Worksheet 1Structure And Bonding PDF

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Worksheet 1: Structure and Bonding General Principles Organic compounds are compounds based on carbon. There are several ways to represent organic compounds this includes Lewis structures, condensed structures, kekule structures, skeletal structures, and resonance structures. 1. Lewis Structures Lewis structures indicate which atoms are bonded together and show lone pairs and formal charges. Number of covalent bonds an atom forms depends on the number of additional valence electrons it needs to reach a stable octet  Hydrogen has one valence electron (1s1), and thus, forms one bond.  Carbon has four valence electrons (2s2 2p2), and thus, forms four bonds.  Nitrogen has five valence electrons (2s2 2p3), and thus, forms three bonds.  Oxygen has six valence electrons (2s2 2p4), and thus, forms two bonds.  Halogens have seven valence electrons (2s2 2p5), and thus, form one bond. a) The number of bonds plus the number of lone pairs equals four

b) If carbon does not form four bonds, it has a charge (or it is a radical).

c) Nitrogen has one lone pair. If nitrogen does not form three bonds, it is charged.

d) Oxygen has two lone pairs. If oxygen does not form two bonds, it is charged.

e) A halogen has three lone pairs. If hydrogen or halogen does not form one bond, it has a charge (or it is a radical).

f)

If the number of electrons used is greater than the number of valence electrons of the atoms, make multiple bonds (double and triple bonds) while keeping octets.

Worksheet 1: Structure and Bonding

g) The charge of an atom in a Lewis structure is calculated using this formula: Formal charge = # of valence electrons- (all the lone-pair electrons and half of the bonding electrons). h) A single bond consists of a sigma bond (), a double bond consists of a one sigma () and one pi bond (), a triple bond consists of one sigma () and two pi bonds (). i) A  bond results from an axial overlap of two orbitals, a  bond results from the side-by-side overlap of two parallel p orbitals. 2. Hybridization, bond angle and Molecular Geometry The hybridization of C, N, or O depends on the number of  bonds the atom forms: no  bonds = sp3, one  bond = sp2, and two  bonds = sp. Exceptions are carbocations and carbon radicals, which are sp2.  The greater the electron density in the region of orbital overlap, the shorter and stronger the bond.  The more s character in the orbital used to form a bond, the shorter and stronger the bond and the larger the bond angle.  The shorter the bond, the stronger it is.  Triple bonds are shorter and stronger than double bonds, which are shorter and stronger than single bonds.  A  bond is weaker than a  bond.

Worksheet 1: Structure and Bonding 3. Bond Strength and Length

4. Resonance Structures Some molecules have structures that cannot be shown with a single representation. They are represented by structures that contribute to the final structure but differ in the position of the  bond or lone pairs. Such structures are delocalized and are represented by resonance forms. Resonance forms differ only in the placement of their  or nonbonding electrons. Individual resonance forms are imaginary, and the real structure is a hybrid of different forms. Different resonance forms of a substance do not have to be equivalent. When two resonance forms are non-equivalent, the actual structure of the resonance hybrid resembles the more stable form. Resonance hybrid is more stable than any individual resonance form. Resonance leads to stability. When writing resonance structures, we used curved arrow. Curved arrow indicates movement of electrons, not of the atoms. Rules for Drawing Resonance Contributors  Only electrons move. Atoms never move.  Only π electrons (electrons in π bonds) and lone-pair electrons can move. (σ electrons never move).  The total number of electrons in the molecule does not change.  Electrons always move to an sp2 or sp atom; sp3 atoms cannot accept electrons because they already have an octet.

Worksheet 1: Structure and Bonding

Features that decrease the predicted stability of resonance structures  An atom with an incomplete octet.  A negative charge that is not on the most electronegative atom.  A positive charge that is on an electronegative atom.  Separated charges.

B has separated charges

F is less stable (has an incomplete octet) .

C and D have similar stabilities

H has a positive charge on an electronegative atom

J is more stable cause it has a negative charge on the more electronegative atom (O), whereas I has negative charge on carbon. 5. More structures of Organic Compounds Several shorthand methods have been developed to write structures

Worksheet 1: Structure and Bonding Condensed structures C-H or C-C single bonds are not shown, they are understood

Example Skeletal structures Carbon atoms aren’t usually shown. Carbon atom is assumed to be at each intersection of two lines (bonds) and at the end of each line. Hydrogen atoms bonded to carbon aren’t shown. Atoms other than carbon and hydrogen are shown.

Worksheet 1: Structure and Bonding In this exercise, you will write Lewis structures, determine formal charges, hybridization, and molecular shapes of some organic compounds. You will also write resonance structures, condensed, and skeletal structures. Answer the following questions on your lab notebook and submit the carbon copy. Questions 1. There are two different substances with each of the following formulas: C4H10, C3H6 and C2H6O. Draw the Lewis structure of each formula. (Show all lone pairs) 2.

Oxaloacetic acid, an important intermediate in food metabolism, has the formula C4H4O5 and contains three C=O bonds and two O-H bonds. Propose the two most stable possible Lewis structures. (Show all lone pairs)

3. Draw the structures for the following molecules (Show all lone pairs) : a) Acrylonitrile, C3H3N, which contains a carbon-carbon double bond and a carbon-nitrogen triple bond. b) Ethyl methyl ether, C3H8O, which contains an oxygen atom bonded to two carbons. c) Cyclohexene, C6H10, which contains a ring of six atoms and one carbon-carbon double bond. d) Determine the hybridization, the shape, the bond angle and the polarity of each of the structures.

4. How many sigma and pi bonds are in each of the following compounds? Give the hybridization of each carbon atom.

a) H2C=C=CH2 b) c) 5. Give the hybridization, the bond angle, and the shape around each of the indicated carbons below.

6.

Classify the following bonds from the shortest to the longest. Explain!

7.

Convert the following molecular models into a condensed structure and a skeletal structure.

a)

8. a) Convert the molecules below into skeletal structures.

b)

c)

Worksheet 1: Structure and Bonding

(I)

(II)

b) Indicate the hybridization of each of the central atoms. 9. For each of the indicated atoms in Cipro®. a) Give the molecular shape. b) Determine the bond angle. c) Predict the hybridization

10. Calculate the formal charges on the indicated atoms in each compound below.

(I)

(II)

(III)

11. Draw resonance structures for each of the species below. Show all lone pairs and the arrows. Circle the most stable

(A)

(B)

(C)

(D)...