity difference

Two cutoffs, three boxes, and the calculation is a single subtraction. It is done before any discussion of polarity, since it is what tells you whether there are polar bonds to cancel.

$\Delta EN$	Bond type	What it means
$\Delta EN<0.4$	NONPOLAR covalent	the electrons are shared equally
Example: The C-H bond: $2.55-2.20=0.35$ , so nonpolar. This is why hydrocarbons are insoluble in water.
$0.4\leq\Delta EN<1.7$	POLAR covalent	the electrons lean toward the more electronegative atom
Example: The H-Cl bond: $3.16-2.20=0.96$ . The O-H bond: $3.44-2.20=1.24$ , the most important in all the chemistry of living things.
$\Delta EN\geq 1.7$	IONIC	an electron is transferred, not shared
Example: $NaCl$ : $3.16-0.93=2.23$ . There is no longer a molecule but a lattice of ions, which explains a melting point of $801$ °C.
$\Delta EN=0$	purely nonpolar covalent	perfectly equal sharing
Example: Any bond between two identical atoms: $H_{2}$ , $O_{2}$ , $Cl_{2}$ . The difference is exactly $0$ , there is no pole.

These cutoffs are conventions, not laws of nature: a bond with $\Delta EN=1.68$ is not fundamentally different from one at $1.72$ . The exam uses them as sharp cutoffs all the same, and expects the written calculation before the conclusion.

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. Confusing the geometry of the domains with the geometry of the molecule

the whole question, 5 marks, and the wrong polarity afterwards

What not to write

“Water has four domains around the oxygen, so it is tetrahedral.”

What to write

“Its ELECTRONIC geometry is tetrahedral, but two domains are lone pairs: its MOLECULAR geometry is BENT, with an angle of $104.5°$ .”

The four domains of oxygen form a tetrahedron, but you only name what the atoms draw: water is BENT.

Why: You only draw and name the atoms. Lone pairs take up room and push the bonds apart, but they are not part of the observed shape.

2. Declaring a molecule polar because its bonds are

3 marks, and a wrong ranking of boiling points

What not to write

“Each C=O bond is very polar, so carbon dioxide is a polar molecule.”

What to write

“The two bonds are polar but opposite and of equal magnitude: the moments cancel, and $CO_{2}$ is NONPOLAR.”

Why: The dipole moment of a molecule is a VECTOR SUM. A symmetric geometry with identical substituents always cancels this sum, whatever the polarity of each bond.

3. Granting an expanded octet to a second-period element

the whole structure, 5 marks

What not to write

“In the nitrate ion, I put a double bond on each oxygen: nitrogen then has twelve electrons.”

What to write

“ $N$ , $O$ , $C$ and $F$ NEVER exceed eight electrons: they have no accessible $d$ orbital. Only the third period and beyond can expand the octet.”

Three double bonds carry nitrogen beyond the octet, which a second-period element cannot do; the right structure keeps eight electrons and formal charges.

Why: An expanded octet requires empty $d$ orbitals close in energy, which exist only from $n=3$ . This is also why $PCl_{5}$ exists and $NCl_{5}$ does not.

4. Placing a lone pair in an axial position in a bipyramid

the whole question, 5 marks

What not to write

“ $SF_{4}$ has five domains: I put the lone pair on top, in an axial position, and the molecule is square pyramidal.”

What to write

“The lone pair ALWAYS goes EQUATORIAL, where it has only two neighbours at $90°$ instead of three. $SF_{4}$ therefore has a SEESAW shape.”

Axial, the lone pair would have three neighbours at

90°

, equatorial only two: it goes to the equator and

SF_{4}

takes the seesaw shape.

Why: A lone pair is bulkier than a bond, so it seeks the least crowded position. Equatorial, it suffers two repulsions at $90°$ ; axial, it would suffer three.

5. Counting the bonding electrons in full in the formal charge

2 marks per atom, often 6 marks on the question

What not to write

“The nitrogen of ammonia has one lone pair and three bonds, so $FC=5-2-6=-3$ .”

What to write

“Bonding electrons count for HALF: $FC=5-2-\dfrac{6}{2}=0$ .”

