(c) doc bDoc 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.  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.

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

SHAPES OF MOLECULES INDEX: introduction * diatomic molecules * Shapes based on groups of electron pairs (bonding/non-bonding) : 2  3  4  5  6 * some more complex inorganic/organic molecules/ions * transition metal complexes * Some other molecules/ions of carbon, nitrogen, sulfur and chlorine * shapes and bond angles of organic molecules * GCSE/IGCSE/AS Science-CHEMISTRY bonding notes

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.
    • This effectively means to produce as wide as possible bond angle e.g. X-A-X angle
  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 hydrogen and A is nitrogen or oxygen, because larger X atoms or more bulky X groups might override the valence shell electron pair rule of 2. This tends to be overlooked by exam boards 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.

These relative 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, simple molecules and ions with three to six electron pairs (including lone pairs of electrons) surrounding the central atom.

Introduction - electron pair repulsion theory and bond angle

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.

This is known as The VALENCE SHELL ELECTRON PAIR REPULSION THEORY MODEL (VSEPR theory, valence shell electron pair repulsion).

It 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 on 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)

underdeveloped test! on shapes and angles
 

 

Diatomic molecules

These are not considered to have a 'shape' or a 'bond angle', but they are useful dot and cross diagram revision based on the outer valence electrons and help you to construct Lewis dot and cross diagrams for molecules with >2 atoms.

(c) doc b H-H e.g. hydrogen H2

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H-Cl e.g. hydrogen chloride HCl, HX in general where X = halogen

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(c) doc b Cl-Cl e.g. chlorine Cl2, iodine(I) chloride ICl (iodine monochloride)

(c) doc b Q and X are both halogen atoms from group 7

  O=O (c) doc b oxygen molecule

  

Two groups of electrons around the central atom

two bonding pairs of electrons or two double bond pairs give a linear shape and bond angle of 180o

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H-Be-H

gaseous beryllium hydride BeH2 (Q = H, X = Be)

linear shaped molecule (VSEPR theory argument)

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Cl--Be-Cl

gaseous beryllium halides BeCl2 (X = Be, Q = F, Br, Cl, I)

Similarly the VSEPR theory argument gives a linear shaped molecule for gaseous beryllium fluoride, beryllium chloride, beryllium bromide and gaseous beryllium iodide too I presume?

valence bond dot and cross diagrams

O=C=O

(c) doc b carbon dioxide CO2

full dot and cross electron structure

Here you have two groups of bonding electrons either side of a central atom in the carbon dioxide triatomic molecule producing a linear shape. Double bond pairs count just the same as single bond pairs in determining the shape of a molecule.

[H3N-Ag-NH3]+ [Ag(NH3)2]+

transition metal complex of co-ordination number 2: e.g. the diamminesilver(I) ion, [Ag(NH3)2]+, where the :NH3 ammonia molecule acts as an electron pair donor to form the dative covalent bond. Its not a true linear shaped complex ion because of the pyramidal ammonia ligands but the N-Ag-N bond is linear i.e. bond angle 180o.

[F-Cl-F]-  

Can you show this has two bond pairs and three lone pairs around the central chlorine atom and is a linear molecule with an F-Cl-F bond angle of 180o!

  

Three groups of electrons around the central atom

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electrons: two bond pairs, one lone pair

shape of molecule BENT, bond angle approximately 120o

Does anyone know of any example?

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electrons: two bond pairs, one lone pair

molecule shape BENT, bond angle approximately 120o

Does anyone know of an example? but for X=Q double bonds see sulfur dioxide

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electrons: 3 bond pairs, no lone pairs

(molecular compounds/ions involving hydrogen)

(c) doc bThe VSEPR theory argument gives the shape TRIGONAL PLANAR: Q-X-Q bond angle exactly 120o: 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.

Carbon in group 4/14 has four valence electrons, one of which is lost in the formation of carbocation (carbonium ion). 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 120o. 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.

