1 Bond and Lone Pairs Valence electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS.Valence electrons are distributed as shared.

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Presentation transcript:

1 Bond and Lone Pairs Valence electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS.Valence electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS. HCl lone pair (LP) shared or bond pair This is called a LEWIS structure.

2 Bond Formation A bond can result from an overlap of atomic orbitals on neighboring atoms. Cl HH + Overlap of H (1s) and Cl (2p) Note that each atom has a single, unpaired electron.

3 Review of Valence Electrons Remember from the electron chapter that valence electrons are the electrons in the OUTERMOST energy level… that’s why we did all those electron configurations!Remember from the electron chapter that valence electrons are the electrons in the OUTERMOST energy level… that’s why we did all those electron configurations! B is 1s 2 2s 2 2p 1 ; so the outer energy level is 2, and there are 2+1 = 3 electrons in level 2. These are the valence electrons!B is 1s 2 2s 2 2p 1 ; so the outer energy level is 2, and there are 2+1 = 3 electrons in level 2. These are the valence electrons! Br is [Ar] 4s 2 3d 10 4p 5 How many valence electrons are present?Br is [Ar] 4s 2 3d 10 4p 5 How many valence electrons are present?

4 Steps for Building a Dot Structure Ammonia, NH 3 1. Decide on the central atom; never H. Why? If there is a choice, the central atom is atom of lowest affinity for electrons. (Most of the time, this is the least electronegative atom…in advanced chemistry we use a thing called formal charge to determine the central atom. But that’s another story!) Therefore, N is central on this one If there is a choice, the central atom is atom of lowest affinity for electrons. (Most of the time, this is the least electronegative atom…in advanced chemistry we use a thing called formal charge to determine the central atom. But that’s another story!) Therefore, N is central on this one 2. Add up the number of valence electrons that can be used. H = 1 and N = 5 H = 1 and N = 5 Total = (3 x 1) + 5 Total = (3 x 1) + 5 = 8 electrons / 4 pairs = 8 electrons / 4 pairs

5 3.Form a single bond between the central atom and each surrounding atom (each bond takes 2 electrons!) H H H N Building a Dot Structure H H H N 4.Remaining electrons form LONE PAIRS to complete the octet as needed (or duet in the case of H). 3 BOND PAIRS and 1 LONE PAIR. Note that N has a share in 4 pairs (8 electrons), while H shares 1 pair.

6 5.Check to make sure there are 8 electrons around each atom except H. H should only have 2 electrons. This includes SHARED pairs. Building a Dot Structure 6. Also, check the number of electrons in your drawing with the number of electrons from step 2. If you have more electrons in the drawing than in step 2, you must make double or triple bonds. If you have less electrons in the drawing than in step 2, you made a mistake! H H H N

7 Carbon Dioxide, CO 2 1. Central atom = 2. Valence electrons = 3. Form bonds. 4. Place lone pairs on outer atoms. This leaves 12 electrons (6 pair). 5. Check to see that all atoms have 8 electrons around it except for H, which can have 2. C 4 e- O 6 e- X 2 O’s = 12 e- Total: 16 valence electrons

8 Carbon Dioxide, CO 2 6. There are too many electrons in our drawing. We must form DOUBLE BONDS between C and O. Instead of sharing only 1 pair, a double bond shares 2 pairs. So one pair is taken away from each atom and replaced with another bond. C 4 e- O 6 e- X 2 O’s = 12 e- Total: 16 valence electrons How many are in the drawing?

9 Double and even triple bonds are commonly observed for C, N, P, O, and S H 2 CO SO 3 C2F4C2F4C2F4C2F4

10 Now You Try One! Draw PH3

11 Sigma (  ) and Pi (  ) bond sigma bond - direct orbital overlap between the two nuclei. pi bond has orbital overlap off to the sides of the line joining the two nuclei. Sigma bonds are stronger than pi.

