Doc Brown's Advanced A Level Chemistry

Theoretical Physical Chemistry Revision Notes

The Shapes of Molecules and Ions and bond angles related to their Electronic Structure - mainly inorganic molecules on this page

Part 1 from diatomic molecules to polyatomic molecules

The shapes and bond angles of a variety of molecules are described, explained and discussed using valence shell electron pair repulsion theory (VSEPR theory) and patterns of shapes deduced for 2, 3, 4, 5 and 6 groups of bonding electrons or non-bonding electrons in the valence shell of the central atom of the molecule or ion. 

So, this page is all about how to work out molecule shapes and work out bond angles is described and explained! i.e. using the electron pair repulsion theory to deduce the shapes of molecules and work out bond angles from the various repulsions i.e lone pair - lone pair, lone pair - bond pair and bond pair - bond pair repulsions.

Shapes can be worked out from dot & cross diagrams and bond angles deduced from the established shape, all such deductions are essentially based on electron pair(s) repulsion theory.

A summary of the shapes of molecules is given in APPENDIX 1.

If you think I've missed a molecule/ion shape that I should have covered for pre-university level, email it to me!

Sub-index for 'shapes of molecules' pages

Introduction to VSEPR theory

No-shape diatomic molecules!

Molecule shapes based on total groups of 2  3  4  5  6  electron pairs (bonding or non-bonding)

Some more complex inorganic/organic molecules/ions

transition metal complexes

Some other molecules/ions of carbon, nitrogen, sulfur and chlorine (separate page)

shapes and bond angles of organic molecules (separate page)

Introduction to VSEPR theory

You could be given familiar and unfamiliar examples of species and asked to deduce the shape according to the valence shell electron pair repulsion (acronym VSEPR) theory principles.

1. Consider bonding pairs and lone (non-bonding) pairs of electrons as charge clouds that repel each other (remember that like electrical charges repel).
   • In deducing the shape of free radicals, a single electron can be treated as a lone pair to a reasonable approximation.
2. These pairs of bonding electrons or non-bonding electrons in the outer shell of atoms arrange themselves as far apart as possible to minimise this repulsion.
   • We are talking about the electron clouds or more precisely, the orbitals that these bonding and non-bonding pairs of electrons occupy.
   • This effectively means to produce a range of as wide as possible angles between adjacent orbitals.
   • These maybe lone pair - lone pair, bond pair - lone pair or bond pair - bond pairs of electrons.
   • Note that the electron cloud orbitals could involve double or triple bonds - but that makes no difference in deducing the shape of a molecule using VSEPR theory.
3. Know the general rule: lone pair–lone pair repulsion is greater than lone pair–bond pair repulsion, which is greater than bond pair–bond pair repulsion.
   • Generally speaking this doesn't affect the shape, BUT, it can make small differences to the expected 'perfect' bond angle. An excellent example is the comparison of the H-X-H angle for water, ammonia and methane, which I've discussed in detail on this page.
   • You also have to be careful in predicting bond angles using the rule expressed in 2. which works well if X is nitrogen or oxygen.
   • BUT, larger X atoms or more bulky X groups might override the valence shell electron pair rule of 2.
     • This can be overlooked by exam boards in their marking schemes and 'stuff' you find on the internet!
   • Frequently in the text I've used the abbreviated term 'lone pair' meaning a pair of non-bonding electrons - in exams its a good idea to initially define any abbreviations you may use several times and then you shouldn't lose any marks - take care!
4. Last introduction point: Be very careful when describing your deduction of a molecular shape to distinguish between the electron pair geometry and the actual molecular geometry of the molecule i.e. the actual shape! If no lone pairs are involved, they are the same, if at least one lone pair is involved, they cannot be the same!
   • The electron pair geometry is the arrangement of the (usually) pairs of bonding and non-bonding electrons around the central atom of the neutral molecule (or molecular ion).
   • The molecular geometry shows the actual shape of the molecule.
   • Check out the extensive summary table in APPENDIX 1, it covers all shapes needed for UK A level chemistry.
   • I haven't introduced this APPENDIX 1 summary here at the start - its too much in one go, but when you have studied a variety of examples of deducing molecule shapes and bond angles take a good look at it and you should then fully appreciate point 4.