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 molecule unless X is a single atom and the three X's identical as in the case of hydrogen.

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electrons: 3 bond pairs, no lone pairs

(molecular compounds/ions NOT involving hydrogen)

The VSEPR theory argument gives the molecule shape TRIGONAL PLANAR: bond angle, 120o: e.g. gaseous boron trifluoride BF3, boron trichloride BCl3 and aluminium fluoride AlF3 (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 or Al for F)

but for X=Q double bonds see sulfur trioxide

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 ~120o.

     

 

Four groups of electrons around the central atom

(see also sulfate and sulfite ions)

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electrons: two bond pairs and two lone pairs

(molecular compounds/ions involving hydrogen)

Oxygen and sulfur in group 6/16 have six outer shell valence electrons and by pairing with two hydrogens you get the stable shell octet.

(c) doc b 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 109o (actually 104.5o for water) and similarly ions like the amide ion NH2- (Q = H, X = O, S etc. in group 6)

(c) doc b  (c) doc b 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 109o?  

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

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 ~109o, 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 92o. (this great reduction from 109o, 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 ~109o, 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 104.7o, almost identical with water (104.5o), 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.

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electrons: two bond pairs and two lone pairs

(molecular compounds/ions NOT involving hydrogen)

Oxygen and sulfur in group 6/16 have six outer shell valence electrons and by pairing with two halogens you get the stable shell octet.

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

(c) doc b F2O with F-O-F bond angle of approximately 109o

The actual bond angle is 103o, 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 109o to 103o, but read on ....!

(c) doc bCl2O [Chlorine (I) oxide, dichlorine monoxide] is also V or bent shaped with a Cl-O-Cl bond angle of ~111o, 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 <109o. 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 ~110o, very close to the perfect 109.5o, but this is by coincidence! Lots of X2O molecules on the internet are quoted as having a bond angle of 104.5o, but unless X is hydrogen, they are usually wrong!

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 112o in Br2O (dibromine monoxide, bromine(I) oxide). Another nice pattern in a changing bond angle trend (but do pre-university exam boards realise this?.

(c) doc bSulfur(II) chloride (sulphur dichloride) SCl2, will have the same dot and cross diagram (two bond pairs and two lone pairs of electrons) as F2O and Cl2O and a predicted bent or V shaped molecule. The predicted Cl-S-Cl bond angle would be ~109o. but the extra lone pair - lone pair repulsion lowers this to 103o. SF2 is similar with a bond angle of 98o 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!

[ (c) doc b ]+ type of electronic structure

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 109o not allowed but 107-108 allowed. BUT, it might well be ~109o because the F - F repulsion might compensate for the extra lone pair - lone pair repulsion OR it might be ~104.5o 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 ~109o from Google books, but this maybe just an assumption

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electrons: three bond pairs and one lone pair

(molecular compounds/ions involving hydrogen)

Nitrogen and phosphorus in group 5/15 have five outer valence shell electrons, when three pair up with hydrogen you get three bond pairs and one lone pair.

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

(c) doc b (c) doc b 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 104.5o due to the extra repulsion of two lone pairs, the H-N-H is 107.5o in NH3 (one lone pair) and H-C-H is 109o (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 ~109o 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 <109o.

Theoretically calculated value in the liquid is 106.7o, 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 109.5o, exactly as in ammonia (107.5o).

(c) doc b 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 93.5o, 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.

(c) doc b (c) doc b

electrons: three bond pairs and one lone pair

(molecular compounds/ions NOT involving hydrogen)

Nitrogen and phosphorus in group 5/15 have five outer valence shell electrons, when three pair up with a halogen you get three bond pairs and one lone pair.

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 ~109o (Q = F, Cl etc. X = N, P etc.)

The bond angles maybe <109o due to the larger lone pair - bond pair > bond pair - bond pair repulsion BUT the halogen atoms are more bulky than hydrogen. This partly explains the following 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.

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electrons: 4 bond pairs

(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.

