The Shapes of Molecules Chemistry II
Lewis Structures DEFINITIN: A structure of a molecule showing how the valence electrons are arranged. 1) nly the valence electrons appear in a Lewis structure.
2) The line joining two atoms represents a pair of electrons shared between two atoms. single bond - two shared electrons, one line double triple bond - six shared electrons, three lines bond - four shared electrons, two lines
3) Dots placed next to an atom represent nonbonding electrons. Procedure Step 1: Place the atoms relative to each other. If the formula is in the form AB n, place the atom with the lower group number in the center since it needs more e - to form an octet.
If atoms have the same group number, place the one with the higher period number (Lowest EN) in the center. can form only 1 bond so it is never in the center. Step 2: Determine the total number of valence electrons available.
Step 3: Draw a single bond from each surrounding atom to the central atom. then subtract 2 from the total number of valence electrons for each bond drawn. Step 4: Distribute the remaining electrons in pairs so that each atom obtains eight electrons. (2 for )
Start by placing pair of electrons around the more EN atoms to give them the stable octet. If any electrons remain, place them around the central atom. Lewis structures do not indicate shape.
Writing Lewis Structures for Molecules with ne Central Atom Write a Lewis structure for the molecule CCl 3, chloroform. Step 1: Cl C Cl Cl
Step 2: Count valence electrons. [1xC(4e - )] [1x(1e - )] [3xCl(7e - )] = 26 electrons Step 3: Draw single bonds between the atoms, and subtract 2 electrons per bond. Cl C Cl Cl
26 e - - 8 e- = 18 e-. Step 4: Distribute the remaining electron in pairs beginning with the surrounding atoms. Cl C Cl Cl
Writing Lewis Structures for Molecules with More than ne Central Atom! Write the Lewis structure for hydrogen peroxide (molecular formula, 2 2 ) an important household bleach.
Step 1. Place the atoms in the best geometry, with the hydrogen atoms having only one bond, they are on the ends or outside, and oxygen can have up to two bonds so put them in the middle.
Step 2. Find the sum of electrons: [2 x (1e - )] + [2 x (6e - )] = 14e - Step 3. Add single bonds and subtract 2e - for each bond: - - - 14e - -6e - = 8e -
Step 4. Add the remaining electron in pairs around the oxygen atoms as hydrogen can only have two! - - -
Writing Lewis Structure for Molecules with More than ne Central Atom Write the Lewis structure for methanol (molecular formula C 4 ), an important industrial alcohol that is being used as a gasoline alternative in car engines. C and must be next to each other with filling in the bonds. 4(1) + 4 + 6 = 14 valence e -. : C :
Writing Lewis Structures for Molecules with Multiple Bonds Write the Lewis structure for acetylene(c 2 2 ): Use the first 4 steps: placing atoms, counting electrons, placing single bonds, and completing octets.
-C -C - Neither of the carbon atoms has an octet, or if they are placed around one atom, the other has only 4! So no octet! Place both pairs into forming multiple bonds, a triple bond! - C C -
Lewis Structures of Simple Molecules C C 4 Methane Cl Cl Cl 2 Chlorine C C C 2 4 2 Acetic Acid C Cl Cl C Cl C C 2 Carbon Dioxide C Carbon Monoxide Cl CCl 4 Carbon Tetrachloride 2 Water ydrogen xide
.. N N.. C N.. ydrocyanic acid ydrogen Cyanide : CN Nitrogen N 2 Molecular xygen : 2 Ethylene : C 2 4................ C C
C C K + Ethyl Alcohol (Ethanol) Cl KCl 3 Potassium Chlorate Note brackets and charge CF 4 F F C F F Carbon Tetrafluoride
. N. Ammonia.. N C N.. +.... N Ammonium Ion Urea
Resonance: Symmetrical Molecules zone : 3 I II Which is the correct Lewis Structure?
Resonance ybrid Structure ne pair of electron s resonances between the two locations! The bonding pairs create what appears to be 1.5 bonds.
In single, double, and triple bonds, the e - pairs are attracted by nuclei of two bonded atoms. The e- pairs are localized. Multiple Lewis Structures are caused by e- pair delocalization.
Write resonance structures for the nitrate ion, N 3-. Nitrate has 1(5) + 3(6) + 1 = 24 valence e - N N N N does not have an octet; a pair of e - will move in to form a double bond. N N N
Benzene C C C C C C C C C C C C C C C C C C
The octet rule can be used as a guide. Some central atoms have fewer than 8 e - around them and some have more.
