Bonding : General Concepts Chapter 8 Valence electrons
Bonding : General Concepts Chapter 8 Valence electrons are the outer shell electrons of an atom. The valence electrons are the electrons that participate in chemical bonding. Group e- configuration # of valence e- 1A ns1 1 2A
ns2 2 3A ns2np1 3 4A ns2np2 4 5A ns2np3
5 6A ns2np4 6 7A ns2np5 7 Lewis Electron-Dot Symbols A Lewis electron-dot symbol is a symbol in which the electrons in the valence shell of an atom or ion are represented by dots placed around the letter symbol of the element.
Group I Group II Group III Group IV Group V Group VI Group VII Group VIII : : number of valence electrons. : : :
: . . Si Na. . Mg . .Al P S Cl Ar . . . . . : . : . : : . number . .indicates the Note that the group Lewis symbol (Lewis structure) - a way to represent atoms (and their bonds) using the element symbol and valence electrons as dots.
Chemical Bonding Chemical Bond - the force of attraction between any two atoms in a compound. Interactions involving valence electrons are responsible for the chemical bond. The Ionic Bond Ionic bond: the electrostatic force that holds ions together in an ionic compound. Li+ F Li + F 1s22s1 LiF 1s22s22p5 Li+ + e- Li
e- + Li+ + 1s2 1s22s22p6 [He] [Ne] F F - F - Li+ F - 6 Essential Features of Ionic Bonding Atoms with low I.E. and low E.A. tend to form positive ions.
Atoms with high I.E. and high E.A. tend to form negative ions. Ion formation takes place by electron transfer. The ions are held together by the electrostatic force of the opposite charges. Reactions between metals and nonmetals (representative) tend to be ionic. Describing Ionic Bonds Noble gas configurations are stable. The atoms of metals loses electron(s) to become a cation (positive). eg., Na Na+ [Ne]3s1 [Ne] + 1 e(Na has low I.E, easy to loose an e-) The atom that gains the electron becomes an anion (negative), eg., Cl, Cl + eCl- (Cl has high E.A) 3s23p5 + e- 3s23p6 or [Ar]
Another example The magnesium has two electrons to give, whereas the fluorines have only one vacancy each. F ] 2+ Mg F . [
: : : : [ - : : : : : Mg . . F. :
F ] - : : : : Consequently, magnesium can accommodate two fluorine atoms. 9.1 Use Lewis dot symbols to show the formation of aluminum oxide (Al2O3). The mineral corundum (Al2O3). 9.1
Thus, the simplest neutralizing ratio of Al3+ to O2 is 2:3; two Al3+ ions have a total charge of +6, and three O 2 ions have a total charge of 6. So the empirical formula of aluminum oxide is Al2O3, and the reaction is Ionic bonds MgO magnesium oxide 2Mg(s)+O2(g) 2MgO Isoelectronic Series A series of ions/atoms containing the same number of electrons. O2-, F-, Ne, Na+, Mg2+, and Al3+ Copyright Cengage Learning. All rights reserved 13 CONCEPT CHECK! Choose an alkali metal, an alkaline earth metal, a noble
gas, and a halogen so that they constitute an isoelectronic series when the metals and halogen are written as their most stable ions. What is the electron configuration for each species? Determine the number of electrons for each species. Determine the number of protons for each species. Copyright Cengage Learning. All rights reserved 14 Electrostatic (Lattice) Energy Lattice energy (U) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions. E = k Q+Qr
Lattice energy increases as Q increases and/or as r decreases. Q+ is the charge on the cation Q- is the charge on the anion r is the distance between the ions Compound Lattice Energy (kJ/mol) Q: +2,-1 MgF2 2957 Q: +2,-2 MgO 3938 E is negative for ionic bond formation, but lattice energy is the LiF reverse, energy to break a bond, LiCl
which is +ve. 1036 853 r F- < r Cl15 Formation of an Ionic Solid 1.Sublimation of the solid metal. M(s) M(g) [endothermic] 2. Ionization of the metal atoms. M(g) M+(g) + e- [endothermic] 3. Dissociation of the nonmetal. 1/2X2(g) X(g) [endothermic] Copyright Cengage Learning. All rights reserved 16 Formation of an Ionic Solid
(continued) 4 Formation of nonmetal ions in the gas phase. X(g) + e- X-(g) [exothermic] 5.Formation of the solid ionic compound. M+(g) + X-(g) MX(s) [quite exothermic] Copyright Cengage Learning. All rights reserved 17 Born-Haber Cycle for Determining Lattice Energy Lattice energy = +1017 o DHoverall = DH1o+ DH2o+ DH3o+ DH4o+ DH5 o
Covalent Bond A covalent bond is a chemical bond in which two or more electrons are shared by two atoms. Why should two atoms share electrons? F + 7e- F F F 7e- 8e- 8e- Noble gas
config. Lewis structure of F2 single covalent bond lone pairs F F lone pairs lone pairs F F lone pairs single covalent bond
20 Features of Covalent Bonds The diatomic elements have totally covalent bonds (totally equal sharing.) H2, N2, O2, F2, Cl2, Br2, I2 Covalent compounds are usually formed from nonmetals. Molecules - compounds characterized by covalent bonding. not a part of a massive three dimensional crystal structure. Lewis structure of water H + O +
H single covalent bonds H O H or 2e-8e-2e- Hydrogen molecule, diatomic H H H : H H O H
Carbon dioxide and nitrogen molecule Double bond two atoms share two pairs of electrons O C O or O O C double bonds 8e- 8e- 8e- Triple bond two atoms share three pairs of electrons N N
8e-8e- or N N triple bond 23 Lengths of Covalent Bonds Bond Lengths Triple bond < Double Bond < Single Bond Polar Covalent Bonds A polar covalent bond is one in which the bonding electrons spend more time near one of the two atoms involved.
