Bonding – General Concepts

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Presentation transcript:

Bonding – General Concepts

Chemical Bonds bonds form when the potential energy of the system will be lowered chemical bond – mutual electrical attraction between the nuclei and valence electrons of different atoms that binds the atoms together

Sodium Chloride Crystal Lattice Ionic compounds form solids at ordinary temperatures. Ionic compounds organize in a characteristic crystal lattice of alternating positive and negative ions. Solid sodium chloride has a melting point ~ 800oC WOW!

Electronegativity: The ability of an atom in a molecule to attract shared electrons to itself.

Do we get it???? Let us see! Order the following bonds according to their Polarity (lowest to highest): H-H, O-H, Cl-H, S-H and F-H. (Table on pg. 334) H-H < S-H < Cl-H < O-H < F-H 0 0.4 0.9 1.4 1.9 POLARITY INCREASES

Ionic Bonds Electrons are transferred Electronegativity differences are generally greater than 1.7 The formation of ionic bonds is always exothermic!

Determination of Ionic Character Electronegativity difference is not the final determination of ionic character Compounds are ionic if they conduct electricity in their molten state

Coulomb’s Law “The energy of interaction between a pair of ions is proportional to the product of their charges, divided by the distance between their centers”

Sodium Chloride Crystal Lattice

Estimate Hf for Sodium Chloride Na(s) + ½ Cl2(g)  NaCl(s) Lattice Energy -786 kJ/mol Ionization Energy for Na 495 kJ/mol Electron Affinity for Cl -349 kJ/mol Bond energy of Cl2 239 kJ/mol Enthalpy of sublimation for Na 109 kJ/mol Na(s)  Na(g) + 109 kJ Na(g)  Na+(g) + e- + 495 kJ ½ Cl2(g)  Cl(g) + ½(239 kJ) Cl(g) + e-  Cl-(g) - 349 kJ Na+(g) + Cl-(g)  NaCl(s) -786 kJ Na(s) + ½ Cl2(g)  NaCl(s) -412 kJ/mol

Bond Length Diagram

Covalent Bonding Forces Electron – electron repulsive forces Electron – proton attractive forces Proton – proton repulsive forces

Covalent Bonds Polar-Covalent bonds Nonpolar-Covalent bonds Electrons are unequally shared Electronegativity difference between .3 and 1.7 Nonpolar-Covalent bonds Electrons are equally shared Electronegativity difference of 0 to 0.3 Electrostatic Potential Diagram

Polarity Dipole Moment uneven distribution of charge a molecule with a imbalance in electrons has a dipole moment polyatomic molecules have dipole moments if their polar bonds don’t cancel out

Practice Determine whether each of the following bonds will be: ionic, polar covalent, OR nonpolar covalent Cl and O 3.5-3.0=0.5 polar covalent Cl and Br 3.0-2.8 nonpolar covalent S and H 2.5-2.1=0.4 polar covalent S and Cs 2.5-0.7=1.8 ionic

Practice For each of the following molecules, indicate which ones have a dipole moment. HCl Cl2 H: 2.1 < Cl: 3.0 Dipole moment - + Cl: 3.0 = Cl: 3.0 No dipole moment

Practice CH4 C: 2.5 > H: 2.1 no dipole moment H2S + 4- + + + + + 2-

Ions: Electron Configurations and Sizes nonmetals sharing to make covalent bonds taking electrons from metals to form anions metals giving electrons to nonmetals to form cations Sizes of Ions trends are most important cation < parent atom more protons attracting electrons in anion > parent atom more electrons without additional protons

Sizes of Ions isoelectronic ions: ions with the same number of electrons Ex: S2- Cl- K+ Ca2+ electrons have greater attraction to nucleus as # of protons increases so the size decreases as nuclear charge (Z) increases

Table of Ion Sizes

Example Give six ions that are isoelectronic with neon. N3-, O2-, F-, Na+, Mg2+, Al3+ 7 > 8 > 9 > 11 > 12 > 13 Place them in order of decreasing size. (Largest to smallest)

Example for each of the following groups, place atoms in order of decreasing size Cu, Cu+, Cu2+ all have 29 p, lower #e- mean smaller ion Cu > Cu+ > Cu2+ Ni2+, Pd2+, Pt2+ different amounts of e- and p-energy levels Pt2+ > Pd2+ > Ni2+ O, O-, O2- same #p, more electrons means larger ion O2- > O- > O

4. Which of the following would represent a non-polar molecule containing polar bonds: I2 CO2 PF3 SO2 H2O

Which of the following would represent a non-polar molecule containing polar bonds: I2 CO2 PF3 SO2 H2O Answer B You tell me why???

