concepts of chemical bonding brown, lemay ch 8 ap chemistry
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Concepts of Chemical Bonding
Brown, LeMay Ch 8AP Chemistry
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8.1: Types of “Inter-Atomic” Bonds
1. Ionic: electrostatic attraction between oppositely charged ions
2. Covalent: sharing of e- between two atoms (typically between nonmetals)
3. Metallic: “sea of e-”; bonding e- are relatively free to move throughout the 3D structure
Co va le n t M e ta llic
Io n ic
IncreasingDiff. of EN
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Lewis symbolsValence e-: e- in highest energy level and involved in
bonding; all elements within a group on P.T. have same # of valence e-
Lewis symbol (or electron-dot symbol):
Shows a dot only for valence e- of an atom or ion.
Place dots at top, bottom, right, and left sides and in pairs only when necessary (Hund’s rule).
Primarily used for representative elements only (Groups 1A – 8A)Ex: Draw the Lewis symbols of C and N.•
• C ••
•
: N ••
Gilbert N. Lewis
(1875 – 1946)
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The Octet Rule Atoms tend to gain, lose, or share e- until
they are surrounded by 8 valence e- (have filled s and p subshells) and are thus energetically stable.
Exceptions do occur (and will be discussed later.)
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8.2: Ionic Bonding Results as atoms lose or gain e- to achieve a
noble gas e- configuration; is typically exothermic. The bonded state is lower in energy (and therefore
more stable). Electrostatic attraction results from the opposite
charges. Occurs when diff. of EN of atoms is > 1.7
(maximum is 3.3: CsF) Can lead to interesting crystal structures (Ch.
11). Use brackets when writing Lewis symbols of ions.
Ex: Draw the Lewis symbol of sodium fluoride.••
: F :••
[ ]1-Na[ ]1+
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Lattice Energy Measurement of the energy of stabilization present
in ionic solidsHlattice = energy required to completely separate 1 mole of solid ionic compound into its gaseous ions
rr
QQH lattice
Electrostatic attraction (and thus lattice energy) increases as ionic charges increase and as ionic radii decrease. Ex: Which has a greater lattice energy?
NaCl or KCl NaCl or MgS
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Transition metals typically form +1, +2, and +3 ions. It is observed that transition metal atoms first
lose both “s” e-, even though it is a higher energy subshell.
Most lose e- to end up with a filled or a half-filled subshell.
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8.4 - 8.5: Covalent Bonding Atoms share e- to achieve noble gas
configuration that is lower in energy (and therefore more stable).
Occurs when diff. of EN of atoms is ≤ 1.7 Polar covalent:
0.3 < diff. of EN ≤ 1.7 (e- pulled closer to more EN atom)
Nonpolar covalent:≤diff. of EN ≤ 0.3 (e- shared equally)
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8.6: Drawing Lewis Diagrams1. Add up valence e- from all atoms in formula.
If there is a charge, add e- (if an anion) or subtract e- (if a cation).
2. Draw the “molecular skeleton”: Place the least EN atom(s) in the center. Array the remaining elements around the center
and connect them with a single bond. (When in doubt, put the element written first in the formula in the center of the molecule.)
3. Complete the octets of the outer (more EN) atoms first.
4. Place leftover e- on the central atom, even if it violates the octet rule.
5. If the central atom does not have an octet, create multiple bonds by sharing e- with the outer atoms.
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Ex: Draw the Lewis structure, and name the molecule.
SO42- HCN
H2O2 CNS1-
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8.8: Exceptions to the Octet Rule Odd-electron molecules:
Ex: NO or NO2 (involved in breaking down ozone in the upper atmosphere)
Incomplete octet:H2 He BeF2 BF3
NH3 + BF3 → NH3BF3 (Lewis acid/base rxn)
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Expanded octet: occurs in molecules when the central atom is in or beyond the third period, because the empty 3d subshell is used in hybridization (Ch. 9)PCl5 SF6
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8.6: Formal Charge For each atom, the numerical difference between
# of valence e- in the isolated atom and # of e- assigned to that atom in the Lewis structure.
To calculate formal charge:1. Assign unshared e- (usually in pairs) to the atom
on which they are found.2. Assign one e- from each bonding pair to each
atom in the bond. (Split the electrons in a bond.)