When the atoms are alike, as in the H-H bond of H2 , the bonding electrons are shared equally (a nonpolar covalent bond). When the two atoms are of different elements, the bonding electrons need not be shared equally, resulting in a polar bond. Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms electron poor region H electron rich region
F e- poor H d+ e- rich F d- Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond. Electron Affinity - measurable, Cl is highest X (g) + e- X-(g)
Electronegativity - relative, F is highest 28 Dipole Moment Property of a molecule whose charge distribution can be represented by a center of positive charge and a center of negative charge. Use an arrow to represent a dipole moment. Point to the negative charge center with the tail of the arrow indicating the positive center of charge. Copyright Cengage Learning. All rights reserved 29 Dipole Moment 30
No Net Dipole Moment (Dipoles Cancel) Copyright Cengage Learning. All rights reserved 31 Polar Covalent Bonds Electronegativity is a measure of the ability of an atom in a molecule to draw bonding electrons to itself. In general, electronegativity increases from the lower-left corner to the upper-right corner of the periodic table. The current electronegativity scale, developed by Linus Pauling, assigns a value of 4.0 to fluorine and a value of 0.7 to cesium. The Electronegativities of Common Elements
Variation of Electronegativity with Atomic Number Classification of bonds by difference in electronegativity Difference Bond Type 0 Covalent 2 0 < and <2 Ionic Polar Covalent Increasing difference in electronegativity Covalent
Polar Covalent share e- partial transfer of e- Ionic transfer e- The greater the difference in electronegativity between two atoms, the greater the polarity of a bond. Which would be more polar, a H-F bond or a HCl bond? H-F 4.0 - 2.1 = 1.9 H-Cl 3.0 - 2.1 = 0.9 HF bond is more polar than the HCl bond.
Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H2S; and the NN bond in H2NNH2. Cs 0.7 Cl 3.0 3.0 0.7 = 2.3 Ionic H 2.1 S 2.5 2.5 2.1 = 0.4 Polar Covalent N 3.0
N 3.0 3.0 3.0 = 0 Covalent Chemical bonding (cont.) Classify the following bonds as ionic, polar covalent, or covalent: E.N. diff (a)the bond in HCl 3.0 2.1 = 0.9 polar covalent (b)The bond in KF 4.0 0.8 = 3.2 Ionic (c)the CC bond in H3CCH3 2.5 2.5 = 0 covalent Writing Lewis Structures
1. Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center. 2. Count total number of valence e-. 3. Complete an octet for all atoms except hydrogen. 4. If structure contains too many electrons, form double and triple bonds on central atom as needed. 39 9.3 Write the Lewis structure for nitrogen trifluoride (NF3) in which all three F atoms are bonded to the N atom. NF3 is a colorless, odorless, unreactive gas. 9.3 Solution We follow the preceding procedure for writing Lewis
structures. Step 1: The N atom is less electronegative than F, so the skeletal structure of NF3 is Step 2: The outer-shell electron configurations of N and F are 2s22p3 and 2s22p5, respectively. Thus, there are 5 + (3 7), or 26, valence electrons to account for in NF3. 9.3 Step 3: We draw a single covalent bond between N and each F, and complete the octets for the F atoms. We place the remaining two electrons on N: lone pair of electrons
bonding pair Because this structure satisfies the octet rule for all the atoms, step 4 is not required. Check Count the valence electrons in NF3 (in bonds and in lone pairs). The result is 26, the same as the total number of valence electrons on three F atoms (3 7 = 21) and one N atom (5). How about NH3? Similar to NF3, Difference for H ? Multiple Bonds In the molecules described so far, each of the bonds has been a single bond, that is, a covalent bond in which a single pair of electrons is shared. It is possible to share more than one pair. A double bond involves the sharing of two pairs between atoms. : :
H : :C : C H : H H H or
H C H C H Multiple Bonds Triple bonds are covalent bonds in which three pairs of electrons are shared between atoms. C : C :::
: H H or H C C H Formal Charge of an atom An atoms formal charge is the difference between the number of valence electrons in an isolated atom and the
number of electrons assigned to that atom in a Lewis structure. formal charge on an atom in a Lewis structure = total number of valence electrons in the free atom - total number of nonbonding electrons
- 1 2 ( total number of bonding electrons ) The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion. 45 H H
0 C formal charge on an atom in a Lewis structure 0 O = Formaldehyde, CH2O C 4 eO 6 e2H 2x1 e12 e- total number of valence electrons in
the free atom - 2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 total number of nonbonding electrons formal charge on C = 4 - 0 - x 8 = 0 formal charge on O
= 6 -4 - x 4 = 0 - 1 2 ( total number of bonding electrons ) 46 Lewis structure and formal charge of carbonate ion
The Lewis structure for the carbonate ion is, formal charge on an atom in a Lewis structure = total number of valence electrons in the free atom - total number of nonbonding electrons
- 1 2 ( total number of bonding electrons ) Lewis structure and formal charge of carbonate ion The Lewis structure for the carbonate ion is, formal charge on an atom in
a Lewis structure = total number of valence electrons in the free atom - total number of nonbonding electrons C: 4-0-4 =0 O(a): 6-4-2 =0 O (b):6-6-1 = -1 O : 6-6-1 = -1, Total of -2 for carbonate ion.
