Periodic Table of the ElementsIA IIA IIIA IVA VA VIA VIIA VIIIA H He 2 Li Be B C N O F Ne 3 Na Mg IIIB IVB VB VI VIIB VIIIB IB IIB Al Si P S Cl Ar 4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 6 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 1. Methane, CH4
- Looking at the Periodic Table, carbon and hydrogen are both Intermediate, therefore covalent;
- 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=
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- Considering C as the central atom, have 4 bonded atoms and no lone-pairs, therefore
- steric number = 4, so tetrahedral electronic geometry,
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- 4 atoms so tetrahedral molecular geometry
rotated to a different angle =
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Periodic Table of the ElementsIA IIA IIIA IVA VA VIA VIIA VIIIA H He 2 Li Be B C N O F Ne 3 Na Mg IIIB IVB VB VI VIIB VIIIB IB IIB Al Si P S Cl Ar 4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 6 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 2. Hydrogen sulfide
- Looking at the Periodic Table, H is Intermediate and S is Hi, therefore covalent;
- 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 =
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- Considering S as the central atom, have 2 bonded atoms and 2 lone-pairs, therefore
- steric number = 4, so tetrahedral electronic geometry,
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- two lone pairs, so bent molecular geometry
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Periodic Table of the ElementsIA IIA IIIA IVA VA VIA VIIA VIIIA H He 2 Li Be B C N O F Ne 3 Na Mg IIIB IVB VB VI VIIB VIIIB IB IIB Al Si P S Cl Ar 4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 6 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 3. Carbon monoxide, CO
- Looking at the Periodic Table, carbon is Intermediate and oxygen is Hi, therefore covalent;
- 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
Polarity:
- C and O have significantly different Electronegativity values (2.5& 3.5), so the C-O bond will be polar, with O partially negative.
- Carbon monoxide is polar as shown with the dipole arrow in the image:
Periodic Table of the ElementsIA IIA IIIA IVA VA VIA VIIA VIIIA H He 2 Li Be B C N O F Ne 3 Na Mg IIIB IVB VB VI VIIB VIIIB IB IIB Al Si P S Cl Ar 4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 6 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 4. Ammonia, NH3
- Looking at the Periodic Table, hydrogen is Intermediate and nitrogen is Hi, therefore covalent;
- 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=
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- Considering N as the central atom, have 3 bonded atoms and one lone-pair, therefore
- steric number = 4, so tetrahedral electronic geometry,
- only 3 atoms so trigonal pyramidal molecular geometry
rotated to a different angle =
Polarity:
- H and N have significantly different Electronegativity values (2.1& 3.0), so the H-N bond will be polar, with N partially negative.
- Ammonia is polar as shown with the dipole arrow in the image below:
- The polar contributions of the angled polar H-O bonds along the central axis add up so overall the molecule is polar.
In addition to these exercises you should familiarize yourself with the text materials.
VSEPR Theory & Molecular Geometry
Engravings of geometric solids from Encyclopedia Britannica 11th edition (1910) vol 7.
© R A Paselk
Last modified 24 October 2009