Simple Molecules

 

 

To determine the type of molecule (polar/ non-polar) and IMF, you will need these four steps:

 

 

1) Draw dot-cross diagram:

 

1 Determine the central atom and surrounding atoms.

Central atom is commonly the:
- Element with the lesser no. of atoms
- First element in chemical formula (except H)
- Least electronegative element
2 For polyatomic ions:
Anion: add the e– to the most electro-ve atom (one e– per atom)
Cation: remove the e– from the least electro-ve element Ignore this step for neutral molecules
3 Determine no. of bonds each surrounding atom can form:
no. of valence e– on atom [n(e)] no. of bonds the atom can form [n(b)]
0 – 4 n(e)
5 – 8 8 – n(e)
For H atom: n(b) = 2 – n(e)
4 Draw all the bonds to the central atom.
For atoms with 8 e–: form dative bond from central atom to surrounding atom
For atoms with 0 e–: form dative bond from surrounding atom to central atom
5 Assign remaining e– on central atom as lone pairs.
If central atom is in Period 2 and no. of e– around the atom is > 8, convert a double bond to a dative bond from the central atom.

 

Examples:

 

SO2

1 Central atom: S
Surrounding atoms: O
2 N.A.
3
4
5 N.A. (S is in P3; can expand beyond octet)

 

HCN

1 Central atom: C
Surrounding atoms: H and N
2 N.A.
3
4
5 N.A.

 

NH3

1 Central atom: N
Surrounding atoms: H
2 N.A.
3
4
5 N has 5 e– and 3 bonds > it has 2 remaining e–:

 

NO3–

1 Central atom: N
Surrounding atoms: O
2 Add one e– to one O atom.
3
4
5 N has 10e– > convert one double bond to dative bond from N to O:

 

AlH4–

1 Central atom: Al
Surrounding atoms: H
2 Add one e– to one H atom.
3
4
5 N.A.

 

NH4+

1 Central atom: N
Surrounding atoms: H
2 Remove one e– from one H atom.
3
4
5 N.A.

 

2) Shape (VSEPR)

  • Electron pairs arrange themselves as fair apart as possible to minimise repulsion
  • Multiple bonds are treated as single bond pairs
  • Repulsion: lp-lp > lp-bp > bp-bp

Approach:

  1. Determine no. of e– pairs (bp+lp)
  2. Find parent shape and bond angle
  3. (For molecules with lp) Erase lp and name resultant shape
  4. For each lp, reduce bond angle by 2°

No. of e– pairs

 Shape Bond Angle

2

 Linear 180°

3

 Trigonal Planar 120°

4

 Tetrahedral 109.5°

5

 Trigonal Bipyramidal equatorial (120°)
axial (90°)

6

 Octahedral 90°

 

Example: NH3

  • 4 e– pairs (3 bp + 1lp)
  • Parent shape: Tetrahedral (109.5°)

  • Resultant shape: Trigonal Pyramidal
  • Angle: 107°

 

3) Polarity

 

Draw the dipole moments for each bond in molecule (from less electro-ve atom to more electro-ve atom)

  • Net dipole → Polar
  • No net dipole → Non-Polar

If you have problems determining if there is a net dipole, visualise the dipole moments as forces acting on the central atom. If the central atom moves → a net dipole exists.

 

Convenient Generalisations:

  • All hydrocarbons are non-polar since C–H bond is non-polar.
  • Molecules with lone pairs on central atom are polar (Exception: XeF4)

 

4) Intermolecular Forces

 

Type Exist between:
Permanent dipole – permanent dipole (pd-pd) polar molecules
Induced dipole – induced dipole (id-id)
(Dispersion)

between all molecules/ atoms

non-polar molecules only have dispersion
Hydrogen bonds between H attached to F,O,N and lone pair on F,O,N on another molecule