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Chapter 9
Chemical Bonding 1: The Lewis Model
| Term | Definition |
|---|---|
| Why Do Chemical Bonds Form? | Chemical bonds form because they lower the potential energy between the charged particles that compose atoms |
| 3 Types of Bonds | 1. Ionic.. 2. Covalent.. 3. Metallic |
| Ionic | Between metal and nonmetal. Electrons transferred |
| Covalent | Between 2 nonmetals or a nonmetal and a mettalloid. Electrons shared |
| Metallic | Between metal and metal. Electrons pooled |
| Octet | An atom has an octet when it has eight valence electrons. These are particularly stable. Have eight dots for Lewis structure |
| Exception Helium(He) | He always has two paired dots for Lewis structure. The number of valence electrons for helium is two even though it is in group 8 |
| Chemical Bond | The sharing or transfer of electrons to attain stable electron configuration for the bonding atoms |
| Octet Rule | The tendency for most bonded atom to possess or share eight electrons in their outer shell to obtain stable electron configurations and lower their potential energy |
| Lattice Energy | The energy associated with the formation of a crystalline lattice of alternating cations and anions from the gaseous ions |
| Born-Haber Cycle | A hypothetical series of steps that represent the formation of an ionic compound from its constituent elements |
| Step 1 Born-Haber Cycle Steps for the Formation of NaCl(s) from Na(s) and Cl2(g) | The first step is the formation of gaseous sodium from solid sodium... Na(s) ---> Na(g) ÎH*step1(sublimation energy of Na) = +108 kJ |
| Step 2 Born-Haber Cycle Steps for the Formation of NaCl(s) from Na(s) and Cl2(g) | The second step is the formation of a chlorine atom from a chlorine molecule... 1/2Cl2(g) ---> Cl(g) ΔH*step2(bond energy of Cl2 x 1/2) = +122 kJ |
| Step 3 Born-Haber Cycle Steps for the Formation of NaCl(s) from Na(s) and Cl2(g) | The third step is the ionization of gaseous sodium. The enthalpy change for this step is the ionization energy of sodium... Na(g) ---> Na^+(g) + e^- ΔH*step3(ionization energy of Na) = +496 kJ |
| Step 4 Born-Haber Cycle Steps for the Formation of NaCl(s) from Na(s) and Cl2(g) | The fourth step is the addition of an electron to gaseous chlorine. The enthalpy change for this step is the electron affininty of chlorine... Cl(g) + e^- ---> Cl^-(g) ΔH*step4(electron affininty of Cl) = -349 kJ |
| Step 5 Born-Haber Cycle Steps for the Formation of NaCl(s) from Na(s) and Cl2(g) | The fifth and final step is the formation of the crystalline solid from the gaseous ions. The enthalpy change for this step is the lattice energy, the unknown quantity... Na^+(g) + Cl^-(g) ---> NaCl(s) ΔH*step5 = ΔH*lattice = ? |
| Increase in Ionic Radius Causes Decrease in Lattice Energy | As the ionic radii increase as we move down the column, the ions cannot get as close to each other and therefore do not release as much energy when the lattice forms(decrease ΔH*lattice) |
| Couloumb's Law | The magnitude of the potential energy of two interacting charges depends not only on the distance between the charges, but also on the product of the charges... E = (q1 x q2)/r |
| Relationship Between Magnitude of Ionic Charge in Chemical Bonds, and Lattice Energy | As Magnitude of Ionic Charges in a Chemical Bond Increase, Lattice Energy Increase(increase ΔH*lattice) |
| Summarizing Trends in lattice Energies | 1. Lattice energies( ΔH*lattice) become less exothermic(less negative) with increasing ionic radius... 2. Lattice energies( ΔH*lattice) become more exothermic(more negative) with increasing magnitude of ionic charge |
| Bonding Pair | A shared pair of electrons |
| Lone Pair | A pair of electrons that is associated with only one atom in a Lewis structure and is therefore not involved in bonding |
| Nonbonding Electrons | Lone pair electrons in the Lewis structure. Electrons associated with only one atom and is therefore not associated with bonding |
| Double Bond | When two electron pairs are shared between two atoms. Double bonds are shorter and stronger than single bonds |
| Triple Bond | The bond that form when three electron pairs are shared between two atoms, Triple bonds are even shorter and stronger than double bonds |
| Polar | Having a positive pole and a negative pole |
| Polar Covalent Bond | Is intermediate in nature between a pure covalent bond and an ionic bond |
| Electronegativity | The ability of an atom to attract electrons to itself in a chemical bond(which result in polar and ionic bond) |
| Periodic Trends for Electronegativity(EN) | 1. EN increases across a row in the periodic table.. 2. EN decreases down a column in the periodic table.. 3. Fluorine is the most EN element.. 4. Francium is the least EN element.. 5. Increases from bottom left to top right.. 6. Noble gases excluded |
| Electronegativity with Relation to Atomic Size | Electronegativity is inversely related to atomic size. The larger the atom, the less ability it has to attract electrons to itself in a chemical bond |
| Dipole Moment(μ) | Occurs any time there is a separation of positive and negative charge |
| Equation for Dipole Moment | The magnitude of a dipole moment created by separating two particles of equal but opposite charges of magnitude q by a distance r is given by the equation... μ = qr |
| Debye(D) | The unit commonly used for reporting dipole moments...1 D = 3.34 x 10^-30 C * m |