These relative electron pair repulsions have a profound effect on bond angles and molecule shape, both of which you need to be able to deduce.

For most pre-university courses you should be able to explain the shapes of, and bond angles in, of simple molecules and ions with two bonding pairs of electrons OR three to six electron pairs (including both bonding and lone pairs of electrons) surrounding the central atom.

The shape of a molecule is determined by the number of groups of electrons around the central atom. The 'groups' might be a non-bonding single electron, a non-bonding or bonding pair of electrons, a double pair of bonding electrons or triple pair of bonding electrons etc. The electron 'groupings' repel to minimise the potential energy of the system i.e. to make the A-B-C angle as wide as possible.

The dot and cross diagrams (ox) are presented in 'Lewis style'

In the diagrams the central atom is denoted by X and attached surrounding bonded atoms by Q.

The bond angle is therefore based on angle between the atoms Q-X-Q.

The phrase lone pair, usually refers to a pair of non-bonding electrons in the outer valence shell of the central atom of the molecule.

As already mentioned, this has an important 'sub-rule' which affects the precise bond angle.

Any lone pairs of non-bonding electrons on the central atom X, are closer to X than bond pairs because there is no Q atom attracting/sharing the lone pair electron charge.

This will increase the repulsion between a lone pair of non-bonding electrons on X and any other bonding/non-bonding pairs of electrons in the outer shell of the central atom X.

The result is two-fold:

In terms of electron pair repulsion: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair.

As the lone pair - 'other pair' repulsion increases, the angle between these pairs increases, so the Q-X-Q angle will be slightly reduced compared to what might be expected from the 'simple' geometry of the shape (this is best illustrated by the sequence H₂O, NH₃ and CH₄, see below)

Three groups of electrons around the central atom

trigonal planar electron pair geometry

electrons: two bond pairs, one lone pair

shape of molecule V or BENT, bond angle approximately 120^o

Does anyone know of any example? Q would have to be H.

the Lewis dot and cross electronic diagram used to predict the shape - trigonal planar electron pair geometry

V or bent shape molecule

electrons: two bond pairs, one lone pair

molecule shape V or BENT, bond angle approximately 120^o

Does anyone know of an example?

For X=Q double bonds see sulfur dioxide

SO₂, NO₂^- and O₃ involve 2 groups of bonding electrons and lone pair and VSEPR correctly predicts their V or bent shape and a Q-X-Q bond angle of ~120^o.

Electron pair geometry different from the molecular geometry.

the Lewis dot and cross electronic diagram used to predict the shape of boron hydride - trigonal planar electron pair geometry

trigonal planar shaped molecule

electrons: 3 bond pairs, no lone pairs

Electron pair geometry same as the molecular geometry.

(molecular compounds/ions involving hydrogen)

The VSEPR theory argument gives the shape TRIGONAL PLANAR: Q-X-Q bond angle exactly 120^o:

e.g. X = B and Q = H for gaseous boron hydride BH₃. Boron in Group 3/13 has three outer valence electrons each of which pairs up with a hydrogen electron - dot and cross diagram on the right.

Boron is in group 3/13 with three outer valence electrons and can therefore form three single covalent bonds with hydrogen, but no 'octet' of valence electrons on the central atom.

(X = C, Q = H) Carbon in group 4/14 has four valence electrons, one of which is lost in the formation of carbocation. When the remaining three electrons are paired with those from hydrogens you get the methyl carbocation CH₃⁺. This will also have a perfect trigonal planar shape with a perfect H-C-H bond angle of 120^o. Prior to bonding, the black dots of the hydrogen electrons and the black crosses of the carbon electrons and not a black cross has gone from the carbon atom, one of its four outer valence electrons has been lost to give he overall single positive charge.

DO NOT assume examination questions will always be based on neutral molecules!

the Lewis dot and cross electronic diagram used to predict the shape of some group III trihalides - trigonal planar electron pair geometry

trigonal planar shaped molecule/ion

electrons: 3 bond pairs, no lone pairs

Electron pair geometry same as the molecular geometry.