Electron Deficient Molecules Gaseous molecules containing Be or B often have fewer than 8 e - around them. Cl Cl Be Cl B Cl Cl
Lewis Structures for ctet Rule Exceptions F F Cl F Each fluorine atom has 8 electrons associated. Chlorine has 10 electrons! Unpaired e- is called a free radical N 2 is an odd electron atom. The nitrogen has 7 electrons.. N
Free radicals are paramagnetic and extremely reactive. Thought to be dangerous to body tissue.
Expanded Valence Shells Sometimes an atom may expand its valence shell in a process that releases energy. This is done using empty outer level d orbitals as well as the occupied p orbitals.
Expanded valence shells only occur in central atoms that are Period 3 elements or higher in which d orbitals are available.
Resonance Structures - Expanded Valence F F F S F F F Shells S = 12e - p = 10e - F F P F F Sulfur hexafluoride Phosphorous pentafluoride F S Resonance Structures Sulfuric acid S S = 12e -
Writing Lewis Structures for Exceptions to the ctet Rule. Write Lewis structures for (a) 3 P 4 and (b) BFCl 2. In (a) decide on the most likely structure. Draw the Lewis structures for the molecule and determine if there is an element which can be an exception to the octet rule. Note that (a) contains P which is a Period-3 element and can have an expanded valence shell. 0 0 0-1 P 0 (a) 3 P 4 has two resonance forms and formal charges indicate the more important form. +1 0 0 0 0 0 P 0 0 0 0 0 more stable lower formal charges (b) BFCl 2 will have only 1 Lewis structure. Cl F B Cl
Valence Shell Electron Pair Repulsion Model Lewis-Dot structures that give a reasonable picture of how the valence electrons are distributed in a molecule. Lewis-Dot structures do not indicate the shape of the molecule.
XeF 4 and CF 4 have different shapes even though the central atom is coordinated by four F atoms. Shape can be resolved by using the Valence Shell Electron Pair Repulsion Model. (VSEPR model.)
VSEPR Definition: Groups of valence electrons around a central atom stay as far away from each other as possible to minimize repulsion.
What do groups mean? X X : X= X X Triple Bond Double Bond Lone Pair Single Bond Unpaired e -
Each is a group and will repel other groups. Each will try to occupy as much space as possible around the central atom. Minimizing e - pair repulsions. The bonding and non-bonding pairs around a given atom will be positioned as far apart as possible.
There are 5 electron arrangements possible about a central atom and these e - groups will repel each other giving rise to 5 different geometries created by the groups.
Electron Groups Around Central Atom 2 Groups Linear
3 Groups Trigonal planar 4 Groups Tetrahedral
5 Groups Trigonal bipyramidal 6 Groups ctahedral
Electron-group repulsions and the five basic molecular shapes. Ideal Bond Angles
These electron group geometries give rise to identical molecular geometries if each electron pair group is a bond pair to the same atom. If some of the bonding pairs are nonbonding pairs, different molecular geometries exist for all shapes except for the linear geometry.
All molecules considered will be of three atoms or more. Simple two atomed molecules are easily understood with Lewis structures. They must be linear.
Classification Notation A = central atom X m = bonded atom E n = lone (unbonded) pair m and n are whole numbers
VSEPR Rules for Determining Molecular Shapes Draw the Lewis Structure Determine the number of lone and bonding e - pairs around the central atom. Determine the type structure.axe etc and hence the molecular geometry. Determine the bond angles.
Example (1) AX 3 E - geometry = Examples: S 3, BF 3, N 3-, C 3 2- trigonal planar X Molecular geometry = A trigonal planar X X All Bond Angles = 120 o
Example Examples: (2) AX 2 E S 2, 3, E - geometry = trigonal planar PbCl 2, SnBr 2 E Molecular geometry A = bent X X Why? Bond Angle < 120 o
Bond Angle < 120 o The lone pair is only attracted by the central nucleus and not by another nuclei.
Therefore the lone pair can expand (taking up more room) and exert a greater repulsion, squeezing the bond angle between the other 2 bonding atoms down from 120 o. This lp-bp repulsion is greater than bp-bp repulsion.