1. The compound most likely to be ionic is: KF CCl4 CO2 ICl CS2

2. Ranking the ions S2-, Ca2+, K+, Cl- from smallest to largest gives the order as: S2-, Cl-, K+, Ca2+ Ca2+, Cl-, K+,, S2- K+, Ca2+, Cl-, S2- Cl-, S2-, K+, Ca2+ Ca2+, K+, Cl-, S2-

3. The type of bonding within a water molecule is: ionic bonding polar covalent bonding nonpolar covalent bonding metallic bonding hydrogen bonding

4. What type of bond would you expect in CsI?: Ionic Covalent Hydrogen Metallic van der walls

PLEASE PASS YOUR RESPONSES RIGHT TO LEFT LAST PERSON…..BRING THEM UP. THAT’S ALL FOLKS! PLEASE PASS YOUR RESPONSES RIGHT TO LEFT LAST PERSON…..BRING THEM UP.

The compound most likely to be ionic is: KF CCl4 CO2 ICl CS2 Answer: A Bonds formed from elements with greater differences in electronegativity are most likely ionic in nature. (remember trends!)

Ranking the ions S2-, Ca2+, K+, Cl- from smallest to largest gives the order as: S2-, Cl-, K+, Ca2+ Ca2+, Cl-, K+,, S2- K+, Ca2+, Cl-, S2- Cl-, S2-, K+, Ca2+ Ca2+, K+, Cl-, S2- Answer E Note that all ions in this isoelectronic series have an Ar electron configuration; therefore the nuclear charge determines the size.

The type of bonding within a water molecule is: ionic bonding polar covalent bonding nonpolar covalent bonding metallic bonding hydrogen bonding Answer B Consider the nature of the bonding between hydrogen and oxygen in water.

What type of bond would you expect in CsI?: Ionic Covalent Hydrogen Metallic van der walls Answer A You tell me why???

Bond Length and Energy Bond Bond type Bond length (pm) Bond Energy (kJ/mol) C - C Single 154 347 C = C Double 134 614 C  C Triple 120 839 C - O 143 358 C = O 123 745 C - N 305 C = N 138 615 C  N 116 891 Bonds between elements become shorter and stronger as multiplicity increases.

Bond Energy and Enthalpy Energy required Energy released D = Bond energy per mole of bonds Breaking bonds always requires energy Breaking = endothermic (+DH) Forming bonds always releases energy Forming = exothermic (-DH)

Example Calculate the change in energy for the following reaction: (use table 8.4 on pg 351 in your textbook) H2(g) + F2(g)  2HF(g) ∆H = ∑ D (bonds broken) – ∑ D (bonds formed) ∆H = (DH-H + DF-F) – 2DH-F ∆H = (432 kJ + 154 kJ) – 2 (565 kJ) ∆H = -544 kJ

Example Using bond energies, calculate ∆H for the reaction: CH4(g) + 2Cl2(g) + 2F2(g)  CF2Cl2(g) + 2HF(g) + 2HCl(g) ∆H = ∑ D (bonds broken) – ∑ D (bonds formed) ∑ D (bonds broken) = 4 (413 kJ) + 2 (239 kJ) + 2 (154 kJ) ∑ D (bonds formed) = 2 (485 kJ) + 2 (339 kJ) + 2 (565 kJ) + 2 (427 kJ) ∆H = 2438 kJ – 3632 kJ = -1194 kJ

The Octet Rule Combinations of elements tend to form so that each atom, by gaining, losing, or sharing electrons, has an octet of electrons in its highest occupied energy level. Diatomic Fluorine

Formation of Water by the Octet Rule

Comments About the Octet Rule 2nd row elements C, N, O, F observe the octet rule. 2nd row elements B and Be often have fewer than 8 electrons around themselves - they are very reactive. 3rd row and heavier elements CAN exceed the octet rule using empty valence d orbitals. Below are three examples of compounds containing elements not obeying the Octet Rule:

Lewis Structures Shows how valence electrons are arranged among atoms in a molecule. Reflects central idea that stability of a compound relates to noble gas electron configuration.

Completing a Lewis Structure -CH3Cl Make carbon the central atom Add up available valence electrons: C = 4, H = (3)(1), Cl = 7 Total = 14 Join peripheral atoms to the central atom with electron pairs. H .. .. Complete octets on atoms other than hydrogen with remaining electrons H .. C .. Cl .. .. .. H

H H C Cl CH3Cl Example methyl chloride C: 1 x 4e- = 4e- H: 3 x 1e- = 3e- Cl: 1 x 7e- = 7e- 14 e- Carbon is central atom H H C Cl duet octet

CH3Cl Methyl Chloride

Multiple Covalent Bonds: Double bonds Ethene Two pairs of shared electrons

Multiple Covalent Bonds: Triple bonds Ethyne Three pairs of shared electrons

.. .. .. Example CH2O O H – C - H Formaldehyde C: 1 x 4e- = 4 e- H: 2 x 1e- = 2 e- O: 1 x 6e- = 6 e- 12 O H – C - H

Resonance Resonance is invoked when more than one valid Lewis structure can be written for a particular molecule. Benzene, C6H6 The actual structure is an average of the resonance structures. The bond lengths in the ring are identical, and between those of single and double bonds.