Why: The formal charge shares each bonding pair equally between the two atoms. The foolproof check: the sum of the formal charges must equal the charge of the species, here zero.

6. Describing resonance as an oscillation between two structures

3 marks on the reasoning question

What not to write

“Ozone keeps flipping between a single bond on the left and a double bond on the right.”

What to write

“The real molecule is a HYBRID: the two bonds are identical, intermediate between single and double, at all times. No resonance structure exists separately.”

The two drawings are only pictures: the real molecule has two identical bonds of

128

pm, between the single and the double bond.

Why: Experiment measures two EQUAL bond lengths in ozone, intermediate between those of a single and a double bond. An oscillation would give two different lengths.

7. Attributing hydrogen bonding to a hydrogen carried by a carbon

3 marks, and a completely wrong ranking of boiling points

What not to write

“Methane contains hydrogen, so it forms hydrogen bonds and boils at a high temperature.”

What to write

“Hydrogen bonding requires H bonded to N, O or F. Methane has only London forces and boils at $-161$ °C.”

The group 14 hydrides rise steadily with the period, London forces only. Water boils about

190

degrees above the trend of its group: hydrogen bonding.

Why: You need both a hydrogen that is strongly depleted of electrons and a small, very electronegative acceptor carrying a lone pair. Carbon is neither electronegative enough nor carrying a lone pair.

8. Forgetting to adjust the number of valence electrons for an ion

the whole structure, 5 marks

What not to write

“For the nitrate ion $NO_{3}^{-}$ : $5+3\times 6=23$ valence electrons.”

What to write

“You must ADD one electron per negative charge: $5+18+1=24$ electrons, that is $12$ pairs.”

Why: An odd number of electrons signals a radical, which nitrate is not. Checking the parity of the total is therefore an instant check on common ions.

9. Announcing $109.5°$ for a molecule that carries lone pairs

2 marks, but the question comes up on almost every exam

What not to write

“Ammonia is derived from the tetrahedron, so its angles are $109.5°$ .”

What to write

“A lone pair is bulkier than a bond: it tightens the angles to about $107°$ in ammonia and $104.5°$ in water.”

Same tetrahedron of domains for the three molecules, but each lone pair tightens the angles:

109.5°

, then

107°

, then

104.5°

Why: A lone pair is held by only one nucleus, so it spreads out more than a bonding pair held by two. The more lone pairs, the more the angle closes.

Which method to choose

From the Lewis structure to the shape

The number of domains around the central atom, then the number of lone pairs

With five domains, the lone pairs occupy the equator first; with six domains, the second pair goes opposite the first, hence the square plane.

If 2 domains → linear, $180°$ , nonpolar if the two substituents are identical

Example: $CO_{2}$ , $BeCl_{2}$
If 3 domains, no lone pair → trigonal planar, $120°$

Example: $BF_{3}$ , $NO_{3}^{-}$
If 3 domains of which 1 lone pair → bent, about $118°$ , POLAR

Example: $SO_{2}$ , $O_{3}$
If 4 domains → tetrahedral ( $109.5°$ ), trigonal pyramidal with 1 lone pair ( $107°$ ), bent with 2 ( $104.5°$ )

Example: $CH_{4}$ , $NH_{3}$ , $H_{2}O$
If 5 domains → bipyramid; the lone pair or pairs go EQUATORIAL: seesaw, T-shape, then linear

Example: $PCl_{5}$ , $SF_{4}$ , $ClF_{3}$ , $I_{3}^{-}$
If 6 domains → octahedron; 1 lone pair gives the square pyramid, 2 OPPOSITE pairs give the square plane

Example: $SF_{6}$ , $BrF_{5}$ , $XeF_{4}$

A double or triple bond counts as one domain only. This is why carbon dioxide, which nevertheless has four bonding pairs, is linear and not tetrahedral.

Which intermolecular force dominates

The composition of the molecule, in this order

Hydrogen bonding requires H on N, O or F; a polar molecule without that H has only dipole-dipole; a nonpolar one, only London forces.