Deviations from ideal occur with the following because their electron clouds take different amounts of space around the central atom. Lone pairs Multiple bonds Different surrounding atom Unpaired electron (requires less space rare)
Effect of Double Bonds Ideal 120 0 120 0 C
Real 122 0 116 0 C Larger EN Greater Electron Density
If there is a double bond, it should affect the bond angle, however N 3 - has three equivalent resonance structures, hence bond angle remains at 120 o. The double bond N could be in any of three positions.
Example (3) AX 4 E - geometry = Examples: C 4, SiCl 4, S 4 2-, Cl 4 - tetrahedral Molecular geometry = tetrahedral All Bond Angles = 109 o
Example Examples: (4) AX 3 E N 3, PF 3 E - geometry = Cl 3, 3 + tetrahedral Molecular geometry = trigonal pyramidal Bond Angles = 107. 3 o
Example (5) AX 2 E 2 E - geometry = Examples: 2, F 2 SCl 2 tetrahedral Molecular geometry = Bent V shaped Bond Angles = 104.5
Example (6) AX 5 E - geometry = trigonal Examples: PF 5, AsF 5 SF 4 bipyramidal Molecular geometry =trigonal bipyramidal Bond Angles = Axial = 180 o Equatorial = 120 o
Example Examples: (7) AX 4 E SF 4,Xe 2 F 2 E - geometry = trigonal IF 4+, I 2 F 2 - bipyramidal Molecular geometry = see saw Bond Angles reduce to 86.8 o and 101.5 o respectively
Example (8) AX 3 E 2 Examples: ClF 3, BrF 3 E - geometry = trigonal bipyramidal Molecular geometry = T - Shaped Bond Angles: Two 90 0 and one 180 o become two @ 86.2 o and 172.4 o
Example (9) AX 3 E 2 Examples: XeF 2, I 3-, IF 2 - E - geometry = trigonal bipyramidal Molecular geometry = Linear Bond Angles: 180 o
Example (10) AX 6 Examples: SF 6, IF 5 E - geometry = ctahedral Molecular geometry = ctahedral Bond Angles: six 90 o and three 180 o
Example Examples: (11) AX 5 E BrF 5, TeF 5-, XeF 4 E - geometry = ctahedral Molecular geometry = Square Pyramidal Bond Angles: Many 90 o and 180 o are reduced to less than ideal.
Example (12) AX 4 E 2 Examples: XeF 4. ICl 4 - E - geometry = ctahedral Molecular geometry = Square Planar Bond Angles: 2 180 o and 4 90 o
Predicting Molecular Shapes with Two, Three, or Four E - Groups Draw the molecular shape and predict the bond angles (relative to the ideal bond angles) of PF 3 and CCl 2. For PF 3 - there are 26 F P F valence electrons, 1 F nonbonding pair.
The shape is based upon the tetrahedral arrangement. The F-P-F bond angles should be <109.5 0 due to F P F F the repulsion of the <109.5 0 nonbonding electron pair. The type of shape is AX 3 E. The final shape is trigonal pyramidal.
For CCl 2, C has the lowest EN and will be the center atom. There are 24 valence e -, 3 atoms attached to the center atom. Cl C Cl
C does not have an octet; a pair of nonbonding electrons will move in C from the to make a double bond. Cl Cl The shape for an atom with three atom attachments and no nonbonding pairs on the central atom is trigonal planar.
Type AX 3 The Cl-C-Cl bond angle will be less than 120 0 due to the electron density of the C=. 124.5 0 Cl C 111 0 Cl
Predicting Molecular Shapes with Five or Six Electron Groups Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of SbF 5 and BrF 5. SbF 5-40 valence e - ; all electrons around central atom will be in bonding pairs.
F F F Sb F F Shape is AX 5 - trigonal bipyramidal.
BrF 5-42 valence e - ; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX 5 E, square pyramidal. F F F Br F F
Molecular Geometry CCl 4 S 2 gbr 2 C 2 S 3-2 BF 3 Xe 4 AX 4 AX 2 E AX 2 AX 2 AX 3 E AX 3 AX 4 Tetrahedral Bent Linear Linear Trig. Pyramidal Trig. Planar Tetrahedral
Molecular Geometry ICl 3 SeF 6 IF 4 + AsCl 5 I 3-1 I 2 F 2-1 XeF 4 AX 3 E 2 AX 6 AX 4 E AX 5 AX 2 E 3 AX 4 E AX 4 E 2 T-shaped ctahedral See-saw Trig. Pyramidal T-shaped See-saw Square Planar
The End