(molecular compounds/ions NOT involving hydrogen)

The VSEPR theory argument gives the molecule shape TRIGONAL PLANAR: bond angle, 120^o:

Boron and aluminium are in Group 3/13 with three outer valence electrons and can form three single covalent bonds with halogens (ignoring ionic compounds) - but no 'octet' of electrons on the central atom.

e.g. gaseous boron trifluoride BF₃, boron trichloride BCl₃ and aluminium fluoride AlF₃ (the latter is ionic in solid). Boron and aluminium are in group 3/13 with three outer shell valence electrons each of which is paired with an outer valence shell electron of the halogen.

(others in the gaseous state  e.g. if Q = F or Cl, then X = B, for Al for F only)

but for X=Q double bonds see sulfur trioxide

COCl₂ (below) and the carbonate ion CO₃^2- are also trigonal planar, no lone pairs, but involve single and double bonds

Another mention of carbocations: There will always be a trigonal planar arrangement of the three C-X bonds around the positive carbon atom of a carbocation e.g. (CH₃)₃C⁺, but it will NOT be a planar molecular ion unless X is a single atom and the three Q's identical as in the case of hydrogen.

Does [CCl₃]⁺ exist?, if so, it will be trigonal planar in shape (diagram to do)

COCl₂ (g) carbonyl dichloride

(many other commonly used names! e.g. carbonyl chloride, carbon oxychloride, carbon dichloride oxide, phosgene-gas warfare agent).

The dot and cross diagram shows this is another example of three groups of bonding electrons (two single C-Cl bonds and a C=O double bond) giving a trigonal planar structure with Cl-C-Cl and O=C-Cl bond angles of ~120^o. The VSEPR theory predicts the correct shape based on three groups of bonding electrons and no lone pairs on electrons on the central carbon atom.

Four groups of electrons around the central atom

(see also sulfate and sulfite ions for four groups of bonding electrons involving single and double bonds)

In all the examples in these groups there are 8 electrons in the valence shell of the central atom.

the Lewis dot and cross electronic diagram used to predict the shape of water, hydrogen sulfide and the amide ion - tetrahedral electron pair geometry

the 3D shape

V or bent shaped molecule

electrons: two bond pairs and two lone pairs

Electron pair geometry different from the molecular geometry.

(molecular compounds/ions involving hydrogen)

Oxygen and sulfur in group 6/16 have six outer shell valence electrons and by pairing with two hydrogen atoms you get the stable octet shell of electrons around the central atom.

The VSEPR theory argument gives an ANGULAR or BENT or V shape: e.g. hydrogen sulfide, H₂S, or water H₂O, i.e. H₂X with H-X-H bond angle of approximately 109^o (actually 104.5^o for water) and similarly ions like the amide ion NH₂^- (Q = H, X = O, S etc. in group 6)

  The two pairs of double dots represent the lone pairs of non-bonding electrons not involved in the covalent bonding in water. Similar diagram for hydrogen sulfide H₂S.

Why isn't the H-O-H angle 109^o?  

The exact H-O-H angle in H₂O is reduced from 109.5^o to 104.5^o due to the extra repulsion of two lone pairs, the H-N-H bond angle is 107.5^o in NH₃ (one lone pair) and the H-C-H bond angle is 109^o (no lone pairs). This trend results from the 'repulsion order' lone pair-lone pair > lone pair-bond pair > bond pair-bond pair trend - details on these other molecules later.

All electrons shown on the left dot and cross diagram.

You would expect hydrogen sulfide H₂S, to have the same bent V shape and a H-S-H bond angle of ~109^o, but due to the greater repulsion of the non-bonding pairs of electrons (lone pairs) the actual H-S-H bond angle is reduced to 92^o. (this great reduction from 109^o, isn't just about the repulsion rule, but also involves the subtle bonding behaviour of the s and p orbitals - of no concern to us here at pre-university level!)

An example of an ion with a bent V shape is the negative ion, the amide ion HN₂^-.

The expected bond angle would be ~109^o, but, just as with the water molecule, the lone pair - lone pair repulsion is greater than the bond pair - bond pair repulsion giving a bond angle of 104.7^o, almost identical with water (104.5^o), and note that oxygen and nitrogen have similar atomic radii.