12 Violations of the Octet Rule Usually occurs with B and elements of higher periods. Common exceptions are: Be, B, P, S, and Xe. BF 3 SF 4 Be: 4 B: 6 P: 8 OR 10 S: 8, 10, OR 12 Xe: 8, 10, OR 12

13 MOLECULAR GEOMETRY

14 VSEPR VSEPR V alence S hell E lectron P air R epulsion theory.V alence S hell E lectron P air R epulsion theory. Most important factor in determining geometry is relative repulsion between electron pairs.Most important factor in determining geometry is relative repulsion between electron pairs. Molecule adopts the shape that minimizes the electron pair repulsions. MOLECULAR GEOMETRY

15 Some Common Geometries Linear Trigonal Planar Tetrahedral

16 VSEPR charts Use the Lewis structure to determine the geometry of the moleculeUse the Lewis structure to determine the geometry of the molecule Electron arrangement establishes the bond anglesElectron arrangement establishes the bond angles Molecule takes the shape of that portion of the electron arrangementMolecule takes the shape of that portion of the electron arrangement Charts look at the CENTRAL atom for all data!Charts look at the CENTRAL atom for all data! Think REGIONS OF ELECTRON DENSITY rather than bonds (for instance, a double bond would only be 1 region)Think REGIONS OF ELECTRON DENSITY rather than bonds (for instance, a double bond would only be 1 region)

17

18 Other VSEPR charts

19 Structure Determination by VSEPR Water, H 2 O The electron pair geometry is TETRAHEDRAL The molecular geometry is BENT. 2 bond pairs 2 lone pairs

20 Structure Determination by VSEPR Ammonia, NH 3 The electron pair geometry is tetrahedral. The MOLECULAR GEOMETRY — the positions of the atoms — is TRIGONAL PYRAMID.

21 Hybridization The orbitals mix to give hybrid orbitals The number of atomic orbitals that mix and form the hybrid orbitals are equal to the total number of electron pairs 2 electron pairs=sp hybrid 3 electron pairs=sp 2 hybrid (one s, 2 p orbitals) 4 electron pairs =sp 3 hybrid (one s, 3 p) 5 electron pairs= sp 3 d hybrid 6 electron pairs = sp 3 d 2 hybrid

22 Molecular Models Lab Chem ical form ulas Lewis Structur e Effe ctiv e pair s Bon d pair s Lone pairs Molecular Geometry(Dra w) Hybridi zation Name of Shape H2H2 O2O2 N2N2 H2OH2O PH 3 CF 4 CO 2 PCl 5 HCN CO

23 Bond Polarity HCl is POLAR because it has a positive end and a negative end. (difference in electronegativity- Cl more elctronegative) Cl has a greater share in bonding electrons than does H. Cl has slight negative charge (-  ) and H has slight positive charge (+  ) - polar covalent bond

24 Polar and Non-polar molecules Bond can be polar but because of symmetry of molecule, polarity cancels out. Ex- CCl 4 Bond is polar but molecule is not symmetrical- then molecule stays polar. Ex- H 2 O Bonds are not polar due to same electronegativity- molecule is nonpolar. Ex- CH 4.

25 Electronegativity difference and bond polarity Electronegativity difference Bond character >1.7Mostly ionic Polar covalent <0.4Mostly covalent 0Nonpolar covalent

26 Like dissolves like Polar solvents dissolve polar solutes- Example: Salt in water= Na+ Cl- Non-polar solutes dissolve in nonpolar solvents Example: Oil in CCl 4

27 This is why oil and water will not mix! Oil is nonpolar, and water is polar.This is why oil and water will not mix! Oil is nonpolar, and water is polar. The two will repel each other, and so you can not dissolve one in the otherThe two will repel each other, and so you can not dissolve one in the other Bond Polarity

28 Bond Polarity “Like Dissolves Like”“Like Dissolves Like” –Polar dissolves Polar –Nonpolar dissolves Nonpolar

29 Polarity Practice Worksheet For each of the following pairs, determine which is most polar based on Lewis structure and electronegativity. 1. CHCl 3 or CHBr 3 2. H 2 O or H 2 S 3. HCl or HI 4. C 2 HBr or C 2 HCl 5. CH 3 OH or CH 3 OCH 3 6. CH 3 C=OCH 3 or C 3 H 8 O

30 Molecular Bonding Practice For each of the following molecules: »1. H 2 S2. NCl 3 3. CHF 3 »a. Draw the Lewis dot structure and the geometry of the molecule »B. Draw the dipoles and determine the overall polarity of the molecule »C. What is the hybridization of the central atom? »D. Give the number of sigma and pi bonds »E. Build the molecule using your model kits.

31 Diatomic Elements These elements do not exist as a single atom; they always appear as pairs When atoms turn into ions, this NO LONGER HAPPENS! –Hydrogen –Nitrogen –Oxygen –Fluorine –Chlorine –Bromine –Iodine Remember: BrINClHOF