# Shapes of Molecules – Answers to Practice Questions

1: Predict the shapes and bond angles in:

i) sulphur dichloride (SCl2)     non linear    104.5°

ii) silane (SiH4)  tetrahedral   109.5°

iii) carbon disulphide (CS2)  linear   180°

iv) sulphur trioxide  trigonal planar   120°

v) ozone  non-linear   117.5°

2: State and explain the shape of the ammonia molecule, NH3

The central N atom in NHhas 3 bonding pairs and 1 lone pair around it. The electron pairs all repel to get as far apart as possible, and the lone pair repels more strongly than the bonding pairs. As a result the bond angles are 107º and the shape is PYRAMIDAL.

3:  Compare the shapes of the H2O molecule and the H3Oion, and use this to explain why the bond angle in H2O is 104.5º while the bond angle in H3O+ is 107º.

The central O atom in H2O has 2 bonding pairs and 2 lone pairs around it giving it a NON-LINEAR shape, while the central O atom in H3Ohas 3 bonding pairs and 1 lone pair around it giving it a PYRAMIDAL shape. In both cases the four electron pairs repel to get as far apart as possible, but because lone pairs repel more than bonding pairs the bond angle in H2O closes to 109.5 – (2 x 2.5) = 104.5º while the bond angleH3Oin only closes to 109.5 – 2.5 = 107º.

4:  Draw the 3D shape of the following molecules

i) chloromethane, CH3Cl

ii) phosphine, PH3

iii) phosphorus pentachloride, PCl5

5: State the shape around each of the three carbon atoms in propene, CH2CHCH3

The 3D shape can be drawn as:

The carbon on the left has 3 bonding pairs and no lone pairs (remember a double bond counts as 1 bonding pair for the purposes of VSEPR Theory). It thus has TRIGONAL PLANAR shape.

The carbon in the middle is the same as the one on the left, with 3 bonding pairs and no lone pairs, and therefore TRIGONAL PLANAR shape.

The carbon on the right has 4 bonding pairs and no lone pairs, so it has TETRAHEDRAL shape.