If the molecule contains H bonded directly to N, O or F → hydrogen bonding, the strongest of the three, and an abnormally high boiling point

Example: $H_{2}O$ boils at $100$ °C against $-60$ °C for $H_{2}S$
If the molecule is polar without H on N, O or F → dipole-dipole, plus London

Example: $HCl$ , $CH_{3}Cl$ , $SO_{2}$
If the molecule is nonpolar → London only, growing with molar mass and chain length

Example: liquid $Br_{2}$ and gaseous $Cl_{2}$ at room temperature
If you must compare two nonpolar molecules → the heavier or the less branched one boils higher

a linear chain offers more contact surface than a branched chain of the same mass
If you must compare an ionic solid with a molecular solid → the ionic one always wins, and between two ionic ones you compare $\dfrac{Q_{1}Q_{2}}{r}$

MgO melts far higher than NaCl because the charges are double

London forces are present in ALL molecules, including those that form hydrogen bonds. The other forces add to London forces, they do not replace them.

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.

Writing a Lewis structure and deducing the shape

When to use it: The question gives a formula and asks for the structure, the geometry, the angles or the polarity.

1 Count the valence electrons: the sum of the valence electrons of each atom, PLUS one per negative charge, MINUS one per positive charge. Divide by two to get the number of pairs.
2 Choose the central atom, the least electronegative apart from hydrogen, and link it to each outer atom with a single bond.
3 Complete the octets of the outer atoms first, then place the remaining pairs on the central atom.
4 If the central atom is short of electrons, form double or triple bonds by moving a lone pair from a neighbour, unless the central atom is in the second period and already full.
5 Count the DOMAINS around the central atom, counting each multiple bond as one, then name the electronic geometry and the molecular geometry.
6 Conclude on the angles and on the polarity by reasoning on the vector sum of the bond moments.

Concluding sentence

“With $34$ valence electrons, sulfur carries four bonds and one lone pair, that is five domains: the electronic geometry is a trigonal bipyramid, the lone pair occupies an equatorial position, so the molecule has a seesaw shape and it is POLAR.”

The trap: Skipping the counting step and drawing from memory. Counting is never wasted time: it decides the number of lone pairs, hence the shape and the polarity.

Marking: Typically 2 marks for the count, 3 for the correct structure, 2 for the two geometries named, 1 for the angle, 2 for the justified polarity.

Choosing the best resonance structure

When to use it: Several Lewis structures are possible and the question asks which one contributes most.

1 Write all the conceivable structures, first respecting the octet of second-period elements, which is not negotiable.
2 Calculate the formal charge of EACH atom in each structure, and check that the sum equals the charge of the species.
3 Keep the structure whose formal charges are closest to zero.
4 In a tie, place the negative charge on the most electronegative atom, and rule out any structure carrying two charges of the same sign on neighbouring atoms.

Concluding sentence

“Both structures respect the octet, but the one that places the negative charge on oxygen rather than on carbon has smaller and better distributed formal charges: it is the one that contributes most to the hybrid.”

The trap: Choosing a structure with zero formal charges at the price of an expanded octet on a second-period atom. The octet always comes before the formal charge.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The sum of the formal charges

Add the formal charges of all the atoms: the total must equal exactly the charge of the molecule or ion. A gap signals a misplaced bond or pair.

A sum of $-2$ on a neutral molecule: two electrons too many have been drawn.
The total count of electrons drawn

Count the electrons actually present in the drawing, two per bond and two per lone pair, and compare with the total calculated in the first step.

$34$ expected and $36$ drawn for $SF_{4}$ : one lone pair too many on sulfur.
Consistency between shape and polarity

A perfectly symmetric geometry with identical substituents always gives a nonpolar molecule. Any contrary conclusion is wrong without further examination.

A $CCl_{4}$ declared polar: the tetrahedral geometry cancels the four moments.
The parity of the number of valence electrons

An odd number signals a radical. If the species is not one, a charge adjustment has been forgotten.

$23$ electrons for the nitrate ion: the negative charge has not been counted.
The ranking of boiling points reread through the forces

Check that every species forming hydrogen bonds is placed above a comparable species that does not, at a similar molar mass.

An $H_{2}O$ ranked below $H_{2}S$ : hydrogen bonding has been ignored.