The VSEPR theory argument gives a TETRAHEDRAL shape: e.g. methane CH4, silicon hydride SiH4 with H-X-H bond angle of the perfect 109.5o 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).

(c) doc b alkanes structure and naming (c) doc b (Q = H, X = C, Si, Ge etc. in group 4)

Note: the exact H-O-H angle in H2O is 104.5o due to the extra repulsion of two lone pairs, the H-N-H is 107.5o in NH3 (one lone pair) and H-C-H is the perfect tetrahedral angle of 109.5o (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 is109o, 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 !!

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electrons: 4 bond pairs

(molecular compounds/ions NOT involving hydrogen)

Carbon is in group 4/14 with four outer valence shell electrons which can pair up with group 7/17 halogens which are one electron short of the stable octet shell electron arrangement.

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 109.5o.

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 109.5o.

X can be C, Si etc. and Q can be F, Cl, Br etc. e.g. CF4, CBr4, CI4, SiF4, SiCl4 etc.

All of which will be perfect tetrahedral shaped molecules with all Q-X-Q bond angles of 109.5o.

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 'dot' has been removed 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.

 

Five groups of electrons around the central atom

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valence bond dot and cross diagram

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electrons: 5 bond pairs

Phosphorus in group 5/15 has five outer shell valence electrons, all of which can pair up with a halogen atom such as fluorine and chlorine.

Note that this expands the valence shell electrons of bonded phosphorus to ten, two beyond the 'octet rule'.

The VSEPR theory argument gives a TRIGONAL BIPYRAMID shape: e.g. for phosphorus(V) fluoride (phosphorus pentafluoride) PF5, gaseous phosphorus(V) chloride, PCl5, with bond angles on 90o and 180o based on the vertical and right-angled Q-X-Q bonds and bond angles of 120o based on the central trigonal planar arrangement.

In the dot and cross diagram on the left, X = P or As (but NOT N, can't use 3d level as a valence shell) and Q = F, Cl and Br (not I).

Note that solid PCl5 has an ionic structure and is not a trigonal bipyramidal (bipyramid) molecule - a 4 bond pair tetrahedral [PCl4]+ ion and a 6 bond pair octahedral [PCl6]- ion.

  

Six groups of electrons around the central atom

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electrons: 6 bond pairs

The VSEPR theory argument gives a OCTAHEDRAL SHAPE: e.g. sulfur(VI) fluoride (sulfur hexafluoride molecule) SF6

or the [PCl6]- ion and many transition metal complexes (see below), with Q-X-Q bond angles of 90o and 180o.

In the dot and cross diagram X = S or P and Q = F or Cl

Prior to bonding, for the SF6 molecule the six dots represent the six valence shell electrons of sulfur (group 6/16) and the six sets of seven crosses represent the seven valence electrons of the fluorine atoms.

For the [PCl6]- ion the six dots represent the original five valence electrons of phosphorus (group 5/15) plus one extra dot for the extra electron that gives the ion its single overall negative charge. Similarly to above, the six sets of seven crosses represent the seven valence electrons of the chlorine atoms. You should put the dot and cross diagram in big square brackets and a superscript minus sign.

Note that this expands the valence shell electrons of bonded phosphorus or sulfur to twelve, four beyond the 'octet rule'.

  

More complex inorganic molecules/ions and organic molecules

These often are not given a particular shape name, but never-the-less, an appreciation of the 3D spatial arrangement is expected e.g.

(c) doc b

Ethane consists of two joined 'tetrahedral halves', with all C-C-H and H-C-H bond angles of 109o.

See other page for more shape and bond angle analysis of organic molecules

 

H3N:=>BF3

would be like ethane above

Boron trifluoride with 3 bonding pairs acts as a lone pair acceptor (Lewis acid) and ammonia with a lone pair of electrons enables it to act as a Lewis base - a an electron pair donor. It donates the lone pair to the 4th 'vacant' boron orbital to form a sort of 'adduct' compound. This gives a stable 'octet' of electrons around both the nitrogen and boron atoms.