Note for the amide ion in the o and x diagram, prior to bonding, the two black dots of the hydrogen electrons, the five black crosses of nitrogen's valence electrons and the extra purple cross of the excess electron which gives the amide group its overall single negative charge to form the amide anion.

the Lewis dot and cross electronic diagram used to predict the shape of fluorine oxide, bromine oxide, sulfur dichloride - tetrahedral electron pair geometry

V or bent shaped molecule

electrons: two bond pairs and two lone pairs

Electron pair geometry different from the molecular geometry.

(molecular compounds/ions NOT involving hydrogen)

Oxygen and sulfur in group 6/16 have six outer shell valence electrons and by pairing with two halogen atoms you get the stable octet shell of electrons around the central atom.

The VSEPR theory argument gives a BENT or V angular shape: e.g. fluorine oxide (oxygen(II) fluoride) X = O, Q = F

OF₂ (F₂O) with F-O-F bond angle of approximately 109^o

The actual bond angle is 103^o, and this does fit in with the argument for water where the greater lone pair - lone pair > bond pair - bond pair electron repulsion reduces the F-O-F bond angle from 109^o to 103^o, but read on ....!

(X = O, Q = Cl) Cl₂O [Chlorine (I) oxide, dichlorine monoxide] is also V or bent shaped with a Cl-O-Cl bond angle of ~111^o, close to the perfect 'tetrahedral' bond angle for 4 pairs of valence shell electrons. Although lone pair - lone pair > lone pair - bond pair > bond pair - bond pair repulsion, do NOT assume, as with water, the bond angle is <109^o. The reason being that chlorine is a much more bulky atom than hydrogen (radii Cl >> H) and so the Cl - Cl electron cloud repulsion compensates for the extra lone pair - lone pair repulsion effect and the bond angle is actually ~110^o, very close to the perfect 109.5^o, but this is by coincidence! Lots of X₂O molecules on the internet are quoted as having a bond angle of 104.5^o, but unless X is hydrogen, they are usually wrong!

(X = O, Q = Br) With the even more bulky bromine atom, the Br - Br repulsion completely overrides the lone pair - lone pair repulsion effect giving a Br-O-Br bond angle of 112^o in Br₂O (dibromine monoxide, bromine(I) oxide). Another nice pattern in a changing bond angle trend (but do pre-university exam boards realise this?.

(X = S, Q = Cl) Sulfur(II) chloride (sulphur dichloride) SCl₂, will have the same dot and cross diagram (two bond pairs and two lone pairs of electrons) as OF₂ (F₂O) and Cl₂O and a predicted bent or V shaped molecule. The predicted Cl-S-Cl bond angle would be ~109^o. but the extra lone pair - lone pair repulsion lowers this to 103^o. SF₂ is similar with a bond angle of 98^o but this greatly lowered angle is not simply to do with the effect of lone pair - lone pair repulsion. There are complications due to differences in electronegativity and the way the valence orbitals are distributed and discussion of these points is well beyond pre-university chemistry!

[]⁺ is the outer electronic structure of the [F-Cl-F]⁺ ion.

Chlorine is in group 7/17 with seven outer valence shell electrons. Remove an electron and pair up two of the remaining six electrons with fluorine and you get a dot and cross diagram with two bond pairs and two pairs of non-bonding electrons. e.g. the difluorochlorinium(III) cation [ClF₂]⁺ ion is also bent shaped. You expect the bond angle to be ~109.

I've seen values of the F-Cl-F bond angle of 104.5 quoted on the internet as is if this ion is analogous to water? not sure, because F atoms are more bulky than H atoms (bigger group of electron clouds) giving increased F - F repulsion. I've also seen in a (Salters A level?) marking scheme 109^o not allowed but 107-108^o allowed. BUT, it might well be ~109^o because the F - F repulsion might compensate for the extra lone pair - lone pair repulsion OR it might be ~104.5^o like F₂O but I can't actually find an authentic value on the internet for the F-Cl-F bond angle?

Can anybody help on this one with a clear argument for a particular value?