The typical problem, taken apart

Five domains against six: sulfur tetrafluoride and xenon tetrafluoride

For $SF_{4}$ and then for $XeF_{4}$ , establish the Lewis structure, determine the number of domains around the central atom, name the electronic geometry and the molecular geometry, then conclude on the polarity.

These two molecules have the same apparent formula $AX_{4}$ and yet opposite shapes and polarities: explain why.

Step 1

$SF_{4}$ : valence electrons $=6+4\times 7=34$ , that is $17$ pairs. Four S-F bonds use $8$ electrons, the four fluorines carry $4\times 6=24$ nonbonding electrons, which leaves $34-8-24=2$ electrons, that is ONE lone pair on sulfur.

Why

The count decides everything that follows. What remains after serving the outer atoms is exactly what gives the number of lone pairs on the centre, hence the shape.

Step 2

Sulfur therefore carries $4$ bonds and $1$ lone pair: FIVE domains, type $AX_{4}E$ . Electronic geometry: trigonal bipyramid.

Why

You name the electronic geometry first, because it fixes the available positions, axial and equatorial, between which you will have to choose.

Step 3

The lone pair takes the EQUATORIAL position, where it faces only two neighbours at $90°$ . The molecular geometry is therefore a SEESAW, with angles tightened to about $102°$ and $173°$ .

Why

The equatorial choice is the only truly physical decision of the question, and it is the one that student papers miss. It is justified by counting the repulsions at $90°$ , not by invoking habit.

Step 4

Polarity of $SF_{4}$ : the four S-F bonds are very polar and the seesaw is not symmetric, so the moments do not cancel. The molecule is POLAR.

Why

The presence of an uncompensated lone pair is almost always enough to conclude: it breaks the symmetry that could have cancelled the moments.

Step 5

$XeF_{4}$ : valence electrons $=8+4\times 7=36$ , that is $18$ pairs. Four bonds use $8$ electrons, the fluorines $24$ , which leaves $36-8-24=4$ electrons, that is TWO lone pairs on xenon.

Why

Two more valence electrons on the central atom, and everything changes: the same count, carried through, gives one more lone pair and hence another geometric family.

Step 6

Xenon carries $4$ bonds and $2$ lone pairs: SIX domains, type $AX_{4}E_{2}$ . Octahedral electronic geometry; the two lone pairs take OPPOSITE positions, at $180°$ from each other.

Why

Two lone pairs repel each other more than anything else: they go as far apart as possible, which fixes their position and therefore that of the four fluorines.

Step 7

The molecular geometry is therefore SQUARE PLANAR, with angles of $90°$ . The four bond moments, coplanar and at $90°$ , cancel in pairs: $XeF_{4}$ is NONPOLAR.

Why

This is the most useful counterexample of the chapter: a molecule with two lone pairs that is nonetheless nonpolar, because those pairs are opposite and compensate each other.

Step 8

Checking the counts: $8+24+2=34$ for $SF_{4}$ and $8+24+4=36$ for $XeF_{4}$ , matching the starting totals. Formal charges all zero in both cases.

Why

Recounting the electrons drawn and checking that the formal charges are zero validates the structure without having to redraw it: two independent checks in ten seconds.

The conclusion, written out

“ $SF_{4}$ has five domains, including one equatorial lone pair: its shape is a seesaw and it is polar. $XeF_{4}$ has six domains, including two opposite lone pairs: its shape is square planar and it is nonpolar. The same formula $AX_{4}$ therefore covers two geometries and two polarities, which only the electron count can tell apart.”

The classic mistake on this problem: Treating the two molecules as tetrahedral $AX_{4}$ because they have four outer atoms. The number of atoms bonded is never enough: it is the DOMAINS, lone pairs included, that decide.

Learn by heart

• $\Delta EN$ : $<0.4$ nonpolar, $0.4$ to $1.7$ polar, $\ge 1.7$ ionic.
• $FC=N_{v}-N_{nb}-\dfrac{N_{b}}{2}$ , and the sum of the $FC$ equals the charge of the species.
• Ion: ADD one electron per negative charge, REMOVE one per positive charge.
• A multiple bond counts as ONE domain.
• Lone pair: always EQUATORIAL with five domains, always OPPOSITE with six domains when there are two.
• Angles: $109.5°$ with no lone pair, $107°$ with one, $104.5°$ with two.
• Polar = polar bonds AND a non-symmetric geometry. $CO_{2}$ , $CCl_{4}$ , $BF_{3}$ , $SF_{6}$ , $XeF_{4}$ are nonpolar.
• Hydrogen bonding only if H is bonded to N, O or F. London forces exist everywhere.