| Percent Ionic Character | The ratio of a bond's actual dipole moment to the dipole moment it would have if the electron were completely transferred from on atom to the other, multiplied by 100... |
| Equation for Percent Ionic Character | Percent ionic character = ((measured dipole moment of bond) / (dipole moment if electron were completely transferred)) x 100 |
| Electronegativity Difference(ΔEN) for Classifying Bond Types | 1. ΔEN(small(0 - 0.4))...Bond Type(Covalent)...Example(Cl2)... 2. ΔEN(intermediate(0.4 - 2.0))...Bond Type(Polar Covalent)...Example(HCl)... 3. ΔEN(Large(2.0+))...Bond Type(Ionic)...Example(NaCl) |
| 1st Step for Drawing Lewis Structures for Molecular Compounds | Determine the total number of valence electrons available for compound. Done by adding the individual valence electrons of each atom in compound.. Charges are adjusted for ions. Subtract charge for cation(+). Add charge for anions(-) |
| 2nd Step for Drawing Lewis Structures for Molecular Compounds | Determine the central atom of the structure. Central atom is the... 1. Atom that is single.. 2. Atom with lowest number of valence electrons.. 3. Atom with the lowest EN... NEVER Hydrogen(max of 2 electrons) |
| 3rd Step for Drawing Lewis Structures for Molecular Compounds | Assembly the structure. Attach the remaining atoms to the central atom using single bonds. The outer atoms are attached such that the molecule has the greatest symmetry possible |
| 4th Step for Drawing Lewis Structures for Molecular Compounds | Reduce electron pool. For each bond used to assemble the structure, subtract 2 electrons from the total number of electrons available for bonding, the electron pool |
| 5th Step for Drawing Lewis Structures for Molecular Compounds | Complete octets of outer atoms. Each outer atom in structure is sharing one pair of e^- 's with central atom; therefore, each outer atom still requires six electrons to complete octet. Electrons are shown as pairs at each remaining side of the symbol |
| 6th Step for Drawing Lewis Structures for Molecular Compounds | Reduce electrons pool. Subtract number of electrons used to complete the octets of the outer atoms |
| 7th Step for Drawing Lewis Structures for Molecular Compounds | Check central atom(CE). If e^- 's remain, place in pairs at CE. CE MUST HAVE AT LEAST 8 ELECTRONS. Exception is Boron(B). The CE can have more than 8 e^- 's. Can occur with CE that have available d orbitals or of 3rd principle energy level or higher |
| 8th Step for Drawing Lewis Structures for Molecular Compounds | If any atoms lack an octet, form double or triple bonds as necessary to give them octets |
| Drawing Lewis Structure for Polyatomic Ions | When calculating number of valence electrons, add one electron for each negative charge(anion), and subtract on electron for each positive charge(cation). Drawn with the charge of the ion in the upper right-hand corner, outside the bracket |
| Resonance Structures | One of two or more Lewis structures that have the same skeletal formula(the atoms are in the same location), but different electron arrangements. Shown with double--headed arrow between the structures |
| Resonance Hybrid | The actual structure of the molecule which is intermediate between the two(or more) resonance structures |
| Formal Charge(FC) | A fictitious charge assigned to each atom that determines the best way to draw a Lewis Structure. The charge an atom would have if all bonding electrons were shared equally between the bonded atoms |
| Calculating Formal Charge(FC) of an Atom | FC = number of valence electrons - number of bonds(1/2 number of bonded electrons) - number nonbonded electrons |
| Rules For Formal Charge(FC) | 1. Sum of all FCs in neutral molecule must be 0.. 2. Sum of all FCs in ion must equal charge of the ion.. 3. Small(or zero) FCs on individual atoms are better than larger ones.. 4. When FCs cannot be avoided, negative FCs should reside on most En atom |
| Free Radicals(or simply radicals) | Molecules and ions with an odd number of electrons in their Lewis structures. Ex: NO |
| Incomplete Octets | Molecules of ions with fewer than eight electrons around an atom. Ex: sometimes B and Be compounds |
| Expanded Octets | Molecules or ions with more than eight electrons around an atom. Elements in the third row of the periodic table and beyond often exhibit expanded octets of up to 12(and occasionally 14) electrons. Never occur in second-period(row) elements |
| Bond Energy | The energy required to break 1 mole of the bond in the gas phase. Always positive(endothermic) because it always takes energy to break a bond... Cl2(g) ---> 2Cl(g) ΔH = 243 kJ.... HCl(g) ---> H(g) + Cl(g) ΔH = 431 kJ |
| Bond Energy For Breaking Bonds | When bonds break, the process is endothermic(positive bond energy, +ΔH)... Cl2(g) ---> 2Cl(g) ΔH = 243 kJ |
| Bond Energy for Forming Bonds | When bonds form, the process is exothermic(negative bond energy, -ΔH) It is always the forming of bonds that releases energy ... 2Cl(g) ---> Cl2(g) ΔH = -243 kJ |
| Calculating ΔH with Bond Energy | ΔHrxn = Σ(reactant ΔH's bonds broken) - Σ(product ΔH's bonds formed).... 1. A reaction is exothermic when weak bonds break and strong bonds form.. 2. A reaction is endothermic when strong bonds break and weak bonds form |
| Bond Length | The average length of a bond between two particular atoms in a variety of compounds |
| Bond Length with Increasing Energy, Decreasing Length | Single --> Double --> Triple |
| Best Insulators | Most electronegative elements make the best insulators |
Created by:
TimChemistry1
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