5.111 Principles of Chemical Science

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1 MIT penourseware Principles of hemical Science Fall 2008 For information about citing these materials or our Terms of Use, visit:

2 5.111 Lecture Summary #12 Readings for today: Section 2.9 (2.10 in 3 rd ed), Section 2.10 (2.11 in 3 rd ed), Section 2.11 (2.12 in 3 rd ed), Section 2.3 (2.1 in 3 rd ed), Section 2.12 (2.13 in 3 rd ed). Read for Lecture #13: Section 3.1 (3 rd or 4 th ed) The Basic VSEPR Model, Section 3.2 (3 rd or 4 th ed) Molecules with Lone Pairs on the entral Atom. Topics: I. Breakdown of the octet rule ase 1. dd number of valence electrons ase 2. ctet deficient molecules ase 3. Valence shell expansion II. Ionic bonds III. Polar covalent bonds and polar molecules I. BREAKDW F TE TET RULE ase 1. dd number of valence electrons For molecules with an odd number of valence electrons, it is not possible for each atom in the molecule to have an octet, since the octet rule works by e - s. Example: 3 2) 3(1) + 4 = valence electrons 3) 3(2) + 8 = electrons needed for octet 4) 14 7 = bonding electrons 3 Radical species: molecule with an electron. Radicals are usually very reactive. The reactivity of radical species leads to interesting (and sometimes harmful) biological activity. Free radicals in biology: a paradox Free radical species damage Free radicals DA. in biology: a Free paradox radicals are essential for life. Free radical ive species oxygen damage radicals are DA. a byproduct Free radicals Free are radicals essential are involved for life. in cell ighly reactive oxygen radicals are a byproduct Free radicals are involved in cell of metabolism and cause DA damage. signaling Some enzymes (see use example free radicals on p. 2). to igarette smoke contains free radicals that can carry out essential reactions in the damage DA in lung cells and lead antioxidant-rich to cancer. Some body. enzymes use free radicals to carry out essential reactions in the antioxidant-rich For example, a pr bo otein dy. called ribonucleotide reductase oxidants (such as vitamin b A, lueberries, and E) reduce (RR) catalyzes an essential step in DA synthesis For and example, repair a using protein a free called radical ribonucleotide species. reductase Antioxidants (such as vitamin A,, and E) reduce (RR) catalyzes an essential step in DA synthesis DA damage in the body by trapping radicals. and repair using a free radical species.

3 Some radicals are more stable. For example, 1) Draw skeletal structure 2) = 11 valence electrons 3) = 16 electrons needed for octet 4) = bonding electrons itric oxide, vasodilation: the widening of blood vessels Figure by MIT penourseware. an important cell-signaling molecule in humans. diffuses freely across cell membranes and signals for the smooth muscle in blood vessels to relax, resulting in vasodilation and increased blood flow. a radical species, has a short lifetime in the body, which makes it an ideal messenger molecule between adjacent cells. You may be familiar with a drug that inhibits the breakdown of an binding partner (an enzyme), leading to increased blood flow:. ow let s think about molecular oxygen, 2. What we expect: 2) valence electrons 3) electrons needed for octet 4) bonding electrons 5) Add two electrons per bond. 6) 2 bonding electrons remaining. Make double bond. 7) valence electrons make lone pairs Lewis method seems to work here, but in reality 2 is a! We need molecular orbital (M) theory (Lecture #14). 2

4 ase 2. ctet deficient molecules Some molecules are stable with an incomplete octet. Group 13 elements and have this property. onsider BF 3 F F B F First, let s write the Lewis structure that achieves octets on every atom. 2) 3 + 3(7) = valence electrons 3) 8 + 3(8) = electrons needed for octet 4) = bonding e - s 5) assign two electrons per bond. 6) 8 6 = 2 extra bonding electrons 7) 24 8 = 16 lone pair electrons 8) calculate formal charges: F B = 3 0 (½)(8) = -1 F FDB = 7 (½)( ) = F F = 7 (½)( ) = But experiments suggest that all three B-F bonds have the same length, that of a bond. F F B = 3 0 (½)(6) = 0 F B F F F = 7 6 (½)(2) = 0 The formal charges are more favorable for this structure. ase 3. Valence shell expansion Elements with n = or > 3 have empty - orbitals, which means more than eight electrons can fit around the central atom. Expanded valence shells are more common when the central atom is and is bonded to small, highly electronegative atoms such as, F, and l. 3

