# Shapes of Molecules and Bonding

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1 Shapes of Molecules and onding Molecular geometry is governed by energy. Molecules receive such geometry as to minimize their potential energy. A striking example is DA. Lewis Dot Structures 1. VAL (total number of valence electrons) = sum of the valence electrons of each element (sum of the group numbers, G) charge (of an ion) 2. STA (stable noble gas configuration) = sum of the electrons in the noble gas configuration for each element (2 for, 8 for and beyond). Since an atom achieves its greatest stability and lowest energy when it has a filled shell, the noble gas configuration of ns 2 np 6, we assume that an atom in a molecule or ion will also be most stable when it has its noble gas complement of electrons around it the ubiquitous octet rule. 3. D (total number of bonding electrons) = STA VAL and P (number of bond pairs) = D/2. Since electrons are shared between atoms that are bonded together, the octet around each atom over counts the actual numbers of electrons involved. The amount of over counting is the number of bonding electrons. 4. LE (total number of lone pair electrons) = VAL D and LP (number of lone pairs) = LE/2. Valence electrons are of two varieties: they are bonding electrons (D) or nonbonding electrons, lone electrons (LE). 5. Draw the Lewis structure using your calculated P and LP. Remember that nature likes symmetry, put the odd atom in the center. Draw all resonance structures. 6. () formal charge on atom) = G LE D/2. The sum of the formal charges on all atoms is equal to the charge on the molecule or ion. 7. Valence shell expands to reduce and maximize resonance if atom is in the third period or beyond. Example 1

2 3 VAL: 5 () + (3x7) () = 26 e STA: 8 () + 3 (8) = 32 e D: STA VAL: = 6 e [/2 = 3 P] LE: VAL D: 26 6 = 2 e.... : :.. : : : :.. Example Write a Lewis structure for 2 2, one of the compounds responsible for the depletion of stratospheric ozone VAL: 4() + 2 (7) () + 2 (7) () = 32 e STA: 5(8) = 4 e D: STA VAL: 4 32 = 8 e [/2 = 4 P] LE: VAL D: 32 8 = 16 e arbon has the lowest E and is the central atom. The other atoms are placed around it. Make bonds and fill in remaining valence electrons placing 8 e around each atom Make 4 bonds and fill in remaining valence electrons placing 8e- around each atom. Example Write the Lewis structure for methanol, (molecular formula 4 ), an important industrial alcohol that is being used as a gasoline alternative in car engines. 2

3 VAL: 4(1)() + 4() + 6 () = 14 valence e STA: 4(2)+2(8) = 24 e D: STA VAL: = 1 e [/2 = 5 P] LE: VAL D: 14 1 = 4 e ydrogen can have only one bond so and must be next to each other with filling in the bonds..... In reality this is what do in order to convert a molecular formula to a Lewis dot structure 3

4 Writing Lewis Structures for Molecules with Multiple onds PRLEM: Write Lewis structures for the following: (a) Ethylene ( 2 4 ), the most important reactant in the manufacture of polymers (b) itrogen ( 2 ), the most abundant atmospheric gas PLA: or molecules with multiple bonds, there is a Step 5 which follows the other steps in Lewis structure construction. If a central atom does not have 8e -, an octet, then two e - (either single or nonbonded pair) can be moved in to form a multiple bond. SLUTI: (a) There are 2(4) + 4(1) = 12 valence e -. can have only one bond per atom. (b) 2 has 2(5) = 1 valence e -. Therefore, a triple bond is required to make the octet around each. :.. : : :.... : : Resonance: Delocalized Electron-Pair onding 3 can be drawn in 2 ways - A A either structure depicts 3 accurately but can be drawn to represent a structure which is a hybrid of the two - a resonance structure. A Resonance structures have the same relative atom placement but a difference in the locations of bonding and nonbonding electron pairs. is used to indicate that resonance A 4

5 Writing Resonance Structures PRLEM: Write resonance structures for the nitrate ion, 3 -. itrate has 1(5) + 3(6) + 1 = 24 valence e - does not have an octet; a pair of e - will move in to form a double bond. 5

6 ormal harge: Selecting the est Resonance Structure An atom owns all of its nonbonding electrons and half of its bonding electrons. ormal charge is the charge an atom would have if the bonding electrons were shared equally. ormal charge of atom = # of valence e - (# of unshared electrons + 1 # of shared electrons) 2 or A or # valence e - = 6 # valence e - = 6 # nonbonding e - = 4 A # nonbonding e - = 6 or # bonding e - # bonding e = 2 x = 4 x = # valence e - = 6 ormal charge = -1 ormal charge = = 1 # nonbonding e - = 2 # bonding e - 1 = 6 x 2 = 3 ormal charge = +1 Selecting the est Resonance Structure Three criteria for choosing the more important resonance structure: Smaller formal charges (either positive or negative) are preferable to larger charges. Avoid like charges (+ + or - -) on adjacent atoms. A more negative formal charge should exist on a more electronegative atom (an atom with a larger E value). 6

