2. Shapes

L2CHEM

---

Determining the shape of molecule is crucial for this standard.

Shapes can be determined by looking at how many bonding pairs, and lone pairs there are in the lewis structure.

For example, when we’re given the lewis structure for $H_{2}O$ we should first note that the lewis structure is complete, since hydrogen has 2, and oxygen has 8 valence electrons. Now, we need to look at the central atom, which in this case is oxygen. A central atom is very self-explanatory, and is obvious. As we can see in the lewis diagram, the central atom, oxygen, has 4 regions of ‘electron density’, which refers to the groups of electrons around the central atom. We can see 4 groups of electrons (grouped into 2s) — 2 regions bonded to nothing, and the other 2 regions bonded to hydrogen each.

When we know that the central atom has 4 regions of electron density, we know that the parent shape of this molecule is ‘tetrahedral’ (there ain’t such a way to explain this — remember it). From the 4 regions of electron density, 2 regions are bonding pairs, since 2 regions are attached to 1 hydrogen each. Therefore, I can say that the central atom has 2 bonding pairs.

When there are 4 regions of electron density, and 2 of the regions are bonding pairs, while the other 2 regions are lone pairs (regions bonded to nothing), it has the overall shape of ‘bent’ (again, there ain’t such a way to explain this — remember it).

Here is a good flowchart you can follow:

4EDR:
parent shape: tetrahedral
2BP, 2LP:
overall shape: bent
4BP, 0LP:
overall shape: tetrahedral
3BP, 1LP:
overall shape: trigonal pyramidal

3EDR:
parent shape: trigonal planar
3BP, 0LP:
overall shape: trigonal planar
2BP, 1LP:
overall shape: bent
1BP, 2LP:
overall shape: linear

2EDR:
parent shape: linear
2BP, 0LP:
overall shape: linear
1BP, 1LP:
overall shape: linear

1EDR:
parent shape: linear
1BP, 0LP:
overall shape: linear

---

You should also remember the angles that each shape has.

Tetrahedral: $109.5^o$,
Trigonal Planar: $120^o$,
Trigonal Pyramidal: $109.5^o$,
Bent: $109.5^o$,
Linear: $180^o$.

---

There will most likely be a question that asks to compare and contrast the shape and bond angles or something — follow the CRABS structure:
C: Central Atom
R: Repulsion
A: Arrangement
B: BP and LP (NBP)
S: Shape

e.g.
[C] $H_{2}O$ has a central atom of $O$, which has 4 regions of electron density. [R] All regions of electron density repel each other to maximize separation and minimize overall repulsion. [A] This ends up with a tetrahedral arrangement with a bond angle of $109.5^o$. [B] Since $H_{2}O$ has 2 bonding pairs, and 2 lone pairs, [S] this results in an overall shape of bent.

Free excellence using the CRABS structure.