3. Then, subtract the e- assigned from the original number of valence e-. #VALENCE e- in free atom
– #NON-BONDING e-– ½(#BONDING e-) FC
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Used to select most stable (and therefore most likely structure) when more than one structure are reasonable according to “the rules”.
The most stable: Has FC on all atoms closest to zero Has all negative FC on most EN atoms.
FC does not represent real charges; it is simply a useful tool for selecting the most stable Lewis structure.
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Examples: Draw at least 2 Lewis structures for each, then calculate the FC of each atom.
SCN1-
N2O BF3
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8.7: Resonance Structures Equivalent Lewis structures that describe a
molecule with more than one likely arrangement of e-
Notation: use double-headed arrow between all resonance structures. Ex: O3
Note: one structure is not “better” than the others. In fact, all resonance structures are wrong, because none truly represent the e- structure of the molecule. The “real” e- structure is an “average” of all resonance structures.
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Bond Order An indication of bond strength and bond length
Single bond: 1 pair of e- sharedEx: F2 •• ••
:F-F:•• ••
O=O:
:
:
:
:N ≡ N:
Longest,
weakest
Shortest, stronges
t
Double bond: 2 pairs of e- shared
Ex: O2
Triple bond: 3 pairs of e- shared
Ex: N2
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Bond Order & Resonance Structures To determine bond order with resonance
structures: Pick one bond and add up the integer
bond order in one resonance structure to the same bond position in all other resonance structures.
Divide the sum by the number of resonance structures to find bond order.
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Examples
SO3 C6H6
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8.9: Bond enthalpy: H/mol to break a particular bond of substance
(g)Ex: CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g) Hrxn =
? 1 C-H & 1 Cl-Cl bond are broken (per mole) 1 C-Cl & 1 H-Cl bond are formed (per mole)
Hrxn ≈ (Hbonds broken) - (Hbonds formed)
Note: this is the “opposite” of Hess’ Law whereHrxn = Hproducts – Hreactants
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Ex: CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g) Hrxn = ?Bond Ave H/mol Bond Ave H/molC-H 413 Cl-Cl 242H-Cl 431 C-Cl 328
C-C 348 C=C 614
Hrxn ≈ (Hbonds broken) - (Hbonds formed)
Hrxn ≈ [(1(413) + 1(242)] – [1(328) + 1(431)]
Hrxn ≈ -104 kJ/mol
Hrxn = -99.8 kJ/mol (actual)
Note: 2 C-C ≠ 1 C=C2(348) = 696 kJ ≠ 614 kJ
H
Absorb E, break 1 C-H and 1 Cl-Cl
bond
Release E, form 1 C-
Cl and 1 H-Cl bondCH4(g) + Cl2(g)
CH3Cl (g) + HCl (g)Hrxn
*CH3(g) + H(g) + 2 Cl(g)
Hrxn = (Hbonds broken) + (- Hbonds formed)Hrxn = (Hbonds broken) - (Hbonds
formed)
Ex: CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) Hrxn=?
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23.5: Metallic bonding Metallic elements have low I.E.; this means
valence e- are held “loosely”. A metallic bond forms between metal atoms
because of the movement of valence e- from atom to atom to atom in a “sea of electrons”. The metal thus consists of cations held together by negatively-charged e- "glue.“ This results in excellent
thermal & electrical conductivity, ductility, and malleability.
A combination of 2 metals is called an alloy.
http://www.uwgb.edu/dutchs/EarthSC202Notes/minerals.htm
Free e- move rapidly in response to electric fields, thus metals are excellent conductors of electricity.
Free e- transmit kinetic energy rapidly, thus metals are excellent conductors of heat.
Layers of metal atoms are difficult to pull apart because of the movement of valence e-, so metals are durable.However, individual atoms are held loosely to
other atoms, so atoms slip easily past one another, so metals are ductile.