7 ormal harge: Selecting the est Resonance Structure EXAMPLE: - has 3 possible resonance forms: A formal charges orms and have negative formal charges on and ; this makes them more preferred than form A. orm has a negative charge on which is the more electronegative element, therefore, contributes the most to the resonance hybrid. Write Lewis structures for (a) 3 P 4 (pick the most likely structure); (b) 2. a) VAL: 3(1)() + 5(P) + (4)(6)() = 32 valence e STA: 3(2)+5(8) = 46 e D: STA VAL: = 14 e [/2 = 7 P] LE: VAL D: = 18 e P is a Period-3 element and can have an expanded valence shell. Lower -1 P +1 P 7

8 b) 2 will have only 1 Lewis structure. Valence Shell Electron Pair Repulsion Geometries (VSEPR Model) This model is based on the idea that bond and lone paira in the valence shell of an element repel each other and seek to be as far apart as possible, It is remarkably successful to predict structures of molecules and ions of main group elements It is less effective and it is used seldom to predict structures of compounds containing transition metal elements. 1. Repulsions P/P < P/LP < LP/LP 2. umber of electron pairs (steric number) electronic geometry 3. Molecular geometry (shape) bond angles, distortions 4. Dipole moment, polarity 8

9 umber of electron pairs Geometry (bond angle) umber of lone pairs 2 Linear (18 o ) 3 Trigoanal planar (12 o ) ent 1 4 Tetrahedral (19.47 o ) Trigonal pyramidal 1 ent 2 5 Triginal bipyramidal (9 o, 12 o ) seesaw 1 T-shaped 2 Linear 3 6 ctahedral (9 o ) Square Pyramidal 1 Square Planar 2 A - entral atom X - Surrounding atom E - onbonding valence electron-group 9

10 The single molecular shape of the linear electron-group arrangement The two molecular shapes of the trigonal planar electron-group arrangement. 1

11 actors Affecting Actual ond Angles ond angles are consistent with theoretical angles when the atoms attached to the central atom are the same and when all electrons are bonding electrons of the same order. Effect of Double onds ideal 12 o Effect of onbonding (Lone) Pairs 12 o larger E greater electron density 116 o 122 o real Lone pairs repel bonding pairs more strongly than bonding pairs repel each other. Sn The three molecular shapes of the tetrahedral electron-group arrangement 95 o Lewis structures and molecular shapes. 11

12 The four molecular shapes of the trigonal bipyramidal electron-group arrangement The three molecular shapes of the octahedral electrongroup arrangement 12

13 The steps in determining a molecular shape Molecular formula Lewis Dot Structure electron group arrangement note positions of lone pairs and double bonds count bonding and non-bonding electrons molecular shape Predicting Molecular Shapes with Two, Three, or our Electron Groups PRLEM: Draw the molecular shape and predict the bond angles (relative to the ideal angles) of (a) P 3 and (b) 2. SLUTI: (a) or P 3 - there are 26 valence electrons, 1 nonbonding pair P The shape is based upon the tetrahedral arrangement. P <19.5 o The -P- bond angles should be <19.5 o due to the repulsion of the nonbonding electron pair. The final shape is trigonal pyramidal. The type of shape is AX 3 E (b) or 2, has the lowest E and will be the center atom. There are 24 valence e -, 3 atoms attached to the center atom. does not have an octet; a pair of nonbonding electrons will move in from the to make a double bond. Type AX 3 The shape for a molecule with a central atom having three attachments and no nonbonding pairs is trigonal planar. The -- bond angle will be less than 12 o due to the electron density of the = o 111 o 13

14 Predicting Molecular Shapes with ive or Six Electron Groups PRLEM: SLUTI: Sb Determine the molecular shape and predict the bond angles (relative to the ideal angles) of (a) Sb 5 and (b) r 5. (a) Sb 5-4 valence e - ; all electrons around central atom will be in bonding pairs; shape is AX 5, trigonal bipyramidal. Sb (b) r 5-42 valence e - ; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX 5 E, square pyramidal. The orientation of polar molecules in an electric field r 14

15 Predicting the Polarity of Molecules PRLEM: rom electronegativity (E) values and their periodic trends predict whether each of the following molecules is polar and show the direction of bond dipoles and the overall molecular dipole when applicable: (a) Ammonia, 3 (b) oron trifluoride, 3 (c) arbonyl sulfide, S (atom sequence S) PLA: Draw the shape, find the E values and combine the concepts to determine the polarity. SLUTI: (a) 3 E = 3. E = 2.1 bond dipoles molecular dipole The dipoles reinforce each other, so the overall molecule is definitely polar. (b) 3 has 24 valence e - and all electrons around the will be involved in bonds. The shape is AX 3, trigonal planar. 12 o (E 4.) is more electronegative than (E 2.) and all of the dipoles will be directed from to. ecause all are at the same angle and of the same magnitude, the molecule is nonpolar. (c) S is linear. and S have the same E (2.) but the = bond is quite polar ( E) so the molecule is polar overall. S 15

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