Applications of VSEPR Theory: MDCAT Chemistry notes
Applications of VSEPR Theory for MDCAT: shapes and bond angles of AB2, AB3 and AB4 molecules with lone pairs, and planar versus non-planar species.
How to predict a shape
- Find the central atom and count its bond pairs. A multiple bond counts as one.
- Count its lone pairs: (valence electrons − electrons used in bonding) ÷ 2, adjusted for any ionic charge.
- The total number of pairs gives the electron-pair arrangement.
- The shape is named from the positions of the atoms only. Lone pairs are there but are not "seen" in the name.
Shapes table
| Type | bp, lp | Shape | Angle | Examples |
|---|---|---|---|---|
| $\mathrm{AB_2}$ | 2, 0 | Linear | 180° | $\mathrm{BeCl_2}$, $\mathrm{CO_2}$, $\mathrm{CS_2}$, HCN |
| $\mathrm{AB_3}$ | 3, 0 | Trigonal planar | 120° | $\mathrm{BF_3}$, $\mathrm{BCl_3}$, $\mathrm{AlCl_3}$, $\mathrm{SO_3}$, $\mathrm{NO_3^-}$ |
| $\mathrm{AB_2E}$ | 2, 1 | Bent (angular) | less than 120° | $\mathrm{SnCl_2}$, $\mathrm{SO_2}$ |
| $\mathrm{AB_4}$ | 4, 0 | Regular tetrahedral | 109.5° | $\mathrm{CH_4}$, $\mathrm{CCl_4}$, $\mathrm{SiCl_4}$, $\mathrm{NH_4^+}$, $\mathrm{MnO_4^-}$ |
| $\mathrm{AB_3E}$ | 3, 1 | Trigonal pyramidal | 107° ($\mathrm{NH_3}$) | $\mathrm{NH_3}$, $\mathrm{PH_3}$, $\mathrm{NF_3}$, $\mathrm{PCl_3}$ |
| $\mathrm{AB_2E_2}$ | 2, 2 | Bent (angular) | 104.5° ($\mathrm{H_2O}$) | $\mathrm{H_2O}$, $\mathrm{H_2S}$ |
Four electron pairs always take a tetrahedral arrangement. With no lone pair the shape is tetrahedral, with one lone pair it is pyramidal, and with two it is bent.
Worked reasoning
- $\mathrm{SnCl_2}$: Sn has 4 valence electrons. Two are used for the two Sn–Cl bonds and one lone pair remains. So the molecule is bent, not linear, even though it has two bond pairs like $\mathrm{CO_2}$.
- $\mathrm{H_2O}$ and $\mathrm{SnCl_2}$ are therefore both bent. $\mathrm{NH_3}$ (pyramidal) and $\mathrm{AlCl_3}$ (planar) are not alike, and neither are $\mathrm{AlCl_3}$ and $\mathrm{PCl_3}$.
- $\mathrm{BF_3}$: B has 3 bond pairs and no lone pair, so the molecule is planar with an exact angle of 120°.
- Lone pairs shrink the angle below the ideal. $\mathrm{PH_3}$ is pyramidal with an angle well below 109.5° (about 93°), so it is not a 109.5° molecule.
Planar and non-planar species
- Planar: $\mathrm{BF_3}$, $\mathrm{BCl_3}$, $\mathrm{SO_3}$, $\mathrm{NO_3^-}$, $\mathrm{SO_2}$, $\mathrm{H_2O}$ (three atoms always lie in a plane), ethene, benzene and linear molecules.
- Non-planar (3-D): $\mathrm{CH_4}$, $\mathrm{NH_4^+}$, $\mathrm{MnO_4^-}$, $\mathrm{NH_3}$, $\mathrm{NF_3}$, $\mathrm{PH_4^+}$.
- Exactly 109.5°: only the regular tetrahedra, such as $\mathrm{CH_4}$, $\mathrm{CCl_4}$, $\mathrm{SiCl_4}$ and $\mathrm{NH_4^+}$. $\mathrm{NH_3}$, $\mathrm{H_2O}$ and $\mathrm{PH_3}$ are smaller.
Common MDCAT traps
- $\mathrm{SnCl_2}$ is the odd one out among two-bond-pair molecules because its lone pair bends it.
- $\mathrm{NH_3}$ is not planar: its lone pair makes it pyramidal. $\mathrm{BF_3}$ is the planar one.
- $\mathrm{AlCl_3}$ monomer is trigonal planar like $\mathrm{BF_3}$. Do not confuse it with $\mathrm{PCl_3}$, which is pyramidal.
- The BF3 angle is exactly 120°, not 119.5° or 109.5°.
- In a "set" question, one wrong member rules out the whole option. For example, $\mathrm{SO_3}$ and benzene are planar.
Quick revision
- $\mathrm{AB_4}$ with no lone pairs is a regular tetrahedron at 109.5°.
- Three bond pairs and one lone pair give a trigonal pyramid.
- Two bond pairs and two lone pairs give a bent shape at 104.5° in water.
- Four electron pairs always give a tetrahedral arrangement.
- $\mathrm{CO_2}$, $\mathrm{CS_2}$ and HCN are linear, while $\mathrm{SnCl_2}$ is bent.