Humboldt State University ® Department of Chemistry

Richard A. Paselk

	Science 331
Fall 2004	Lecture/Activity	Office: SA560a
Notes: 13 September		Phone: x 5719 Home: 822-1116
		e-mail: rap1

Molecular Geometry

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:

• Linear (carbon dioxide, 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
  • 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= :O::C::O:
    • 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
    • 
• Trigonal planar (carbocation, CH₃⁺; formaldehyde, CH₂O)
  • carbocation, CH₃⁺
    • valence electrons = 4 + 3x1 - 1 = 6
    • three bonds possible, since only three 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 3 bonded atoms and no lone-pairs, therefore
    • steric number = 3, so trigonal planar electronic geometry, and 3 atoms so
    • trigonal planar molecular geometry
    • 
  • 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
    • or, showing the double bond
• 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
  • rotated to a different angle =
• 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
  • rotated to a different angle =
• 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
  • 

Other geometries occur when there are more than eight electrons (as frequently occurs with transition metals), but we won't worry about them.

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© R A Paselk

Last modified 15 September 2004