Electrochemistry

Determine Type of Bonding for Simple Molecules

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

1) Draw dot-cross diagram:

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

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

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 Hydrogen: n(b) = 0 for n(e) = 0 or 2

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

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]

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:

Determine no. of e^– pairs (bp+lp)
Find parent shape and bond angle
(For molecules with lp) Erase lp and name resultant shape
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

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Electrochemistry