I've found one quote of ~109^o from Google books, but this maybe just an assumption

the Lewis dot and cross electronic diagram used to predict the shape - tetrahedral electron pair geometry

pyramidal shaped molecule/ion

electrons: three bond pairs and one lone pair

Electron pair geometry different from the molecular geometry.

(molecular compounds/ions involving hydrogen)

Nitrogen and phosphorus in group 5/15 have five outer valence shell electrons, so when three electrons pair up with hydrogen you get three bond pairs and one lone pair and the central atom has a stable octet of electrons around it.

The VSEPR theory argument gives a TRIGONAL PYRAMIDAL or TRIGONAL PYRAMID shape: e.g. ammonia NH₃ with bond angle of approximately 109^o. Why isn't the H-N-H angle 109^o? The exact H-N-H angle is 107^o due to the extra repulsion of one lone pair (see below).

The double dots represent a lone pair of non-bonding electrons not involved in the covalent bonding in ammonia.

Note: the exact H-O-H angle in H₂O is 104.5^o due to the extra repulsion of two lone pairs, the H-N-H is 107.5^o in NH₃ (one lone pair) and H-C-H is 109^o (no lone pairs) because of the 'repulsion order' lone pair - bond pair > bond pair - bond pair repulsion,

 and similar for the oxonium ion H₃O⁺, H-O-H bond angle ~109^o shown below where the addition of a hydrogen ion produces three bond pairs and a lone pair of electrons.

one lone pair of electrons left on protonation, the H-O-H bond angle should be <109^o.

Theoretically calculated value in the liquid is 106.7^o, but I can't find an experimental value on the internet? This value fits in with the lone pair - bond pair repulsion > bond pair - bond pair repulsion reducing the bond angle from 109.5^o, exactly as in ammonia (107.5^o).

The shape of phosphine PH₃, would be expected to be pyramidal (trigonal pyramid) like ammonia with a bond angle of ~109., BUT, in fact the experimental H-P-H bond angle is 93.5^o, but this is only partly explained by the lone pair - bond pair repulsion > bond pair - bond pair repulsion, and, as with hydrogen sulfide, it probably involves the subtle bonding behaviour of the s and p orbitals (of no concern to us here at pre-university level!)

The methyl carbanion ion CH₃^- (negative methyl ion) will also have a trigonal pyramid shape based on three bond pairs and a lone pair. Prior to bonding the 4 black dots represent carbon's valence electrons and the crosses the electrons from the hydrogen atoms. Then, the purple dot represents the extra electron added to give the overall single negatively charged ion.

The methyl radical CH₃^. is also a pyramidal shape because the lone electron effectively acts as a 4th electron cloud (orbital) along with three bond pairs associated with carbon's valence shell electrons combining with the hydrogen electrons.  Prior to bonding the 4 black dots represent carbon's valence electrons and the crosses the electrons from the hydrogen atoms. With only 3 bond pairs, this leaves a lone electron in its own orbital of the electrically neutral free radical.

the Lewis dot and cross electronic diagram used to predict the shape - tetrahedral electron pair geometry

pyramidal shaped molecule

electrons: three bond pairs and one lone pair

Electron pair geometry different from the molecular geometry.

(molecular compounds/ions NOT involving hydrogen)

Nitrogen and phosphorus in group 5/15 have five outer valence shell electrons, so when three electrons pair up with halogen atoms you get three bond pairs and one lone pair and the central atom has a stable octet of electrons around it.

The VSEPR theory argument gives a TRIGONAL PYRAMIDAL or TRIGONAL PYRAMID shape. e.g. nitrogen trifluoride NF₃, nitrogen trichloride NCl₃, or phosphorus(III) fluoride PF₃, or phosphorus(III) chloride PCl₃ (phosphorus trifluoride/trichloride).

NF₃, NCl₃, PF₃ and PCl₃ should have bond angles Q-X-Q of approximately ~109^o (Q = F, Cl etc. X = N, P etc.)

The bond angles maybe <109^o due to the larger lone pair - bond pair > bond pair - bond pair repulsion BUT the halogen atoms are more bulky than hydrogen. Is the bond angle trend: Br-P-Br > Cl-P-Cl > F-P-F (but I can't find actual values yet?)