Frequently asked questions

What is the difference between electronic geometry and molecular geometry?

Electronic geometry describes the arrangement of all the domains around the central atom, lone pairs included, and gives the starting angles. Molecular geometry describes only the atoms, so it ignores the lone pairs while still being shaped by their bulk. Water is tetrahedral from the first point of view and bent from the second.

Why is carbon dioxide nonpolar when its bonds are polar?

Because the dipole moment of a molecule is a vector sum. The two carbon-oxygen bonds are strongly polar, but they point in exactly opposite directions and have the same magnitude: they cancel. The molecule is therefore nonpolar overall, despite very polar bonds.

Where does a lone pair go in a trigonal bipyramid?

Always in an equatorial position. There, the lone pair has only two neighbours at ninety degrees, against three if it were axial. Since a lone pair is bulkier than a bond, it chooses the least crowded position. This is what gives a molecule with four bonds and one lone pair the seesaw shape.

Which atoms can exceed the octet?

Only those of the third period and beyond, such as phosphorus, sulfur, chlorine or xenon, because they have accessible d orbitals. Carbon, nitrogen, oxygen and fluorine never exceed eight electrons. This is why phosphorus pentachloride exists and nitrogen pentachloride does not.

When does a molecule form hydrogen bonds?

Only when a hydrogen atom is bonded directly to a nitrogen, an oxygen or a fluorine. You need both a hydrogen strongly depleted of electrons and a small, very electronegative acceptor carrying a lone pair. A hydrogen carried by a carbon meets neither condition.

Why are the angles of ammonia not equal to one hundred and nine point five degrees?

Because nitrogen carries a lone pair in addition to its three bonds. A lone pair is held by only one nucleus, so it takes up more room than a bonding pair and pushes the bonds harder. The angles close to about one hundred and seven degrees, and to one hundred and four and a half in water, which has two.

Practise it

Corrected exercises: Chemical bonding and molecular geometry, 202-SN1-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-chem-sn1-bonding-vsepr. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Get in touch for a first session. On chemical bonding, we work through the complete chain, from counting valence electrons to polarity: it is always at the same place that the chain breaks.

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Revision sheet: chemical bonding and molecular geometry (202-SN1-RE)

The essentials

The cutoffs and formulas to know by heart

The VSEPR table, in the only usable form

Polarity: two conditions, not one

The rules in table form

The complete VSEPR table, one molecule per row

The bond type is read from the electronegativity difference

The mistakes that cost marks

1. Confusing the geometry of the domains with the geometry of the molecule

2. Declaring a molecule polar because its bonds are

3. Granting an expanded octet to a second-period element

4. Placing a lone pair in an axial position in a bipyramid

5. Counting the bonding electrons in full in the formal charge

6. Describing resonance as an oscillation between two structures

7. Attributing hydrogen bonding to a hydrogen carried by a carbon

8. Forgetting to adjust the number of valence electrons for an ion

9. Announcing 109.5°109.5°109.5° for a molecule that carries lone pairs

Which method to choose

From the Lewis structure to the shape

Which intermolecular force dominates

How the answer is expected to be written

Writing a Lewis structure and deducing the shape

Choosing the best resonance structure

Check before you hand in

The typical problem, taken apart

Five domains against six: sulfur tetrafluoride and xenon tetrafluoride

Learn by heart

Frequently asked questions

What is the difference between electronic geometry and molecular geometry?

Why is carbon dioxide nonpolar when its bonds are polar?

Where does a lone pair go in a trigonal bipyramid?

Which atoms can exceed the octet?

When does a molecule form hydrogen bonds?

Why are the angles of ammonia not equal to one hundred and nine point five degrees?

Corrected exercises: Chemical bonding and molecular geometry, 202-SN1-RE

See also

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9. Announcing $109.5°$ for a molecule that carries lone pairs