Resonance Bond Length and Bond Energy Resonance bonds are shorter and stronger than single bonds. Resonance bonds are longer and weaker than double bonds.

Resonance in Ozone, O3 Neither structure is correct. Oxygen bond lengths are identical, and intermediate to single and double bonds

Resonance in Polyatomic Ions Resonance in a carbonate ion: Resonance in an acetate ion:

Resonance in Polyatomic Ions How about the nitrate ion(NO3-) ?: How about the nitrite ion (NO2-) ?: Lewis Structure Nitrate ion Lewis Structure Nitrite ion

Resonance The lewis structure of which molecule requires resonance structures? MgCl2 PCl5 SiO2 SO2 OCl2 Answer: D In this case there is a single bond between sulfur and one of the oxygens, and a double bond between sulfur and the other oxygen.

Formal Charge Helps you to compare various Lewis Structures and choose the best or most likely structure FC = (# valence e-) – (#e- assigned to it) Assigned: all of the lone pairs belong to that atom plus half of the electrons involved in bonding Valence: from periodic table

Formal Charge Not REAL: but provides less extreme charges than oxidation numbers sum of the FC on molecule must equal overall charge on molecule atoms try to achieve a FC as close to zero as possible any negative FC must be on most electronegative atom

Example SO42- sulfate ion 6e- + 4(6e-)= 30 e- + 2- = 32 electrons -1 +2 -1 -1 -1

Example NO2 nitrogen dioxide 17 electrons

Formal Charge (FC) Dinitrogen oxide, N2O has two double bonds. The general structure is N=N=O. The formal charge on the oxygen atom in this molecule is? zero Positive 1 (+1) Positive 2 (+2) Negative one (-1) Negative two (-2) Answer: A Use the formula! If necessary, draw the lewis structure.

Localized Electron Bonding Model (LEM) LEM is composed of three parts: Description of the valence electrons arrangement in the molecule (Lewis structure) B. Prediction of the geometry of the molecule (VSEPR) C. Description of the type of atomic orbitals used by the atoms (Chapter 9 – Hybridization)

VSEPR Are you ready for 3-Dimensions?

VSEPR – Valence Shell Electron Pair Repulsion Steric # Overall Structure Forms 2 Linear AX2 3 Trigonal Planar AX3, AX2E 4 Tetrahedral AX4, AX3E, AX2E2 5 Trigonal bipyramidal AX5, AX4E, AX3E2, AX2E3 6 Octahedral AX6, AX5E, AX4E2 Steric # = X+E A = central atom X = atoms bonded to A E = nonbonding electron pairs on A

O O Cl B Molecular Shapes VSEPR Theory (Valence Shell Electron Pair Repulsion) pairs of electrons are arranged as far apart from each other as possible. Linear: 180o between bonds Ex: O2 O O 2. Trigonal Planar: “flat triangle” – one central atom surrounded by three atoms with NO LONE PAIRS. Ex: BCl3 B Cl

C H N H 3. Tetrahedral: four atoms bonded to a central atom. 4. Trigonal pyramidal: three atoms bonded to a central atom with one lone pair of electrons about the central atom. N H

5. Bent: a central atom bonded to two other atoms AND one or two LONE PAIRS.

VSEPR: Linear AX2 CO2 Shape: Linear

VSEPR: Trigonal Planar AX3 BF3 Shape: Trigonal Planar AX2E SnCl2 Shape: Bent

AX4 CCl4 AX3E PCl3 AX2E2 Cl2O Shape: Tetrahedral VSEPR: Tetrahedral AX4 CCl4 Shape: Tetrahedral AX3E PCl3 Shape: Trigonal Pyramidal AX2E2 Cl2O Shape: Bent

VSEPR: Trigonal Bi-pyramidal AX5 PCl5 Shape: Trigonal Bipyramidal AX4E SF4 Shape: Seesaw AX3E2 ClF3 Shape: T-shaped AX2E3 I3- Shape: Linear

AX6 SF6 AX5E BrF5 AX4E2 ICl4- Shape: Octahedral VSEPR: Octahedral AX6 SF6 Shape: Octahedral AX5E BrF5 Shape: Square Pyramidal AX4E2 ICl4- Shape: Square planar

The predicted geometry (shape) of PH3 according VSEPR The predicted geometry (shape) of PH3 according to the VSEPR theory is? linear bent Trigonal pyramidal tetrahedral Trigonal planar Answer: C Easy one!

State the steric number, electronic famly and Predict the molecular structure for each of the following: Steric# E-Family M-Structure PCl3 SCl2 SiF4 4 Tetrahedral Trigonal Pyramidal AX3 4 Tetrahedral Bent AX2E2 4 Tetrahedral Tetrahedral AX4