Its shape is essentially the same as ethane, a sort of double tetrahedral with H-N-H, N-B-F and F-B-F bond angles of ~109o.

(c) doc b or (c) doc b

Benzene is a completely planar molecule, with all C-C-C or C-C-H bond angles of 120o.

See other page for more on shape and bond angle analysis of organic molecules

 

TRANSITION METAL COMPLEX IONS

COMPLEXES(c) doc b

The three examples below show cis/trans isomerism

(c) doc bcis/trans octahedral

(c) doc bcis/trans octahedral

(c) doc b cis/trans square planar

more details and examples on the "Transition Metals" pages

All the bonds shown,__ or ...., are dative covalent, with lone electron pair donation by the ligand L, to the central metal ion i.e. L: (c) doc b Mn+ etc.


The shapes and bond angles of BeH2 BeCl2 CO2 [Ag(NH3)2]+ BH3 BF3 BCl3 AlF3 COCl2 H2O H2S NH3 F2O PF3 PF5 PCl3 PCl5 H3O+ NCl3 CH4 CCl4 PCl4+ PCl6- SF6 H3NBF3 NH3BF3 dot and cross diagrams bond angles H-B-H VSEPR molecule shape of BH3 bond angles H-C-H VSEPR molecule shape of CH3+ bond angles F-B-F VSEPR molecule shape of BF3 bond angles Cl-B-Cl VSEPR molecule shape of BCl3 bond angles F-Al-F VSEPR molecule shape of AlF3 bond angles VSEPR molecule shape of COCl2 bond angles VSEPR molecule shape of H2S bond angles H-O-H VSEPR molecule shape of H2O bond angles H-N-H VSEPR molecule shape of NH2- bond angles VSEPR molecule shape of [NH2]- bond angles F-O-F VSEPR molecule shape of F2O bond angles Cl-O-Cl VSEPR molecule shape of Cl2O bond angles Br-O-Br VSEPR molecule shape of Br2O bond angles Cl-S-Cl VSEPR molecule shape of SCl2 bond angles F-Cl-F VSEPR molecule shape of ClF2+ bond angles VSEPR molecule shape of [ClF2]+ bond angles H-N-H VSEPR molecule shape of NH3 bond angles H-P-H VSEPR molecule shape of PH3 bond angles H-O-H VSEPR molecule shape of H3O+ bond angles VSEPR molecule shape of [H3O]+ bond angles H-C-H VSEPR molecule shape of CH3- bond angles VSEPR molecule shape of [CH3]- bond angles VSEPR molecule shape of CH3. radical bond angles Cl-N-Cl VSEPR molecule shape of NCl3 bond angles F-N-F VSEPR molecule shape of NF3 bond angles F-P-F VSEPR molecule shape of PF3 bond angles Cl-P-Cl VSEPR molecule shape of PCl3 bond angles H-C-H VSEPR molecule shape of CH4 bond angles H-Si-H VSEPR molecule shape of SiH4 bond angles H-N-H VSEPR molecule shape of NH4+ bond angles VSEPR molecule shape of [NH4]+ bond angles Cl-C-Cl VSEPR molecule shape of CCl4 bond angles F-C-F VSEPR molecule shape of CF4 bond angles Br-C-Br VSEPR molecule shape of CBr4 bond angles I-C-I VSEPR molecule shape of CI4 bond angles Cl-P-Cl VSEPR molecule shape of PCl4+ bond angles Cl-P-Cl VSEPR molecule shape of [PCl4]+ bond angles F-P-F VSEPR molecule shape of PF5 bond angles Cl-P-Cl VSEPR molecule shape of PCl5 bond angles F-S-F VSEPR molecule shape of SF6 bond angles Cl-P-Cl VSEPR molecule shape of PCl6- bond angles VSEPR molecule shape of [PCl6]- dot and cross diagrams

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