The Cl-N-Cl bond angle is ~107.5, similar to ammonia, presumably due to lone pair - bond pair > bond pair - bond pair repulsion, but the Cl atoms are much more bulky than H atoms, so I'm surprised it is lowered?

Not sure at all on these last two points.

the Lewis dot and cross electronic diagram used to predict the shape of methane - tetrahedral electron pair geometry

tetrahedral shaped molecule/ion

electrons: 4 bond pairs, no lone pairs

Electron pair geometry same as the molecular geometry.

(molecular compounds involving hydrogen)

Carbon and silicon are in group 4/14 with four outer valence shell electrons, each of which can pair up with the electron of a hydrogen atom to give the stable octet arrangement around the central atom. The same applies to all the other group 4/14 elements, even if they are metallic in nature.

The VSEPR theory argument gives a TETRAHEDRAL shape: e.g. methane CH₄, silicon hydride SiH₄ with H-X-H bond angle of the perfect 109.5^o and similarly ions like the ammonium ion NH₄⁺. Note: No lone pair, no extra repulsion, no reduction in angle, therefore perfect tetrahedral angle (for H-X-H angles:  CH₄ > NH₃ > H₂O, see below).

(Q = H, X = C, Si, Ge etc. in group 4)

Note: the exact H-O-H angle in H₂O is 104.5^o due to the extra repulsion of two lone pairs, the H-N-H is 107.5^o in NH₃ (one lone pair) and H-C-H is the perfect tetrahedral angle of 109.5^o (no lone pairs) because of the 'repulsion order' lone pair-lone pair > lone pair-bond pair > bond pair-bond pair.

The ammonium ion is also a perfect tetrahedral shape, so the H-N-H bond angle is109^o, identical to methane.

No lone pair of electrons left on protonation of the ammonia molecule.

Prior to protonation, you have five crosses of nitrogen's valence electrons and three dots for the bonding hydrogen electrons. The lone pair of ammonia accepts the proton in the formation of a dative covalent bond, BUT on proton bonding, all 4 N-H bonds are identical, even if you represent the molecule as H₃N:H !!

the Lewis dot and cross electronic diagram used to predict the shape of tetrachloromethane etc. - tetrahedral electron pair geometry

tetrahedral shaped molecule/ion

electrons: 4 bond pairs, no lone pairs

Electron pair geometry same as the molecular geometry.

(molecular compounds/ions NOT involving hydrogen)

Carbon and silicon are in group 4/14 with four outer valence shell electrons, each of which can pair up with the electron of a halogen atom to give the stable octet arrangement around the central atom. The same applies to all the other group 4/14 elements, even if they are metallic in nature.

The VSEPR theory argument gives a TETRAHEDRAL shape: e.g. tetrachloromethane CCl₄ or [PCl₄]⁺ with exact Cl-C-Cl and Cl-P-Cl bond angles of 109.5^o.

For CCl₄ in the dot and cross diagram the four crosses represent carbon's outer shell valence electrons and the four sets of seven dots represent the original chlorine outer shell valence electrons.

If one or more of the atoms are not identical, then there will be small deviations from the perfect Q-X-Q bond angle of 109.5^o.

X can be C, Si etc. and Q can be H, F, Cl, Br etc. e.g. CF₄, CBr₄, CI₄, SiF₄, SiCl₄ but also 'mixed' substituents e.g. CBr₂Cl₂, CHCl₃, CH₂F₂, CBr₂F₂ etc. etc. !!!

All of which will be nearly or perfect tetrahedral shaped molecules with all Q-X-Q bond angles of ~109.5^o. All involve four pairs of bonding electrons around the central C, Si or any other group 4/14 atom,

You can get a tetrahedral molecule-ion from a group 5/15 element, if one of the five outer electrons is effectively lost to make a positive ion.  In the case of the [PCl₄]⁺ ion in the dot and cross diagram the four crosses represent the remaining phosphorus outer shell valence electrons. P is in group 5/15 with five outer electrons, BUT a 'dot' has been removed from the dot and cross Lewis diagram to give the single positive charge on the ion. The four sets of seven dots represent the original chlorine outer shell valence electrons as in CCl₄ above. The phosphorous supplies the four electrons needed to produce the four P-Cl covalent bonds.