Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 107	Fundamentals of Chemistry	Fall 2008
Lecture Notes: 7 October	© R. Paselk 2005
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Chemical Bonds, cont.

Bond Type

So how do we predict whether a strong chemical bond is covalent or ionic?

• For numbers, use a difference of 1.7 to distinguish ionic (D EN> 1.7) and covalent (D EN< 1.7).
• Easier to use the hi, lo, intermediate system.
  • First recall Hi-Lo-Intermediate assignment:
    • all metals are low
    • the most electronegative of the non-metals are high (N, O, F, S, Cl, Br, I)
    • other elements are intermediate
  • Then apply rules:
    • If combine hi + lo, then ionic
    • Otherwise, covalent (hi + hi, lo + lo, hi + inter., lo + inter., inter. + inter.)
• Lewis Structure Examples:
• Barium iodide
• Carbon disulfide
• Arsenic triiodide
• Hydrogen selenide
• Oxygen difluoride

Multiple Bonds

Recall we must show an octet (or duet for Period I) in the outer-most shell (valence electrons). When this does not occur with single electron pairs (bonds) between atoms can sometimes make it happen with multiple bonds. You might find "Clark's Method" useful for determining the bonding patterns of various molecules:

Clark's Method for determining bonding in covalent Lewis Structures

• Add up all of the valence electrons in the structure (remember to add one electron for each negative charge, or subtract one for each positive charge)
  • If e^- = 6y + 2 where y = # atoms other than H, then octet rule is followed with single bonds only.
  • If e^- < 6y + 2 then probably have multiple bonding with the number of multiple bonds = /2 (remember a triple bond is 2 multiple bonds!).
• If you can draw more than one structure, then chose the most symmetrical.

Examples:

• carbon dioxide
• carbon monoxide

• If two or more structures are equally symmetrical, then you probably have resonance and should show all structures connected by double arrows.

Resonance example:

• Carbonate ion

 

Molecular Geometry

The importance of molecular shape: recognition at the molecular level in organisms. Shape and electron density are extraordinarily important to the interaction of biomolecules - Examples

• Sarin (nerve gas)
• "drugs"
• estrogen mimics ("feminization" of various animal populations - birth control complication)

Lewis Structures enable us to predict bonding patterns for compounds of the representative elements, but how can we predict their shapes? We will add another tool, VSEPR Theory, to our chemical toolbox - a simple way to predict the geometry of bonds around a central atom (for larger molecules predict one center at a time).

VSEPR Theory

VSEPR (Valence Shell Electron Pair Repulsion) Theory is based on three assumptions (there are more advanced versions, but unnecessary for us):

• Electron pairs will orient around a central point to minimize repulsion.
• Lone-pairs of electrons will have greater repulsion than bonded pairs of electrons (note that the atoms are ignored in terms of repulsion).
• Repulsion is strong at 90° and weaker at 120° (weakest at 180°).

VSEPR predicts geometry based on these assumptions in a few simple, sequential, steps:

1. Draw a correct Lewis Structure.
2. Determine the Steric Number = the number of bonded atoms + the number of lone pairs = "electron clouds" in valance shell of central atom.
3. Maximize the angles between electron pairs, placing the lone (unbonded) pairs at the extremes.

For central atoms with eight outer electrons (octets) there are three possible electron pair geometries:

• Linear with angles of 180° ( a single pair and a triple bond, or two double bonds).
• Trigonal planar with angles of 120° (one double bond and two single pairs).

• Tetrahedral with angles of 109.5° (four single pairs).

These three electron pair geometries can lead to five molecular geometries.

Five molecular geometries

Linear

• Carbon monoxide, CO
  • valence electrons = 4 + 6 = 10
  • 6y + 2 = 14, thus 4 fewer electrons than required for all single bonds, 4/2 = 2 multi-bonds (2 double or 1 triple)
  • LS = :C:::O:
  • Considering C as the central atom, have one bonded atom and one lone-pair, therefore
  • steric number = 2, so linear electronic geometry, and two atoms so
  • linear molecular geometry
• Carbon dioxide, CO₂
  • valence electrons = 4 + 2x6 = 16
  • 6y + 2 = 20, thus 4 fewer electrons than required for all single bonds, 4/2 = 2 multi-bonds (2 double or 1 triple)
  • LS: from symmetry C will be central atom, therefore=
  • Considering C as the central atom, have 2 bonded atoms and no lone-pairs, therefore
  • steric number = 2, so linear electronic geometry, and
  • linear molecular geometry

    or

Trigonal planar

• Formaldehyde, CH₂O
  • valence electrons = 4 + 2x1 + 6 = 12
  • 6y + 2 = 6 x 2 + 2 = 14; so molecule has 2 fewer electrons than required for all single bonds, 1 double bond
  • LS: from symmetry C will be central atom, therefore=
  • Considering C as the central atom, have 3 bonded atoms and no lone-pairs, therefore
  • steric number = 3, so trigonal planar electronic geometry, and 3 atoms so
  • trigonal planar molecular geometry =

Tetrahedral

• Methane, CH₄
• valence electrons = 4 + 4x1= 8
• four bonds possible, since only 4 pairs, single bonds because only have H's bound to C.
• LS: from symmetry C will be central atom, therefore=
• Considering C as the central atom, have 4 bonded atoms and no lone-pairs, therefore
• steric number = 4, so tetrahedral electronic geometry, and 4 atoms so
• tetrahedral molecular geometry = another view =

Trigonal pyramidal

• Ammonia, NH₃
  • valence electrons = 5 + 3x1= 8
  • only 4 pairs, single bonds because only have H's bound to N, 3 bonds, since only 3 H's
  • LS: from symmetry N will be central atom, therefore=
  • Considering N as the central atom, have 3 bonded atoms and one lone-pair, therefore
  • steric number = 4, so tetrahedral electronic geometry, but only 3 atoms so
  • trigonal pyramidal molecular geometry = view from beneath N=

Bent

• Water, H₂O
  • valence electrons = 6 + 2x1= 8
  • only 4 pairs, single bonds because only have H's bound to O, 2 bonds, since only 2 H's
  • LS: from symmetry O will be central atom, therefore=
  • Considering O as the central atom, have 2 bonded atoms and 2 lone-pairs, therefore
  • steric number = 4, so tetrahedral electronic geometry,
  • but only 2 atoms, so bent molecular geometry =

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