5 onsider Pl 5 l l P l 2) 5 + 5(7) = valence electrons l l 3) 8 + 5(8) = electrons needed for octet 4) = bonding e - s To make five P-l bonds, need shared electrons. So = 30 lone-pair electrons. onsider r 4 2 r -2 2) 6 + 4(6) + = valence electrons 3) 8 + 4(8) = electrons needed for octet 4) = bonding e - s 7) 32 8 = 24 lone-pair electrons. 8) calculate formal charges: F r = 6 0 (½)(8) = +2 F = 6 6 (½)(2) = -1 Total charge = 2 + 4(-1) = -2 But experimentally, r- bond length and strength are between that of a single and double bond! -2-2 r r plus other resonance structures. F r = 6 0 (½)12 = 0 F DB = 6 4 (½)4 = 0 F = 6 6 (½)2 = -1 4

6 Valence shell expansion around r results in formal charge separation. More stable Lewis structure. II. II BDS Ionic bonds involve the complete of (one or more) electrons from one atom to another with a bond resulting from the electrostatic attraction between the cation and anion. onsider the formation of al from the neutral atoms, a and l. a (g) a + (g) + e E = = 494 kj/mol l + e l (g) E = EA = kj/mol a (g) + l (g) a + (g) + l (g) E = + kj/mol E = IE a + ( EA l ) = 145 kj/mol Problem: a (g) + l (g) a + (g) + l (g) has a positive E. It energy. Solution: oulomb attraction. a + (g) + l (g) al (g) E = 589 kj/mol et change in energy for the overall process: a (g) + l (g) al (g) E = kj/mol The mutual attraction between the oppositely-charged ions releases energy. The net energy change for the formation of al is a decrease in energy. We can calculate the oulomb attraction based on the distance between the two ions (assume here that the ions are point charges): U(r) = z 1 z 2 e 2 for 2 unlike charges, z = charge numbers of the ions and 4πε 0 r e = absolute value of the charge of an e - (1.602 X ) alculate U(r) for a + and l -. al has a bond length (r) = 2.36Å. 5

7 U(r) = ( )( )( ) = 4π(8.854 x J -1 m -1 )( ) onvert to kj/mol U(r) = x J x x = Simple ionic model predicts: Experiments measure: E = - E d = kj/mol E = - E d = kj/mol The discrepancy results from the following approximations: ignored repulsive interactions. Result: E d than experimental value. treated a + and l as. ignored quantum mechanics. This simple model is applicable only to very ionic bonds. III. PLAR VALET BDS Perfectly-ionic and perfectly-covalent bonds are the two extremes of bonding. In reality, most bonds fall somewhere in the middle. A polar covalent bond is an sharing of electrons between two atoms with different electronegativities (χ). onsider -l versus - (Pauling electronegativity values are given): l χ = 2.2 χ l = 3.2 δ+ l δ where δ is fraction of a full charge (e) that is asymmetrically distributed. 2 is a perfectly covalent bond, δ = 0. 6

8 Dipole moment Asymmetric charge distribution results in an electric dipole, two unlike charges separated by a finite distance. We can quantify charge separation by defining a dipole moment, µ. δ+ δ r μ m μ = Q r Magnitude of charge Separation where Q = δe Dipole moment is measured in m or in debye 1 debye = 1 D = x m In chemistry, the arrow points toward the negative charge in a polar bond. Polar molecules have a non-zero net dipole moment. χ = molecule 2 molecule χ = 2.6 χ = 3.0 χ = 3.4 In large organic molecules and in biomolecules, such as proteins, we often consider the number of polar groups within the molecule. For example, let s compare vitamin A to vitamin B9 Which vitamin contains a higher number of polar bonds? vitamin Vitamin A soluble Vitamin B9 ( ) soluble 7