- 1 2 ( total number of bonding electrons ) Resonanace A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure. +
- - + O O (+Oand O areOthe formal charges) O O-O bond in O3 should be longer than the double bond, but both oxygen bonds are equal in length because of resonance as shown. Delocalized Bonding: Resonance According to theory, one pair of bonding electrons is spread over the region of all
three atoms. O O O This is called delocalized bonding, in which a bonding pair of electrons is spread over a number of atoms. . What are the resonance structures of the Benzene? carbonate (CO32-) ion? benzene Carbonate ion -
O C O O - O C O O - -
- O C O O - Lewis Structures and Exceptions to the Octet Rule 1. Incomplete Octet - less then eight electrons around an atom other than H. Lets look at BF3 2. Odd Electron - if there is an odd number of valence electrons it isnt possible to give every atom eight electrons.
Lets look at NO 3. Expanded Octet - elements in 3rd period and beyond may have 10 and 12 electrons around it. Exceptions to the Octet Rule The Incomplete Octet BeH2 BF3 H F B F F
Be H Exceptions to the Octet Rule Odd-Electron Molecules NO N 5eO 6e11e- N O The Expanded Octet (central atom with principal quantum number n > 2) Electrons can be accomodated in the d orbitals SF6
S 6e6F 42e48e- F F F S F F F 6 single bonds (6x2) = 12 18 lone pairs (18x2) = 36 Total = 48 Exceptions to the Octet Rule
For example, the bonding in phosphorus pentafluoride, PF5, shows ten electrons surrounding the phosphorus. P : : :F: : : : : :
:F :F: F: F: : Exceptions to the Octet Rule In xenon tetrafluoride, XeF4, the xenon atom must accommodate two extra lone pairs. Xe F: : : :
: : : F: : : :F : : :F The enthalpy change required to break a particular bond in one
mole of gaseous molecules is the bond energy. Bond Energy DH0 = 436.4 kJ H2 (g) H (g) + H (g) Cl2 (g) Cl (g) + Cl (g) DH0 = 242.7 kJ HCl (g) H (g) + Cl (g) DH0 = 431.9 kJ O2 (g) O (g) + O (g) DH0 = 498.7 kJ
O O N2 (g) N (g) + N (g) DH0 = 941.4 kJ N N Bond Energies Single bond < Double bond < Triple bond Bond Energies (BE) and Enthalpy changes in reactions Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products. (Examples 9.13,9.14)
DH0 = total energy input total energy released = SBE(reactants) SBE(products) H2 (g) + Cl2 (g) 2HCl (g) 2H2 (g) + O2 (g) 2H2O (g) Use bond energies to calculate the enthalpy change for: H2 (g) + F2 (g) 2HF (g) DH0 = SBE(reactants) SBE(products) Type of bonds broken H
H F F Type of bonds formed H F Number of bonds broken Bond energy (kJ/mol) Energy
change (kJ) 1 1 436.4 156.9 436.4 156.9 Number of bonds formed Bond energy (kJ/mol) Energy change (kJ)
2 568.2 1136.4 DH0 = 436.4 + 156.9 2 x 568.2 = -543.1 kJ Bond Energy To illustrate, lets estimate the DH for the following reaction. CH 4 ( g ) Cl 2 ( g ) CH 3Cl( g ) HCl ( g ) In this reaction, one C-H bond and one ClCl bond must be broken. In turn, one C-Cl bond and one H-Cl bond are formed. Bond Energy simple arithmetic yields DH.
CH 4 ( g ) Cl 2 ( g ) CH 3Cl( g ) HCl( g ) DH0 = [BE (C-H) + BE (Cl-Cl)] [BE (C-Cl) + BE (H-Cl)] =[(411+240)] [(327+428)] kJ = -104 kJ
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